Numerical reasoning I
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Transcript of Numerical reasoning I
![Page 1: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/1.jpg)
Numerical Reasoning
![Page 2: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/2.jpg)
Problem on Numbers
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Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d Sum of n terms of A.P S = n/2 *[2a+(n-1)d)]
Geometrical Progression:
Tn = arn – 1.
Sn = a(rn – 1) / (r-1)
Problems on Numbers
![Page 4: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/4.jpg)
Basic Formulae
1. ( a+b)2 = a2 + b2 + 2ab
2. (a-b)2 = a2 +b2 -2ab
3. ( a+b)2 - (a – b)2 = 4ab
4. (a+b)2 + (a – b)2 = 2 (a2 +b2)
5. (a2 – b2) = (a+b) (a-b)
6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca)
7. (a3 +b3) = ( a+b) (a2 –ab +b2)
8. (a3 –b3) = (a-b) (a2 +ab + b2)
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Problem - 1
A 2 digit number is 3 times the sum of its digits if 45 is added to the number. Its digits are interchanged. The sum of digits of the number is?
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Solution
The number is 3 times the sum of its digits
45 is added = 4 +5 = 9
So, common numbers in 3 and 9th table.
9, 18, 27, 36, 45….
27 + 45 = 72
2 + 7 = 9 or 4 + 5 = 9
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Problem - 2
A number when divided by 119 leaves a remainder of 19. If it is divided by 17. It will leave a remainder of?
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Solution
= 19/17 = 2 remainder
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Problem - 3A boy was asked to find the value of 3/8 of sum of money instead of multiplying the sum by 3/8 he divided it by 8/3 and then his answer by Rs.55. Find the correct answer?
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Solution
8/3 – 3/8 = 55/24
= 55/55/24
= 24
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Problem - 4
A man spends 2/5rd of his earning. 1/4th of the expenditure goes to food, 1/5th on rent, 2/5th on travel and rest on donations. If his total earning is Rs.5000, find his expenditure on donations?
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Solution
5000*2/5 = 2000
Remaining amount has given as donation
2000* (1/5 + 2/5 + ¼)
Total amount = 200*17/20 = 1700
2000 – 1700 = 300
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Problem - 5
From a group of boys and girls 15 girls leave. There are then left, 2 boys for each girl. After this 45 boys leave, there are then left 5 girls for each boy, find the number of girls in the beginning?
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Solution15 girls leave = 2 boys for each girl
45 boys leave = 5 girls fro 1 boy
Let the boys be x; Girls = x/2 +15
After the boys have left,
No.of boys = x – 45 and girls = 5(x-45)
x/2 = 5(x-45)
X = 2(5x-22)
X = 10x – 450
X =50
50/2 +15 =40
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Problem - 6
An organization purchased 80 chairs fro Rs.9700. For chairs of better quality they paid Rs.140 each and for each of the lower grade chair they paid Rs.50 less. How many better quality chairs did the organization buy?
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SolutionBetter quality chairs = x;
Lower quality = 80 –x
Price of better quality = Rs.140, Lower quality = 140-50 = 90
140*x + 90(80-x) = 9700
140x + 7200 – 90x = 9700
50x = 9700 – 7200;
50x = 2500
X = 50
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Problem - 7
A labour is engaged for 30 days, on the condition that Rs.50 will be paid for everyday he works and Rs.15 will be deducted from his wages for everyday he is absent from work. At the end of 30 days he received Rs.850 in all. For how many days did he wanted?
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Solution
Total wages = 30*50 = 1500 (without Absent)Wages received in 30 days = 850 (with Absent)Let the labourer work for x daysAbsent = 30 – x50x – (30-x)15 = 85050x -450 +15x = 85065x = 1300X = 1300/65 = 20 days
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Problem - 8
The rent is charged at Rs.50 per day for first 3 days Rs.100 per day next 5 days, and 300 per day thereafter. Registration fee is 50 at the beginning. If a person had paid Rs.1300 for his stay how many days did he stay?
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Solution
3 days = 150 + 50 = 200
5 days = 100*5 = 500
= 200 + 500 = 700
1300 – 700 = 600
2 days = 300*2 = 600
= 5 + 3 + 2 = 10 days
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Problem - 9
In a school 20% of students are under the age of 8 years. The number of girls above the age of 8 years is 2/3 of the number of boys above the age of 8 years and amount to 48. What is the total number of students in the school?
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SolutionGirls above 8 yrs = 48Boys above 8 yrs = 48 / 2/380% of students above 8 yrs = 48 + 72 = 120
80 120 20 x80x = 120*20X = 120*20/80 = 30Total No.of students = 120+30 = 150
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Ratio and Proportion
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Ratio and Proportion
Ratio : Relationship between two variables.
= a : b
Proportion : Relationship between two ratios.
= a : b : : c : d
Proportion Calculation = a*d : b*c
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Problem - 1
The ratio of number of boys to that of girls in a school is 3:2. If 20% boys and 25% of girls are scholarship holders, find the percentage of the school students who are not scholarship holders?
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Solution
Let the total number of students be 100
Boys = 100*3/5 = 60
Girls = 100*2/5 = 40
S. holders = 60*20/100 = 12, non S. holders = 60 -12 = 48
Girls s. holders = 40*25/100 = 10,
Non s. holders = 40 – 10 = 30
Students who do not have scholarship = 48 + 30 = 78
78/100*100 = 78%
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Problem - 2
The cost of diamond varies as the square of its weight. A diamond weighing 10 decigrams costs Rs. 32000. Find the loss incurred when it breaks into two pieces whose weights are in the ratio 2:3?
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Solution
1st piece = 10*2/5 = 4
2nd piece = 10*3/5 = 6
Cost of the diamond varies as square of its weight
42 : 62 102 = 100k
16k : 36k
100k – 52 k = 48k(loss)
100k = 32000; k = 320
48*320 = 15360
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Problem - 3
The ratio of the first and second class fares between two railway stations 4 : 1 and the ratio of the number of passengers traveling by first and second class is 1:40. If the total of Rs.1100 is collected as fare from passengers of both classes what was the amount collected from first class passengers?
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Solution
Fare = 4 : 1
Passengers traveling = 1 : 40
Amount = No. pas * fare = 4*1 :10*1 = 4 : 40
= 1:10
Total amount = 1100.
First class passengers’ amount = 1*1100/11
= 100
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Problem - 4
A vessel contains a mixture of water and milk in the ratio 1:2 and another vessel contains the mixture in the ratio 3:4. Taking 1 kg each from both mixtures a new mixture is prepared. What will be the ratio of water and milk in the new mixture?
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Solution
1st vessel = water = 1/3 , milk = 2/3
2nd vessel = water = 3/7, milk = 4/7
Water = 1/3 + 3/7 = 16/21
Milk = 2/3 + 4/7 = 26/21
16 : 26 = 8:13
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Problem - 5
Ratio of the income of A, B, C last year 3 : 4 : 5. The ratio of their individual incomes of last year and this year are 4:5, 2:3 and 3:4 respectively. If the sum of their present income is Rs.78,800. Find the present individual income of A, B and C.
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SolutionA’s Present Income = 5/4*3x = 15x/4B’s Present Income = 3/2*4x = 12x/2C’s Present Income = 4/7*5x = 20x/715x/4 + 6x+20x/3 = 78,800197x/12 = 78,800X = 945600/197X = 4,800A’s Present income = 15x/4 = 15*4800/4 = 18,000B’s Present income = 6*x = 6*4,800 = 28,800C’s Present income = 20x/3 = 20*4800/3 = 32,000
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Problem - 6
Of the three numbers, the ratio of the first and the second is 8:9 and that of the second and third is 3:4. If the product of the first and third numbers is 2,400, then find the second number?
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Solution a : b = 8 : 9 b : c = 3 : 4
b : c = 3*3 : 4*3 = 9 : 12 a : b : c = 8 : 9 : 12Product of first and third = 8k * 12k = 240096k2 = 2400; k2 = 2400/96 = 25 k = 5Second number = 9 * 5 = 45
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Problem - 7
Annual income of A and B are in the ratio of 4 : 3 and their annual expenses are in the ratio 3 : 2. If each of them saves Rs.600 at the end of the year, what is the annual income of A?
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Solution
Income = 4 : 3, Expenses = 3 : 2Savings 600 eachA’s income = 4x, expenses = 3x, savings = x i.e 600
Income = 4*600 : 3*600A : B = 2400 : 1800A income = 2400
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Problem - 8
The property of a man was divided among his wife, son and daughter according to his will as follows. Wife’s hare is equal to 6/7th of son’s share and daughter share is equal of 4/7th of Son’s. If the son and daughter together receives Rs.1,02,300. How much does his wife get?
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Solution
Let the Son’s share be x.
Daughter’s share = x*4/7 = 4x/7
Wife’s share = x* 6/7 = 6x/7
X + 4x/7 = 1,02,300
7x + 4x = 1,02,300
X = 1,02,300 /11 = 65,100
Wife Share = 65,100 *6/7 = Rs. 55, 800
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Problem - 9
A pot containing 81 litres of pure milk of the milk 1/3 is replaced by the same amount of water. Again 1/3 of the mixture is replaced by the same amount of water. Find the ratio of milk to water in the new mixture?
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Solution
Milk : Water
Initial = 81 : 0
1/3 removed = 54 : 27
1/3 mixture = 36 : 45
Ratio of Milk and Water = 4 : 5
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Problem - 10
729 ml of mixture contains milk and water are in the ratio 7 : 2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7 : 3.
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Solution
Water = 729 * 2/9 = 162
Ratio Water
2 162
3 x
2x = 3*162/2 = 243
243 – 161 = 81 ml water is to be added
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Problem - 11
Price of a scooter and a television set are in the ratio 3 : 2. If the scooter costs Rs.600 more than the television set, then find the price of television?
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Solution
Diff. in ratio = 3 – 2 = 1
1 ratio is 600 means, the television cost is 2 ratio so, cost of television = 1200
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Problem - 12
The annual income and expenditure of man and his wife are in the ratio of 5:3 and 3:1 respectively, if they decide to save equally and find their balance is 4000. Find their income at the end of the year?
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Solution
Man and Wife income = 5 : 3 = 2 (diff)
Man and Wife Expenses = 3 : 1 = 2 (diff)
so, both of them are saving ratio of 2
Total saving of Man and Women = 4000, individual saving 2000
So, Man income = 5000 and Women income = 3000
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Problem - 13
In a class room, ¾ of the boys are above 160 cm in height and they are in 18 number. Also out of the total strength, the boys are only 2/3 and the rest are girls. Find the total number of girls in a class?
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Solution
¾ of the boys in 18 numbers means, ¼ of the boys = 6
Total number of boys = 18+6 = 24Ratio Number2/3 241/3 x2/3*x = 24*1/3 x = 24/2 = 12 Girls
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Problem - 14
Rs. 770 was divided among A, B and C such that
A receives 2/ 9th of what B and C together
receive. Find A’s share?
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Solution
A = 2/9 (B+C)
B+C =9A/2
A+B+C = 770
A + 9A/2 = 770
11A = 770*2
A = 140
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Problem - 15A sporting goods store ordered an equal
number of white and yellow balls. The tennis
ball company delivered 45 extra white balls
making the ratio of white balls to yellow balls
1/5 : 1/6. How many white tennis balls did the
store originally order for?
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Solution
Let the number of yellow balls be x
(x + 45) : x = 1/5 : 1/6
Solving the above equation,
The number of white balls originally ordered
would be = 225 balls
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Alligation and Mixture
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Alligation and Mixture
Alligation : It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.
(Quantity of cheaper / Quantity of costlier)
(C.P. of costlier) – (Mean price)
= --------------------------------------
(Mean price) – (C.P. of cheaper)
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Alligation or Mixture
Cost of Cheaper Cost of costlier c d
Cost of Mixture m
d-m m-c
(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
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Problem -1
Three glasses of size 3 lit, 4 lit and 5 lit contain mixture of milk and water in the ratio of 2:3, 3:7 and 4:11 respectively. The content of all the three glasses are poured into a single vessel. Find the ratio of milk and water in the resulting mixture.
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Solution1st Vessel
= Milk = 3*2/5 = 6/5 = Water = 3*3/5 = 9/5
2nd Vessel:
= Milk = 4*3/10 = 12/10= Water = 4*7/10 = 28/10
3rd Vessel:= Milk = 5*4/15 = 20/15= Water = 5*11/15 = 55/15
Milk : Water = 6/5 +12/10 + 20/15 : 9/5 + 28/10 + 55/15 = 18/15 + 18/15 +20/15 : 27/15 + 42/15+55/15 = 56 : 124 (or) 14:31
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Problem - 2
How many kg of tea worth Rs. 25 per kg must be blended with 30 kg tea worth Rs. 30 per kg, so that by selling the blended variety at Rs.30 per kg there should be a gain of 10%?
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Solution
30*100/110 = 300/1125 30
300/11
30/11 25/1130 : 256 : 536 : 30kg
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Problem - 3
A man buys cows for Rs. 1350 and sells one so as to lose 6% and the other so as to gain 7.5% and on the whole he neither gains nor loses. How much does each cow cost?
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Solution
6 7.5
0
7.5 6
15 12
5 : 4
1350*5/9 = 750
1350 *4/9 = 600
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Problem - 4
There are 65 students in a class, 39 rupees are distributed among them so that each boy gets 80p and each girl gets 30p. Find the number of boys and girls in a class.
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SolutionGirls Boys
30 80
60
20 30
2 : 3
65*2/5 = 26
65*3/5 = 39
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Problem - 5
A person covers a distance 100 kms in 10 hr Partly by walking at 7 km per hour and rest by running at 12 km per hour. Find the distance covered in each part.
![Page 67: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/67.jpg)
Solution
Speed = Distance / Time = 100 / 10 = 107 12
102 : 3
Time taken in 7 km/hr = 10 * 2/5 = 4 4*7 = 28 kmTime taken in 12 km/hours = 10*3/5 = 6 12*6 = 72 km
![Page 68: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/68.jpg)
A merchant has 100 kg of salt, part of which
he sells at 7% profit and the rest at 17% profit.
He gains 10% on the whole. Find the quantity
sold at 17% profit?
Problem - 6
![Page 69: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/69.jpg)
7 17 10 (17-10) (10-7) 7 : 3The quantity of 2nd kind = 3/10 of 100kg = 30kg
Solution
![Page 70: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/70.jpg)
In what ratio two varieties of tea one costing Rs. 27 per kg and the other costing Rs. 32 per kg should be blended to produce a blended variety of tea worth Rs. 30 per kg. How much should be the quantity of second variety of tea, if the first variety is 60 kg?
Problem - 7
![Page 71: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/71.jpg)
27 32
30
2 3
Quantity of cheaper tea = 2
Quantity of superior tea 3
Quantity of cheaper tea =2*x/5 = 60 , x=150
Quantity of superior tea = 3 * 150/5 = 90 kg
Solution
![Page 72: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/72.jpg)
A 3-gallon mixture contains one part of S and
two parts of R. In order to change it to mixture
containing 25% S how much R should be
added?
Problem - 8
![Page 73: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/73.jpg)
R : S
2 : 1
75% : 25%
3 : 1
1 gallon of R should be added.
Solution
![Page 74: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/74.jpg)
Three types of tea A,B,C costs Rs. 95/kg, Rs.
100/kg. and Rs 70/kg respectively. How many
kg of each should be blended to produce 100
kg of mixture worth Rs.90/kg given that the
quantities of B and C are equal?
Problem - 9
![Page 75: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/75.jpg)
B+C/2 A85 95
905 5
Ratio is 1:1 so A = 50 , B + C = 50
The quantity would be 50 : 25 : 25
Solution
![Page 76: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/76.jpg)
In what proportion water must be added to
spirit to gain 20% by selling it at the cost price?
Problem - 10
![Page 77: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/77.jpg)
Profit%=20%
Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3
0 10
25/3
5/3 25/3
The ratio is 1: 5
Solution
![Page 78: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/78.jpg)
In an examination out of 480 students 85% of
the girls and 70% of the boys passed. How
many boys appeared in the examination if total
pass percentage was 75%
Problem - 11
![Page 79: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/79.jpg)
Solution
Solution:
70 85
75
10 5
Number of Boys = 480 * 10/15
Number of Boys = 320
•
![Page 80: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/80.jpg)
Problem - 12
A painter mixes blue paint with white paint so that the mixture contains 10% blue paint. In a mixture of 40 litre paint how many litre of blue paint should be added, so that the mixture contains 20% of blue paint?
![Page 81: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/81.jpg)
Solution
Quantity of blue paint in the mixture = 10% of 40
40*10/100 = 4
40 – 4 = 36 litre
Let x litre blur paint can be mixed
4+x/30 = 20/80 = 4+x = 9
x = 5
![Page 82: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/82.jpg)
Problem - 13
From a 100 litre mixture containing water and milk equal proportion, 10 litres of mixture is replaced by 10 litres of water in succession twice. At the end, what is the ratio of milk and water?
![Page 83: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/83.jpg)
Solution
Milk Water10 lit(1st) 50 : 50
45 : 45 45 : 55
2nd 10 lit 40.5 : 49.5Add Water 40.5 : 59.5
81 : 119
![Page 84: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/84.jpg)
Problem - 14
In a mixture of 400 gms, 80% is copper, sliver is 20%. How much copper is to be added, so that the new mixture has 84% copper?
![Page 85: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/85.jpg)
Solution400*80/100 = 320 Copper400*20/100 = 80 Sliver
Percen Mixture 80 320 84 x= 320*84/80 = 336(320+x) = (400+x) 84/100320+x = 400+84/100 + 84x/10016x/100 = 336 – 320; 16x/100 = 16; x = 100
![Page 86: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/86.jpg)
Problem - 15
A jar full of whisky contains 50% alcohol. A part of this whisky is replaced by another containing 30% alcohol and now the percentage of alcohol was found to be 35%. Find the quantity of whisky replaced?
![Page 87: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/87.jpg)
Solution
50 30
35
5 : 15
5 : 15 = 1 : 3
Replaced = 3/4
![Page 88: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/88.jpg)
Partnership
![Page 89: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/89.jpg)
Type - 1A invest = 10000B invest = 15000Profit = 5000Find their Individual Share ?A : B = 10000 : 15000 = 2 : 3A’s Share = 5000*2/5 = 2000B’s Share = 5000*3/5 = 3000This is a first and basic step for any Partnership
Problem.
![Page 90: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/90.jpg)
Type - 2
A invest = 5000,
After 3 months B joined A, with an investment of 3000
Profit at the end of the year = 3500
Find their Share ?
Any thing happen after a month, like a person joining a business, or withdraw from business or withdraw some amount means given amount is for month.
Cont…
![Page 91: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/91.jpg)
Type - 2
A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9
A’s share = 3500*20/29 = 2413.7
B’s Share = 3500*9/29 = 1086.3
![Page 92: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/92.jpg)
Type - 3
A invest 5000B invest 6000After 3 months A withdraw amount 1000, after 5
months a withdraw amount 1000 again.Profit at the end of the Year = 5000Find their Share ?A = 5*3 + 4*5 + 3*4 = 15 +20 + 12 = 47B = 6*12 = 72
![Page 93: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/93.jpg)
Type - 3
A’s share = 5000* 47/119 = 1974.8
B’s share = 5000*72/119 = 3025.2
![Page 94: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/94.jpg)
Type - 4
A invest twice as much as B, B invest 1/3rd of C. At the end of the year their Profit is 6000. Find their Share?
A = 2BB = 1/3CC = xA : B : C = 2x/3 : x/3 : xA : B : C = 2x/3 : x/3 : 3x /3A : B : C = 3 : 2 : 6A’s Share = 6000*3/11 = 1636B’s Share = 6000*2/11 = 1091C’s Share = 6000*6/11 = 3273
![Page 95: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/95.jpg)
Problem - 1
A, B and C started a business in partnership by investing Rs.12000 each. After 6 months, C left and after 4 months D joined with his capital of Rs.24,000. At the end of a year, a profit of Rs.8,500 shared among all the partners. Find B’s share?
![Page 96: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/96.jpg)
These are all the basic types remaining we will see when we solve problems.
![Page 97: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/97.jpg)
Solution
A : B : C : D
12000 : 12000 : 12000 : 24000
1 : 1 : 1 : 2
1*12 : 1*12 : 1*6 : 2*2
12 : 12: 6 : 4
6 : 6 : 3 : 2
B’s share = 6/17*8500 = 3000
![Page 98: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/98.jpg)
Problem - 2
A, B and C enter into partnership. A contributes one third of the capital while B contributes as much as A and C together contributed. If the profit at the end of the year amounted to Rs.840. What would be B’s share?
![Page 99: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/99.jpg)
SolutionA’s share = 1/3 of the capital
A’s share = 1/3*840 = 280
B’s share = A + C = 280 + x
A + B + C = 840
280 + 280 + x + x = 840
560 + 2x = 840
2x = 840 – 560
X = 140
B’s share = 280+140 = 420
![Page 100: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/100.jpg)
Problem - 3
Akilesh and Jaga enter into a partnership. Akilesh contributing Rs.8000 and Jaga contributing Rs.10000. At the end of 6 months they introduce Prakash, who contributes Rs.6000. After the lapse of 3 years, they find that he firm has made a profit of Rs.9660. Find Prakash’s share?
![Page 101: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/101.jpg)
Solution
Akilesh : Jaga : Prakash
8 : 10 : 6
4 : 5 : 3
4*36 : 5836 : 3*30
144 : 180 : 90
8 : 10 : 5
Prakash’s share = 9660*5/23 = 2100
![Page 102: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/102.jpg)
Problem - 4
Priya and Vijay enter into partnership. Priya supplies whole of the capital amounting to Rs.45,000 with the conditions that the profit are to be equally divided and that Vijay pays Priya interest on half of the capital of 10% p.a. but receives Rs.120 per month for carrying on the concern. Find their total yearly profit. When Vijay’s income is one half of Priya’s income?
![Page 103: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/103.jpg)
Solution45,000 *1/2 = 22,50022,500 *10//100 = 2250 (interest p.a)Vijay receives Rs.120 per month = 120*12 = 1440Total profit be xRatio of Profit sharing = 1 : 1Priya’s income = x/2+2250Vijay’s income = x/2 – 2250 +14401 Priya = ½ VijayPriya Income = Twice of Vijay incomex/2 + 2250 = 2(x/2 – 2250 +1440)X+4500/2 = 2(x/2 – 810)X+4500/2 = x – 1620 = x +4500 = 2x – 3240X = 7740Total Profit of the year = 7740+1440 = 9,180
![Page 104: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/104.jpg)
Problem - 5
Revathy and Shiva are partners sharing profits in the ratio of 2:1. They admit Pooja into partnership giving her 1/5th share in profits which she acquires from Revathy and Shiva in the ratio of 1:2. Calculate the new profit sharing ratio?
![Page 105: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/105.jpg)
Solution
Pooja gets her share of 1/5th of total share of Profit from Revathy and Shiva in the ratio 1 : 2
From Revathy = 1/3*1/5 = 1/15From Shiva = 2/3*1/5 = 2/15Total Pooja share = 1/15+2/15 = 3/15 = 1/15Revathy share = 2/3 – 1/15 = 9/15Shiva share = 1/3 – 2/15 = 3/15Shares = Revathy : Shiva : Pooja = 3 : 1 : 1
![Page 106: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/106.jpg)
Problem - 6
A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined them after six months with an amount equal to that of B. In what proportion should the profit at the end of 1 year be distributed among A, B and C?
![Page 107: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/107.jpg)
Solution
Let the investment,
3 : 5 : 5
3*12: 5*12 : 5*6
36 : 60 : 30
6 : 10 : 5
![Page 108: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/108.jpg)
Problem - 7
If 4(A’s capital) = 6(B’s capital) = 10 (C’s capital) then out of a profit of rs.4650. Find C’s share?
![Page 109: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/109.jpg)
Solution
Let the unknown value be x
x/4 : x/6 : x/10
15x/60 : 10x/60 :6x/60
15 : 10 : 6
C’s share = 6/31*4650 = Rs. 900
![Page 110: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/110.jpg)
Problem - 8
A, B, C subscribe Rs.50,000 fro business. A subscribes Rs.4000 more than B and B Rs.5000 more than C. Out of total profit of Rs.35,000. Find A’s share?
![Page 111: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/111.jpg)
SolutionC = x, B = x + 5000A = x+5000+4000 = x + 9000 x +x+5000 +x+9000 = 500003x+14000 = 500003x = 50000 – 140003x = 36000,x = 12000C : B : A12000 : 17000 : 21000A = 35000*21/50 = 14,700
![Page 112: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/112.jpg)
Problem - 9
A and B are partners in a business, A contributes ¼ of he capital for 15 months and B received 2/3 of the profit. For how long B’s money was used?
![Page 113: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/113.jpg)
Solution
B = 2/3A = 1/3A : B = 1/3 : 2/3 = 1 : 2Investment 1/4x+15 : 3/4x*y15x/4 : 3xy/415x/4 : 3xy/4 : : 1 : 230x/4 = 3xy/4Y = 30x/4 * 4/3x = 10 months
![Page 114: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/114.jpg)
Problem - 10
A, B and C invests Rs.4,000, Rs.5,000 and Rs.6,000 respectively in a business and A gets 25% of profit for managing the business and the rest of the profit is divided by A, B and C in proportion to their investment. If in a year, A gets Rs.200 less than B and C together, what was the total profit for the year?
![Page 115: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/115.jpg)
SolutionTotal Profit = 10025% for managing the business = 100 – 25 = 75%A : B : C 4000 : 5000 : 6000 4 : 5 : 64x : 5x : 6x = 25x100*15x/75 = 20xA gets 4x + 25% of 20x
= 4x + 20x *25/100 = 9xB = 5x, C = 6x(5x + 6x) – 9x = 20011x – 9x = 2002x = 200; x = 100Total Profit 20x = 20*100 = 2000
![Page 116: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/116.jpg)
Problem - 11
A and B entered into partnership with capitals in the ratio of 4 : 5. After 3 months, A withdraw ¼ of his capital and B withdraw 1/5 of his capital. The gain at the end of 10 months was Rs.760. Find the share of B?
![Page 117: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/117.jpg)
SolutionA : B
4 : 5
4000 : 5000
A’s share = 4000*1/4 = 4000 – 1000 = 3000
B’s share = 5000*1/5 = 5000 – 1000 = 4000
A : B
3*4+3*7 : 5*3 +4*7
12 + 21 : 15+28
33 : 43
60*43/76 = 430
![Page 118: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/118.jpg)
Problem - 12
Rs. 1290 is divided between A, B and C. So, that A’s share is 1 ½ times B’s and B’s share is 1 ¾ times C. What is C’s share?
![Page 119: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/119.jpg)
Solution
A : B = 1 ½ : 1 = 3/2 : 1 = 3 : 2
B : C = 1 ¾ : 1 = 7/4 : 1 = 7 : 4
A:B :C =3*7(A) : 2*7(B) : 7*2(B) : 4*2(C)
= 21 : 14 : 8
B = 1290*8/43 = Rs.240
![Page 120: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/120.jpg)
Problem - 13
A man starts a business with a capital of Rs.90000 and employs an assistant. From the yearly profit he keeps an amount equal to 4 ½ of his capital and pay 35% of the remainder of the profits. Find how much the assistant receives in a year, in which profit is Rs.30,000.
![Page 121: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/121.jpg)
Solution
Investment = 90,000
4 ½ of investment = 9/2/100*90000 = Rs.4050
Profit = 30,000 – 4050 = 25,950
35/100*25,950 =9082.50
![Page 122: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/122.jpg)
Problem - 14
A and B invest in a business in the ratio 3 : 2. If
5% of the total profit goes to charity and A’s
share is Rs. 855, what is the total profit %?
![Page 123: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/123.jpg)
Solution
Let the total profit be Rs. 100
After paying charity A’s share = 3/5 *95 = 57
If A’s share is Rs. 57, the total profit is 100
If A’s share is Rs. 855, the total profit is
100 * 855/57 = Rs. 1500
The total profit = Rs. 1500
![Page 124: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/124.jpg)
Problem - 15
A,B,C entered into a partnership by making an
investment in the ratio of 3 : 5 : 7. After a year
C invested another Rs. 337600 while A withdrew
Rs. 45600. The ratio of investments then
changed into 24: 59 : 167. How much did A
invest initially?
![Page 125: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/125.jpg)
Solution
Solution:
Let the investments of A, B, and C be 3x, 5x, 7x
(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167
(3x – 45600)/5x = 24/59
x = 47200
Initial investment of A = 47200 * 3 = Rs. 141600
![Page 126: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/126.jpg)
Problems on Age
![Page 127: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/127.jpg)
Problem - 1
The age of the Father is 4 times the age of his Son. If 5 years ago, Father’s age was 7 times the age of his Son, what is the Father’s present age?
![Page 128: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/128.jpg)
Solution
F = 4S
F - 5 = 7(S - 5)
4S – 5 = 7S – 35
3S = 30
S = 10
Father’s age = 4* 10 = 40 years
![Page 129: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/129.jpg)
Problem - 2
The age of Mr. Gupta is four times the age of his Son. After Ten years, the age of Mr. Gupta will be only twice the age of his Son. Find the present age of Mr. Gupta’s Son.
![Page 130: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/130.jpg)
Solution
G = 4S
G + 10 = 2 ( S + 10)
4S + 10 = 2S + 20
2S = 10
S = 5
Son’s Age = 5 years
![Page 131: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/131.jpg)
Problem - 3
10 years ago Anu’s mother was 4 times older than her daughter. After 10 years, the mother will be twice as old as her daughter. Find the present age of Anu.
![Page 132: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/132.jpg)
SolutionTen years before:M – 10 = 4(A – 10 )M – 10 = 4A – 40M = 4A – 40 + 10M = 4A – 30 Ten Years After:M + 10 = 2(A + 10)M + 10 = 2A + 20M = 2A + 20 – 10 M = 2A + 104A – 30 = 2A + 102A = 10 + 302A = 40: Anu’s Age = 20
![Page 133: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/133.jpg)
Problem – 4
The sum of the ages of A and B is 42 years. 3 years back, the ages of A was 5 times the age of B. Find the difference between the present ages of A and B?
![Page 134: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/134.jpg)
SolutionA + B = 42 A = 42 – BA – 3 = 5 ( B – 3)A – 3 = 5B – 1542 – B – 3 = 5B – 1542 – 3 + 15 = 5B + B54 = 6BB = 54 /6 = 9A = 42 – B; A = 42 – 9 = 33Difference in their ages = 33 – 9 = 24 Years
![Page 135: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/135.jpg)
Problem - 5
The sum of the ages of a son and father is 56 years. After 4 years, the age of the father will be 3 times that of the son. Find their respective ages?
![Page 136: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/136.jpg)
SolutionF + S= 56S = 56 – FF + 4 = 3 (S + 4)F + 4 = 3 (56 – F + 4)F + 4 = 168 – 3F + 124F = 168 + 12 – 44F = 176 ; F = 44S = 56 – F ; S = 56 – 44 = 12Father Age = 44; Son Age = 12
![Page 137: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/137.jpg)
Problem – 6
The ratio of the ages of father and son at present is 6:1. After 5 years, the ratio will become 7:2. Find the Present age of the son.
![Page 138: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/138.jpg)
Solution
6x + 5/x + 5 = 7/2
12x + 10 = 7x + 35
12x – 7x = 35 – 10
5x = 25
x = 25 / 5
x = 5 years
Son age = 1* 5 = 5 years
![Page 139: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/139.jpg)
Problem - 7
The ages of Ram and Shyam differ by 16 years. Six years ago, Shyam’s age was thrice as that of Ram’s. Find their present ages?
![Page 140: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/140.jpg)
Solution
S = R + 16
S – 6 = 3(R – 6)
S – 6 = 3R – 18
R + 16 – 6 = 3R – 18
R + 10 = 3R – 18
2R = 28 ; R = 14
Shyam’s Age = 14 + 16 = 30.
![Page 141: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/141.jpg)
Problem - 8
A man’s age is 125% of what it was 10 years ago, 83 1/3% of what it will be after 10 years. What is his present age?
![Page 142: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/142.jpg)
Solution
Let the age be x
125% of (x – 10) = 83 1/3 % of (x +10)
125/100 * x – 10 = 250/ 300 * x +10
5/4 x – 10 = 5/6 x – 10
5x / 4 – 5x / 6 = 50/6 + 50/4
5x /12 = 250/12
5x = 250 ; x = 50 years
![Page 143: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/143.jpg)
Problem - 9
3 years ago, the average age of a family of 5 members was 17. A baby having born, the average age of the family is the same today. What is the age of the child?
![Page 144: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/144.jpg)
Solution
Average age of 5 members = 17Total age of 5 members = 17*5 = 85 3 years later, the age of 5 members will be = 85 + 15 = 100100 + x / 6 = 17100 + x = 17*6100 + x = 102x = 102 – 100 = 2 years
![Page 145: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/145.jpg)
Problem - 10
The sum of the age of father and his son is 100 years now. 5 years ago their ages were in the ratio of 2 : 1. The ratio of the ages of father and his son after 10 years will be?
![Page 146: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/146.jpg)
Solution
F + S = 1005 years ago 2 : 15 years agoF + S = 100 – 10 = 9090*2/3 = 60 : 30Present age = 65 : 35 10 years ago = 75 : 45 = 5 : 3
![Page 147: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/147.jpg)
Problem - 11
Six years ago, Sushil’s age was triple the age of Snehal. Six years later, Sushil’s age will be 5/3 of the age of Snehal. What is the present age of Snehal?
![Page 148: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/148.jpg)
Solution
Six years ago,Snehal = x; Sushil = 3xSix years later,3x + 6+6 = 5/3(x+6+6)9x +36 = 5x+604x = 60 – 36X = 6Present Age of Snehal = 6+6 = 12 years
![Page 149: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/149.jpg)
Problem - 12
Susan got married 6 years ago. Today her age is 1¼ times that at the time of her marriage. Her son is 1/6 as old as she today. What is the age of her son?
![Page 150: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/150.jpg)
Solution
6 years ago Susan got married.
So her son’s age will be less than 6 years.
Let as consider, her son’s age is 5 years.
Susan’s Age is 5*6 = 30 yrs, since the son is 1/6th of Susan’s age.
6 years ago her age must have been 24 yrs
24*1 ¼ = 24*5/4 = 30 yrs
As it satisfies the conditions her son’s age is 5 years
![Page 151: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/151.jpg)
Problem - 13
My brother is 3 years elder to me. My father was 28 years of age when my sister was born, while my mother was 26 years of age, when I was born. If my sister was 4 years of age when my brother was born, then, what was the age of my father and mother respectively when my brother was born?
![Page 152: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/152.jpg)
Solution
My brother was born 3 years before I was born and 4 years after my sister was born
Father’s age when brother was born
= 28 + 4 = 32 years
Mother’s age when brother was born
= 26 – 3 = 23 years
![Page 153: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/153.jpg)
Problem - 14
If 6 years are subtracted from the present age of Gagan and the reminder is divided by 18, then the present age of his grandson Aunp is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then what is Gagan’s present age?
![Page 154: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/154.jpg)
Solution
Anup’s age = 5 – 2 = 3 years
Let Gagan’s age be x
= x – 6 / 18 = 3
x – 6 = 3*18 ; x – 6 = 54
x = 54 + 6
Gagan’s age = 60
![Page 155: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/155.jpg)
Problem - 15
Ramu’s grandfather says, “ Ram, I am now 30 years older than your father. 15 years ago, I was 2½ times as old as your father”. How old is the grandfather now?
![Page 156: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/156.jpg)
SolutionLet the father’s age be x.Grandfather’s age will be 30 + x15 years ago,X + 30 – 15 = 5/2 (x – 15)X + 15 = 5/2 (x – 15)2x + 30 = 5x – 75105 = 3xX = 105 / 3 = 35Grandfather’s age = 35 + 30 = 65
![Page 157: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/157.jpg)
Average
![Page 158: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/158.jpg)
Average
Average = Sum of Quantities Number of Quantities
Sum of quantities= Average*Number of
Quantities.
Number of quantities = Sum of Quantities
Average
![Page 159: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/159.jpg)
Problem - 1
The average age of a class of 22 students is 21 years. The average increases by 1 when the teacher’s age is also included. What is the age of the teacher?
![Page 160: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/160.jpg)
Solution
Total age of the students be xx/22 = 21; x = 21*22= 462Teacher’s age is also includedx/23 = 22; x = 22*23 = 506Total age of 23 people – Total age of 22 peoplewill be the age of teacher506 – 462 = 44 yearsThe age of teacher = 44
![Page 161: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/161.jpg)
Problem - 2
The average of 7 numbers is 25. The average of first three of them is 20 while the last three is 28. What must be the remaining number?
![Page 162: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/162.jpg)
Solution
Average of 7 numbers = 25,
Sum of 7 numbers = 25* 7 = 175
Avg. of first three numbers = 20, 20* 3 = 60
Avg. of last three numbers = 28, 28*3 = 84
The 4th number = 175 – (60+84) = 175 – 144
= 31
![Page 163: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/163.jpg)
Problem - 3
The average age of a team of 10 people remains the same as it was 3 years ago, when a young person replaces one of the member. How much younger was he than the person whose place he took?
![Page 164: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/164.jpg)
Solution
Let Average be x
10 members’ Average = 10x
Average of 10 members (including new one) is same as it was 3 yrs ago.
Now 10*3 = 30 years have increased, so a person of 30 years should have replaced to keep the average as same.
![Page 165: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/165.jpg)
Problem - 4
The average age of a couple was 26 years at that time of their marriage. After 11 years of marriage the average age of the family with 3 children become 19 years. What is the average age of the Children?
![Page 166: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/166.jpg)
Solution
Average of parents ages is 26, sum= 26*2 = 52Parents age after 11 years = 52 +22 = 74Average age of Family = 19, Sum = 19*5 = 95Sum of family’s age – Sum of parents’ age= 95 – 74 = 21Sum of the ages of 3 children = 21,Average Age = 21/3 = 7 yrs
![Page 167: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/167.jpg)
Problem - 5
9 members went to a hotel for taking meals. Eight of them spent Rs. 12 each on their meals and the ninth person spent Rs. 8 more than the average expenditure of all the nine. What was the total money spent by them?
![Page 168: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/168.jpg)
Solution
Average = x/9
Amount Spent by 8 members = 12 * 8 = 96
96 + x/9 + 8 = x
104 = x – x/9
104 = 8x/9
8x = 104 *9 = 936
x = 936/8 = 117
![Page 169: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/169.jpg)
Problem - 7
A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th innings?
![Page 170: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/170.jpg)
Solution
17th innings avg. = x, Runs = 17x
16th innings avg. = x -3, Runs = 16 (x -3)
16 (x-3) + 87 = 17x
16x – 48 +87 = 17x
X = 39
![Page 171: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/171.jpg)
Problem - 7
There are 24 students in a class. One of them, who was 18 yrs old left the class and his place was filled up by the newcomer. If the average of the class thereby was lowered by one month, what is the age of the newcomer?
![Page 172: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/172.jpg)
Solution
Average reduced by 1 month,
24 * 1 = 2 years
So, the newcomer’s age is 18 -2 = 16 years
![Page 173: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/173.jpg)
Problem - 8
The average of marks in mathematics for 5 students was found to be 50. Later, it was discovered that in the case of one student the mark 48 was misread as 84. What is the correct average?
![Page 174: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/174.jpg)
Solution
Difference = 84 – 48 = 36
36 /5 = 7.2 (Increased)
The corrected average = 50 – 7.2 = 42.8
![Page 175: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/175.jpg)
Problem - 9
The average salary of all the workers in a factory is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. What is the total number of workers in the factory?
![Page 176: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/176.jpg)
Solution
Members Avg.
7 12000
X 6000
6x = 7*12
X = 7812/6 = 14
Total no. of workers = 7 + 14 = 21
![Page 177: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/177.jpg)
Problem - 10
Average salary of all the 50 employees including 5 officers of the company is Rs. 850. If the average salary of the officers is 2500, find the average salary of the remaining staff of the company.
![Page 178: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/178.jpg)
Solution
x/50 = 850; x = 42,500
5 officers’ salary = 2500*5 = 12500
50 – 5 members = 42500 – 12500
45 members = 30000
Avg. salary of 45 members = 30000/45
= 667(App)
![Page 179: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/179.jpg)
Problem - 11
Find the average of 8 consecutive odd numbers 21,23,25,27,29,31,33,35
![Page 180: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/180.jpg)
Solution
1st number + last Number /2
= 21 + 35 /2 = 28
![Page 181: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/181.jpg)
Problem - 12
A train covers 50% of the journey at 30 km/hr, 25% of the journey at 25 km/hr, and the remaining at 20 km/hr. Find the average speed of the train during entire journey.
![Page 182: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/182.jpg)
Solution
Total Journey = 100 km
S = Distance / Time = 100 / 5/3 + 1/1 + 5/4
= 100 * 12 /20+12+15
= 1200/47 = 25 25/47 km/hr
![Page 183: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/183.jpg)
Problem - 13
The average of 10 numbers is 7. What will be the new average if each number is multiplied by 8?
![Page 184: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/184.jpg)
Solution
If numbers are multiplied by 8,
Average also to be multiplied by 8
= 7*8 = 56
{or}
x/10 = 7
x = 10*7 = 70
= 70* 5 = 560 /10 = 56
![Page 185: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/185.jpg)
Problem - 14
The mean marks of 10 boys in a class is 70% whereas the mean marks of 15 girls is 60%. What is the mean marks of all 25 students?
![Page 186: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/186.jpg)
Solution
Boys = x/10 = 70 = 700
Girls = x/15 = 60 = 900
10 + 15 = 700 + 900
25 = 1600
1600/25 = 64%
![Page 187: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/187.jpg)
Problem - 15
Of the three numbers the first is twice the second and the second is thrice the third. If the average of the three numbers is 10, what are the numbers?
![Page 188: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/188.jpg)
SolutionA = 2xB = xC = x/32x + x + x/3/3 = 106x + 3x + x /9 = 106x + 3x + x = 9010x = 90 ; x = 9.A = 18, B = 9, C = 3
![Page 189: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/189.jpg)
Percentage
![Page 190: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/190.jpg)
Percentage
• By a certain Percent, we mean that many
hundredths.
• Thus, x Percent means x hundredths, written
as x%
![Page 191: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/191.jpg)
•Finding out of Hundred.
If Length is increased by X% and Breadth is decreased by Y% What is the percentage Increase or Decrease in Area of the rectangle?
Formula: X+Y+ XY/100 %
Decrease 20% means -20
Percentage
![Page 192: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/192.jpg)
Problem -1
When 75% of the Number is added to 75%, the result is the same number. What is the number?
![Page 193: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/193.jpg)
Solution
Percentage Number
75 x+75
100 x
100x + 7500 = 75x
25x = 7500
x = 300
![Page 194: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/194.jpg)
Problem - 2
A tank is full of milk. Half of the milk is sold and the tank is filled with water. Again half of the mixture is sold and the tank is filled with water. This operation is repeated thrice. Find the percentage of milk in the tank after the third operation?
![Page 195: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/195.jpg)
Solution
Milk Water
100 0
50 50(1st)
25 75 (2nd)
12.5 87.5 (3rd)
After 3 operation Milk 12.5%
![Page 196: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/196.jpg)
Problem 3
A large water-melon weighs 20kg with 96% of its weight being water. It is allowed to stand in the sun and some of the water evaporates so that now, only 95% of its weight is water. What will be its reduced weight?
![Page 197: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/197.jpg)
Solution
20 *96/100=19.2kg of water
Let the evaporated water be x
19.2-x=95%(20-x)
19.2-x=95(20-x)/100
1920-100x=1900-95x
5x=20 ;x=4
20-4=16kg.
![Page 198: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/198.jpg)
Problem 4
The population of a city is 155625. For every1000 men, there are 1075 women. If 40% of men and 24% of women be literate, then what is the percentage of literate people in the city?
![Page 199: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/199.jpg)
SolutionRatio of men and women=1000:1075=40:43
Number of men=40*155625/83=75000
Number of women=155625-7500=80625
Number of literate men=75000*40/100=3000
Number of literate women
=80625*24/100=19350
Literate people =30000+19350=49350
Percentage of literate people
=49350/155625*100=2632/83=31 59/83%
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Problem 5
300 grams of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?
![Page 201: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/201.jpg)
Solution
Grams Sugar
300 40%
X 50%
50x = 40*300
x = 40*300/50 = 240
300 – 240 =60 Kg
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Problem - 6
A man lost 12½% of his money and after spending 70% of the remainder, he has Rs. 210 left. How much did the man have at first?
![Page 203: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/203.jpg)
SolutionLet the amount be 100Then, 100.00 – 12.50 = 87.50
70% of 87.50 = 87.50 *70/100 =61.25
The remaining amount will be Rs. 26.25Initial Final100 26.25X 210
26.25x = 21000; x = 21000/26.25 = 800
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Problem - 7
During one year the population of a town increases by 10% and during next year it diminished by 10%. If at the end of the second year, the population was 89,100, what was the Population at the beginning of first year?
![Page 205: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/205.jpg)
Solution
Let the population be 1001st Year = 100 + 10 = 1102nd Year = 110 * 10/100 = 110 -11 = 99Percentage Population99 89100100 x99x = 89100*100;x = 8910000/99 = 90000
![Page 206: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/206.jpg)
Problem - 8
When a number is first increased by 20% and then again 20% by what percent should the increase number be reduced to get back the original number?
![Page 207: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/207.jpg)
Solution
Let the number be 100
20% increase = 100*20/100 = 20
New Value = 120
Again increase by 20% = 120*20/100 = 24
New value = 144
Increased amount = 44/144*100 = 30 5/9%
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Problem - 9
The number of students studying Arts, Commerce and Science in an institute were in the ratio 6 : 5 : 3 respectively. If the number of students in Arts, Commerce and science were increased by 10%, 30% and 15% respectively, what was the new ratio between number of students in the three streams?
![Page 209: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/209.jpg)
Solution
A : C : S
6 : 5 : 3
6x : 5x : 3x
6x*110/100 : 5x*130/100 : 3x*115/100
6x*110 : 5x*130 : 3x*115
660 : 650 : 345
132 : 130 : 69
![Page 210: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/210.jpg)
Problem - 10
In measuring the sides of rectangle errors of 5% and 3% in excess are made. What is the error percent in the calculated area?
![Page 211: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/211.jpg)
Solution
Area = xy
X = 5% Excess = 100* 5/100 = 105
Y = 3% Excess = 100*3/100 = 103
103*105/100 = 10815/100 = 108.15
Error – Actual = 108.15 – 100 = 8.15% Excess
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Problem - 11
In a certain examination there were 2500 candidates. Of them 20% of them were girls and rest were boys. If 5% of boys and 40% of girls failed, what was the Percentage of candidates passed?
![Page 213: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/213.jpg)
Solution
Girls = 2500*20/100 = 500Boys = 2500*80/100 = 2000Students who failed wereBoys = 2000*5/100 = 100Girls = 500*40/100 = 200Total Failed Students = 300Total Pass students = 2500 – 300 = 2200Pass Percentage = 2200/2500*100 = 88%
![Page 214: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/214.jpg)
Problem - 12
A person saves every year 20% of his income. If his income increases every year by 10% then his saving increases by?
![Page 215: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/215.jpg)
Solution
Every year saving, if the income is Rs. 100
= 100 *20/100 =Rs. 20
Salary increases = 110*20/100 = 22
Percentage increase (Savings) = 2/20*100 = 10%
![Page 216: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/216.jpg)
Problem - 13
On a test containing 150 questions carrying 1 mark each, meena answered 80% of the first answers correctly. What percent of the other 75 questions does she need to answer correctly to score 60% on the entire exam?
![Page 217: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/217.jpg)
Solution
Required correct answer = 150*60/100 = 90 Questions need to be correct.
80% of 75 questions = 60 q answered correctly.
Remaining 30 questions need be correct out of 75
= 30/75*100 = 40
![Page 218: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/218.jpg)
Problem - 14
A boy after giving away 80% of his pocket money to one companion and 6% of the remainder to another has 47 paise left with him. How much pocket money did the boy have in the beginning?
![Page 219: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/219.jpg)
SolutionLet the amount be 100To the first companion = 100*80/100 = 80 Remaining = 100 – 80 = 20To the 2nd person = 20*6/100 = 1.20The remaining = Rs.18.80 or 1880 paiseInitial Final100 1880X 471880x = 47*100x = 4700/1880 = 2.5
![Page 220: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/220.jpg)
Problem - 15
The length of a rectangle is increased by 10% and breath decreased by 10%. Then the area of the new rectangle?
![Page 221: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/221.jpg)
Solution
I – D – I*D /100
10 -10 – 10*10/100
0 – 1 = -1
Decrease by 1%
![Page 222: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/222.jpg)
Profit and Loss
![Page 223: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/223.jpg)
• Gain =(S.P.)-(C.P.)• Loss =(C.P.)-(S.P.)• Loss or gain is always reckoned on C.P.• Gain% = [(Gain*100)/C.P.]• Loss% = [(Loss*100)/C.P.]• S.P. = ((100 + Gain%)/100)C.P.• S.P. = ((100 – Loss%)/100)C.P.
Profit and Loss
![Page 224: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/224.jpg)
Problem - 1
A trade man allows two successive discount of 20% and 10%. If he gets Rs.108 for an article. What was its marked price?
![Page 225: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/225.jpg)
Solution
I1 + I2 – I1*I2/100
20 + 10 – 20*10 /100
= 28%
Discount = 28%, 72 Percent Cost is 108
Then 100percent cost = 72 108
100 x
100*108/72 = 150
![Page 226: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/226.jpg)
Problem - 2
A trade man bought 500 metres of electric wire at 75 paise per metre. He sold 60% of it at profit of 8%. At what gain percent should he sell the remainder so tas to gain 12% on the whole
![Page 227: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/227.jpg)
Solution 500* 60/100 = 3008 X
12300 200300 : 200 = 6 : 48 18
126 4Remainder at 18% Profit
![Page 228: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/228.jpg)
Problem - 3
A man purchased a box full of pencils at the rate of 7 for Rs. 9 and sold all of them at the rate of 8 for Rs. 11. in this bargains he gains Rs. 10. How many pencils did the box contains.
![Page 229: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/229.jpg)
Solution
LCM = 7 and 8 = 56
56 pencil cost price = 8*9 = 72
56 Pencil selling price = 7*11
Profit = 77 – 72 = Rs. 5 for 56 pencil
Rs. 5 for 56 pencil means , for Rs. 10 the pencils are 112
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Problem - 4
A cloth merchant decides to sell his material at the cost price, but measures 80cm for a metre. His gain % is?
![Page 231: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/231.jpg)
Solution
100 – 80 = 20 cm difference
Actual = 80
20/80*100 = 25% Gain
![Page 232: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/232.jpg)
Problem - 5
Sales of a book decrease by 2.5% when its price is hiked by 5%. What is the effect on sales?
![Page 233: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/233.jpg)
Solution
Let the sales be 100 – 2.5 = 97.5Profit = 100+5 = 105Sales Profit97.5 105100 X100x = 97.5*105x = 97.5*105/100 = 102.375100 – 102.375 = 2.375 = 2.4 profit (app)
![Page 234: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/234.jpg)
Problem - 6
A dealer buys a table listed at Rs.1500 and gets successive discount of 20% and 10%. He spends Rs. 20 on transportation and sells it at a profit of 10%. Find the selling price of the table.
![Page 235: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/235.jpg)
SolutionDiscount = 20+10 – 20*10/100 = 28%Actual price = 100 – 28 = 72 100 1500 72 x72*1500/100 = 1080Transport = 1080 +20 = 1100 100 1100 110 x1100*110/100 = 1210
![Page 236: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/236.jpg)
Problem - 7
A fridge is listed at Rs. 4000. due to the off season, a shopkeeper announces a discount of 5%. What is the S.P?
![Page 237: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/237.jpg)
Solution
= 4000*95/100 = 3800
![Page 238: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/238.jpg)
Problem - 8
If the cost price of 9 pens is equal to the S.P of 11 pens. What is the gain or loss?
![Page 239: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/239.jpg)
Solution
= 11 – 9 = 2
= 2/11*100 = 18 2/11% loss
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Problem - 9
A machine is sold for Rs.5060 at a gain of 10% what would have been the gain or loss percent if it had been sold Rs.4370?
![Page 241: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/241.jpg)
Solution
S.P = Rs.5060 = Gain = 10%
C.P = 100/110*5060 = 4600
IF S.P = Rs.4370 and C.P = Rs.4600
Loss = 230
Loss % = 230/4600 * 100 = 5% loss
![Page 242: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/242.jpg)
Problem - 10
A person purchased two washing machines each for Rs.9000. he sold one at a loss of 10% and other at a gain of 10%. What is his gain or loss?
![Page 243: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/243.jpg)
Solution
Each Rs.9000. one is 10% profit and other is 10% loss. So No profit and No loss
![Page 244: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/244.jpg)
Problem - 11
Four percent more is gained by selling an article for Rs.180, then by selling if for Rs.175. then its C.P is?
![Page 245: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/245.jpg)
Solution
Let the cost price = Rs. X
4% of x = 180 – 175 = 4x/100 = 5
4x = 500; x = 500/4 = 125
![Page 246: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/246.jpg)
Problem - 12
An article is sold at a profit of 20%. If it had been sold at a profit of 25%. It would have fetched Rs.35% more. The Cost Price of the article is?
![Page 247: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/247.jpg)
Solution
Let C.P = Rs. X
125% of x – 120% of x = 35
5% of x =Rs.35 = x = 35*100/5 = 700
C.P = Rs. 700
![Page 248: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/248.jpg)
Problem - 13
A reduction of 20% in the price of orange enables a man to buy 5 oranges more for Rs. 10. The price of an orange before reduction was,
![Page 249: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/249.jpg)
Solution
20% Rs. 10 = Rs.2
Reduced price of 5 oranges = Rs. 2
Reduced price of 1 oranges = 40 p
Original price = 40/ 1- 0.20 = 400/8 = 50 Paise
![Page 250: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/250.jpg)
Problem - 14
A man sells two horses for Rs.1475. The cost price of the first is equal to the S.P of the second. If the first is sold at 20% loss and the second at 25% gain. What is his total gain or loss? ( in rupees)
![Page 251: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/251.jpg)
Solution
Let cost price of 1st horse = S.P of 2nd = x
C.P of 2nd = S.P of 2nd * 100/125 = x*100/125 = 4x/5
S.P of 1st = C.P of 1st *80/100 = x*80/100 = 4x/5
Neither loss nor gain
![Page 252: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/252.jpg)
Problem - 15
Rekha sold a watch at a profit of 15%. Had he bought it at 10% less and sold it for Rs. 28 less, he would have gained 20%. Find the C.P of the Watch.
![Page 253: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/253.jpg)
Solution
C.P be Rs. XFirst S.P = 115% of x = 23x/20 and second C.P =
90% x = 9x/10Second S.P = 120% of 9x/10 = 120/100 * 9x/10
= 27x/25Given 23x/20 – 27x/25 = 28 = 115x – 108x/100
= 287x/100 = 28 = x = 28*100/7 = 400C.P = Rs.400
![Page 254: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/254.jpg)
Probability
![Page 255: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/255.jpg)
Probability
• Probability:
P(є) = n(є) / n(s)• (Addition theorem on probability:
n(AUB) = n(A) + n(B) - n(AB)• Mutually Exclusive:
P(AUB) = P(A) + P(B)• Independent Events:
P(AB) = P(A) * P(B)
![Page 256: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/256.jpg)
Problem - 1
Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting all face cards?
![Page 257: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/257.jpg)
Solution
n(E) = 52C4
n(S) = 12C4 = 12C4/52C4
![Page 258: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/258.jpg)
Problem - 2
Four persons are to be chosen at random from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man, 1 woman or 2 children?
![Page 259: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/259.jpg)
Solution
Total 3 M + 2 W + 4 C = 9 C 4 = 126
n (E) = 3C1 * 2C1 * 4C2 = 36
36/126 = 2/7
![Page 260: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/260.jpg)
Problem - 3
A word consists of 9 letters, 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability that more than one vowels will be selected?
![Page 261: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/261.jpg)
Solution
n(E) = 9C3 = 84
More than one Vowels. So,
2V +1C or 3 V
4C2 *5C1 + 4C3 = 34
= 34/84 = 17/42
![Page 262: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/262.jpg)
Problem - 4
A bag contains 10 mangoes out of which 4 are rotten. Two mangoes are taken out together. If one of them was found to be good, then what is the probability that the other one is also good?
![Page 263: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/263.jpg)
Solution
10 mangoes – 4 are rotten = 6 good mangoes
Getting good mangoes = 6C1/10C1 = 6/10
Getting second mango to be good = 5/9
1st and 2nd mangoes
6/10 *5/9 = 1/3
![Page 264: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/264.jpg)
Problem - 5
Out of 13 applicants for a job there are 5 women and 8 men. It is desired to select 2 persons for the job. What is the probability that at least one of the selected person will be a woman?
![Page 265: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/265.jpg)
Solution
n(E) = 13C2 = 78
n(S) = 1m and 1 w or 2 w
= 8C1*5C1 + 5C2 = 50
= 50/78 = 25/39
![Page 266: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/266.jpg)
Problem - 6
Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are queen?
![Page 267: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/267.jpg)
Solution
P(A) = Both are Black
P(B) = Both are Queen
P(AnB) = Both are queen and Black
P(A) = 26C2/52C2 = 325/1326
P(B) = 4C2 /52C2 = 6/1326
P(AnB) = 2C2 /52C2 = 1/1326
325/1326 + 6/1326 - 1/ 1326 = 55/221
![Page 268: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/268.jpg)
Problem -7
A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is 1/7 and the probability of wife’s selection is 1/5. Find the probability that only one of them is selected?
![Page 269: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/269.jpg)
Solution
Husband’s Selection = 1/7;
Not getting selected = 1 – 1/7 = 6/7
Wife’s selection = 1/5;
Not getting selected = 1 – 1/5 = 4/5
Only one of them is selected =
(Husband’s Selection + Wife Not getting selected) or (Wife’s selection + Husband’s Not getting selected)
= (1/7*4/5) + 1/5*6/7) = 2/7
![Page 270: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/270.jpg)
Problem - 8
Four persons are chosen at random from a group of 3 men, 2 women and 4 children. What is the chance that exactly 2 of them are children?
![Page 271: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/271.jpg)
Solution
3 + 2 + 4 = 9C4 = 126
4 members 2(M and W) + 2(boy)
5C2 + 4C2 = 60
= 60 / 126 = 10/21
![Page 272: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/272.jpg)
Problem - 9
Prakash can hit a target 3 times in 6 shots, Priya can hit the target 2 times in 6 shots and Akhilesh can hit the target 4 times in 4 shots. What is the probability that at least 2 shots hit the target?
![Page 273: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/273.jpg)
Solution
Prakash hitting = 3/6; not hitting = 3/6
Priya hitting = 2/6; not hitting = 4/6
Akilesh = 4/4 = 1
At least 2 shots hit target
= 3/6*4/6 + 3/6*2/6 = ½
![Page 274: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/274.jpg)
Problem - 10
There are two boxes A and B. A contains 3 white balls and 5 black balls and Box B contains 4 white balls and 6 black balls. One box is taken at random and what is the probability that the ball picked up may be a white one?
![Page 275: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/275.jpg)
Solution
(Box A is selected and a ball is picked up ) or (Box B is selected and a ball is picked up)
½*3/8 + ½*4/10 = 31/80
![Page 276: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/276.jpg)
Problem - 11
A bag contains 6 white balls and 4 black balls. Four balls are successively drawn without replacement. What is the probability that they are alternately of different colour?
![Page 277: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/277.jpg)
Solution
Suppose the balls drawn are in the order white, black, white, black…
= 6/10 *4/9*5/8*3/7 = 360/5040
Suppose the balls drawn are in the order black, white, black, white…
= 4/10*6/9*3/8*5/7 = 360/5040
360/5040 +360/5040 = 1/7
![Page 278: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/278.jpg)
Problem - 12
A problem in statistics is given to four students A, B, C and D. Their chances of solving it are 1/3, ¼, 1/5 and 1/6 respectively. What is the probability that the problem will be solved?
![Page 279: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/279.jpg)
Solution
A is not solving problem = 2/3,
B is not solving problem = ¾
C not solving problem = 4/5
D not solving problem = 5/6
2/3*3/4*4/5*5/6 = 1/3
All together the probability of solving the problem = 1 -1 /3 = 2/3
![Page 280: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/280.jpg)
Problem - 13
There are 8 questions in an examination each having only 2 answers choices ‘Yes’ or ‘No’. All the questions carry equal marks. If a student marks his answer randomly, what is the probability of scoring exacting 50%?
![Page 281: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/281.jpg)
Solution
Each questions having 2 ways of answering,
1 question = 2!........ 8 question = 2!
= 2!*2!*2!*2!*2!*2!*2!*2! = 256
To get 50%, 4 questions need to be correct,
8c4 = 8*7*6*5/1*2*3*4 = 70
= 70/256 = 35/128
![Page 282: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/282.jpg)
A group consists of equal number of men and
women. Of them 10% of men and 45% of
women are unemployed. If a person is randomly
selected from the group find the probability for
the selected person to be an employee.
Problem - 14
![Page 283: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/283.jpg)
Let the number of men is 100 and women be 100
Employed men and women = (100-10)+(100-45)
= 145
Probability = 145 / 200 = 29 / 40
Solution
![Page 284: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/284.jpg)
Problem - 15
The probability of an event A occurring is 0.5 and that of B is 0.3. If A and B are mutually exclusive events. Find the probability that neither A nor B occurs?
![Page 285: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/285.jpg)
Solution
It is Mutually exclusive events P(A n B)=0
Probability = 1 – ( P(A) + P (B) – P(A n B) )
= 1 – (0.5 + 0.3 – 0)
= 0.2
![Page 286: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/286.jpg)
Permutation and Combination
![Page 287: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/287.jpg)
Permutation and Combination
Permutation means Arrangement
Combination means Selection
![Page 288: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/288.jpg)
Permutation and Combination
• Permutations: Each of the arrangements which can be made by
taking some (or) all of a number of items is called permutations.
npr = n(n-1)(n-2)…(n-r+1)=n!/(n-r)!• Combinations: Each of the groups or selections which can be made
by taking some or all of a number of items is called a combination.
nCr = n!/(r!)(n-r)!
![Page 289: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/289.jpg)
Types
1. How many ways of Arrangement possible by using word SOFTWARE?
SOFTWARE = 8!
2. How many ways of arrangement Possible by using word SOFTWARE, vowels should come together.
SFTWR (OAE) = 6! * 3!
![Page 290: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/290.jpg)
Types
3. How many ways of Arrangement Possible by using word SOFTWARE, vowels should not come together?
SFTWR ( ARE)
Not together
= Total arrangement – Vowels together
= 8! – (5! * 3!)
![Page 291: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/291.jpg)
Types4. How many ways of arrangement possible by using
word MACHINE, so that vowels occupy only ODD places.
- - - - - - - (7 places)
MCHN (AIE) 4 Consonant and 3 vowels.
7 places = 4 ODD places, 3 EVEN places
Vowels = 4P3 = 4!
Consonant = 4P4 = 4!
Total Number of arrangement = 4!*4!
![Page 292: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/292.jpg)
Types
5. How many ways of arrangement possible by using word ARRANGEMENT
Letter’s Repetition = 2(A) 2(R) 2 (E) 2 (N)
= 11!/2!*2!*2!*2!
In a given problem, any letter is repeated more than once that should be divided with total number.
![Page 293: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/293.jpg)
Problem - 1
A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways this can be done, when at least 2 ladies are included?
![Page 294: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/294.jpg)
Solution
a. 2 ladies * 3 Gents
4C2 * 6 C3 = 120
b. 3 ladies * 2 Gents
4C3 * 6C2 = 60
c. 4 ladies * 1 Gent
4C4 *6C1 = 1*6 = 6
Total ways = 120 +60 +6 = 186
![Page 295: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/295.jpg)
Problem - 2
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
![Page 296: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/296.jpg)
Solution
Total places = 9Odd places = 5Even places = 4 4 even places occupied by 4 women
= 4P4 = 4! = 245 odd places occupied by 5 men
= 5P4 = 5! = 120Total ways = 120*24 = 2880 ways
![Page 297: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/297.jpg)
Problem - 3
A set of 7 parallel lines is intersected by another set of 5 parallel lines. How many parallelograms are formed by this process?
![Page 298: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/298.jpg)
Solution
Two parallel lines from the first set and any two from the second set will from a parallelogram.
7C2 *5C2 = 21 * 10 = 210
![Page 299: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/299.jpg)
Problem - 4
There are n teams participating in a football championship. Every two teams played one match with each other. There were 171 matches on the whole. What is the value of n?
![Page 300: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/300.jpg)
Solution
Total number of matches played = nC2
nC2 = 171
n(n-1)/2= 171
n2 – n – 342 = 0
(n+18) (n-19) = 0
n = 19
![Page 301: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/301.jpg)
Problem - 5
In an examination, a candidate has to pass in each of the 6 subjects. In how many ways can he fail?
![Page 302: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/302.jpg)
Solution
6C1 + 6C2 + 6C3 + 6C4+6C5+6C6
1 + 6 + 15 + 20 + 15 + 6 = 63 ways
![Page 303: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/303.jpg)
Problem - 6
In how many ways can a pack of 52 cards be distributed to 4 players, 17 cards to each of 3 and one card to the fourth player?
![Page 304: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/304.jpg)
Solution
17 cards can be given to 1st player = 52 C17
2nd player = 35C17
3rd player = 18C17
4th player = 1
= 52C17*35C17*18C17
= 52!/17!35! * 35!/17!*18! * 18!/17!*1!
= 52!/(17!)3
![Page 305: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/305.jpg)
A foot race will be held on Saturday. How many
different arrangements of medal winners are
possible if medals will be for first, second and
third place, if there are 10 runners in the race …
Problem - 7
![Page 306: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/306.jpg)
n = 10
r = 3
n P r = n!/(n-r)!
= 10! / (10-3)!
= 10! / 7!
= 8*9*10
= 720
Number of ways is 720.
Solution
![Page 307: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/307.jpg)
To fill a number of vacancies, an employer must
hire 3 programmers from among 6 applicants,
and two managers from 4 applicants. What is
total number of ways in which she can make her
selection ?
Problem - 8
![Page 308: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/308.jpg)
It is selection so use combination formula
Programmers and managers = 6C3 * 4C2
= 20 * 6 = 120
Total number of ways = 120 ways.
Solution
![Page 309: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/309.jpg)
Problem - 9
A man has 7 friends. In how many ways can
he invite one or more of them to a party?
![Page 310: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/310.jpg)
Solution
In this problem, the person is going to select his friends for party, he can select one or more person, so addition
= 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7
= 127
Number of ways is 127
![Page 311: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/311.jpg)
Problem - 9
Find the number of different 8 letter words
formed from the letters of the word EQUATION
if each word is to start with a vowel
![Page 312: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/312.jpg)
Solution
For the words beginning with a vowel, the first
letter can be any one of the 5 vowels, the
remaining 7 places can be filled by
7P7 = 5040
The number of words = 5 * 5040 = 25200
![Page 313: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/313.jpg)
Problem - 10
In how many different ways can the letters of the
word TRAINER be arranged so that the vowels
always come together?
![Page 314: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/314.jpg)
Solution
A,I,E can be arranged in 3! Ways
(5! * 3!) / 2! = 360 ways
![Page 315: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/315.jpg)
Problem - 11
In how many different ways can the letters of the Word DETAIL be arranged so that the vowels may occupy only the odd positions?
![Page 316: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/316.jpg)
Solution
___ ___ ___ ___ ___ ___
3P3 = 3! = 1*2*3 = 6
3P3 = 3! = 1*2*3 = 6
= 6*6 = 36
![Page 317: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/317.jpg)
Problem - 12
There are 5 red, 4 white and 3 blue marbles in a bag. They are taken out one by one and arranged in a row. Assuming that all the 12 marbles are drawn, find the number of different arrangements?
![Page 318: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/318.jpg)
Solution
Total number of balls = 12
Of these 5 balls are of 1st type (red), 4 balls are the 2nd type and 3 balls are the 3rd type.
Required number of arrangements = 12!/5!*4!*3!
= 27720
![Page 319: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/319.jpg)
Problem - 13
5 men and 5 women sit around a circular table, the en and women alternatively. In how many different ways can the seating arrangements be made?
![Page 320: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/320.jpg)
Solution
5 men can be arranged in a circular table in 4 ways = 24 ways
There are 5 seats available for 5 women they can be arranged in 5 ways
No. of ways = 5!*4! = 2880 ways
![Page 321: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/321.jpg)
Problem - 14
In a chess board there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.
![Page 322: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/322.jpg)
Solution
Solution:
9C2 * 9C2 = 1296
![Page 323: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/323.jpg)
Problem - 15
In how many ways can a cricket team of 11 players be selected out of 16 players, If one particular player is to be excluded?
![Page 324: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/324.jpg)
Solution
Solution:
If one particular player is to be excluded, then selection is to be made of 11 players out of 15.
15C11= 15!/( 11!*4!)=1365 ways
![Page 325: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/325.jpg)
Area and Volume
![Page 326: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/326.jpg)
Area and VolumeCube:
• Let each edge of the cube be of length a. then,
• Volume = a3cubic units
• Surface area= 6a2 sq.units.
• Diagonal = √3 a units.
![Page 327: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/327.jpg)
Cylinder:
• Let each of base = r and height ( or length) = h.
• Volume = πr2h
• Surface area = 2 πr h sq. units
• Total Surface Area = 2 πr ( h+ r) units.
Area and Volume
![Page 328: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/328.jpg)
Cone:
• Let radius of base = r and height=h, then
• Slant height, l = √h2 +r2 units
• Volume = 1/3 πr2h cubic units
• Curved surface area = πr l sq.units
• Total surface area = πr (l +r)
Area and Volume
![Page 329: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/329.jpg)
Sphere:
• Let the radius of the sphere be r. then,
• Volume = 4/3 πr3
• Surface area = 4 π r2sq.units
Area and Volume
![Page 330: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/330.jpg)
Circle: A= π r 2
Circumference = 2 π r
Square: A= a 2
Perimeter = 4a
Rectangle: A= l x b
Perimeter= 2( l + b)
Area and Volume
![Page 331: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/331.jpg)
Triangle:
A = 1/2*base*height
Equilateral = √3/4*(side)2
Area of the Scalene Triangle
S = (a+b+c)/ 2
A = √ s*(s-a) * (s-b)* (s-c)
Area and Volume
![Page 332: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/332.jpg)
Problem - 1
A rectangular sheet of size 88 cm * 35 cm is bent to form a cylindrical shape with height 35 cm. What is the area of the base of the cylindrical shape?
![Page 333: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/333.jpg)
Solution
The circumference of the circular region = 88 cm
2r = 88
r = 88*7/22*2 = 14 cm
Area of the base = r2 = 22/7*14*14 v= 616 cm2
![Page 334: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/334.jpg)
Problem - 2
The radius of the base of a conical tent is 7 metres. If the slant height of the tent is 15 metres, what is the area of the canvas required to make the tent?
![Page 335: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/335.jpg)
Solution
R = 7 m
L = 15 m
Area of Canvas required = Curved Surface Area of cone
rl = 22/7*7*15 = 330 sq.m
![Page 336: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/336.jpg)
Problem - 3
Three spherical balls of radius 1 cm, 2 cm and 3 cm are melted to form a single spherical ball. In the process, the material loss was 25%. What would be the radius of the new ball?
![Page 337: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/337.jpg)
SolutionVol. of sphere = 4/3 r3
Vol. of 3 small spherical balls = 4/3 ( 13+23+33)= 4/3 (1+8+27) = 4/3 (36) = 48Material loss = 25%Vol. of the single spherical ball = 48*75/100
= 48 * ¾ = 36 V = 4/3r3 = 36r3 = 36*3/4 = 27r = 3 cm
![Page 338: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/338.jpg)
Problem - 4
A rectangular room of size 5m(l)*4m(w)*3m(h) is to be painted. If the unit of painting is Rs. 10 per sq.m, what is the total cost of painting?
![Page 339: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/339.jpg)
Solution
Area of 4 walls = 2h(l+b)
The area to be painted includes the 4 walls and the top ceiling.
Area to be painted = 2h (l+b) +lb
= 2*3 (5+4) + 5*4
= 54+20 = 74 sq.m.
Total cost of painting = 74*10 = Rs.740
![Page 340: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/340.jpg)
Problem - 5
The radius of a sphere is r units. Each of the radius of the base and the height of a right circular cylinder is also r units. What is the ratio of the volume of the sphere to that of the cylinder?
![Page 341: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/341.jpg)
Solution
Vol. of sphere = 4/3r3 and Vol. of Cylinder = r2h = r3
Required Ratio = 4/3 r3: r3 = 4/3 : 1
= 4 : 3
![Page 342: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/342.jpg)
Problem - 6
A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each, are dropped in it and they sink down in the water completely. What will be the increase in the level of water in the jar?
![Page 343: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/343.jpg)
Solution
Radius of each ball = 1 cm
Vol. of 4 balls = 4* 4/3 (r)3 = 16/3 cm3
Vol. of water raised in the Jar = Vol. of 4 balls
Let h be the rise in water level, then
Area of the base *h = 16/3 *5*5*h = 16/3 H = 16/3*25 = 16/75 cm
![Page 344: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/344.jpg)
What is the cost of planting the field in the form
of the triangle whose base is 2.8 m and height
3.2 m at the rate of Rs.100 / m2
Problem - 7
![Page 345: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/345.jpg)
Area of triangular field = ½ * 3.2 * 2.8 m2
= 4.48 m2
Cost = Rs.100 * 4.48
= Rs.448..
Solution
![Page 346: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/346.jpg)
Problem - 8
Find the length of the longest pole that can be
placed in a room 14 m long, 12 m broad, and 8
m high.
![Page 347: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/347.jpg)
Solution
Length of the longest pole = Length of the diagonal of the room
= √(142 + 122 + 82)
= √ 404 = 20.09 m
![Page 348: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/348.jpg)
Area of a rhombus is 850 cm2. If one of its
diagonal is 34 cm. Find the length of the other
diagonal.
Problem - 9
![Page 349: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/349.jpg)
850 = ½ * d1 * d2
= ½ * 34 * d2
= 17 d2
d2 = 850 / 17
= 50 cm
Second diagonal = 50cm
Solution
![Page 350: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/350.jpg)
A grocer is storing small cereal boxes in large
cartons that measure 25 inches by 42 inches by 60
inches. If the measurement of each small cereal
box is 7 inches by 6 inches by 5 inches then what
is maximum number of small cereal boxes that can
be placed in each large carton ?
Problem - 10
![Page 351: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/351.jpg)
No. of Boxes = 25*42*60 / 7*6*5 = 300
300 boxes of cereal box can be placed.
Solution
![Page 352: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/352.jpg)
Problem - 11
If the radius of a circle is diminished by 10%,
what is the change in area in percentage?
![Page 353: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/353.jpg)
Solution
= x + y + xy/100
= -10 - 10 + 10*10/100
= -19%
Diminished area = 19%.
![Page 354: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/354.jpg)
Problem - 12
A circular wire of radius 42 cm is bent in the
form of a rectangle whose sides are in the ratio
of 6:5. Find the smaller side of the rectangle?
![Page 355: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/355.jpg)
Solution
length of wire = 2 πr = (22/7*14*14)cm
= 264cm
Perimeter of Rectangle = 2(6x+5x) cm
= 22xcm
22x =264 x = 12 cm
Smaller side = (5*12) cm = 60 cm
![Page 356: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/356.jpg)
Problem - 13
A beam 9m long, 40cm wide and 20cm deep is made up of iron which weights 50 kg per cubic metre. Find the weight of the Beam.
![Page 357: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/357.jpg)
Solution
Vol. of the Beam = lbh = 9*40/100*10/100
= 72 m3
Weight of the iron beam is given as lm3 = 50 kg
72/100 m3 = 72/100*50 = 36 kg
![Page 358: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/358.jpg)
Problem - 14
If the length of a rectangle is reduced by 20%
and breadth is increased by 20%. What is the
percentage change in the area?
![Page 359: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/359.jpg)
Solution
x + y + (xy/100)%
= - 20 + 20 – 400/100
= -4
The area would decrease by 4%
![Page 360: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/360.jpg)
Problem - 15
Find the number of bricks measuring 25 cm in length, 5 cm is breadth and 10 cm in height for a wall 40 m long, 75 cm broad and 5 metres in height?
![Page 361: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/361.jpg)
Solution
Vol. of the wall = 40*72/100*5 = 150 m3
Vol. of 1 bricks = 25/100*5/100*10/100
= 1/80 m3
Number of bricks required = 150/1/800
= 150*800
= 120000
![Page 362: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/362.jpg)
Calendar
![Page 363: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/363.jpg)
CalendarOdd days:
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
![Page 364: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/364.jpg)
CalendarMonth code: Ordinary year
J = 0 F = 3
M = 3 A = 6
M = 1 J = 4
J = 6 A = 2
S = 5 O = 0
N = 3 D = 5
Month code for leap year after Feb. add 1.
![Page 365: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/365.jpg)
Calendar
Ordinary year = (A + B + C + D )-2
-----------------------take remainder
7
Leap year = (A + B + C + D) – 3
------------------------- take remainder
7
![Page 366: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/366.jpg)
Problem - 1
11th January 1997 was a Sunday. What day of the week on 7th January 2000?
![Page 367: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/367.jpg)
Solution
11th Jan 1997 = Sunday
11th Jan 1998 = Monday
11th Jan 1999 = Tuesday
11th Jan 2000 = Wednesday
7th Jan 2000 is on Saturday
![Page 368: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/368.jpg)
Problem - 2
What day of the week was on 5th June 1999?
![Page 369: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/369.jpg)
Solution
A+B+C+D – 2 / 7
A = 1999/7 = 4
B = 1999/4 = 499/7 = 2
C = June = 4
D = 5/7 = 5
= 4+2+4+5 – 2/7 = 13/7 = 5 = Saturday
![Page 370: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/370.jpg)
Problem - 3
On what dates of August 1988 did Friday fall?
![Page 371: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/371.jpg)
Solution
A = 1988 / 7 = 0
B = 1988/4 = 497/7 = 0
C = 3
D = x
0+0+3+x+3/7 = x/7 = 5(Friday)
Friday falls on = 5,12,19,26
![Page 372: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/372.jpg)
Problem - 4
India got independence on 15 August 1947. What was the day of the week?
![Page 373: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/373.jpg)
Solution
A = 1947/7 = 1
B = 1947/4 = 486/7 = 3
C = 15/7 = 1
D = 2
1+3+1+2 – 2 /7 = 5/7 = Friday
![Page 374: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/374.jpg)
Problem - 5
7th January 1992 was Tuesday. Find the day of the week on the same date after 5 years. i.e on 7th January 1997.
![Page 375: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/375.jpg)
Solution
7th January 1992 = Tuesday
7th January 1993 = Thursday (Leap)
7th January 1994 = Friday
7th January 1995 = Saturday
7th January 1996 = Monday ( Leap)
7th January 1997 = Tuesday
![Page 376: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/376.jpg)
Problem - 6
The first Republic day of India was celebrated on 26th January 1950. What was the day of the week on that date?
![Page 377: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/377.jpg)
Solution
A = 1950/7 = 4
B = 1950/4 = 487/7 = 4
C = 0
D = 26/7 = 5
4+4+0+5 – 2/7 = 11/7 = 4 = Thursday
![Page 378: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/378.jpg)
Problem - 7
Find the Number of times 29th day of the month occurs in 400 consecutive year?
![Page 379: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/379.jpg)
Solution
1 year = 1 (Ordinary Year)
1 year = 12 (Leap Year)
400 years = 97 leap year
97 * 12 = 1164
303*11 = 3333
= 1164+3333 = 4497 times
![Page 380: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/380.jpg)
Problem - 8
If 2nd March 1994 was on Wednesday, 25 Jan 1994 was on,
![Page 381: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/381.jpg)
Solution
A = 1994/7 = 6
B = 1994/4 = 498/7 = 1
C = 0
D = 25/7 = 4
= 6 + 1 + 0 + 4 – 2 / 7 = 3 = Tuesday
![Page 382: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/382.jpg)
Problem - 9
Calendar for 2000 will serve also?
![Page 383: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/383.jpg)
Solution
= 2000 + 2001 + 2002 + 2003 + 2004
= 2 + 1 + 1 + 1 + 2 = 7 (Complete Week)
2005
![Page 384: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/384.jpg)
Problem - 10
If Pinky’s 1st birthday fell in Jan 1988 on one of the Monday’s, the day on which are was born is,
![Page 385: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/385.jpg)
Solution
Jan = 1988 = Monday
Jan = 1987 = Sunday
![Page 386: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/386.jpg)
Problem - 11
Akshaya celebrated her 60th birthday on Feb 24, 2000. What was the day?
![Page 387: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/387.jpg)
Solution
A = 2000 /7 = 7
B = 2000/4 = 500/7 = 3
C = 3
D = 24/7 = 0
= 7+3+3+0-3/7 = 10/7 = 3 = Wednesday
![Page 388: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/388.jpg)
Problem - 12
On what dates of April 2008 did Sunday Fall?
![Page 389: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/389.jpg)
Solution
Calculate for 1st April 2008A = 2008/7 = 6B = 2008/4 = 502/7 = 5C = 1/7 = 1D = 0= 6+5+1+0 – 3/ 7 = 2 = Tuesday1st April on Tuesday, then 1st Sunday fall on 6.Sunday falls on 6, 13, 20, 27.
![Page 390: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/390.jpg)
Problem - 13
Today is Friday. After 62 days it will be,
![Page 391: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/391.jpg)
Solution
62 / 7 = 6 days after Friday then it will be Tuesday
![Page 392: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/392.jpg)
Problem - 14
What will be the day of the week on 1st Jan 2010?
![Page 393: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/393.jpg)
Solution
A = 1
B = 5
C = 0
D = 1
= 1+5+0+1 – 2/ 7 = 5/7 = 5 = Friday
![Page 394: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/394.jpg)
What is the day of the week on 30/09/2007?
Problem - 15
![Page 395: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/395.jpg)
Solution:
A = 2007 / 7 = 5
B = 2007 / 4 = 501 / 7 = 4
C = 30 / 7 = 2
D = 5
( A + B + C + D )-2
= -----------------------
7
= ( 5 + 4 + 2 + 5) -2
----------------------- = 14/7 = 0 = Sunday
7
Calendar
![Page 396: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/396.jpg)
Clock
![Page 397: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/397.jpg)
Clocks
Clock:
Angle between hour hand and minute hand = (11m/2) – 30h
Angle between minute hand and hour hand =30h – (11m/2)
![Page 398: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/398.jpg)
Problem - 1
What is the angle between the minute hand and hour hand when the time is 2.15?
![Page 399: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/399.jpg)
Solution
q = 11 m/2 – 30(h)
= 11 15/2 – 30(2)
= 11(7.5) – 60
= 82.5 – 60 = 22 1/2
![Page 400: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/400.jpg)
Problem - 2
At what time between 5 and 6 o’clock the hands of a clock coincide?
![Page 401: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/401.jpg)
Solution
Coinciding Angle = 0
Min. hand to hour hand = 25 min apart
60/55*25 = 12/11 * 25 = 300/11
= 27 3/11min past 5
![Page 402: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/402.jpg)
Problem - 3
At what time between 12 and 1 o’clock both the hands will be at right angles?
![Page 403: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/403.jpg)
Solution
Right angle = 90 degreesq = 30(h) – 11 m/2
90 = 30(12) – 11 m/2
180 = 360 – 11m
11m = 360 – 180
M = 180/11
16 4/11 past 12
![Page 404: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/404.jpg)
Problem - 4
Find at what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but not together?
![Page 405: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/405.jpg)
Solution
Minute hand to hour hand = 35 min apart
Straight line not together = 30 min apart
Difference = 35 – 30 = 5 min
= 60/55*5 = 12/11*5 = 60/11
= 55 5 / 11 past 7
![Page 406: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/406.jpg)
Problem - 5
At what time between 5 and 6 are the hands of the clock 7 minutes apart?
![Page 407: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/407.jpg)
Solution
7 min space behind the hour hand:
25 min – 7 min = 18 min
60/55 *18 = 216/11 = 19 7/11 min past 5
7 Min space ahead the hour hand
25 min + 7 min = 32 min
60/55*32 = 12/11*32 = 384/11
= 34 10/11 min past 5
![Page 408: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/408.jpg)
Problem - 6
A clock strikes 4 and takes 9 seconds. In order to strike 12 at the same rate what will be the time taken?
![Page 409: Numerical reasoning I](https://reader038.fdocuments.in/reader038/viewer/2022102502/55556e9fb4c9055f5f8b4804/html5/thumbnails/409.jpg)
Solution
Strike Sec
3 (interval) 9
11 x
3x = 11*9
X = 11*9/3 = 33 Sec