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In this Chapter following topics will be covered
Introduction
LCM and HCF
Units Digits and Remainders
Problems on Simple equations
This is the most important chapter in
quantitative aptitude for any competitive exams.
From this chapter, questions will come from
simplifications, divisibility rules, finding
remainders, finding unit digits (or last digits),
problems based on LCM and HCF…etc.
Natural numbers: Numbers which are used for
counting, it means all positive integers are natural
numbers. Examples are 1, 2, 3, and so on.
Based on divisibility there are two types of natural
numbers: Prime and Composite.
Prime Numbers: A number which doesn’t have
factors other than one.
Number 6 factors, are 1, 2 and 3. Factors for 8 are 1,
2 and 4 and factors for 7, only 1 and itself.
Therefore, 7 is a prime number.
Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,
and 97 so on. The only even number which is a
prime number is 2.
Note: For every prime number greater than 3,
generally will be in the form of 6 1n or 6 1n ,
where n is integer, but vice versa is not true, i.e.
numbers which are in the form 6 1n or 6 1n need
not be a prime.
To check given number is prime number or not:
Example: 181
Take a square root of the number, for 181, square
root lies between 13 and 14. Generally we round of
the square root to lower integer, i.e. 13.
Now, we note down all the prime numbers which are
less than 13 are 2, 3, 5, 7 and 11. If any of above
prime number divides given number i.e. 191, then
we say given number is not a prime.
If n is a prime number then, 2n – 1 is also a prime
number. 219
– 1 is prime since 19 is a prime number.
Composite Numbers: Any number other than prime
numbers called as composite numbers. Examples are
2, 4, 8, 10.so on.
Divisibility rules:
Divisibility by 2: Any number ends with even
number is divided by 2.
Examples are 298, 302, 1024 etc.
Divisibility by 3: If Sum of the digits in a number is
multiple of 3, then number is divided by 3. Example,
take a number 1236. Sum of the digits is 1 + 2 + 3 +
6 = 12, which is a multiple of 3 is so number is
divided by 3.
Check for 54749 and 987.
Divisibility by 4: A number formed with last two
digits in a number if it is divided by 4 then number
is divided by 4.
Example: 18548. Check if 48 is divided by 4, then
18548 is divided by 4.
Divisibility by 5: If a number ends with zero or five
then number is divided by 5. (eg.15, 20, 1255...).
Divisibility by 6: If a number is divided by both 2
and 3, then number is divided by 6.
Divisibility by 7: If the difference between the
number of tens in the number and twice of the unit
digit is 0 or divisible by 7, then number is divided by
7.
Take a number 987, double the unit digit is7 2 =14difference is98 14 84 , check if 84 is divided by
7. So number 987 is divided by 7.
Divisibility by 8: A number formed with last three
digits in a number if it is divided by 8 then number
is divided by 8.
The number 4816 is divisible by 8 because 816 is
divided by 8.
Divisibility by 9: A number divisible by 9, if sum of
the digits of a number is multiple of 9. Take an
example 60813, sum of the digits is 6 + 0 + 8 + 1 + 3
=18, is a multiple of 9.
Divisibility by 10: A number should end with 0.
Divisibility by 11: If sum of the alternate digits is
same or differ by multiples of 11. Take 1430,
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1 3 4 and sum of the other digits is also 4 (= 4 +
0) so number is divided by 11.
Take 587543, sums are 5 7 4 8 5 3 16
equal here so number is divided by 11. Check for
398849 and 5477319.
Divisibility by 19: If the addition of the number of
tens in the number and twice of the unit digit is
divisible by 19, then number is divided by 19. Take
684, 68 + 24 =76 is divisible by 19. If we have a
big number we continue above process step by step.
Whole numbers: Natural numbers and zero called
as whole numbers which are non-negative integers.
Real numbers: A real number is a value that
represents any quantity along a number line (number
line represents numbers from - to + ). Because
they lie on a number line, their size can be
compared. You can say one is greater or less than
another, and do arithmetic with them.
Rational numbers: A rational number is one that
can be represented as the ratio of two integers. For
example all the numbers below are rational:
1 3, ...
2 5
Irrational numbers: An irrational number cannot
be written as the ratio of two integers. Examples,
Pi = 3.14, e = 2.718, 2 =1.4142…
Simplify the following:
1. 89 + 19 + 39 + 97 + 78 + 101=?
2. 87 95 89 98 99 ?
3. 989 + 992 – 997 + 991 + 979 =?
Solutions:
1. 90 – 1 + 20 – 1 + 40 – 1 + 100 – 3 + 80 – 2
+ 100 +1 = 90 + 20 + 40 + 100 + 80 + 100 –
7 = 423
2. 90 3 90 5 90 1 + 90 + 8 + 90 + 9
= 468 Or 5 100 13 5 11 2 1
= 50032 =468
3. 1000 11 1000 8 (1000 3)
+1000 9 +1000 21 4000 1000 46 = 2954.
Example 1: 7845 + 3479 + 6432 + 5894 =?
A 24655 B 23650 C 23655 D 24660
Solution: Start with unit digits
5 9 2 4 20 0 carry 2
So, options A and C are eliminated.
4 7 3 9 carry 2 = 25
5 carry 2
Last 2 digits of the answer will have 50, option B is
correct. If one of the options is “none of these” then
we need to do addition till the last step.
Example 2: 9548 + 7314 = 8362 +?
A 8230 B 8410 C 8500 D 8600
Solution: Consider unit digits on both the sides.
8 + 4 = 12 = 2 +? so last digit of the answer should
be 0. Going to next step, i.e. adding the tens digits,
Then, 4 + 1 + carry 1 from previous step = 6 = 6 +?
So tens digit of the answer should be 0. Options A
and B are eliminated. Add the digits of the hundreds
place, 5 + 3 = 8 = 3 +?. 5 should be there. So option
C is correct.
Practice Set – 1:
1. What should be the maximum value of B in the
following equation, 5A9 – 7B2 + 9C6 = 823 A
6 B 5 C 7 D 9
2. When 335 is added to 5A7, the result is 8B2
which is divisible by 3. What is largest possible
value of A? A 8 B 2 C 1 D 4
3. If A381 is divisible by 11, find the value of
smallest number A? A 1 B 2 C 5 D 7
4. If x and y are the integers such that (3x + 7y) is a
multiple of 11, which of the following will also
be divisible by 11?
A 4x + 6y B x + y C 9x + 4y D 4x – 9y
5. What least number should be subtracted from
28038, so that the remaining will be divisible by
15? A 8 B 3 C 5 D 6
6. If a boy multiplied 36 by a number and obtains
8674 as his answer which is wrong. What is
correct answer? A 8672 B 8676 C 8678 D 8680
7. In four consecutive prime numbers, there are in
ascending order, product of first three is 385,
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and that of last three is 1001. The largest given
prime number is? A 17 B 19 C 11 D 13
8. Product of three consecutive numbers is whose
sum is 15? A 120 B 150 C 125 D 210
Suppose 33 divided by 5, 3 is the remainder. So we
write as 33 = 56 + 3.
In a division sum, dividend (N) = divisor (d)
quotient (q) + remainder (r).
Number = dq + r.
Example 3: If a number is divided by 4 gives
remainder 3, find the remainder if twice of that
number divided by 4? (S.S.C 2012)
Solution: Let number is n.
n = 4q + 3. 2n/ 4 = (24q + 23) = (8q + 6)/ 4.
Remainder is 2. If square of that number is divided
by 4, n2 = (4q + 3)
2 = 16q
2 + 24q + 9, remainder is 1
(since 9/4 gives remainder 1).
Example 4: Two numbers 19 and 14 are divided by
a number n leaves the same remainder. Find the
value of n? A 5 B 6 C 4 D 7
Solution: This kind of problems you can check with
options. Take the difference between 19 and 14, i.e.
19 – 14 = 5. It means if 14 and 19 divide by 5 or its
factors, leaves the same remainders.
9. In a division sum, the divisor is 3 times the
quotient and 6 times the remainder, if the
remainder is 2, then dividend? A 36 B 28 C 50
D 48
10. A number is divided by 72 gives 12 remainder
as, find the remainder if the same number is
divided by 9? A 5 B 6 C 3 D 4
11. If a number is divided by a divisor remainder is
24, if twice of the number is divided by same
divisor remainder will be 11. Find the divisor?
A 25 B 31 C 37 D 32
12. Two numbers 79 and 54 are divided by a
number n leaves the same remainder. Find the
value of n? A 10 B 5 C 25 D both B and C
13. Two numbers 4794 and 3378 are divided by a
number n leaves the same remainder. Find the
value of n? A 472 B 365 C 452 D none
Let’s take a 2 digit number, 43, reverse the digits
and subtract the number from original number.
43 – 34 = 9 (= 91).
Take one more example 72, 72 – 27 = 45 (= 9 5).
Example 5: Difference between a two digit number
and number obtained by reversing the digits of
actual number is always divisible by? (S.S.C 2013)
A 11 B 10 C 9 D None
Solution: Assume number is XY number obtained by
reversing the digits is YX.
So (10X + Y) – (10Y + X) = 9 (X – Y).
So difference always equals to 9 multiple.
14. Sum of the digits of a two digit number is 10,
while when the digits are reversed, number
decreases by 54. Find the changed number?
(S.S.C 2013) A 28 B 19 C 82 D 46
15. Raju had to do a multiplication, instead of taking
34 as one of the multipliers, he took 43. As a
result, the product went up by 540. What is
correct answer? A 2020 B 2040 C 1860 D None
16. How many two digit numbers are there to give a
perfect square number when they added to
number formed by reversing the digits? A 4 B 8
C 12 D 10
17. The difference between 791 and its reverse is
divided by 99, and then quotient is?
A 0 B 4 C 3 D 6
Important observation is whenever, if take the
difference between a number and number
obtained by reversing the digits of original
number is always 9 multiple.
Difference between a 3 digit number and
number obtained by reversing the digits of
actual number is, like 783 – 387 = 396 (= 99 x
4).
Let take a 3 digit number, XYZ, value of it is
100X + 10Y + Z. Number obtained by
reversing the digits is ZYX, value is 100Z +
10Y + X.
Difference is, (100X + 10Y + Z) – (100Z +
10Y + X) = 99 (X – Z) = 99 multiple.
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Squares:
Try to remember square of numbers from
1to 20. Procedure to find square of numbers ending
with 5, take some examples 15, 25, 35, 75, 95,
375…etc.
Example 6: Find the value of275 ?
Solution:
Step1: The last 2 digits of the answer is, 25 = 25.
Step2: Remaining part of the answer is, take 7 and
multiply it with next number i.e. 8, so it’s 78 = 56.
Join the two parts 25 and 56. 275 5625 .
It works for any number ending with 5. 235 1225 (Since 34 = 12) 295 9025 (Since 910 =90)
2195 38025 (Since 1920 = 380) 2 236 (35 1) 1225 1 2 35 1 = 1296
2 264 (65 1) 4225 1 2 65 1
2 289 (90 1) 8100 1 2 90 1
2 2 2 272 (70 2) ,61 (60 1)
2 287 (90 3) Or
2(85 2) .
Example 7: Find the squares of
A. 104
B. 109
C. 113
D. 98
Solutions:
A. Take 100 as base, since 104 is close to 100.
Difference is 104 – 100 = 4.The last two digits
of the answer is24 16 .Remaining part of
the answer is = 104 + 4 (difference) = 108 so
final answer is 10816.
B. Difference is 109 – 100 = 9.The last two digits
of the answer is 29 81 =109 + 9 = 118 so
final answer is 11881.
C. 113 – 100 = 13213 169 . The last two
digits of the answer is 69 and 1 goes as carry
for the next step. Remaining part is = 113 + 13
= 126 + carry (i.e.1) =127, so final answer is
=12769.
D. Difference is 98 – 100 = – 2. The last digit in
answer is – 22 = 4. = 98 + (–2) = 96 so answer
is 9604.
Square root of a perfect square number
Perfect square number ends with digits 0, 1, 4, 9, 6,
and 5 only. (12 = 1, 2
2 = 4, 3
2 = 9, 4
2 = 16, 5
2 = 25,
62 = 36, 7
2 = 49, 8
2 = 64, 9
2 = 81, 10
2 = 100)
Example 8: 2209 ?
A 43 B 47 C 57 D 53
Solution: Divide 2209 into two parts as 22 and
09. Number ending with 9 so last digit of the answer
is may be 3 or 7 (since23 09 2, 7 49 )
Take remaining part 22 which is 2 24 22 5 so we
can take 4. From above we can say answer is 43 or
47, but 2209 is close to 452 = 2025 so final answer is
47 (2209 > 2025(=452)
).
Get the answers:
A. Find the 7056 =?
B. Find the 4489 =?
C. Find the 2809 5929 ?
D. 17689 1849
?3249 1089
Square root of a non-perfect square number
Example 9:
1. 39 =?
2. 500 =?
3. 3125 =?
4. 93500 ?
5. 38472148 =?
Solutions:
1. We know 26 36 and
27 49 so answer
must be in the range of 6 to 7. Now start with
guessing. Let’s start with 6 divide 39 by 6.
396.5
6 and take averageof 6 and 6.5.
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6.5 66.25
2
. So 39 is 6.25 (Very
close approximate value).
2. We know that, 220 400 and
225 = 625 so
initial guess is 20.
500 20 2525 22.5
20 2
.
3. 2 255 3025,60 3600 , so expected
answer will be in the range of 55 to 60 and
that too close to 55, therefore start with
guessing 55.
3125 56.8181 5556.8181
55 2
55.90
4. Divide this into three parts as 9, 35 and 00.
For 9 take 3, for 35 take 0 and for 00 take 0
therefore our initial guess is 300. Divide
93500 by 300.
93500 311.666 300311.666
300 2
305.833
5. Divide this number into two digit groups as
48, 21, 47 and 38. For 38 take 6, for 47 take 0,
for 21 take 0, for 48 take 0 therefore divide the
given number by 6000.
384721486412.014
6000
6412.014 6000
2
= 6206.0123
Get the answers:
1. 270 ?
2. 380 132 ?
3. 1150 2864 ?
Example: 2 241 40 81 = (41 + 40) (41 40).
18. Difference between squares of two consecutive
numbers is 71, find the numbers?
A 35 and 34 B 36 and 37 C 36 and 35 D None
Example 10: Difference between the squares of two
consecutive odd integers is always divisible by
(MAT 2003) A 3 B 6 C 7 D 8
Solution: Take 5 and 7, 49 – 25 = 24.
Take 3 and 5, 25 – 9 = 16. So 8 is the answer.
Or 2x is even number, 2x + 1 is odd number, next
odd number will be 2x + 3. =2 2(2 3) (2 1)x x =
12x + 9 – 4x – 1= 8 (x 1)
Difference is multiple of 8.
Cubes
Try to remember cubes of numbers only from 1
to 9.
1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
8³ = 512
9³ = 729
10³ = 1000
One interesting property, please note the last digits
(unit digit for each numbers cube).
1³= 1 => last digit is 1
2³= 8 => last digit is 8
3³= 27 => last digit is 7
4³= 64 => last digit is 4
5³= 125 => last digit is 5
6³= 216 => last digit is 6
7³=343 => last digit is 3
8³= 512 => last digit is 2
9³= 729 => last digit is 9
10³= 1000 => last digit is 0
These are very useful in finding cube root of
a perfect cube number. If last or unit digit of the
perfect cube number is 0, then last digit of the cube
root is 0, if last digit of the perfect cube number is 2,
the last digit of the cube root is 8. (83
= 512, cube
ends with 2)
19. The least number which should be multiplied to
243 to get a perfect cube is (S.S.C 2012) A 2 B 3
C 6 D 9
Difference between squares of two
consecutive numbers always equals to sum
of the numbers.
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Key
Set -1: 1 C 2 D 3 D 4 D 5 B 6 B 7 D 8 A 9 C 10 C
11 C 12 D 13 A 14 C 15 B 16 B 17 D 18 C19 B
Finding cuberoot of a perfect cube number.
Example 11:
A. 3 9261 =? A 21 B 31 C 41 D 29
B. 3 12167 =? A 23 B 33 C 27 D 29
C. 3 51653 =? A 37 B 47 C 39 D None
D. 3 912673 =?
Solutions:
A. Divide the number into two parts first part
should be last three digits of the number, as
9 | 261 . Take the first part 261, last digit is 1,
so last digit of the answer should contain 1.
For remaining part of the answer, take second
part i.e. 9 note the highest cube less than
remaining number i.e. 9 here, 9. i.e. 2³ = 8 < 9.
So answer is 21. This method works out only
for perfect cube number. By this method, for 3 11061 will get same answer as 21 which is
wrong. If we have “None” as one of the options
we can’t use this method, we need to cross
check.
B. Divide the number into two parts as 12|167.
Take 167, last digit is 7, so last digit of the
answer must be 3 (since 3³= 27). 2³ = 8 < 12
therefore final answer is 23.
C. By above method, 51|653, take 653, so last
digit of the answer is 7 (since 7³ = 343). Since
3³= 27 < 51 therefore final answer is 37 which
is wrong. 373 = 50653 not 51653.
D. Solution: 912|673, for 673 its 7. For 912 its 9
(since9³ = 729 < 912) therefore answer is 97.
Get the answers:
A. 3 19683 =?
B. 3 39304 =?
C. 3 379507 103823 ?
D. 3 3
3 3
238328 778688?
24389 4913
LCM and HCF Let us write multiples of numbers 6 and 8.
For 6 6,12,18,24,30,36,42,48,54,...etc and
for 8 8,16,24,32,40,48,56,64,72....etc
There are some multiples which are
common for both i.e. 6 and 8. Common multiples for
6 and 8 are 24, 48, 72, 96, 120 or we can also write
them as 24, 224, 324, 4 24…etc. So common
multiples are in the form of k 24, where
1,2,3...k etc .
Here, 24 is the first (least) common multiple.
L.C.M means least common multiple, a common
multiple which occurs first, i.e. 24. Common
multiples are in the form of kLCM.
We can say L.C.M of two or more numbers is the
least number which is divisible by each of these
numbers (remainder is 0).
Write factors for 6 and 8.
61, 2, 3, 6 and for 81, 2, 4, 8.
Common factors are 1 and 2. Highest common
factor is 2.This is called as G.C.D or H.C.F of 6 and
8. H.C.F or G.C.D of two or more numbers is
highest factor or largest divisor of those numbers.
Example 12: Find H.C.F and L.C.M of 8, 12?
Solution: Take the ratio of the two numbers.
Simplify them to as minimum as possible. =
8 2
12 3 , from here we cannot simplify anymore.
To get H.C.F
= 8 12
42 3
or numerator
(numerator
denominator
denominatoror ).
L.C.M is 83 or 122 = 24(Cross multiplication)
Find H.C.F and L.C.M of following
1. 24 and 36?
2. 24 and 64?
3. 54 and 36?
Answers:
1. 24 12 4 2
36 18 6 3
H.C.F is24 36
122 3
or and L.C.M is 24 3 =
72.
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2. 24 12 6 3
64 32 16 8 ,HCF
248
3 , and
LCM is 248=192
3. 36 18 6 2
54 27 9 3 , HCF
3618
2 and
LCM is 542 = 108
Find H.C.F and L.C.M for following questions.
54 and 72?
36 and 42?
48 and 36?
52 and 72?
255 and 195?
Example 13: 1. Find L.C.M and H.C.F of 24, 36 and 42?
2. L.C.M and H.C.F of 150, 210 and 300?
Solution:
1. Method 1: we write above numbers in terms of
prime factors.324 2 3 ,
2 236 2 3 and
42 2 3 7 . To get H.C.F, write common
prime factors for all the three numbers, i.e. =
2 3 = 6. To get L.C.M, note all the prime
factors that appear at least once in any of the
numbers: 2, 3 and 7. Take their highest powers
in above numbers, = 3 22 3 7 = 504.
Method 2: Take any two at a time, 24/36 = 2/3.
H.C.F = 24/2 = 36/3 = 12. Now, 12/42 = 2/7,
H.C.F = 12/2 = 42/7 = 6. L.C.M = 243 = 362 = 72. Now, 72/42 = 12/7. L.C.M is 727 =
4212 = 504.
2. 150/210 = 5/7. H.C.F is 150/5 = 30 and LCM is
1507 = 1050. 30/300 = 1/10, H.C.F is 30.
1050/300 = 7/2, L.C.M is 10502 = 2100.
Example 14: The H.C.F and L.C.M of two numbers
are 12 and 144 respectively. Find the other number if
one of them is 36?
Solution: 12 144 = 36b, b = (12 144)/ 36 = 48.
Example 15: Find the smallest number which when
divided by 5 or 6, leaves a remainder of 2 in each
case and the number should be greater than 6 and 5?
Solution: Here, the L.C.M of 2 numbers is 5 6 = 30.
Hence the required number is 30 + 2 = 32. (Here,
same remainder in each case i.e.2)
Next numbers which satisfies above condition are 302 + 2 = 62, 30 3 + 2 = 92, 30 4 + 2 = 122
…etc. It means they are in the form of
L.C.M k + remainder, where k is 0, 1, 2, 3…etc.
Smallest 3 digit number which satisfies above
condition is 122.
Largest 3 digit number which satisfies above
condition is, first divide 999 by 30 (LCM) remainder
is 10, so number divided by 5 and 6 leaves
remainder zero is 990, and number is 990 + 2 = 992.
Smallest 4 digit number can be
990 30 1020 2 1022 .
To get largest 4 digit number, we need to divide
9999 by 30. If you divide 9999 by 30, 10 is the
remainder, so answer is 9990 + 2 = 9992.
Example 16: Find the smallest number which when
divided by 5 or 8, gives reminders of 3 and 6
respectively?
Solution: Here remainders are different. But the
difference between divisors and their remainders is
same in each case, 5 3 2 8 6 .
So answer for above problem is 40 – 2 = 38.
a and b are two numbers and relationship
between H.C.F and L.C.F of those two
numbers is,
H.C.FL.C.M = ab, i.e. product of
H.C.F and L.C.M of any two numbers
equals to product of those two numbers.
Any number which when divided by
2 or more numbers leaving same remainder
r in each case in the form of (LCM of those
numbers) k + r, where k is 0,1,2,3,…etc.
Any number which when divided
by a and b, gives remainders of c and d,
where a – c = b – d = constant difference
= (say) r, will be in the form of LCM of
those numbers) k – r, where k = 1, 2,
3…etc. (same approach for when more
than 2 numbers involved.)
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Practice set – 2
1. L.C.M of two numbers is 2079 and their H.C.F
is 27. If one of the numbers is 189, the other
number is (S.S.C 2013) A 189 B 216 C 297 D
584
2. The greatest 4 digit number exactly divisible by
10, 15 and 20 is? (S.S.C 2013)A 9990 B 9960 C
9980 D 9995
3. Find the largest 3 digit number which when
divided by 12 or 8, leaves a remainder 3 in each
case? A 987 B 982 C 984 D 997
4. The least multiple of 13 which when divided by
4, 5, 6, 7 leaves remainder 3 in each case is?
(S.S.C 2012) A 3780 B 3783 C 2520 D 2522
5. Find the largest 4 digit number which when
divided by 5 or 8, gives reminders of 3 and 6
respectively? A 9958 B 9996 C 9978 D 9998
6. Two alarm clocks ring their alarm at regular
intervals of 50 seconds and 48 seconds
respectively. When they ring together, if they
ring at 8.00 AM for the first time? A 8.15 am B
8.20 am C 8.25 am D 8.30 am
7. Two Traffic signals changes after 36 and 42
seconds respectively. If the lights are switched at
9.00AM at what time will they change
simultaneously? A 9.04.12 AM B 9.04.16 AM
C 9.06.12 AM D 9.04.20 AM
8. A red light flashes five times in two minutes and
a green light flashes three times in one minute at
regular intervals of time. If both the lights flash
at the same time, how many times flash together
in each hour? A 31 B 29 C 30 D 28
9. Two persons run in circular track takes 48 and
36 seconds to complete one full cycle. When
they meet for the first time on the track at
starting point if they are running in
A. Same direction B. Opposite direction
10. Find the area of smallest square which can be
formed with dimensions 75 cm? A 125 B 35 C
1225 D None
11. If the students of 9th class are arranged in rows
of 6, 8, 12 or 16, no student is left behind. Then
the possible number of students in the class is?
(S.S.C 2013) A 60 B 72 C 80 D 96
12. There is a number X which when divided by 3,
5, 7 and 9 leaves remainder of 1, but leaves a
remainder of 7 when divided by 8. Find the least
such number possible?
(A) 316 (B) 946 (C) 631 (D) 1261
13. To celebrate their victory in the World Cup, the
Germans distributed sweets among themselves.
If the sweets were distributed equally among the
11 players, 2 sweets were left. If the sweets were
distributed equally among the 11 players, 3
extras and 1 coach, then 6 sweets were left.
What could be the number of sweets in the box?
(A) 336 (B) 170 (C) 156 (D) 160
14. If a person makes a row of toys of 20 each, there
would be 15 toys left. If they made to stand in
rows of 25 each, there would be 20 toys left, if
they made to stand in rows of 38 each, there
would be 33 toys left and if they are made to
stand in rows of 40 each, there would be 35 toys
left. What is the minimum number of toys the
person have?
(A) 3805 (B) 3795 (C) 3185 (D)4285
Example 17: Two numbers are in the ratio 8: 5 their
HCF is 3, numbers are?
Solution: If you take ratio between any 2 numbers,
HCF will get cancelled. Suppose take 16 and 18. =
16 2 8 8
18 2 9 9
(Here HCF is 2 got cancelled).
For above problem go in reverse way, numbers are 83 = 24 and 53 = 15.
Example 18: Find the largest number with which
when 38 and 51 are divided gives remainders of 2
and 3 respectively?
Solution: Answer is HCF of (38 – 2) and (51 – 3)
i.e.12. (38 = 12 3 + 2 and 51 = 12 4 + 3).
Example 19: Find the largest number with which if
we divide 1000, 700 and 250, remainders are same?
The largest number with which when
a and b divided, gives remainders of c and d
will be the HCF of (a – c) and (b – d).
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Solution: 1000 – 700 = 300, 700 – 250 = 450
HCF of 300 and 450 is 150.
Example 19: Find LCM and HCF of2
3,
4
5?
Solution: LCM of numerators 2, 4 is 4.
LCM of denominators 3, 5 is 15.
HCF of numerators 2, 4 is 2.
HCF of denominators 3, 5 is 1.
LCM of fractions = 4
1
HCF Of fractions = HCF of numerators
LCM of denominators=
2
15
15. Two equilateral triangles have the sides of
lengths 102 and 51 respectively.
A. The greatest length of tape that can measure
both of them exactly is.
B. How many such equal parts can be
measured?
16. A farmer wants to plant 44 apple, 88 papayas
and 66 banana trees in equal rows. The number
of rows (minimum) that are required are? A 11
B 18 C 9 D 22
17. There are 315 and 420 students. What is the
minimum number of rooms required such that
all the rooms have equal number of students?
A 105 B 35 C 7 D None
18. The greatest number that will divide 19, 35 and
59 to leave the same remainder in each case is:
(S.S.C 2012) A 6 B 7 C 8 D 9
19. HCF of
2 4,
3 5 and
6
7 is? (S.S.C 2012) A
1
105 B
24
105 C
48
105 D
2
105
20. Three gold coins of weight 780gm, 840gm and
960gm are cut into small pieces, all of which
have the equal weight. Each piece must be heavy
as possible. If one such piece is shared by two
persons, then how many persons are needed to
give all the pieces of gold coins?
(A) 43 (B) 72 (C) 86 (D) 100
21. Find the numbers which when divided by 7
gives a remainder of 3 and when divided by 5
gives the remainder of 2.
A. Least number
B. Least 3 digit number
C. Largest 3 digit number
D. Least 4 digit number
E. Largest four digit number
Unit’s digits (Last digits)
Find the unit’s digit in the following expressions.
83 34 44 91 13 37 19
83 31 41 99 37 65
72 99 75 23 48 31 333 456
22. Unit digit of 3 38537 1256? (S.S.C CGL
2013) A 4 B 2 C 6 D 8
23. If the last digit in the product is 53 457 235n is
zero
A. What is the maximum possible value n may
take?
B. How many such possible values for n?
Find the number of zero’s in the expressions
15 33 42 8
2 5 2 6 42 5 4 15 8
1! 2! 3! 4! 5!1 2 3 4 5
24. Find the number of zero’s in the product 1, 3, 5,
7…99, 101 and 128? A 62 B 12 C 26 D 7
Largest number with which a, b and c
are divided gives same remainders will be HCF
of (a – b) and (b – c) or difference between any
two.
LCM of fractions = LCM of numerators
HCF of denominators
HCF of fractions = HCF of numerators
LCM of denominators
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Find the unit’s digits in the below examples.
199 2537 9 and 199 2537 9
199 253 199 1987 9 2 3
123456789123456789 ?
1554 4551(997) (799) ?
Remainders
Find the remainders in following problems
If 1415 divide by 8
If 135243362 by 12
76 78 79
80
1256 1258 1259
1260
Find remainders for following questions.
13562
9 And
13662
9?
153246
13 And
153336
16?
25. The remainder when 1737
+2937
is divided by 23,
is (S.S.C CGL 2012) A 17 B 29 C 0 D 1
26. The remainder when 6767 67 is divided by 68?
(S.S.C CGL 2001) A 67 B 1 C 0 D 66
Problems on Simple Equations
Example 19: Find the following
A. The product of two numbers is 120 and sum of
their squares is 289, find the difference of the
numbers?
Solution: Let numbers are x and y. xy = 120.
x2 + y
2 = 289. (x – y)
2 = 289 – 2120 = 49.
x – y = 7.
B. The sum of squares of two numbers is 80 and
the square of their difference is 36. The product
of the two numbers is?
Solution: x2 + y
2 = 80. (x – y)
2 =36 = 80 – 2xy.
xy = 44/2 = 22.
C. I have rs.500 in my pocket, I spent 2/5th of
money on a book, find the price of that book and
find the money left with me?
Solution: Cost of the book is 2/5 500 = 200.
Remaining money is 500 – 200 = 300 or 1 – 2/5
= 3/5, = 3/5500 = 300.
D. A person spends 1/3rd
of his salary on food, 1/4th
on clothes and 1/8th on house rent; now he left
with rs.1400, find his salary?
Solution: Remaining is 1 – (1/3) – (1/4) – (1/8) =
(24 – 8 – 6 –3)/24 = 7/24. (7/24)Total salary =
1400. Salary = 20 24 = rs.4800
E. If 21 is added to a number it becomes 7 less than
thrice of a number, and then the number is?
(S.S.C 2012) A 14 B 16 C 18 D 19
Solution: Let number is n. 21 + n = 3n7, n =
14.
F. Two pens and three pencils cost rs.86. Four pens
and a pencil cost rs.112. Find the cost of a pen
and that of a pencil?
Solution: Let the cost of pen is x rupees and
pencil cost is y rupees. 2x + 3y = 86. 4x + y =
112. By solving above equations, we get x =
rs.25 and y = rs.12.
27. 4 5
of of15 7
a number is greater than 4 2
of of9 5
of
the same number by 8. What is the half of that
number? A 630 B 540 C 315 D 100
28. A person spends 1/3rd
of his salary on food,
1/4thof remaining on clothes and 1/8
thof
remaining on house rent; now he left with
rs.1400, find his salary? A 3200 B 2400 C 4800
D 4000
29. In a basket, there are 125 flowers. A man goes to
worship and offers as many flowers at each
temple as there are temples in the city. Thus he
needs 5 baskets of flowers. Find the number of
temples in the city (S.S.C CGL 2012) A 27 B 25
C 26 D 24
30. A man divides rs.8600 among 5 sons, 4
daughters and 2 nephews. If each daughter
receives four times as much as each nephew, and
each son receives five times as much as each
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nephew, how much does each daughter receive?
(S.S.C CGL 2000) A 200 B 600 C 800 D 500
31. When an amount was distributed among 14
boys, each of them got rs.80 more than the
amount received by each boy when the same
amount is distributed equally among 18 boys.
What was the amount? A 4080 B 5040 C 1200
D 1040
32. Kiran had 85 currency notes in all, some of
which were of rs.100 denomination and
remaining of rs.50 denomination. The total
amount of all these currency notes was rs.5000.
How much amount did she have in the
denomination of rs.50? A 15 B 60 C 70 D None
33. The sum of a number and its reciprocal is
164/18. Find the sum of that number and its
square root? A 12 B 16 C 8 D 20
34. The product of three consecutive even numbers
when divided by 24 is 96; find the product of
their square roots? A 32 B 36 C 28 D 48
35. A total of 68 chocolates were distributed among
12 children such that each girl gets 5 chocolates
and each boy gets 6 chocolates, find the number
of boys? A 8 B 6 C 9 D 4
36. My friends were organizing a picnic. They
estimated the expenditure to be Rs.500, they
mobilized some more friends. The number of
people who actually went to the picnic increased
by five, the expenditure per head came down by
five rupees. The people went for the picnics
were?
(A) 35 (B) 15 (C) 20 (D) 25
37. If a number multiplied by 33 and the same
number is added to it, then we get a number that
is half the square of that number. Find the
number:
(A) 66 (B) 67 (C) 68 (D) None
38. The fuel in tank contains 1/5 of the total capacity
of the tank. When 22 more liters of fuel are
poured into the tank, the indicator rests at the
three-fourth of the full mark. Find the capacity
of the fuel tank.
(A) 36 L (B) 40 L (C) 44 L (D) 32 L
39. Out of a group of swans, 7/2 times the square
root of total number of swans are playing on the
shore of the pond. The remaining two are inside
the pond. Find the total number of swans?
(A) 12 (B) 36 (C) 16 (D) 15
40. The highest score in an innings is 2/9th of the
total score in the inning and the next highest
score is 2/9 of the remainder score. The
difference of two scores is 8 runs. Find the total
score in the innings?
(A) 160 (B) 162 (C) 186 (D) 196
41. Find the number of even numbers between 102
and 252 if:
A. Both the ends included
B. Only one end is included
C. Neither end is included
42. Find the number of numbers between 301 and
400 (both included), that are not divisible by 3.
(A) 68 (B) 65 (C) 66 (D) 67
43. The sum of the third powers of the first 100
natural numbers will have a unit’s digit of?
(A) 5 (B) 6 (C) 4 (D) 0
44. In a three digit number the sum of the digits is
13. The tens digit is one less than the units digit.
Which of the following numbers cannot be a
digit in the hundreds place?
(A) 4 (B) 2 (C) 5 (D) 6
45. When we multiply a certain two digit number by
the sum of its digits, 238 is achieved. If we
multiply the number written in reverse order of
the same digits by the sum of the digits, we get
301, find the number:
(A) 27 (B) 43 (C) 34 (D) 26
46. The unit’s digit of the expression 2139 137195 477 3333684 is:
(A) 2 (B) 4 (C) 5 (D) 0
47. Which of the following can never be in the
ending of a perfect square?
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(A) 5 (B) 9 (C) 00 (D) 000
48. The difference of 2510 7 and 2410 x is divisible
by 3, then x:
(A) 2 (B) 3 (C) 7 (D) 0
49. Find the two digit number if it is known that the
ratio of the required number and the sum of its
digits is 8 as also quotient of the product of its
digits and that of sum is 14/9:
(A) 42 (B) 72 (C) 86 (D) 91
50. Let x, y and z be distinct integers. x and y are
odd and positive, and z is even and positive.
Which one of the following statements cannot be
true?
(A) 2.x y z is even (B)
2.x z y is odd
(C) 2.x y z is even (D) 2.x y z is odd
Practice set -2 Key:
1- C 2 – B 3 – A 4 – B 5 – D 6 – B 7 – A 8 – C 9 –
Ans: 144; 10 – C 11 – D 12 – C 13 – C 14- B 15-
Ans: 51, 9; 16 – C 17- C 18 – C 19 - D 20 – C 21 –
22 – D 23 – 24 – D 25 – C 26 – D 27 – C 28 – A 29
– B 30 – C 31- B 32 – D 33 – A 34 – D 35- 36 – D
37 – A 38 – B 39 – C 40 –B 41 - Ans: 76, 75, 74
42 – D 43 – D 44 – C 45 – C 46 – D 47 – D 48 – A
49 – B 50 –D
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