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NUMERICAL REASONING SOLUTIONS

Question 4 = A (40days) i.e. 200*70/100 200*50/100= 40

Question 5= E (22554) i.e. number of staffs that left =90*70/100=63

The detail of each staff will take 358 characters

Therefore for 63 people, 358*63 character worth of space would be left.

Question 6= A (8950 characters) i.e. 25 *358 (each staff record requires 358

characters, from question 5)

Question 7 = B (127) i.e. The First names first letter is character 1, and the last

letter for home address 4 is character 126

Therefore, the first letter for post code will be character 127.

Question 8 = C (99), each employee record need 358, therefore 35,700 will be

for 35700/358 number of employee.

Question 1= C (cant tell from data)

Question 9= A(5.5), if 1 litre of gloss paints 14m2, 11 by 71.e 77m2 will need

77/14 litres of gloss paint.

Question 10= D(6litres) 1 litre undercoat covers 16m2, therefore 96m2 will

require 96/16 litres of undercoat paint.

Question 11= F (7.5 litres) 1 litres of emulsion covers 12m 2, 90m2(6*15) willneed 90/12

Question 12=E (10), note, the room is rectangular that means the opposite

sides have the same area, therefore (11*3)2+ (9*3)2= 120 which is the total

area of the room. Since 1litre of emulsion covers 12m 2, 120m2 will need 120/12

litres of emulsion.

Question 13=A (550), if it trebled to reach 600 in the last year, that means it

was 200 b4 the last year, if it quadrupled to reach 200, that means it was 50 b4

then. Since 50 correspond to the 3 rd year, it means that 11 units of the product

will cost 11*50 =550 dollars.

Question 3= F(2), since company Y spends 12days (5%) of 240 days in

conferences, and 1 staff has already spent 10 days, it implies that the remaining

staff have just two days to spend in attending conferences before the year runs

out.

Question 14=B (2,400), price at second year is 100, then price at year

three=2*100, price at year 4=4*200 and finally price at year five

=800*3=2,400$.

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Question 15=C (15), price at year 5=720, therefore price at year

4=720/3=240, price in year 3=240/4=60, price at year 2=60/2=30, therefore at

year 1=30/2=15.

Question 16= F (200), the current price is 600$ and it trebled to get to 600$

therefore the price last year was 200$.

Question 17=E(305), for year 1 sale after 2nd month= 35,after 4th month= 25,

after 6th month=15, after 8th month=15, after 10th month=20, after 12th

month=15.

For year 2 sale after 2nd month= 25, after 4 th month= 30, after 6th month=25,

after 8th month=35, after 10th month=35, after 12th month=30. Therefore total

for the two years= the sum of sales for year1 and 2 which is 30.

Question 18=D(82.5), if sales in year 3 was 50% greater than year 2, it means

for every two months in year 3, sales was 50% greater than the corresponding

period for year 2, that is, if sales after 4 th month in year 2 is 30, sales after 4 th

month in year 3= 45(30+ 30*50/100), also if sales after 6 th months in year 2=

25, sales after 6th in year 3= 37.5(25+25*50/100). For year, total sales for month

2 and 4=45+37.5=82.5. (Note that 4th and 6th month corresponds to 2nd and 3rd

period on X axis of the graph).

Question 19= A(12.5), subtract the sales for each corresponding month in

between year 1 and 2 then sum them and divide by 6.i.e 35-25+30-25+25-

15+35-15+35-20+30-15=75, then 75/6=12.5

Question 20=F (4 only), if you subtract sales between corresponding months

for year 1 and 2 youd discover that period 4 records the highest difference in

sales.

Question 2 = D (96), since 40% of working days out of 240 days is spent

meeting clients in company X. 40% of 240= 96.

SOLUTIONS to Me.mths

2) B (1850000), since people that earn btw 0 and 4000 will pay 1% of theirincome on tax. Since average income of Agor people is 3700, 1% of

3700=37, there 50000 people therefore total income tax for Agor people

= 37*50000=1,850,000.

3) D(125), for earner btw 6000 and 8000, tax paid will be 80+ 3% of money

over 6000 in this case 1500, since the income is 7500. 3% of 1500= 45

therefore tax on income of 7500= 80+45= 125.

4) F (0< X< 4000) Let 40 + 2X=100, this implies that, 2X= 60, IF 2% OF X =

60, What is X?Putting it in mathematical form we have X*2/100=60,

therefore, X=60*100/2=3000, so X is greater than 0 but less than 4000hence F is the answer.

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5) A(down by 25 pounds), total initial price =

386*100+259*500=168100pence, total price change = (386+25)100 +

(259-10)500= 165600pence, Total change in price= 165600-168100=-

2500(note 1= 100pence)therefore 2500pence= 25

7) C (28), initial price of 200 share at 226pence= 200*226=45200, Finalprice after 14pence pou increase per share= (226+14)200=48000,

change in price= 48000-45200=2800pence which implies 28.

8) F (51.5), the initial price for each share was 111 but since it dropped by

8, the price is therefore 103pence. Consequently, 50shares will now cost

103*50=5150pence which implies 51.5.

9) A (18,500), the price of producing a batch of 1000= 10000, total income

from sale of chip running at speed 1, 2 and 3=8400+9300+10800=

28,500, therefore total profit= (selling price) 28500-(cost price) 10000=

18500.

10) B (4200), since the cost of producing 1000 chips = 10000, it follows

therefore that each chip will cost 10 to produce i.e. 10000/1000=10.

Since 42% of the total 1000 chips run at the lowest speed it means 420

out of 1000 chips run at the lowest speed i.e. 1000*42/100=420.If it cost

10 to produce a chip then, 420 chips will cost 420*10=4200.Note that

total income from sale of chips running at the lowest speed= 8400,

therefore profit from sale of lowest speed running chip= 8400-

4200=4200.

11) E(28,500), sum the amount of income derived from the sale chips

running at speed 1, 2 and 3, which is 8400+ 9300+ 10800= 28500

12) C (4200), the lowest speed chips make up 42% of the 1000 chips,

42% of 1000=420, each chip costs 10 pounds to produce(see solution to

question 10), therefore the cost of 420 chips=420*10=4200.

13) A(4:7:3), 12000:21000:9000, note 3000 is common to all three, so

simplify the ratio by dividing each of the values by 3000i.e

12000/3000:21000/3000:9000/3000=4:7:3

14) E (13500), since the cost of machines from supplier r=9000 and the

ratio of cost of machine from supplier t to supplier r i.e. t: r= 3:2, let the

sum of machine cost from suppliers r and t be W.

Therefore, 2/5*W= 9000(note 2/5 is the same as r/r+t)

W= 5*9000/2=22500, it total cost of supplies from t and r=22500, and

cost of supply from r=9000, it follows therefore that cost of supplies form t

= 22500-9000=13500.

15) D(1,125), let warranty cost be W, let purchase cost be C, since W

C , by introducing the proportional constant K we have, W=KC, since

W=2625, K=?, C= 21000, from W=KC, it follows therefore that

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K=2625/2100=0.125, therefore warranty cost from supplier R =

0.125*9000= 1,125

16) F(30000), two machines from supplier R=2*9000, and cost of one

machine from supplier P=12000, total cost of machines=

18000+12000=30000

17) C (120), find volume of larger container, which is 2.5*4*3=30, and

divide it by volume of box which 0.5*0.5*1=0.25i.e. 30/0.25=120

18) A(21), find volume of access area which is =1*2*3=6 and subtract

from volume of smaller container which is=3*3*3=27, therefore 27-6=21

19) D (318.80) not quite sure though.

20) E(30)(see no 17)

1) F (210), i.e. 1070+7% of 1000(since 26000 is just 1000 naira over 25000)which, 1070+7*1000/100=1070+70=1140, with an increase to 29000 tax

then becomes, 1070+7% of 4000(since 29000 is 4000 greater than

25000), so, 1070+ 7*4000/100=1350, the difference btw tax at 26000 and

tax at 29000=1350-1140=210.

SOLUTIONS to Me.Eng

Note the answers are based on what is in the passage, dont use

1) C and D

2) B and D

3) A and C

4) C and D

5) A and C

6) C and D

7) A and B

8) A and D

9) A and D

10) B and C

11) A and D

12) A and D

13) A and B

14) A and D

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15) C and D

16) B and D

17) C and D

18) A and C

19) B and d

20) B and C

SOLUTIONS TO ABSTRACT REASONING

1) D

2) B

3) F

4) D

5) B

6) B

7) C

8) D

9) D

10) D

11) D

12) B

13) C

14) A

15) F

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16)D