# NIACL AO Prelims – Quantitative Aptitude Questions Day- 66

Dear Readers, Bank Exam Race for the Year 2018 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for NIACL AO Prelims 2018 Exams can practice these questions daily and make your preparation effective.

[WpProQuiz 4747]

1) If 5 men or 6 women or 8 boys can finish a work in 120 days working 8 hours a day. The number of days taken by 10 men, 6 women and 12 boys working 12 hours a day will be?

a) 15 2/7 days

b) 14 7/8 days

c) 16 2/5 days

d) 17 7/9 days

e) None of these

2) A 200 liter mixture of milk and water contains milk and water in the ratio of 7 : 3. How much milk should be added to the mixture to get a new mixture of milk and water in the ratio of 13 : 5?

a) 12 litres

b) 14 litres

c) 16 litres

d) 18 litres

e) None of these

3) Find the respective ratio of the volumes of a sphere and a cylinder, both having radius 14 cm and height of the cylinder is 16 cm?

a) 5 : 8

b) 7 : 6

c) 6 : 7

d) 8 : 5

e) None of these

4) The difference between Simple interest and Compound interest on a certain sum at the rate of 20% per annum for two years is Rs. 1680. Find the simple interest on that sum at the rate of 14% per annum after 8 years?

a) Rs. 44040

b) Rs. 47340

c) Rs. 45040

d) Rs. 47040

e) None of these

5) The ratio of present ages of a husband and wife is 7 : 5. After 10 years, the ratio of their ages will be 9 : 7. A child was born after 2 years of their marriage, then find the age of child if the ratio of ages at the time of marriage was 5 : 3?

a) 12 years

b) 10 years

c) 14 years

d) 8 years

e) None of these

Directions (Q. 6 – 10) Study the following information carefully and answer the given questions:

The following table shows the ratio of students passed from different streams in graduation from different city and also the ratio of males and females among them. 6) If the total number of males who passed in B. Sc from City R is 1275, then find the total number of students passed in B. A from City R?

a) 2380

b) 3256

c) 2178

d) 3384

e) None of these

7) If the total number of females who passed in B. A from City P is 1113, then find the difference between the total number of students passed in B.A and that of B. Sc in City P?

a) 1094

b) 2180

c) 1986

d) 1272

e) None of these

8) If the total number of students passed from City S is 2794, then the total number of females passed in B. A from City S is what percentage of total number of males passed in B. Sc from City S?

a) 115 %

b) 72 %

c) 100 %

d) 88 %

e) None of these

9) The number of females who passed in B. A from City Q is approximately more/less than the number of males who passed in B. A from City Q?

a) 173 % less

b) 167 % more

c) 152 % less

d) 173 % more

e) 167 % less

10) The number of students passed in B. A from City T is 1716. Find the ratio between the total number of females who passed in B. A from City T to that of total number of males who passed in B. Sc from City T?

a) 1287 : 572

b) 1331 : 557

c) 1089 : 452

d) 956 : 369

e) None of these

5 men = 6 women

6 women = 8 boys

10 men, 6 women and 12 boys = 12 women + 6 women + 9 women = 27 women

Women                       days                hours

6                                  120                  8

27                                x                      12

(6*120*8) = (27* x days *12)

X = (6*120*8) / (27*12) = 160/9 = 17 7/9 days

Milk in the mixture = 200*(7/10) = 140 liters

Water in the mixture = 200 – 140 = 60 liters

Let the amount of milk be added be x,

(140 + x)/60 = 13/5

700 + 5x = 780

5x = 80

x = 16 liters

Volume of the sphere = (4/3)*πr3 = (4/3)*(22/7)*14*14*14

Volume of the cylinder = πr2h = (22/7)*14*14*16

Required ratio = [(4/3)*(22/7)*14*14*14] : [(22/7)*14*14*16] = 7 : 6

The difference between CI and SI for two years is,

Difference = P*(r/100)2

= > 1680 = P*(20/100)2

= > 1680 = P*(1/5)2

= > 1680 = P*(1/25)

= > P = Rs. 42000

SI = (42000*14*8)/100 = Rs. 47040

Let the ratio of their present age be 7x and 5x respectively,

(7x + 10) / (5x + 10) = 9 / 7

49x + 70 = 45x + 90

4x = 20

X = 5

Age of husband = 35 years

Age of wife = 25 years

Let their marriage be held y years ago,

So,

(35 – y) / (25 – y) = 5 / 3

105 – 3y = 125 – 5y

2y = 20

Y = 10 years

Age of the child = 10 – 2 = 8 years

The total number of males who passed in B. Sc from City R= 1275

3’s = 1275

1’s = 425

Total number of students passed in B. Sc from City R = 7’s = 2975

Total number of students passed in B. A from City R = 2975*(4/5) = 2380

The total number of females who passed in B. A from City P = 1113

7’s = 1113

1’s = 159

Total number of students passed in B. A from City P = 159*12 = 1908

Total number of students passed in B. Sc from City P = 1908*(5/3) = 3180

Required difference = 3180 – 1908 = 1272

The total number of students passed from City S = 2794

11’s = 2794

1’s = 254

The total number of students passed in B. A from City S = 1270

The total number of students passed in B. Sc from City S = 1524

The total number of females passed in B. A from City S

= > 1270*(3/5) = 762

The total number of males passed in B. Sc from City S

= > 1524*(1/2) =762

Required % = (762/762)*100 = 100 %

(Or)

The total number of females passed in B. A from City S = 5x*(3/5) = 3x

The total number of males passed in B. Sc from City S = 6x*(1/2) = 3x

Required % = (3x/3x)*100 = 100 %

The ratio of total number of male to female students passed in B. A in City Q

= > 3 : 8

Required % = [(8x – 3x)/3x]*100 = (5x/3x)*100 = 166.67 % = 167 % more

The number of students passed in B. A from City T = 1716

6’s = 1716

1’s = 286

The number of students passed in B. Sc from City T = 1430

The total number of females who passed in B. A from City T

= > 1716 *(3/4) = 1287

The total number of males who passed in B. Sc from City T

= > 1430 *(2/5) = 572

Required ratio = 1287 : 572 