Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS exams. We have included Some questions that are repeatedly asked in exams.

**A’s age is 4 years more than B’s age. A’s age is also 8 years less than C. 7 years hence ratio of B’s age to C’s age will be 7 : 13. Find the ratio of present ages of A, B, and C?**

A. 11 : 19 : 7

B. 17 : 9 : 11

C. 11 : 7 : 19

D. 11 : 12 : 13**C. 11 : 7 : 19**

Explanation:

A = B+4

A = C-8

B+7 / C+ 7 = 7/13

Or (A-4)+7 / (A+8)+7 = 7/13

A+3 / A+15 = 7/13

A = 11

So B = A – 4 = 11 – 4 = 7

C = A+ 8 = 11 + 8 = 19**The sum of ages of A and B is 50. 7 years ago, A’s age was 3 more than twice B’s age then. Find the difference of ages of A and B.**

A. 12

B. 14

C. 13

D. 15**B. 14**

Explanation:

A+B = 50

A – 7 = 2(B-7) + 3

Solve the two equation to find A and B

A = 32, B = 18

Diff = 32 – 18 = 14**A gave 40% of amount to B. From this B gave 20% to C. From remaining B spends 40% on food and 20% of transport. Now he is left with Rs 912. Find the amount received by B.**

A. Rs 2890

B. Rs 2090

C. Rs 2270

D. Rs 2850**D. Rs 2850**

Explanation:

Let A has Rs x

B gets 40/100 * x = 2x/5

after giving 20% to C, B is left with 80% of 2x/5 = 8x/25

now he spends 40% and 20%, total 60%, now amount left is 40% of 8x/25 = 16x/125

so, 16x/125 = 912

x=7125

B got 40% of x = 40/100 * 7125 = 2850**2 men or 3 women or 4 children can do a work in 52 days. In how many days 1 man and 1 woman and 1 child can complete the same work?**

A. 48

B. 50

C. 42

D. 46**A. 48**

Explanation:

For this when we are given – a men or b women or c children work in D days. Number of days taken by 1M + 1W + 1C = a*b*c*D / (a*b + b*c + c*a)

So, days = 2*3*4*52 / (2*3 +3*4 + 4*2) = 48**Four coins are tossed. Find the probability of at most 2 heads.**

A. 5/19

B. 6/17

C. 5/16

D. 11/16**D. 11/16**

Explanation:

There are total 2^{4}outcomes

For atmost 2 heads (11 pairs):

TTTT, HTTT, THTT, TTHT, TTTH, HHTT, HTHT, HTTH, THHT, THTH, TTHH

So prob. = 11/16**There are 3 red, 4 white and 6 blue balls. 2 balls are drawn at random. Find the probability that at least 1 is blue.**

A. 19/26

B. 19/21

C. 13/24

D. 14/19**A. 19/26**

Explanation:

Total 13 balls, ways of choosing 2 are^{13}C_{2}= 13*12/2*1 = 78

For 1 blue:^{6}C_{1}*^{7}C_{1}

= 42

For both blue:^{6}C_{2}

= 6*5/2 = 15

Total prob. = 42+15 / 78 = 57/78 = 19/26**A travels a distance f 50 km in 2 hrs and 30 min. How much faster, in km/hr muct he travel to make such a trip in 5/6 hr less time?**

A. 10

B. 20

C. 30

D. 40**A. 10**

Explanation:

Time required = 2hrs 30 min – 50 min = 1 hr 40 min = 1 2/3 hrs

Required speed = 50 * 3/5 = 30 km/hr

Original speed = 50 * 2/5 = 20 km/hr

Diff = 30 – 20 = 10 km/hr**A is travelling on his cycle and has calculated to reach point P at 2 PM if he travels at 10 km/hr. He will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach P at 1 PM.**

A. 13 km/hr

B. 12 km/hr

C. 12.5 km/hr

D. 14 km/hr**B. 12 km/hr**

Explanation:

Let distance = d, time to reach at 1 PM = t

Then time to reach at 2 PM = t+1 = d/10

Time to reach at 12 noon = t-1 = d/15

Solve the 2 equations, d = 60 km, t = 5 hrs

So speed to reach at 1 PM = 60/5 = 12**The volume of a cone is 1584 cm**^{3}and its height is 42 cm. Find the curved surface area of the cone.

A. 565.7√2 cm^{2}

B. 565 cm^{2}

C. 560.67√2 cm^{2}

D. 548.34√2 cm^{2}**A. 565.7√2 cm**^{2}

Explanation:

Vo. of cylinder is = 1/3 * ᴨr^{2}h = 1584

h = 42

Solve and find r, r = 6 cm

Now length, l = √(h^{2}+r^{2}

= √ 42^{2}+6^{2}= 30√2

Curved surface area = ᴨrl = 22/7 * 6 * 30√2**A can do a work in 10 days, B in 15 days and C in 12 days. All started the work, but A leaves after 2 days, further 3 days after B left the work. In how much time C completes the remaining work?**

A. 2/5 day

B. 4/5 day

C. 1/5 day

D. 3/5 day**D. 3/5 day**

Explanation:

After 2 days, B left after further 3 days, so B worked for 5 days. Let C worked for total x days.

1/10 * 2 + 1/15 * 5 + 1/12 * x = 1

Solve, x = 28/5

This gives that C worked for a total of 28/5 days, in which 5 days he worked before he was left by B. So remaining days to complete remaining work = 28/5 – 5 = 3/5th day

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