# Quantitative Aptitude Questions for IBPS RRB PO and Clerk Exam 2017

**Dear Students,**

**Quantitative Aptitude Questions for IBPS Exam**

Today we have been including some Arithmetic and Number Series Questions. This is the Quantitative Aptitude Questions and you should try to do it within 16-17 Minutes. If you fail to complete it in the stipulated time, then try again with full force. These quantitative aptitude questions are very important from exam point of view like IBPS RRB Exam, IBPS PO and IBPS Clerk.

**Q1. A father’s age is four times as much as the sum of the ages of his three children but 6 years hence his age will be only double the sum of their ages. Then the age of the father is**

(a) 50

(b) 60

(c) 65

(d) 55

(e) None of these

**S1. Ans.(b)**

**Sol.**

Let age of 3 children = x

Age of father = 4x

According to question,

4x+6=2(x+6×3)

⇒ x=15

∴ Father’s age = 4 × 15 = 60 years.

**Q2. 15 years hence, A will be twice as old as B, but five years ago A was 4 times as old as B. Find the difference of their present ages.**

(a) 20

(b) 25

(c) 30

(d) 35

(e) None of these

**S2. Ans.(c)**

**Sol.**

Let present age of A is x and that of B is y.

⇒x+15=2 (y+15)

⇒x-2y=15 …(i)

&x-5=4(y-5)

⇒x-4y=-15 …(ii)

From (i) and (ii)

2y=30⇒y=15 years.

andx=45 years.

∴ Difference = 30 years.

**Q3. A says to B “I am twice as old as you were when I was as old as you are”. The sum of their ages is 63 years. Find the difference of their ages.**

(a) 9

(b) 11

(c) 13

(d) 8

(e) None of these

**S3. Ans.(a)**

**Sol.**

A + B = 63 …(i)

and A = 2[B – (A - B)]

⇒ 3A = 4B …(ii)

From (i) & (ii)

A = 36, B = 27

∴ Difference = 9 years.

**Q4. Two years age, A was four times as old as B. 8 years hence, A’s age will exceed B’s age by 12 years. The ratio of the present ages of A and B is**

(a) 5:3

(b) 3:2

(c) 2:1

(d) 3:1

(e) None of these

**S4. Ans.(d)**

**Sol.**

(A – 2) = 4(B - 2)

⇒ A – 4B = –6 …(i)

and A + 8 = B + 8 + 12

⇒ A – B = 12 …(ii)

From (i) and (ii)

–3B = –18

⇒ A = 18 years.

∴ Required ratio=A/B=18/6=3/1

**Directions (5-9): Find the missing term in the following series.**

**Q5. 12, 12, 18, 45, 180, 1170, ?**

(a) 12285

(b) 10530

(c) 11700

(d) 12870

(e) None of these

**S5. Ans.(a)**

**Sol.**

The pattern is

×1.0, ×1.5, ×(1 + 1.5), ×(1.5 + 2.5), ×(2.5 + 4), ×(4 + 6.5)

∴ ? = 10.5 × 1170

? = 12285

**Q6. 1548, 516, 129, 43, ?**

(a) 11

(b) 10.75

(c) 9.5

(d) 12

(e) None of these

**S6. Ans.(b)**

**Sol.**

The pattern in the given series is —

×1/3,×1/4,×1/3,×1/4,........

∴ ?=43×1/4

?=10.75

**Q7. 8, 36, 152, 620, 2496, 10004, ?**

(a) 8190

(b) 8187

(c) 40040

(d) 8163

(e) None of these

**S7. Ans.(c)**

**Sol.**

The pattern in the series is —

×4+4, ×4+8, ×4+12, ×4+16, ×4+20, ×4+24

∴ ? = 10004 × 4 + 24

? = 40040

**Q8. 8, 10, 14.5, 22.5, 35, 53, ?**

(a) 36

(b) 54.25

(c) 65

(d) 77.5

(e) 90

**S8. Ans.(d)**

**Sol**.

The pattern in the given series is,

+2, +4.5, +8, +12.5, +18, +24.5

∴ ? = 53 + 24.5

? = 77.5

**Q9. 18, 21, 16, 23, 12, 25, ?**

(a) 24

(b) 8

(c) 14

(d) 10

(e) None of these

**S9. Ans.(b)**

**Sol.**

The pattern will be —

+3, –5, +7, –11, +13, –17 (alternate + & – of prime numbers)

∴ ? = 25 – 17

? = 8

**Q10. A and B can do a piece of work in 72 days. B and C can do it in 120 days. A and C can do it in 90 days. In what time A, B and C together can do it?**

(a) 80 days

(b) 100 days

(c) 120 days

(d) 150 days

(e) None of these

**S10.Ans.(e)**

**Sol.**

1/A+1/B=1/72; 1/B+1/C=1/120 and 1/C+1/A=1/90

∴2(1/A+1/B+1/C)=(1/72+1/120+1/90)

⇒(1/A+1/B+1/C)=1/2 [(5+3+4)/(6×5×12)]=1/60

i.e. A, B and C together can do the work in 60 days.

**Q11. A is thrice as good a workman as B and therefore is able to finish a job in 30 days less than B. How many days will they take to finish the job working together?**

**Q12. 9 men working 7 hours a day can complete a piece of work in 15 day. In how many days can 6 men working for 9 hours a day, complete the same piece of work ?**

(a) 63/4 days

(b) 16 days

(c) 67/4 days

(d) 35/2 days

(e) None of these

**S12. Ans.(d)**

**Sol.**

Let required days are x

Then using M_1 H_1 D_1=M_2 H_2 D_2

9 × 7 × 15 = 6 × 9 × x

⇒x=35/2 days.

**Q13. Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?**

(a) 6

(b) 8

(c) 10

(d) 11

(e) None of these

**S13. Ans.(b)**

**Sol.**

1 minute work of (A + B)both=(1/24+1/32)

=(4+3)/(8×12)

=7/96 minutes

i.e. tank will full in 96/7 minutes.

Let B is closed after x minutes

∴7x/96-(18-x)/24=1

⇒(7x+72-4x)/96=1

⇒x=8 minutes

**Q14. A tank is normally filled in 8 hours but takes 2 hours longer to fill because of a leak in its bottom. If the cistern is full, in how many hrs will the leak empty it?**

(a) 45

(b) 50

(c) 40

(d) 35

(e) None of these

**S.14 Ans.(c)**

**Sol.**

Required time to empty the tank=1/8-1/10=(5-4)/40=1/40

i.e. 40 hours will be required.

**Q15. Three pipes A, B and C are connected to a tank. A and B together can fill the tank in 10 hours, B and C together in 15 hours and C and A together in 12 hrs. In how much time will the three pipes fill the tank together?**

(a) 8

(b) 12

(c) 11

(d) 10

(e) None of these

**S15. Ans.(a)**

**Sol**.

(A + B)’s 1 hour work=1/10

(B+C)’s 1 hour work=1/15

(C+A)’s 1 hour work=1/12

∴(A+B+C)’s hour work=1/2[1/10+1/15+1/12] =1/2 [(6+4+5)/60]=1/8

∴ (A + B + C) can fill the tank in 8 hours.

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ReplyDeletesir pls explain que no. 3 in other way.....

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