IBPS RRB Clerk Mains Quantitative Aptitude Questions 2019 (Day-03)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for IBPS RRB Clerk Mains 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

Are You preparing for IBPS PO 2019? Start your preparation with Free IBPS PO Mock Test 2019 – Take Test Now

Check here for IBPS RRB PO Mains Mock Test 2019

Check here for IBPS RRB Clerk Mains Mock Test 2019

Check here for IBPS PO Prelims Mock Test 2019

Click Here to Subscribe Crack High Level Puzzles & Seating Arrangement Questions PDF 2019 Plan

IBPS RRB Clerk Mains Quantitative Aptitude Questions 2019 (Day-03)

maximum of 10 points
Table is loading

Directions (1 – 5): Find the wrong number in the given series

1) 96, 46, 72, 180, 630, 2835

a) 96

b) 72

c) 46

d) 180

e) 630

2) 42, 86, 348, 2094, 16758

a) 86

b) 348

c) 2094

d) 16758

e) 42

3) 48, 1764, 1632, 2622, 2530

a) 2622

b) 2530

c) 1632

d) 1764

e) 48

4) 1582, 1602, 1634, 1690, 1782, 1920

a) 1690

b) 1782

c) 1634

d) 1920

e) 1602

5) 124, 63, 64, 130, 524, 4200, 67214

a) 124

b) 63

c) 64

d) 524

e) 67214

Directions (6 – 10): Read the following passage carefully and answer the given questions.

P and Q together can do a piece of work in 6 (2/3) days. R and S can do the same work in 10 days. Q and S together can do the same work in 6 days. P is 50% more efficient than R. P and S started working together after 2 days, Q and R joins with them. At the end of 4 days, all of them left and T completes the remaining work in 3 (1/5) days.

6) A is twice efficient as P and B is thrice efficient as R. A and B started working alternatively begins with A, find the number of days taken to complete the whole work?

a) 12 days

b) 15 days

c) 10 days

d) 24 days

e) 8 days

7) Find the number of days taken to complete the whole work working all together?

a) 6

b) 5

c) 3

d) 4

e) 9

8) P and T started working, after 2 days Q joins with them. After another two days, R joins with them. P and Q left 2 and 3 days before the work was completed respectively, then find the number of days required to complete the whole work?

a) 5 2/3

b) 6 1/3

c) 5 1/4

d) 6 1/2

e) 5 4/5

9) All of them working together and P gets wage of Rs. 1200. Find the total wage?

a) Rs. 8000

b) Rs. 5500

c) Rs. 3800

d) Rs. 4600

e) Rs. 6400

10) P, Q and R started working with 2/3rd, 5/6th and 3/2nd of their original efficiency respectively. Find the number of days required to complete the whole work?

a) 5 days

b) 4 days

c) 6 days

d) 8 days

e) 9 days

Answers :

Direction (1-5) :

1) Answer: c)

96 * 0.5 = 48 (not 46)

48 * 1.5 = 72

72 * 2.5 = 180

180 * 3.5 = 630

630 * 4.5 = 2835

2) Answer: d)

42 * 2 + 2 = 86

86 * 4 + 4 = 348

348 * 6 + 6 = 2094

2094 * 8 + 8 =16760 (16758)

3) Answer: b)

4) Answer: d)

1582           1602           1634           1690           1782           1922

          20               32               56               92               140

                   12               24               36               48

5) Answer: e)

124 * 0.5 + 1 =63

63 * 1 + 1 = 64

64 * 2 + 2 = 130

130 * 4 + 4 = 524

524 * 8 + 8 = 4200

4200 * 16 + 16 = 67216 (not 67214)

Directions (6 – 10):

P + Q = 3/20 -> (1)

R + S = 1/10 -> (2)

Q + S = 1/6 -> (3)

Equations, (1) + (2) – (3)

(P+Q+R+S) – (Q+S) = 3/20 + 1/10 – 1/6

P + R = (9+6-10)/60

P+R = 5/60

P + R = 1/12 à (4)

Efficiency ratio of P to R = 150: 100 = 3:2

Time ratio of P to R = 2: 3

Substitute in equation (4)

1/2x + 1/3x = 1/12

5/6x = 1/12

X = 10

P = 20 days, R = 30 days

Q = 3/20 – 1/20 = 1/10

S = 1/10 -1/30 = 2/30 = 1/15

LCM of 15, 10, 30 and 20 = 60 units (Total work)

P = 3 units

Q = 6 units

R = 2 units

S = 4 units

At the end of 4 days,

(3+4)*2 + (15*2) = 14 + 30 = 44

Remaining = 60 – 44 = 16 units

16 units completed by T in 16/5 days then 60 units completed in 12 days

6) Answer: c)

A =20/2 * 1 = 10 days

B = 30/3 * 1 = 10 days

Total work = 10 units

A = 1 units and B = 1 units

1st cycle (2 days) = 2 units

5th cycle = 2*5 = 10 units

Total number of days = 10 days

7) Answer: b)

LCM of 10, 12, 15, 20 and 30 = 60 units (total work)

P = 3 units

Q = 6 units

R = 2 units

S = 4 units

T = 5 units

Required number of days = 60/(3+6+2+4+5)

= 60/20 = 3 days

8) Answer: d)

LCM of 10, 12, 20 and 30 = 60 units (total work)

P = 3 units

Q = 6 units

R = 2 units

T = 5 units

Let us take the total no of days be x

(x-2)*3 + (x-5)*6 + 5*x +(x-4)*2 = 60

3x-6+6x-30+5x+2x-8 =60

16x – 44 =60

16x = 104

X = 104/16 =26/4 = 6 2/4 = 6 ½ days

9) Answer: a)

LCM of 10, 12, 15, 20 and 30 = 60 units (total work)

P = 3 units

Q = 6 units

R = 2 units

S = 4 units

T = 5 units

Efficiency ratio = 3: 6: 2: 4: 5

Total wage = (1200/3) * 20 = Rs. 8000

10) Answer: c)

LCM of 10, 20 and 30 = 60 units (total work)

P = 3 units => P’s 2/3rd efficiency = 3 * 2/3 = 2 units

Q = 6 units => Q’s 5/6th efficiency = 6 * 5/6 = 5 units

R = 2 units=> R’s 3/2nd efficiency = 2 * 3/2 = 3 units

Required number of days = 60/(2 + 5 + 3) = 60/10 = 6 days

/ 5. Reviews