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

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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

Direction (1-5) :

96 * 0.5 = 48 (not 46)

48 * 1.5 = 72

72 * 2.5 = 180

180 * 3.5 = 630

630 * 4.5 = 2835

42 * 2 + 2 = 86

86 * 4 + 4 = 348

348 * 6 + 6 = 2094

2094 * 8 + 8 =16760 (16758)

1582Â Â Â Â Â Â Â Â Â Â  1602Â Â Â Â Â Â Â Â Â Â  1634Â Â Â Â Â Â Â Â Â Â  1690Â Â Â Â Â Â Â Â Â Â  1782Â Â Â Â Â Â Â Â Â Â  1922

Â Â Â Â Â Â Â Â Â  20Â Â Â Â Â Â Â Â Â Â Â Â Â Â  32Â Â Â Â Â Â Â Â Â Â Â Â Â Â  56Â Â Â Â Â Â Â Â Â Â Â Â Â Â  92Â Â Â Â Â Â Â Â Â Â Â Â Â Â  140

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  12Â Â Â Â Â Â Â Â Â Â Â Â Â Â  24Â Â Â Â Â Â Â Â Â Â Â Â Â Â  36Â Â Â Â Â Â Â Â Â Â Â Â Â Â  48

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) â€“ (3)

(P+Q+R+S) â€“ (Q+S) = 3/20 + 1/10 â€“ 1/6

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

P+R = 5/60

P + R = 1/12 Ã  (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 â€“ 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 â€“ 44 = 16 units

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

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

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

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 â€“ 44 =60

16x = 104

X = 104/16 =26/4 = 6 2/4 = 6 Â½ days

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

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

P = 3 units => Pâ€™s 2/3rd efficiency = 3 * 2/3 = 2 units

Q = 6 units => Qâ€™s 5/6th efficiency = 6 * 5/6 = 5 units

R = 2 units=> Râ€™s 3/2nd efficiency = 2 * 3/2 = 3 units

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