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.
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Question 1 of 10
1. Question
Directions (1 – 5): Find the wrong number in the given series
96, 46, 72, 180, 630, 2835
Correct
Answer: c)
96 * 0.5 = 48 (not 46)
48 * 1.5 = 72
72 * 2.5 = 180
180 * 3.5 = 630
630 * 4.5 = 2835
Incorrect
Answer: c)
96 * 0.5 = 48 (not 46)
48 * 1.5 = 72
72 * 2.5 = 180
180 * 3.5 = 630
630 * 4.5 = 2835
Question 2 of 10
2. Question
Directions (1 – 5): Find the wrong number in the given series
42, 86, 348, 2094, 16758
Correct
Answer: d)
42 * 2 + 2 = 86
86 * 4 + 4 = 348
348 * 6 + 6 = 2094
2094 * 8 + 8 =16760 (16758)
Incorrect
Answer: d)
42 * 2 + 2 = 86
86 * 4 + 4 = 348
348 * 6 + 6 = 2094
2094 * 8 + 8 =16760 (16758)
Question 3 of 10
3. Question
Directions (1 – 5): Find the wrong number in the given series
48, 1764, 1632, 2622, 2530
Correct
Answer: b)
Incorrect
Answer: b)
Question 4 of 10
4. Question
Directions (1 – 5): Find the wrong number in the given series
1582, 1602, 1634, 1690, 1782, 1920
Correct
Answer: d)
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
Incorrect
Answer: d)
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
Question 5 of 10
5. Question
Directions (1 – 5): Find the wrong number in the given series
124, 63, 64, 130, 524, 4200, 67214
Correct
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)
Incorrect
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)
Question 6 of 10
6. Question
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.
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?
Correct
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
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
1^{st} cycle (2 days) = 2 units
5^{th} cycle = 2*5 = 10 units
Total number of days = 10 days
Incorrect
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
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
1^{st} cycle (2 days) = 2 units
5^{th} cycle = 2*5 = 10 units
Total number of days = 10 days
Question 7 of 10
7. Question
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.
Find the number of days taken to complete the whole work working all together?
Correct
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
Incorrect
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
Question 8 of 10
8. Question
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.
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?
Correct
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 â€“ 44 =60
16x = 104
X = 104/16 =26/4 = 6 2/4 = 6 Â½ days
Incorrect
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 â€“ 44 =60
16x = 104
X = 104/16 =26/4 = 6 2/4 = 6 Â½ days
Question 9 of 10
9. Question
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.
All of them working together and P gets wage of Rs. 1200. Find the total wage?
Correct
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
Incorrect
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
Question 10 of 10
10. Question
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.
P, Q and R started working with 2/3^{rd}, 5/6^{th} and 3/2^{nd} of their original efficiency respectively. Find the number of days required to complete the whole work?
Correct
Answer: c)
LCM of 10, 20 and 30 = 60 units (total work)
P = 3 units => Pâ€™s 2/3^{rd} efficiency = 3 * 2/3 = 2 units
Q = 6 units => Qâ€™s 5/6^{th} efficiency = 6 * 5/6 = 5 units
R = 2 units=> Râ€™s 3/2^{nd} efficiency = 2 * 3/2 = 3 units
Required number of days = 60/(2 + 5 + 3) = 60/10 = 6 days
Incorrect
Answer: c)
LCM of 10, 20 and 30 = 60 units (total work)
P = 3 units => Pâ€™s 2/3^{rd} efficiency = 3 * 2/3 = 2 units
Q = 6 units => Qâ€™s 5/6^{th} efficiency = 6 * 5/6 = 5 units
R = 2 units=> Râ€™s 3/2^{nd} efficiency = 2 * 3/2 = 3 units
Required number of days = 60/(2 + 5 + 3) = 60/10 = 6 days
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 2/3^{rd}, 5/6^{th} and 3/2^{nd} 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)
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
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) 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
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
1^{st} cycle (2 days) = 2 units
5^{th} 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 â€“ 44 =60
16x = 104
X = 104/16 =26/4 = 6 2/4 = 6 Â½ 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 => Pâ€™s 2/3^{rd} efficiency = 3 * 2/3 = 2 units
Q = 6 units => Qâ€™s 5/6^{th} efficiency = 6 * 5/6 = 5 units
R = 2 units=> Râ€™s 3/2^{nd} efficiency = 2 * 3/2 = 3 units
Required number of days = 60/(2 + 5 + 3) = 60/10 = 6 days
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