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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeQ (15) What value should come in place of the x in the following questions?
6^{x1} = 24 Ã— 108 Ã— 648
Correct
Explanation
24 = 2^{3} * 3^{1}
108 = 2^{2} * 3^{3}
648 = 2^{3} * 3^{4}
So, 24 Ã— 108 Ã— 648 = 2^{3+2+3 }* 3^{1+3+4 }= 2^{8} * 3^{8} = 6^{8}
6^{x1 }= 6^{8}
x â€“ 1 = 8
Solve, x = 9Incorrect
Explanation
24 = 2^{3} * 3^{1}
108 = 2^{2} * 3^{3}
648 = 2^{3} * 3^{4}
So, 24 Ã— 108 Ã— 648 = 2^{3+2+3 }* 3^{1+3+4 }= 2^{8} * 3^{8} = 6^{8}
6^{x1 }= 6^{8}
x â€“ 1 = 8
Solve, x = 9 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeQ (15) What value should come in place of the x in the following questions?
âˆš1849 Ã— 1326 Ã· 51 + x% of 350 â€“ 12% of 950 = 1165
Correct
Explanation
43 Ã— 26 + x% of 350 â€“ 114 = 1165
x = 46Incorrect
Explanation
43 Ã— 26 + x% of 350 â€“ 114 = 1165
x = 46 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeQ (15) What value should come in place of the x in the following questions?
Correct
Explanation
2 + 5 â€“ 3 + 1 â€“ 1 + (3/4 + 1/6 â€“ 5/8 + 7/10 â€“ 5/12)
= 4 + 69/120 = 4 23/40Incorrect
Explanation
2 + 5 â€“ 3 + 1 â€“ 1 + (3/4 + 1/6 â€“ 5/8 + 7/10 â€“ 5/12)
= 4 + 69/120 = 4 23/40 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeQ (15) What value should come in place of the x in the following questions?
143^{2} + 188^{2} + 201^{2} â€“ 142^{2} â€“ 186^{2} â€“ 199^{2}=x
Correct
Explanation
143^{2}Â + 188^{2}Â + 201^{2}Â â€“ 142^{2}Â â€“ 186^{2}Â â€“ 199^{2}
= (143^{2}Â â€“ 142^{2}) + (188^{2}Â â€“ 186^{2}) + (201^{2}Â â€“ 199^{2})
Using a^{2}Â â€“ b^{2}Â = (a+b) (ab), we have:
(143 + 142) (143 â€“ 142) + (188 + 186) (188 â€“ 186) + (201 + 199) (201 â€“ 199)
= 285 x 1 + 374 x 2 + 400 x 2
= 1833Incorrect
Explanation
143^{2}Â + 188^{2}Â + 201^{2}Â â€“ 142^{2}Â â€“ 186^{2}Â â€“ 199^{2}
= (143^{2}Â â€“ 142^{2}) + (188^{2}Â â€“ 186^{2}) + (201^{2}Â â€“ 199^{2})
Using a^{2}Â â€“ b^{2}Â = (a+b) (ab), we have:
(143 + 142) (143 â€“ 142) + (188 + 186) (188 â€“ 186) + (201 + 199) (201 â€“ 199)
= 285 x 1 + 374 x 2 + 400 x 2
= 1833 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeQ (15) What value should come in place of the x in the following questions?
131.01% of 457.87 + 341.005% of 129.95 = 259.99% of x
Correct
Incorrect

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeRoy, VInay, and Peter can do a work in 18, 30, and 45 days respectively. If they start work with Roy works the first day, Peter the second day and Vinay the third and fourth day. If this process continues than find in how many days they will complete the work?
Correct
Explanation
Roy = 18, Vinay = 30, Peter = 45
LCM = 90
A = 90/18 = 5, B = 3, C = 2
1st dayâ€¦â€¦â€¦â€¦2nd dayâ€¦â€¦â€¦â€¦â€¦3rd day and 4th day
Roy5â€¦â€¦â€¦â€¦ Peter=2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦. Vinay=3+3
than
4 daysâ€™ work = 5 + 2 + 6 = 13
Make it near total (90)
4*6â€¦â€¦â€¦â€¦ 13*6
24â€¦â€¦â€¦ 78
Roy > 1 â€¦ 5
Peter > 1â€¦â€¦â€¦ 2
Vinay> 1â€¦â€¦â€¦3
Add
24+1+1+1=27 daysâ€¦â€¦â€¦â€¦ (78+5+2+3) = 88 days
Now 9088 = 2unit work pending
Vinay does 3unit work in 1 day, so 2 in 2/3. So total 27 2/3 daysIncorrect
Explanation
Roy = 18, Vinay = 30, Peter = 45
LCM = 90
A = 90/18 = 5, B = 3, C = 2
1st dayâ€¦â€¦â€¦â€¦2nd dayâ€¦â€¦â€¦â€¦â€¦3rd day and 4th day
Roy5â€¦â€¦â€¦â€¦ Peter=2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦. Vinay=3+3
than
4 daysâ€™ work = 5 + 2 + 6 = 13
Make it near total (90)
4*6â€¦â€¦â€¦â€¦ 13*6
24â€¦â€¦â€¦ 78
Roy > 1 â€¦ 5
Peter > 1â€¦â€¦â€¦ 2
Vinay> 1â€¦â€¦â€¦3
Add
24+1+1+1=27 daysâ€¦â€¦â€¦â€¦ (78+5+2+3) = 88 days
Now 9088 = 2unit work pending
Vinay does 3unit work in 1 day, so 2 in 2/3. So total 27 2/3 days 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeTen years ago Anju was half of veena in age. If the ratio of their present ages is 3 : 4, what will be the total of their present ages
Correct
Explanation
Let Anjuâ€™s age 10 years ago = x years.
Then, Veena’s age 10 years ago = 2x years.
(x + 10) / (2x+ 10) = 3/4
=> x = 5.
So, the total of their present ages = (x + 10 + 2x + 10)
= (3x + 20) = 35 years.Incorrect
Explanation
Let Anjuâ€™s age 10 years ago = x years.
Then, Veena’s age 10 years ago = 2x years.
(x + 10) / (2x+ 10) = 3/4
=> x = 5.
So, the total of their present ages = (x + 10 + 2x + 10)
= (3x + 20) = 35 years. 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeA hollow cylindrical tube is open at both ends is made of iron 4cm thick. If the external diameter be 52cm and the length of the tube be 120cm, find the number of cubic cm of iron in it? approx.
Correct
Explanation
H = 120
external diameter â€“ 52
External radius = 26
Internal radius = 264 =22
Volume of iron = external volume â€“ internal volume
22/7 * 26 * 26 * 120 â€“ 22/7 * 22 * 22 * 120 = 72411Incorrect
Explanation
H = 120
external diameter â€“ 52
External radius = 26
Internal radius = 264 =22
Volume of iron = external volume â€“ internal volume
22/7 * 26 * 26 * 120 â€“ 22/7 * 22 * 22 * 120 = 72411 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe average weight of 18 workers is 50 kg. A new worker replaced an old worker whose weight is 45 kg. Hence, the average weight of the workers increased by 1 kg. What is the ratio between the old workman and new workman?
Correct
Incorrect

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA man can row 7 (1/2) km/h in still water. If in a river running at 1.5 km an hour it takes him 50 minutes to row to a place and back, how far off is the place?
Correct
Explanation
Speed Downstream = (7.5 + 1.5) kmph = 9 kmph
Speed Upstream = (7.5 – 1.5) kmph = 6 kmph
Let the required distance be x km. Then,
x/9 + x/6 = 50/60
2x + 3x = 5/6 Ã— 18 = 5x = 15, x = 3
Hence the required distance is 3 kmIncorrect
Explanation
Speed Downstream = (7.5 + 1.5) kmph = 9 kmph
Speed Upstream = (7.5 – 1.5) kmph = 6 kmph
Let the required distance be x km. Then,
x/9 + x/6 = 50/60
2x + 3x = 5/6 Ã— 18 = 5x = 15, x = 3
Hence the required distance is 3 km
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