Important Quant Miscellaneous Questions for IBPS Clerk Exams
Moderate Quant Miscellaneous Questions. Important Quant Miscellaneous Questions for IBPS Clerk Prelims 2018. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS Clerk exams 201718, you will come across a section on Quantitative Aptitude Section. Here we are providing you “Important Quant Miscellaneous Questions for IBPS Clerk Prelims 2018“ based on the latest pattern for your daily practice.
This “Important Quant Miscellaneous Questions for IBPS Clerk 2018 “ is also important for other banking exams such as SBI PO, IBPS PO, IBPS Clerk, SBI Clerk, IBPS IBPS SO, SBI SO and other competitive exams
Important Quant Miscellaneous Questions for IBPS Clerk Prelims  Set – 33
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Question 1 of 10
1. Question
Manish and Sumit alone can do a work in ‘x + 4’ days and ‘x + 10’ days. Manish started the work and worked for 16 days and then leaves the work. If the remaining work is done by Sumit and the entire work is completed in 26 days, then find the value of ‘x’.
Correct
A. 20 days
Part of the work done by Manish in 16 days = 16/(x + 4)
Part of the work done by Sumit in 10 days = 10/(x + 10)
So, 16/(x + 4) + 10/(x + 10) = 1
16x + 160 + 10x + 40 = x^{2} + 14x + 40
x^{2} – 12x – 160 = 0
x^{2} + 8x – 20x– 160 = 0
x(x + 8) – 20(x + 8) = 0
x = 8, 20
Since, x cannot be negative.
So, the value of x = 20 days
Incorrect
A. 20 days
Part of the work done by Manish in 16 days = 16/(x + 4)
Part of the work done by Sumit in 10 days = 10/(x + 10)
So, 16/(x + 4) + 10/(x + 10) = 1
16x + 160 + 10x + 40 = x^{2} + 14x + 40
x^{2} – 12x – 160 = 0
x^{2} + 8x – 20x– 160 = 0
x(x + 8) – 20(x + 8) = 0
x = 8, 20
Since, x cannot be negative.
So, the value of x = 20 days

Question 2 of 10
2. Question
A and B together started a business with initial investments of Rs. 10000 and Rs. 15000 respectively. After six months, A invested an additional amount equal to 40% of his initial investment, B withdrew 20% of his initial investment and C joined the business with initial investment of Rs. 30000. Find the profit share of B out of the total profit of Rs. 6324 earned in the business in one year.
Correct
D. Rs. 2108
Amount invested by A = Rs. 10000
Amount invested by B = Rs. 15000
Additional investment made by A after 6 months = 0.40 × 10000 = Rs. 4000
Amount withdrew by B after 6 months = 0.20 × 15000 = Rs. 3000
So, ratio of profit share of A:B:C in one year = (10000 × 6 + 14000 × 6):(15000 × 6 + 12000 × 6):(30000 × 6)
= 144000:162000:180000
= 8:9:10
So, required profit share of B = (9/27) × 6324 = Rs. 2108
Incorrect
D. Rs. 2108
Amount invested by A = Rs. 10000
Amount invested by B = Rs. 15000
Additional investment made by A after 6 months = 0.40 × 10000 = Rs. 4000
Amount withdrew by B after 6 months = 0.20 × 15000 = Rs. 3000
So, ratio of profit share of A:B:C in one year = (10000 × 6 + 14000 × 6):(15000 × 6 + 12000 × 6):(30000 × 6)
= 144000:162000:180000
= 8:9:10
So, required profit share of B = (9/27) × 6324 = Rs. 2108

Question 3 of 10
3. Question
There are two right circular cones on each end of the right circular cylinder in such a way that the radius of the cylinder is equal to the radius of cones and heights of the two cones are in the ratio 9:16. Diameter of the cylinder is 168 cm and its height is 15 cm. Find the curved surface area of the solid structure, if the difference between the volumes of two cones is 362208 cm^{3}.
Correct
B. 72600 cm^{2}
Let the heights of the cones be‘9x’ cm and ‘16x’ cm respectively.
Given, difference in the volumes of cones = 1/3 × 22/7 × 84^{2} × (16x – 9x) = 362208
7x = 49
x = 7
So, the heights of the cones are 112 cm and 63 cm.
Slant height of larger cone = √(112^{2} + 84^{2}) = √(12544 + 7056) = √19600 = 140 cm
Slant height of smaller cone = √(63^{2} + 84^{2}) = √(3969 + 7056) = √11025 = 105 cm
Curved surface area of cones = 22/7 × 84 × (140 + 105) = 64680 cm^{2}
Curved surface area of cylinder = 2 × 22/7 × 84 × 15 = 7920 cm^{2}
Hence curved surface area of the given solid = 64680 + 7920 = 72600 cm^{2}
Incorrect
B. 72600 cm^{2}
Let the heights of the cones be‘9x’ cm and ‘16x’ cm respectively.
Given, difference in the volumes of cones = 1/3 × 22/7 × 84^{2} × (16x – 9x) = 362208
7x = 49
x = 7
So, the heights of the cones are 112 cm and 63 cm.
Slant height of larger cone = √(112^{2} + 84^{2}) = √(12544 + 7056) = √19600 = 140 cm
Slant height of smaller cone = √(63^{2} + 84^{2}) = √(3969 + 7056) = √11025 = 105 cm
Curved surface area of cones = 22/7 × 84 × (140 + 105) = 64680 cm^{2}
Curved surface area of cylinder = 2 × 22/7 × 84 × 15 = 7920 cm^{2}
Hence curved surface area of the given solid = 64680 + 7920 = 72600 cm^{2}

Question 4 of 10
4. Question
The yearly income of Raju was Rs. ‘y’ in 2015 and his income increases by 10% every year with respect to the income of previous year. He deposits 20% of his income every year in a saving account offering compound interest of 10% pa compounded annually. If the amount in his saving account in 2018 is Rs. 85,184, then find the value of y.
Correct
A. Rs. 80000
Income of Raju in 2015 = Rs. y
Amount deposited by him in saving account in 2015 = y × 0.20 = Rs. 0.2y
Income of Raju in 2016 = y × 1.10 = Rs. 1.1y
Amount deposited by him in saving account in 2016 = 1.1y × 0.20 = Rs. 0.22y
Income of Raju in 2017 = 1.1y × 1.10 = Rs. 1.21y
Amount deposited by him in saving account in 2017 = 1.21y × 0.20 = Rs. 0.242y
Income of Raju in 2018 = 1.21y × 1.10 = Rs. 1.331y
Amount deposited by him in saving account in 2018 = 1.331y × 0.20 = Rs. 0.2662y
Amount in the saving account in 2018 = 0.2y(1 + 0.10)^{3} + 0.22y(1+ 0.10)^{2} + 0.242y(1+ 0.10) + 0.2662y = 85184
0.2y × 1.331 + 0.22y × 1.21 + 0.242y × 1.1 + 0.2662y = 85184
0.2662y + 0.2662y + 0.2662y + 0.2662y = 85184
1.0648y = 85184
y = 80000
So, the value of y = Rs. 80000
Incorrect
A. Rs. 80000
Income of Raju in 2015 = Rs. y
Amount deposited by him in saving account in 2015 = y × 0.20 = Rs. 0.2y
Income of Raju in 2016 = y × 1.10 = Rs. 1.1y
Amount deposited by him in saving account in 2016 = 1.1y × 0.20 = Rs. 0.22y
Income of Raju in 2017 = 1.1y × 1.10 = Rs. 1.21y
Amount deposited by him in saving account in 2017 = 1.21y × 0.20 = Rs. 0.242y
Income of Raju in 2018 = 1.21y × 1.10 = Rs. 1.331y
Amount deposited by him in saving account in 2018 = 1.331y × 0.20 = Rs. 0.2662y
Amount in the saving account in 2018 = 0.2y(1 + 0.10)^{3} + 0.22y(1+ 0.10)^{2} + 0.242y(1+ 0.10) + 0.2662y = 85184
0.2y × 1.331 + 0.22y × 1.21 + 0.242y × 1.1 + 0.2662y = 85184
0.2662y + 0.2662y + 0.2662y + 0.2662y = 85184
1.0648y = 85184
y = 80000
So, the value of y = Rs. 80000

Question 5 of 10
5. Question
The monthly expenditure of Rahul in house rent is 12% more than his monthly expenditure in transportation. The monthly expenditure of Rahul in food is 150% of his monthly expenditure in house rent. Monthly expenditure of Rahul on others is 32% of his total income. If monthly savings of Rahul is 17% of his income, then find his monthly income.
Correct
D. Data insufficient
Let the monthly income of Rahul be Rs. ‘a’.
And, let the monthly expenditure of Rahul in transportation = x% of a
Monthly expenditure of Rahul in house rent = 1.12 × (x% of a)
Monthly expenditure of Rahul in food = 1.5×1.12 × (x% of a)
Monthly expenditure of Rahul in others = 32% of a
So, a – (x% of a + 1.12 × (x% of a) + 1.5 × 1.12 × (x% of a) + 32% of a) = 17% of a
Since, we don’t know the value of ‘x’.
Monthly income of Rahul cannot be calculated as data is insufficient.
Incorrect
D. Data insufficient
Let the monthly income of Rahul be Rs. ‘a’.
And, let the monthly expenditure of Rahul in transportation = x% of a
Monthly expenditure of Rahul in house rent = 1.12 × (x% of a)
Monthly expenditure of Rahul in food = 1.5×1.12 × (x% of a)
Monthly expenditure of Rahul in others = 32% of a
So, a – (x% of a + 1.12 × (x% of a) + 1.5 × 1.12 × (x% of a) + 32% of a) = 17% of a
Since, we don’t know the value of ‘x’.
Monthly income of Rahul cannot be calculated as data is insufficient.

Question 6 of 10
6. Question
Mohan sold an article of cost price Rs. 800 at 16% loss. What would be the profit or loss percent if the same article was sold at Rs. 280 more?
Correct
D. 19%
Initially, the selling price of the article was = 84% of 800 = Rs. 672
New selling price of the article = 672 + 280 = Rs. 952
Required profit % = {(952 – 800)/800} × 100 = 19%
Incorrect
D. 19%
Initially, the selling price of the article was = 84% of 800 = Rs. 672
New selling price of the article = 672 + 280 = Rs. 952
Required profit % = {(952 – 800)/800} × 100 = 19%

Question 7 of 10
7. Question
A car takes 6 hours to reach point A from point B. If the speed was decreased by 10% after travelling 120 km then it takes 30 more minutes to reach point A from point B. Find the distance between point A and point B.
Correct
D. 480 km
Let the speed of car be ‘x’ km/h.
So, x × 6 = 120 + (9x/10) ×[6.5 – {120/x}]
6x = 120 + (9/10) × (6.5x – 120)
1.5x = 120
x = 80
Required distance = x × 6 = 480 km
Incorrect
D. 480 km
Let the speed of car be ‘x’ km/h.
So, x × 6 = 120 + (9x/10) ×[6.5 – {120/x}]
6x = 120 + (9/10) × (6.5x – 120)
1.5x = 120
x = 80
Required distance = x × 6 = 480 km

Question 8 of 10
8. Question
The ratio of number of hosteller to number of day scholar in a school is 6:17 respectively, while the ratio of number of girls to number of boys in the school is 13:15, respectively. If (3/8) of total number of students who live in hostel are girls and number of boys who are day scholar is 1200, then find the number of girls in the school.
Correct
D. 1495
Let, number of hosteller and day scholar be ‘6x’ and ‘17x’, respectively.
And, number of girls and boys be in the school be ‘13y’ and ‘15y’, respectively.
So, 6x + 17x = 13y + 15y
23x = 28y
x = 28y/23
Also, 15y – (5/8) × 6x = 1200
15y – (5/8) × 6 × (28y/23) = 1200
15y – (105y/23) = 1200
(345y – 105y)/23 = 1200
240y = 1200 × 23
y = 115
Number of girls in school = 13y = 1495
Incorrect
D. 1495
Let, number of hosteller and day scholar be ‘6x’ and ‘17x’, respectively.
And, number of girls and boys be in the school be ‘13y’ and ‘15y’, respectively.
So, 6x + 17x = 13y + 15y
23x = 28y
x = 28y/23
Also, 15y – (5/8) × 6x = 1200
15y – (5/8) × 6 × (28y/23) = 1200
15y – (105y/23) = 1200
(345y – 105y)/23 = 1200
240y = 1200 × 23
y = 115
Number of girls in school = 13y = 1495

Question 9 of 10
9. Question
A container contains mixture of water and milk in the ratio 9:23, respectively. The milkman sells 320 liters of mixture and then mixes 81 liters of water in the container. After mixing, the ratio of water and milk in the resultant mixture becomes 9:14, respectively. What was the initial quantity of mixture in the container?
Correct
A. 768 liters
Let the amount of water and milk present in the container initially be ‘9x’ liters and ‘23x’ liters, respectively.
So, (9x – 90 + 81)/(23x – 230) = 9/14
(9x – 9)/(23x – 230) = 9/14
126x – 126 = 207x – 2070
81x = 1944
x = 24
Total quantity of mixture in the container = 32x = 32 × 24 = 768 liters
Incorrect
A. 768 liters
Let the amount of water and milk present in the container initially be ‘9x’ liters and ‘23x’ liters, respectively.
So, (9x – 90 + 81)/(23x – 230) = 9/14
(9x – 9)/(23x – 230) = 9/14
126x – 126 = 207x – 2070
81x = 1944
x = 24
Total quantity of mixture in the container = 32x = 32 × 24 = 768 liters

Question 10 of 10
10. Question
There are 5 red, 3 blue, and 4 black balls in a bag. A person draws 2 balls at random from the bag. What will be the probability that both balls are of different colours?
Correct
A. 47/66
Required probability = {(^{5}C_{1}×^{3}C_{1}) + (^{5}C_{1}×^{4}C_{1}) + (^{3}C_{1}×^{4}C_{1})}/^{12}C_{2} =
= {(5 × 3) + (5 × 4) + (3 × 4)}/66
= (15 + 20 + 12)/66
= 47/66
Incorrect
A. 47/66
Required probability = {(^{5}C_{1}×^{3}C_{1}) + (^{5}C_{1}×^{4}C_{1}) + (^{3}C_{1}×^{4}C_{1})}/^{12}C_{2} =
= {(5 × 3) + (5 × 4) + (3 × 4)}/66
= (15 + 20 + 12)/66
= 47/66
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