Hello and welcome to exampundit. Here is a set of Quantitative Aptitude Quiz for SBI PO mains 2019. The following set is based on Quadratic Equations & Misc Questions.
Quantitative Aptitude Quiz for SBI PO Mains Day 2
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Question 1 of 10
1. Question
The flow of water through a drainage pipe was monitored for a 3hour period. In the second hour, the rate of flow was 21 gallons per hour, which was 50 percent faster than the rate of flow for the first hour. If 25 percent more water flowed through the pipe in the third hour than it did in the second, how many gallons of water flowed through the pipe during the entire three hours?
Correct
Solution:
Rate of flow of water in second hour = 21 gallons per hour
Rate of flow of water in first hour = 21/(3/2) = 14 gallons per hour
Rate of flow of water in third hour = (125/100)*21 = (5/4) * 15 = 26.25 gallons per hour
Number of gallons of water that flowed through the pipe during the entire 3 hours = 14+21+26.25 = 61.25 gallonsIncorrect
Solution:
Rate of flow of water in second hour = 21 gallons per hour
Rate of flow of water in first hour = 21/(3/2) = 14 gallons per hour
Rate of flow of water in third hour = (125/100)*21 = (5/4) * 15 = 26.25 gallons per hour
Number of gallons of water that flowed through the pipe during the entire 3 hours = 14+21+26.25 = 61.25 gallons 
Question 2 of 10
2. Question
While working individually, each of A, B, C and D can produce 60 units of a certain item in a, b, c and d hours respectively, where a, b, c, d are consecutive integers in increasing order. If A and C together can produce 70 units in more than 3 hours, then which of the following can be the time they will take to produce 70 units, if all are working together?
Correct
Solution:
so a, b, c and d can be written as a, a+1, a+2 and a+3…
If A and C together can produce 70 units in more than 3 hours =1/a+1/c<1/3 Just substitute a a as 1.. c is 3.. 1+(1/3)=4/3>1/3
a as 3.. c is 5…(1/3)+(1/5)=8/15>1/3
So, minimum value of a is 5( if we substitute a as 5 and c as 8 then their value will be less then 1/3)
Thus the least time all four will take is if a, b, c, and d are5,6,7 and 8
1/5 + 1/6 + 1/7 + 1/8 = 6*7*8 + 5*7*8 + 5*6*8 + 5*6/ 5*6*7*8= 634/1000 hours
= 634*60/1000 minutes
= 38.04 minutesIncorrect
Solution:
so a, b, c and d can be written as a, a+1, a+2 and a+3…
If A and C together can produce 70 units in more than 3 hours =1/a+1/c<1/3 Just substitute a a as 1.. c is 3.. 1+(1/3)=4/3>1/3
a as 3.. c is 5…(1/3)+(1/5)=8/15>1/3
So, minimum value of a is 5( if we substitute a as 5 and c as 8 then their value will be less then 1/3)
Thus the least time all four will take is if a, b, c, and d are5,6,7 and 8
1/5 + 1/6 + 1/7 + 1/8 = 6*7*8 + 5*7*8 + 5*6*8 + 5*6/ 5*6*7*8= 634/1000 hours
= 634*60/1000 minutes
= 38.04 minutes 
Question 3 of 10
3. Question
A can build a wall in 10 days, working alone, B can build the same wall in 20 days, working alone and C can break the entire built wall in 8 days, working alone. The three of them work alone on the wall on successive days with A working on first day, B on second day and C on third day and the cycle then repeats. In how many days will the wall be built for the first time?
Correct
Solution:
The work in three days = (1/10)+(1/20)(1/8)=1/40
Most of us will go wrong in taking 40*3=120 as the answer. That will be the time when the one wall will always be there. But there will be many times when the wall gets completed but some part gets broken by C
Now, our answer will never be a multiple of 3, since the work would complete before C comes into action
Now in one day C breaks 1/8 wall, so we look for the work less this 1(1/8)=7/8 = 35/40
35/40 : 34/40 will be done by 34 days of all three.. next day it will be 1/10, so 34/40+1/10=38/40
Next day it will be 1/20, so 38/40+(1/20)=40/40=1
So our answer is 35 days of A and B and 34 days of C = 35+35+34=104Incorrect
Solution:
The work in three days = (1/10)+(1/20)(1/8)=1/40
Most of us will go wrong in taking 40*3=120 as the answer. That will be the time when the one wall will always be there. But there will be many times when the wall gets completed but some part gets broken by C
Now, our answer will never be a multiple of 3, since the work would complete before C comes into action
Now in one day C breaks 1/8 wall, so we look for the work less this 1(1/8)=7/8 = 35/40
35/40 : 34/40 will be done by 34 days of all three.. next day it will be 1/10, so 34/40+1/10=38/40
Next day it will be 1/20, so 38/40+(1/20)=40/40=1
So our answer is 35 days of A and B and 34 days of C = 35+35+34=104 
Question 4 of 10
4. Question
Two pipes can separately fill a tank in 15 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
Correct
Solution:
Let’s assume the size of the tank to be LCM(15,30) = 30 units.
The individual rates of the two pipes filling are 2 units &1 units respectively.
Together the tanks fill 3 units in an hour. When the tank is 1/ 3rd full(10 units have been filled). The leak develops
causing the tank to fill at 2/3rds of its usual rate. The time taken to fill the first 10 units is 10/3 = 3.33 hour
For the remaining 20 units, the tanks will fill (3*2/3) = 2 units in an hour.
Because of the leak. the pipe takes 20/2 = 10 hours to fill the remaining tank.
Therefore, the total time taken for the two pipes to fill the tank is 3.33 + 10 = 13.33 hourIncorrect
Solution:
Let’s assume the size of the tank to be LCM(15,30) = 30 units.
The individual rates of the two pipes filling are 2 units &1 units respectively.
Together the tanks fill 3 units in an hour. When the tank is 1/ 3rd full(10 units have been filled). The leak develops
causing the tank to fill at 2/3rds of its usual rate. The time taken to fill the first 10 units is 10/3 = 3.33 hour
For the remaining 20 units, the tanks will fill (3*2/3) = 2 units in an hour.
Because of the leak. the pipe takes 20/2 = 10 hours to fill the remaining tank.
Therefore, the total time taken for the two pipes to fill the tank is 3.33 + 10 = 13.33 hour 
Question 5 of 10
5. Question
Together, 15 type A machines and 7 type B machines can complete a certain job in 4 hours. Together 8 type B machines and 15 type C machines can complete the same job in 11 hours. How many hours would it take one type A machine, one type B machine, and one type C machine working together to complete the job (assuming constant rates for each machine)?
Correct
Solution:
Let a, b, and c be the times, in hours, it takes for 1 type A, B, and C machine to finish the job by itself, respectively. So we can create the equations (notice that, for example, 1/a will be the work done by 1 type A machine per hour and n/a will be the work done by n type A machines per hour):
4(15/a) + 4(7/b) = 1
and 11(8/b) + 11(15/c) = 1
If we divide the first equation by 4 and the second by 11, we have:
15/a + 7/b = 1/4
And
8/b + 15/c = 1/11
Now, adding the two new equations, we have:
15/a + 15/b + 15/c = 1/4 + 1/11
15(1/a + 1/b + 1/c) = 15/44
1/a + 1/b + 1/c = 1/44
Thus, the combined rate of 1 type A, B and C machine is 1/44. That is, together they finish 1/44 of the job in one hour. Therefore, it will take them 1/(1/44) = 44 hoursIncorrect
Solution:
Let a, b, and c be the times, in hours, it takes for 1 type A, B, and C machine to finish the job by itself, respectively. So we can create the equations (notice that, for example, 1/a will be the work done by 1 type A machine per hour and n/a will be the work done by n type A machines per hour):
4(15/a) + 4(7/b) = 1
and 11(8/b) + 11(15/c) = 1
If we divide the first equation by 4 and the second by 11, we have:
15/a + 7/b = 1/4
And
8/b + 15/c = 1/11
Now, adding the two new equations, we have:
15/a + 15/b + 15/c = 1/4 + 1/11
15(1/a + 1/b + 1/c) = 15/44
1/a + 1/b + 1/c = 1/44
Thus, the combined rate of 1 type A, B and C machine is 1/44. That is, together they finish 1/44 of the job in one hour. Therefore, it will take them 1/(1/44) = 44 hours 
Question 6 of 10
6. Question
The radius of a cylindrical water tank is reduced by 50%. However, the speed by which water is filled into the tank is also decreased by 50%. How much more or less time will it take to fill the tank now?
Correct
Solution:
(VC)Volume of the cylinder is directly proportional to R^2.
So if radius is 50% less volume will be 1/4th of the original volume(VC/4)
Now if with velocity V tank can be filled in T1 time of volume VC
So now Velocity is 50% less i..e V/2
So time taken to fill the capacity VC/4 by V/2 velocity is T2.
VT1 = VC
V/2*T2 = VC/4
So T1/T2 = 1/2
So Tank will be filled in less time. that is 50 % less.Incorrect
Solution:
(VC)Volume of the cylinder is directly proportional to R^2.
So if radius is 50% less volume will be 1/4th of the original volume(VC/4)
Now if with velocity V tank can be filled in T1 time of volume VC
So now Velocity is 50% less i..e V/2
So time taken to fill the capacity VC/4 by V/2 velocity is T2.
VT1 = VC
V/2*T2 = VC/4
So T1/T2 = 1/2
So Tank will be filled in less time. that is 50 % less. 
Question 7 of 10
7. Question
A flat, rectangular bed with an area of 2,400 square feet is bordered by a fence on three sides and by a walkway on the fourth side. If the entire length of the fence is 140 feet, which of the following could be the length, in feet, of one of the sides of the bed?
I. 30
II. 60
III. 80A. I only
B. II onlyCorrect
Solution:
Area of bed = 2400
Fence covers 3 sides
Let sides be L and B
2(L+B)=140
L*B =2400
Take help from options L= 80 and B =30Incorrect
Solution:
Area of bed = 2400
Fence covers 3 sides
Let sides be L and B
2(L+B)=140
L*B =2400
Take help from options L= 80 and B =30 
Question 8 of 10
8. Question
A right circular cylinder having the radius of its base as 2 cm is filled with water upto a height of 2 cm. This water is then poured into an empty rectangular container the dimensions of whose base are 2π by 3 cm. If the volume of water in the rectangular container is increased by 50 percent by adding extra water, what is the final height, in cm, of the water level in cm in the rectangular container?
Correct
Solution:
Volume of the Cylinder = πr2h
Volume of water in the Cylinder = π∗22∗2=8π
Area of Rectangle container = lb(h) because of 3d figure
Given, lb = 2π*3
Area = 6πh
Volume of water remains the same when it is poured it the rectangular container.
8π = 6πh
h = 4/3 cm
Now, the volume of water is increased by 50%. Means 8π = 12π
12π = 6πh
h = 2cmIncorrect
Solution:
Volume of the Cylinder = πr2h
Volume of water in the Cylinder = π∗22∗2=8π
Area of Rectangle container = lb(h) because of 3d figure
Given, lb = 2π*3
Area = 6πh
Volume of water remains the same when it is poured it the rectangular container.
8π = 6πh
h = 4/3 cm
Now, the volume of water is increased by 50%. Means 8π = 12π
12π = 6πh
h = 2cm 
Question 9 of 10
9. Question
A circle is inscribed in a square with an area of 121 square units. If the diameter of a marble ball is 1.5π, then how many different combinations of 9 distinct marbles can fit around the circumference of the circle?
Correct
Solution:
Area of square = 121 sq units. Hence side = sqrt(121) = 11 units.
As circle is inscribed in square, so the diameter = 11.
Circumference = 2π x 11/2 = 11π
Diameter of each marble = 1.5π
So maximum possible marbles possible on circumference = 1.5π x 7 = 10.5π < 11π
Number of ways to select 7 out of 9 possible marbles = 9C7 = 36.
To arrange these 7 marbles around circumference = (7 – 1 )! = 720.
Total arrangements = 36 x 720 = 25920.Incorrect
Solution:
Area of square = 121 sq units. Hence side = sqrt(121) = 11 units.
As circle is inscribed in square, so the diameter = 11.
Circumference = 2π x 11/2 = 11π
Diameter of each marble = 1.5π
So maximum possible marbles possible on circumference = 1.5π x 7 = 10.5π < 11π
Number of ways to select 7 out of 9 possible marbles = 9C7 = 36.
To arrange these 7 marbles around circumference = (7 – 1 )! = 720.
Total arrangements = 36 x 720 = 25920. 
Question 10 of 10
10. Question
Two right circular cylinders have the same volume. The radius of the second cylinder is 10% more than the radius of the first. What is the relationship between the heights of the two cylinders?
Correct
Solution:
π*r12*h1=π*r22*h2
given
r1=1.1r2
so
we can say
1.21*h1=h2(21% more than second)Incorrect
Solution:
π*r12*h1=π*r22*h2
given
r1=1.1r2
so
we can say
1.21*h1=h2(21% more than second)
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