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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeThe length of a rectangle is 30% more than its breadth, the side of a square is equal to the length of the rectangle. Difference between the areas of square and rectangle is 624m^{2}. Find the length of the rectangle.
Correct
Answer: 3) 52m
Explanation:
Let the breadth of the rectangle = 10x
And length of rectangle = 130% of breadth = 130% of 10x = 13x
Area of rectangle = 13x*10x = 130x^{2}
Also, the side of square = length of rectangle = 13x
Therefore, area of square = 13x*13x = 169x^{2}
ATQ,
169x^{2} – 130x^{2} = 624
39x^{2} = 624
x^{2} = 624/39
x^{2} = 16
x = 4
Therefore, length of rectangle = 13x = 13*4 = 52mIncorrect
Answer: 3) 52m
Explanation:
Let the breadth of the rectangle = 10x
And length of rectangle = 130% of breadth = 130% of 10x = 13x
Area of rectangle = 13x*10x = 130x^{2}
Also, the side of square = length of rectangle = 13x
Therefore, area of square = 13x*13x = 169x^{2}
ATQ,
169x^{2} – 130x^{2} = 624
39x^{2} = 624
x^{2} = 624/39
x^{2} = 16
x = 4
Therefore, length of rectangle = 13x = 13*4 = 52m 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA boat travels upstream distance of 24km in 3 hrs and downstream a distance of 48km in 4 hrs. Find the speed of the stream.
Correct
Answer: 4) 2 km/hr
Explanation:
Let the speed of boat in still water = x km/hr
And speed of stream = y km/hr
Upstream speed of boat = (xy) km/hr
And downstream speed of boat = (x+y) km/hr
ATQ,
24/(xy) = 3
xy = 8 (equation 1)
and 48/(x+y) = 4
x+y = 12 (equation 2)
Subtracting equation 1 from equation 2,
(x+y) – (xy) = 128
2y = 4
y = 2km/hrIncorrect
Answer: 4) 2 km/hr
Explanation:
Let the speed of boat in still water = x km/hr
And speed of stream = y km/hr
Upstream speed of boat = (xy) km/hr
And downstream speed of boat = (x+y) km/hr
ATQ,
24/(xy) = 3
xy = 8 (equation 1)
and 48/(x+y) = 4
x+y = 12 (equation 2)
Subtracting equation 1 from equation 2,
(x+y) – (xy) = 128
2y = 4
y = 2km/hr 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudePipe A can empty a full tank in 50 min and pipe B can fill an empty tank in 36 min. If both the pipes are opened then in how much time can they fill 14/15th of the tank?
Correct
Answer: 1) 2 hours
Explanation:
Let the capacity of the tank is LCM of (36 and 50) = 900 units
Tank filled by B in one hour = 900/36 = 25 units
And tank emptied by A in one hour = 900/50 = 18units
Let the time taken by A and B to fill 14/15th of the tank is x minutes
x*(25 – 18) = (14/15)*900
7x = 840
x = 840/7 = 120 minutes = 2 hoursIncorrect
Answer: 1) 2 hours
Explanation:
Let the capacity of the tank is LCM of (36 and 50) = 900 units
Tank filled by B in one hour = 900/36 = 25 units
And tank emptied by A in one hour = 900/50 = 18units
Let the time taken by A and B to fill 14/15th of the tank is x minutes
x*(25 – 18) = (14/15)*900
7x = 840
x = 840/7 = 120 minutes = 2 hours 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeTwo trains start from the same point and at the same time. Both go in the opposite direction that is one goes in the east direction and the other goes in the west direction. Their speeds are 25 m/s and 15m/s. In how much time the distance between them will become 1036.8km.
Correct
Answer: 2) 432 minutes
Explanation:
Speed of first train = 25m/s = 25*(18/5) km/h = 90 km/h
And speed of second train = 15m/s = 15*(18/5) km/h = 54 km/hr
When both the trains are moving in opposite direction then there relative speed we be the addition of their individual speeds
Relation speed = (90+54) = 144 km/h
Distance between them will become 1036.8 km in = 1036.8/144 = 7.2 hour = 7 hour and 12 minutes = 432 minutesIncorrect
Answer: 2) 432 minutes
Explanation:
Speed of first train = 25m/s = 25*(18/5) km/h = 90 km/h
And speed of second train = 15m/s = 15*(18/5) km/h = 54 km/hr
When both the trains are moving in opposite direction then there relative speed we be the addition of their individual speeds
Relation speed = (90+54) = 144 km/h
Distance between them will become 1036.8 km in = 1036.8/144 = 7.2 hour = 7 hour and 12 minutes = 432 minutes 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA is 8 years younger than C. The ration between the present ages of B and C is 3:11. If the present age of A is thrice the present of B then find the present age of C.
Correct
Answer: 4) 44 years
Explanation:
Let the present age of C = x years and present age of A = x8 years
ATQ,
Age of B : Age of C = 3:11
Age of B : (x) = 3/11
Age of B = 3x/11
Also,
x8 = 3*(3x/11)
x8 = 9x/11
x 9x/11 = 8
(11x9x)/11 = 8
2x/11 = 8
x = 44Incorrect
Answer: 4) 44 years
Explanation:
Let the present age of C = x years and present age of A = x8 years
ATQ,
Age of B : Age of C = 3:11
Age of B : (x) = 3/11
Age of B = 3x/11
Also,
x8 = 3*(3x/11)
x8 = 9x/11
x 9x/11 = 8
(11x9x)/11 = 8
2x/11 = 8
x = 44 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeSum of the price of x bats and (x+5) balls is Rs. 19350. If the price of a bat is Rs. 1250 and the price of a ball is Rs. 350, then what will be the price of (x+3) bats and (x3) balls.
Correct
Answer: 5) Rs. 20300
Explanation:
Price of x bats = x*1250 and price of (x+5) balls = (x+5)*350
Total price of x bats and (x+5) balls is Rs. 19350
x*1250 + (x+5)350 = 19350
1250x + 350x + 1750 = 19350
1600x = 17600
x = 17600/1600
x = 11
Therefore, the price of (11+3) bats and (113) balls we be
(11+3)*1250 + (113)*350
14*1250 + 8*350
17500 + 2800
Rs. 20300Incorrect
Answer: 5) Rs. 20300
Explanation:
Price of x bats = x*1250 and price of (x+5) balls = (x+5)*350
Total price of x bats and (x+5) balls is Rs. 19350
x*1250 + (x+5)350 = 19350
1250x + 350x + 1750 = 19350
1600x = 17600
x = 17600/1600
x = 11
Therefore, the price of (11+3) bats and (113) balls we be
(11+3)*1250 + (113)*350
14*1250 + 8*350
17500 + 2800
Rs. 20300 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe Discount offered on a product is 20% and the profit earned on that product is 12%. If the marked price of that product is Rs. 1568, then what is the cost price of that product?
Correct
Answer: 3) Rs. 1120
Explanation:
Let the cost price of the product is 100x
Selling price = {(100+12)/100}*100x = 112x
MRP*{(10020)/100= 112x
MRP*(80/100) = 112x
MRP = 112x*(100/80)
MRP = 140x
140x = 1568
x = 1568/140
x = 11.2
Cost price of the product = 100x = 100*11.2 = 1120Incorrect
Answer: 3) Rs. 1120
Explanation:
Let the cost price of the product is 100x
Selling price = {(100+12)/100}*100x = 112x
MRP*{(10020)/100= 112x
MRP*(80/100) = 112x
MRP = 112x*(100/80)
MRP = 140x
140x = 1568
x = 1568/140
x = 11.2
Cost price of the product = 100x = 100*11.2 = 1120 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe ratio of the price of a car and a bike is 15:8. The price of the car increases by 10% and the price of the bike decreases by 8%. If the price of the car becomes Rs. 495000, then the price of the bike will become?
Correct
Answer: 1) 220800
Explanation:
Let the initial price of car is 15x and price of bike is 8x
Now, the price of car becomes
15x*{(100+10)/100}
15x*(110/100)
15x*(11/10)
16.5x
16.5x = 495000
x = 495000/16.5
x = 30000
And the price of bike becomes
8x*{(1008)/100}
8x*(92/100)
8x*(23/25)
7.36x
7.36*30000
220800Incorrect
Answer: 1) 220800
Explanation:
Let the initial price of car is 15x and price of bike is 8x
Now, the price of car becomes
15x*{(100+10)/100}
15x*(110/100)
15x*(11/10)
16.5x
16.5x = 495000
x = 495000/16.5
x = 30000
And the price of bike becomes
8x*{(1008)/100}
8x*(92/100)
8x*(23/25)
7.36x
7.36*30000
220800 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe marked price of a product is Rs. 255 more than the cost price. If 15% discount is offered on the marked price and the profit percent on that product is 36%, then find the cost price of the product?
Correct
Answer: 5) Rs. 425
Explanation:
Let the cost price of the product = x and marked price of the product = x+255
ATQ,
Selling price after discount on marked price = Selling price after profit on CP
(x+255){(10015)/100} = x*{(100+36)/100}
(x+255)*{(85)/100} = x*{136/100}
(x+255)*(85) = x*(136)
85x+255*85 = 136x
136x85x = 255*85
51x = 255*85
x = 425
Cost price of the product = Rs. 425Incorrect
Answer: 5) Rs. 425
Explanation:
Let the cost price of the product = x and marked price of the product = x+255
ATQ,
Selling price after discount on marked price = Selling price after profit on CP
(x+255){(10015)/100} = x*{(100+36)/100}
(x+255)*{(85)/100} = x*{136/100}
(x+255)*(85) = x*(136)
85x+255*85 = 136x
136x85x = 255*85
51x = 255*85
x = 425
Cost price of the product = Rs. 425 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe area of the circle (A) is 2464 cm^{2} and that of the circle (B) is 7546 cm^{2}, what is the difference between the circumference of the two circles (in cm)?
Correct
Answer: 4) 132 cm
Explanation:
Let the radius of circle (A) = r_{a }and radius of circle (B) = r_{b}
Area of circle (A) = 2464 cm^{2}
(22/7)*( r_{a})^{2} = 2464
( r_{a})^{2} = 2464*(7/22)
( r_{a})^{2} = 784
r_{a }= 28
Area of circle (B) = 7546 cm^{2}
(22/7)*( r_{b})^{2} = 7546
( r_{b})^{2} = 7546*(7/22)
( r_{b})^{2} = 2401
r_{b } = 49
Difference of circumference
= 2*(22/7)* r_{b } – 2*(22/7)*r_{a}
= 2*(22/7)*49 – 2*(22/7)*28
= 2*(22/7)(4928)
=2*(22/7)(21)
= 2*22*3
= 132 cmIncorrect
Answer: 4) 132 cm
Explanation:
Let the radius of circle (A) = r_{a }and radius of circle (B) = r_{b}
Area of circle (A) = 2464 cm^{2}
(22/7)*( r_{a})^{2} = 2464
( r_{a})^{2} = 2464*(7/22)
( r_{a})^{2} = 784
r_{a }= 28
Area of circle (B) = 7546 cm^{2}
(22/7)*( r_{b})^{2} = 7546
( r_{b})^{2} = 7546*(7/22)
( r_{b})^{2} = 2401
r_{b } = 49
Difference of circumference
= 2*(22/7)* r_{b } – 2*(22/7)*r_{a}
= 2*(22/7)*49 – 2*(22/7)*28
= 2*(22/7)(4928)
=2*(22/7)(21)
= 2*22*3
= 132 cm
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