QUICK TEST QUANT FOR IBPS
If the price of sugar is raised by 60%, by how much per cent must a person reduce his
consumption so as not to increase his expenditure on sugar?
1) 30%
2) 37.5%
3) 60%
4) 40%
5) 62.5%
2.
Ramesh and Mohan can do a piece of work in 15 days. Ramesh is 3 times as efficient as
Mohan. In how many days can Ramesh alone finish the work?
1) 21 days
2) 24 days
3) 28 days
4) 30 days
5) None of these
3.
The sum of ages of three persons A, B and C together is 167 years. B is
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1.
5
A, and C is 44 years older than A. What is the age of B?
1) 45 years
2) 48 years
3) 51 years
4) 60 years
17
times as old as
12
5) 68 years
4.
Mixture A contains 40% milk and Mixture B contains 30%. A container is filled with seven
parts of Mixture A and thirteen parts of Mixture B. Find the percentage of milk in the container?
1) 32.5%
2) 33.5%
3) 35%
4) 37.5%
5) 36%
5.
Two stations A and B are 333 km apart. A train leaves station A for station B and at the same
time another train leaves B for A. Both trains meet 4.5 hours after they start moving. If the train
which starts from station A is 14 km/hr faster than the other one, what is the ratio of the speeds
of both trains?
1) 3 : 2
2) 5 : 3
3) 4 : 3
4) 7 : 4
5) None of these
6.
A shopkeeper has 120 kg sugar. He sells a certain amount of sugar at 20% profit and the rest
at 55% loss. In the whole process his loss is 5%. Find the difference between the amount of
sugar he sold at 20% profit and the amount he sold at 55% loss.
1) 10 kg
2) 20 kg
3) 40 kg
4) 60 kg
5) 80 kg
7.
The length and breadth of a rectangular park are 100m and 400m respectively. If the area of a
path which runs all along the perimeter inside the park is 1056 sqm, what is the width of the
path?
1) 4 metres
2) 5 metres
3) 6 metres
4) 7.5 metres
5) 8 metres
8.
A person can row 11 kmh–1 in still water. If he takes thrice as much to row up as to row down
a river, what is the speed of the stream?
1) 3 kmh–1
2) 4 kmh–1
3) 4.5 kmh–1
4) 5 kmh–1
5) 5.5 kmh–1
9.
A train after covering 1200 km meets with an accident and due to this it proceeds at
2
of its
3
original speed and arrives at the station in 5 hours. Had the accident taken place after covering
a distance of 300 km further, it would have arrived 2.5 hours earlier. What is the speed of the
train?
1) 30 kmh–1
2) 45 kmh–1
3) 48 kmh–1
4) 54 kmh–1
5) 60 kmh–1
10. A person drives a certain distance on a car at the rate of 25 kmh–1 and returns to its starting
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point on a bicycle at the rate of 10 kmh–1. The whole journey took 11 hours and 12 minutes.
Find the distance travelled by him on car.
1) 40 km
2) 50 km
3) 60 km
4) 70 km
5) 80 km
ANSWERS WITH EXPLANATIONS:
1. 2; New price of sugar =
160
8
× old price = × old price
100
5
 Old price of 1 kg = New price of
So, required reduction = 1 –
5
kg
8
5 3
 kg
8 8
3
 100  37.5%
8
2. 2; Let Ramesh’s one day’s work = 5x and Mohan’s one day’s work = 3x
 % reduction =
(Ramesh + Mohan)’s one day’s work =
or 5x + 3x =
1
15
1
15
1
1

15  8 120
 Ramesh’s one day’s work = 5x
x=
1
1

120 24
 Ramesh can finish it in 24 days.
= 5
3. 3; Let the age of A be ‘x’.
So, B’s age =
or, x 
17 x
and C = x + 44
12
17 x
 x  44  167
12
12 x  17 x  12x
 167 – 44  123
12
or, 41x = 12 × 123
or,
x 
12 123
 36
41
 Age of B = 36 
17
 51years
12
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4. 2; Percentage of milk 
40
30
 13 
100
100  100
7  13
28 39

10
10  100  67  100  33.5%

20
10 20
5. 5; Let the speed of the first train be x and that of the second train be y.
 x – y = 14 ...(I)
333
 74
...(II)
4. 5
Adding these two equations,
2x = 14 + 74 = 88
 x = 44 and y = 30
x+y=
 Ratio =
44 22

30 15
6. 3; Let the quantity he sold at 20% profit be x.
 The quantity he sold at 55% loss is (120 – x).
Let the price of sugar per kg be `1
 20% of x – 55% of (120 – x)
= – 120 
 20 
or,
5
 –6
100
(–ve for loss)
x
55
5
–
(120 – x )  –120
 –6
100 100
100
x 11
– (120 – x )  –6
5 20
4x – 1320  11x
 –6
20
or, 15x – 1320 = –120
or,
1200
 80kg
15
 Rest = 120 – 80 = 40kg
 Difference = 80 – 40 = 40kg
x 
7. 1; Let the width of the path be x metres.
 Area of path = 2{100 × 4 + (40 – 2x)x} = 1056
 400 + 40x – 2x2 = 528
 2x2 – 40x + 128 = 0
 x2 – 20x + 64 = 0
 Solving the eqn, x = 4 metres
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8. 3; Let the speed of the stream be ‘x’ kmph.
Let the distance covered by him down stream and upstream be D.
3 D
D

11  x (11 – x )
 4x = 33 – 11 = 22
33 – 3x = 11 + x
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
x 
22
 5.5 kmph
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9. 5; Let the distance be ‘x’ and the original speed be ‘y’.
1200 (x – 1200) 3 x

 5
y
2y
y
 x – 10y – 1200 = 0 ...(I)
1500 ( x – 1500)  3 x 5

 
y
2y
y 2
 x – 5y – 1500 = 0
...(II)
From eqn (I) and (II), x = 1800 km
y = 60 kmph
10. 5; Let the distance be ‘x’ km.

x x
  11.2
25 10
2x  5x
 11.2
50
or 7x = 11.2 × 50 = 560
 x = 80 ie distance = 80 km
or
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