Quantitative Aptitude - Arithmetic Ability Questions

Q:

A car runs at the speed of 50 kms per hour when not serviced and runs at 60 km./hr. when serviced. After servicing the car covers a certain distance in 6 hours. How much time will the car take to cover the same distance when not serviced?

A) 8.2 hrs B) 6.5 hrs
C) 8 hrs D) 7.2 hrs
 
Answer & Explanation Answer: D) 7.2 hrs

Explanation:

After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km Without servicing, speed of car = 50 kmph => Required time = Distance/speed = 360/50 = 7.2 hrs.

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Filed Under: Time and Distance

2 5230
Q:

78 of 448 + 67 of 3374 = ?

A) 3084 B) 3184
C) 3584 D) None of these
 
Answer & Explanation Answer: D) None of these

Explanation:

78 x 448 + 67 x 3374 => 392 + 2892=> 3284

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Filed Under: Simplification
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3 5229
Q:

The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?

A) 12 B) 13
C) 14 D) 15
 
Answer & Explanation Answer: A) 12

Explanation:

(A+B) - (B-C) = 12 <=> A - C = 12.

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Filed Under: Problems on Ages

3 5225
Q:

Two parallel chords on the same side of the centre of a circle are 5 cm apart. If the chords are 20 and 28 cm long, what is the radius of the circle? 

A) 14.69 cm B) 15.69 cm
C) 18.65 cm D) 16.42 cm
 
Answer & Explanation Answer: B) 15.69 cm

Explanation:

circle11487663359.jpg image

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

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Filed Under: Simplification
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4 5219
Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

A) 6/19 B) 3/10
C) 7/10 D) 6/17
 
Answer & Explanation Answer: B) 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

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Filed Under: Probability
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6 5201
Q:

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?

A) 35/36 B) 17/36
C) 15/36 D) 1/36
 
Answer & Explanation Answer: A) 35/36

Explanation:

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

dice_thrown_simulataneously1532668754.png image

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.

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20 5201
Q:

Rishi bought 30 kg of sugar at the rate of Rs. 9 per kg and 45 kg of sugar at the rate of Rs. 10 per kg. He mixed both type of sugar and sold the mixture at the rate of Rs. 9.75 per kg. What was his gain or loss in the selling of whole sugar?

A) Rs. 10.75 B) Rs. 11.25
C) Rs. 12 D) Rs. 13.50
 
Answer & Explanation Answer: B) Rs. 11.25

Explanation:

Total quantity of sugar = 45 + 30 = 75

Gain or loss can be calculated as

9.75 x 75 - (30 x 9 + 45 x 10)

= 731.25 - 720

= 11.25

 

Hence, in the overall transaction, Rishi got Rs. 11.25 gain.

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Filed Under: Profit and Loss
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7 5197
Q:

A 200 m long train passes a motorcycler, running in the same direction at 12 km/hr, in 15 second and a jeep travelling in the same direction in 20 s. At what speed is the car travelling (length of both the motorcycler and jeep is negligible)?

A) 36 kmph B) 32 kmph
C) 29 kmph D) 24 kmph
 
Answer & Explanation Answer: D) 24 kmph

Explanation:

Let the speed of the train be 'S' kmph

From the given data,

Distance = Length of train = D = 200 mts = 200 x 18/5 kms

Time = 10 sec

given speed of the motorcycler = 12 kmph

Relative speed as they are moving in the same direction = (S - 12) kmph

(S- 12) = 200 x 18515S - 12 = 48S = 60 kmph 

Hence, the speed of the train = S = 60 kmph

Now,

Let the speed of the jeep be 'x' kmph

60 - x = 200 x 1852060 - x = 36x = 60 - 36x = 24 kmph

 

Therefore, the speed of the jeep = x = 24 kmph.

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Filed Under: Time and Distance
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10 5196