Quantitative Aptitude - Arithmetic Ability Questions

Q:

What percent is 120 of 85 ?

A) 133.33 % B) 141.17 %
C) 145.54 % D) 139.29 %
 
Answer & Explanation Answer: B) 141.17 %

Explanation:

Px85100=120

 

=> p = 12000/85 = 141.17 %

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2 3512
Q:

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

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

Explanation:

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

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Filed Under: Permutations and Combinations
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9 3511
Q:

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

A) 720 B) 360
C) 120 D) 24
 
Answer & Explanation Answer: D) 24

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

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6 3510
Q:

Shiva purchased 280 kg of Rice at the rate of 15.60/kg and mixed it with 120 kg of rice purchased at the rate of 14.40/kg. He wants to earn a profit of Rs. 10.45 per kg by selling it. What should be the selling price of the mix per kg?

A) Rs. 22.18 B) Rs. 25.69
C) Rs. 26.94 D) Rs. 27.54
 
Answer & Explanation Answer: B) Rs. 25.69

Explanation:

Rate of rice of quantity 280 kg = Rs. 15.60/kg

Rate of rice of quantity 120 kg = Rs. 14.40/kg

He want to earn a profit of Rs. 10.45/kg

Rate of Mix to sell to get profit of 10.45 =  

 

280 x 15.60 + 120 x 14.40280 + 120 + 10.454368 + 1728400 + 10.45=> 15.24 + 10.45= 25.69 

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12 3510
Q:

A seller charges 28% more from a customer as GST on an article. If customer paid Rs. 4544, what is the cost price of the article?

A) Rs. 3550 B) Rs. 2650
C) Rs. 3840 D) Rs. 2450
 
Answer & Explanation Answer: A) Rs. 3550

Explanation:

Let the cost price of the article be 'P'

Then from given data,

P + 28P100 = 4544=> 128P = 454400=> P = 3550

Hence, the cost price of the article = P = Rs. 3550

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15 3510
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 3510
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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4 3503
Q:

The population of a city is 415600.It increased by 25% in the first year and decreased by 30% in the second year.What is the population of the city at the end of second year?

Answer

As per given data, P = 415600, R1= 25% incresed, R2 = 30% decreased


Population of the city at the end of the second year= 



363650.

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Subject: Percentage Exam Prep: CAT
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12 3501