Arithmetic(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

ARITHMETIC MCQs

Total Questions : 147 | Page 7 of 15 pages

Question 61. A tub is filled with a mixture of 2 miscible fluids P and Q in the proportion 5:3. When 16 litres of the mixture is drawn off and the tub is filled with same amonut of fluid Q, the proportion becomes P:Q=3:5. How many litres does the tub hold___

Discuss Question

:
Let the tub contain 5x litres of P and 3x litres of Q. total = 8x litres of mixture. When are taken out, (8x-16) litres will be left
The mixture will now contain

\frac{5}{8} (8 x - 16) = 5 (x - 2)

litres of P and 3(x-2) litres of Q.
Given that 16 litresof Q are added.
Ratio is given as

\frac{5 (x - 2)}{3 (x - 2) + 16} = \frac{3}{5}

Solving we get x-2=3 ; x=5
Tub will hold 8x = 40 litres

Question 62.

1200
1600
2000
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
OPTION D
Given that

\frac{P N R}{100} = 900

3600 \times 5 \times \frac{r}{100} = 900

R = 5 %

He will split the amount exactly in half. Therefore

\frac{3600}{2} = 1800

will be the amount invested

Question 63. A bottle contains

100 %

phenyl.

\frac{1}{3}

of it is replaced with water. The operation is repeated 4 times. What is the ratio of phenyl: water after the last operation?

16:81
1:5
14:82
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Option (d)

\frac{L i q u i d A l e f t t a k e n n^{t h} o p e r a t i o n}{I n i t i a l q u a n t i t y o f A i n t h e v e s s e l} = \frac{(a - b)^{n}}{a^{n}}

Amount of phenyl remaining

= {(1 - \frac{1}{3})}^{4} = \frac{16}{81}

Therefore, ratio or phenyl:water = 16:65. (81-16=65)

Question 64. Dexter is coming up with a chemical complex using 3 different soluble substances in the volume ratio 3:4:7. The densities of these substances are in the ratio 5:2:6 respectively. The mass of the

2^{n d}

substance contained in 130 litres of the complex is ___

Discuss Question

D e n s i t y = \frac{M a s s}{V o l u m e}

Individual volume ratio = volume of component
Total volume
Ratio of volumes

= \frac{3}{14} : \frac{4}{14} : \frac{7}{14}

Ratio of densitites

= \frac{5}{13} : \frac{2}{13} : \frac{6}{13}

Ratio of masses

= \frac{3}{14} \times \frac{5}{13} : \frac{4}{14} \times \frac{2}{13} : \frac{7}{14} \times \frac{6}{13} = 15 : 8 : 42

. mass in

2^{n d}

substance

= \frac{8}{65} \times 130 = 16

litres.

Question 65. In an online poll for the title of “Miss photogenic”, the winner gets 40 votes more than the one who comes second and 75 votes more than the person who comes

3^{r d}

. there are only 3 candidates. The winner and her nearest competitor polled an average of 90 votes together. How many votes were totally cast online, assuming that there are no invalid votes?

Discuss Question

Answer: Option B. -> 215
:
B
Winner gets 40 votes more the person who came second and the average is 90

\Rightarrow

winner = 110 votes
second person = 70 votes
Total votes = 110 + 70 + 35
= 215

Question 66. In Little flowers school, for a school drill, the students are divided into 2 groups of lilies and daisies.The ratio of lilies: daisies = 8:3. The ratio of boys: girls is 7:4.

60 %

of the daisy group is boys.
What is the ratio of the number of girls in the lily group and the number of boys in the Daisy group ?

20 :13
11:20
14:9
Cannot be determined

Discuss Question

Answer: Option C. -> 14:9
:
C

In Little Flowers School, For A School Drill, The Students A...

Ratio between boys and girls are 7 : 4.

\frac{(1.8 x + b)}{(9.2 x - b)} = \frac{7}{4}

b = 5.2
8x - b = 2.8
GIRLS IN LILLY : BOYS IN DAISY = 28 :18 = 14:9

Question 67. There is a bottle of coke.

25 %

is replaced by pepsi. Again

25 %

is replaced by Pepsi and finally another

25 %

is replaced by Pepsi. What is the approximate final

%

of coke in the bottle?

Discuss Question

Answer: Option C. -> 42
:
C
Option (c)
Let the bottle contain 100 units of liquid.
Liquid taken out in each case is

25 %

The final quantity of component from the original mixture that is not being replaced, if this operation is repeated 3 times, is

100 (1 - \frac{25}{100})^{3}

100 (0.75)^{3} = 42.18 %

Question 68. At a gold shop’s annual promotion, gold coins are distributed with each purchase.The higher the purchase, the higher the value of the gold coin. The value of the gold coin is directly proportional to the thickness and square of the diameter of the coin.Two gold coins are given to a customer. The value of coin1: coin 2 = 4:1. The diameter of the coins are in a ratio 4:3. Find the ratio of the thickness of the coins?

9 : 6
11 : 7
9 : 4
12 : 5

Discuss Question

Answer: Option C. -> 9 : 4
:
C

\frac{T h i c k n e s s_{1} \times d i a_{1}^{2}}{T h i c k n e s s_{2} \times d i a_{2}^{2}} = \frac{4}{1} \frac{T h i c k n e s s_{1}}{T h i c k n e s s_{2}} = \frac{4}{1} \times \frac{9}{16} = \frac{9}{4}

Question 69. At the end of year 1998, Ramesh bought 108 shares. Henceforth, every year he added

p %

of the shares at the beginning of the year and sold

q %

of the shares at the end of the year where p

>

0 and q

>

0. If Ramesh had 108 shares at the end of year 2002, after making the sales for that year, which of the following is true?

p = q
p
p>q
p=q2

Discuss Question

Answer: Option C. -> p>q
:
C
If the numbers of shares remain constant, then p has to be greater than q. Because, we are taking

p %

on base price after getting increased we take

q %

of that, which is greater than the base price. Answer is option (c).

Question 70. C and D are the children of A

&

B, who are married to each other. D is 5 years younger than C. Five years ago, A was twice as old as C. The current age of B is the average of A and C’s ages. If the average age of all of them is 34 years, find the current age of D?

Discuss Question

Answer: Option C. -> 22
:
C
Let the present age of each of them be the letters themselves.
Present age of D= C-5
Presnt age of A will be 2(C-5)+5
=2C-5 and the present age of B will be