Free Arithmetic Miscellaneous - 01 Practice Test - CAT

Question 1

Ram and Shyam invest Rs.15,000 and Rs.25,000, respectively, in a business and at the end of the year , make a profit of Rs.10,000. They agree to re-invest $12 %$ of profit. Out of the remaining profit, each of them takes Rs.1000 and rest is divided in the ratio of their original investment. Find Ram’s total share in the profit.

A. Rs. 4250

B. Rs.3000

C. Rs.2550

D. Rs.3550

SOLUTION

Solution : D

$12 %$ of profit = 1200

Remaining = 8800

After taking 1000 each, amount remaining = 6800

Ratio of shares of Ram and Shyam = 15,000 : 25,000 = 3 : 5

Ram’s share $= \frac{3}{8} \times 6800 = 2550$

Ram’s profit = 2550 + 1000 = 3550.

Question 2

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?

A. 16:81

B. 1:5

C. 14:82

D. None of these

SOLUTION

Solution : 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 3

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

$p < q$

$p > q$

$p = \frac{q}{2}$

SOLUTION

Solution : 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 4

A man purchased 40 kg of cotton at a rate of Rs 65 per kg. then he extracted $20 %$ waste(by weight) from it so that the quality of the cotton improved and he was able to sell them at Rs.80 per kg. Also the waste had $60 %$ cotton seed by weight which he was able to sell Rs.20 per kg. what $%$ profit did he make?

8.65 %

1.24 %

1.47 %

2.15 %

SOLUTION

Solution : D

wt = 40 kg
cost = Rs.65/kg
Total cost $= 40 \times 65$
= Rs 2600
SP $\to$
wt after $20 %$ is extracted
$\Rightarrow$ 32 kg
Selling cost = 32 $\times$ 80
= Rs 2560
$60 %$ of waste $= \frac{60}{100} \times 8 = 4.8 k g$
selling price $= 4.8 \times 20$
= 96
Total price = 2656
Profit $= \frac{56}{2600} \times 100 = 2.15 %$

Question 5

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?

A. 350

B. 215

C. 355

D. 245

SOLUTION

Solution : 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 6

How many kilograms of rice costing 3.6 per kg should be mixed with 8 kgs of rice costing 4.20 per kg, such that the mixture yields a profit of $10 %$ when it is sold at 4.40 per kg?

A. 5 kgs

B. 6 kgs

C. 3 kgs

D. 4 kgs

SOLUTION

Solution : D

For all alligation questions, remember that we need to consider only the cost price

Given that the Selling Price = .

$C o s t p r i c e \times 1.1 = S e l l i n g p r i c e$

$C o s t p r i c e = \frac{S e l l i n g p r i c e}{1.1} = \frac{4.4}{1.1} = 4 r s$

Ratio of quantity of rice = 2:4 or 1:2. Given that the quantity of the second variety is 8 kg, the first variety should be $\frac{8}{2}$ =4 kgs

Question 7

A. 1200

B. 1600

C. 2000

D. None of these

SOLUTION

Solution : 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 8

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?

A. 56

B. 52

C. 42

D. 33

SOLUTION

Solution : 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 9

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___

SOLUTION

Solution :
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 litres of 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 10

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?

A. 18

B. 21

C. 22

D. 27

SOLUTION

Solution : 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 $\frac{C + 2 C - 5}{2} = \frac{(3 C - 5)}{2}$

Average age of members =34

$C + C - 5 + 2 C - 5 + \frac{(3 C - 5)}{2} = 34 \times 4$

$4 C + \frac{(3 C - 5)}{2} = 146$

11C= 297

C=27

D= 27-5=22

Option (C)

Question 11

Fresh grapes contain $90 %$ water by weight and dry grapes contain $20 %$ water by weight. The weight of dry grapes obtained from 20 kg of fresh grapes is (in kg., in decimal, up to 2 decimal places) ___

SOLUTION

Solution :
20 kg $\to 10 %$ (non water content) = 2kg
2 = $80 %$ (non water content dry grapes)
$\Rightarrow$ wt = 2.50 kg

Question 12

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 ___

SOLUTION

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

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 13

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?

A. 9 : 6

B. 11 : 7

C. 9 : 4

D. 12 : 5

SOLUTION

Solution : 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 14

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 ?

A. 20 :13

B. 11:20

C. 14:9

D. Cannot be determined

SOLUTION

Solution : C

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 15

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 difference in the number of boys who are in the lily group and the number of girls in the daisy group, given that there are 48 girls in the daisy group?

A. 200

B. 240

C. 160

D. None of these

SOLUTION

Solution : C

1.2X is given as 48 hence x will be 40.

So the number of boys in lilies will be $5.2 \times 40 = 208$

So the difference will be 208 - 48 = 160

Free Arithmetic Miscellaneous - 01 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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