## Fractions Questions for CAT PDF

Download important CAT Fractions Problems with Solutions PDF based on previously asked questions in CAT exam. Practice Fractions Problems with Solutions for CAT exam.

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**Question 1: **There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

a) 251 : 163

b) 239 : 161

c) 220 : 149

d) 229 : 141

**Question 2: **A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

a) 30.3

b) 35.2

c) 25.4

d) 20.5

**Question 3: **If decreasing 70 by X percent yields the same result as increasing 60 by X percent, then X percent of 50 is

a) 3.84

b) 4.82

c) 7.10

d) The data is insufficient to answer the question

**Question 4: **If $\frac{x}{y}=\frac{7}{4}$, find the value of $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

a) $\frac{27}{49}$

b) $\frac{43}{72}$

c) $\frac{33}{65}$

d) None of the above

**Question 5: **Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

a) $\frac{1}{5}$

b) $\frac{6}{19}$

c) $\frac{1}{4}$

d) $\frac{7}{33}$

**Question 6: **Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

a) 17 : 25

b) 18 : 25

c) 19 : 24

d) 21 : 25

**Question 7: **Consider three mixtures — the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a) The same amount of water and liquid B

b) The same amount of liquids B and C

c) More water than liquid B

d) More water than liquid A

**Question 8: **X and Y are the two alloys which were made by mixing Zinc and Copper in the ratio 6 : 9 and 7 : 11 respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form another alloy Z, what is the ratio of Zinc and Copper in the new alloy Z?

a) 6 : 9

b) 59 : 91

c) 5 : 9

d) 59 : 90

**Question 9: **Consider the formula, $S=\frac{\alpha\times\omega}{\tau+\rho\times\omega}$ positive integers. If ⍵ is increased and ⍺, τ and ρ are kept constant, then S:

a) increases

b) decreases

c) increases and then decreases

d) decreases and then increases

e) cannot be determined

**Question 10: **A manufacturer has 200 litres of acid solution which has 15% acid content. How many litres of acid solution with 30% acid content may be added so that acid content in the resulting mixture will be more than 20% but less than 25%?

a) More than 100 litres but less than 300 litres

b) More than 120 litres but less than 400 litres

c) More than 100 litres but less than 400 litres

d) More than 120 litres but less than 300 litres

e) None of the above

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**Question 11: **Three classes X, Y and Z take an algebra test.

The average score in class X is 83.

The average score in class Y is 76.

The average score in class Z is 85.

The average score of all students in classes X and Y together is 79.

The average score of all students in classes Y and Z together is 81.

What is the average for all the three classes?

a) 81

b) 81.5

c) 82

d) 84.5

**Question 12: **A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a) 2

b) 3

c) 4

d) 5

**Question 13: **A medical practitioner has created different potencies of a commonly used medicine by dissolving tables in water and using the resultant solution. Potency 1 solution: When 1 tablet is dissolved in 50

ml, the entire 50 ml is equivalent to one dose. Potency 2 solution: When 2 tablets are dissolved in

50 ml, the entire 50 ml of this solution is equivalent to 2 doses, … and so on. This way he can give fractions of tablets based on the intensity of infection and the age of the patient. For particular patient, he administers 10 ml of potency 1, 15 ml of potency 2 and 30 ml of potency 4. The dosage administered to the patient is equivalent to

a) > 2 and ≤ 3 tablets

b) > 3 and ≤ 3.25 tablets

c) > 3.25 and ≤ 3.5 tablets

d) > 3.5 and ≤ 3.75 tablets

e) > 3.75 and ≤ 4 tablets

**Question 14: **There are two alloys P and Q made up of silver, copper and aluminium. Alloy P contains 45% silver and rest aluminum. Alloy Q contains 30% silver, 35% copper and rest aluminium. Alloys P and Q are mixed in the ratio of 1 : 4 . 5. The approximate percentages of silver and copper in the newly formed alloy is:

a) 33% and 29%

b) 29% and 26%

c) 35% and 30%

d) None of the above

**Question 15: **The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘A’ is 3 :4. In alloy ‘B’ same metals are mixed in the ratio 5:8. If 26 kg of alloy ‘B’ and 14 kg of alloy ‘A’ are mixed then find out the ratio of ‘metal 1’ and ‘metal 2’ in the new alloy.

a) 3:2

b) 2:5

c) 2:3

d) None of the above

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**Answers & Solutions:**

**1) Answer (B)**

It is given that in drum 1, A and B are in the ratio 18 : 7.

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given that drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A

$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$

$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$

$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$

$\Rightarrow$ $x = \dfrac{239}{161}$

Therefore, we can say that in drum 2, A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

**2) Answer (B)**

Final quantity of alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml

Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %

**3) Answer (A)**

Given that,

$70-\dfrac{70x}{100} = 60+\dfrac{60x}{100}$

Thus, $7000-70x = 6000+60x$

=> $1000 = 130x$

Thus, $x\approx 7.7$

Thus, $7.7\% of 50 = $dfrac{7.7*50}{100} \approx 3.84$

Hence, option A is the correct answer.

**4) Answer (C)**

Given that If $\frac{x}{y}=\frac{7}{4}$

Therefore, $(\frac{x}{y})^2=\frac{49}{16}$ … (1)

$\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}$ this can be written as,

$\Rightarrow$ $\dfrac{(\frac{x}{y})^2-1}{(\frac{x}{y})^2+1}$

$\Rightarrow$ $\dfrac{\frac{49}{16}-1}{\frac{49}{16}+1}$

$\Rightarrow$ $\dfrac{49-16}{49+16}$

$\Rightarrow$ $\dfrac{33}{65}$

Hence, option C is the correct answer.

**5) Answer (D)**

Let the number of marbles with Raju and Lalitha initially be 4x and 9x.

Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6

24x + 6a = 45x – 5a

11a = 21x

a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).

a/9x = 21/99

= 7/33.

Therefore, option D is the right answer.

**6) Answer (C)**

The selling price of the mixture is Rs.40/kg.

Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.

It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2

Let the cost price of the mixture be x.

It has been given that 1.1x = 40

x = 40/1.1

Price per kg of the mixture in ratio 3:2 = $\frac{3a+2b}{5} $

$\frac{3a+2b}{5} = \frac{40}{1.1}$

$3.3a+2.2b=200$ ——–(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.

Price per kg of the mixture in ratio 2:3 = $\frac{2a+3b}{5}$

$\frac{2a+3b}{5} = \frac{40}{1.05}$

$2.1a+3.15b=200$ ——(2)

Equating (1) and (2), we get,

$3.3a+2.2b = 2.1a+3.15b$

$1.2a=0.95b$

$\frac{a}{b} = \frac{0.95}{1.2}$

$\frac{a}{b} = \frac{19}{24}$

Therefore, option C is the right answer.

**7) Answer (C)**

The proportion of water in the first mixture is $\frac{1}{3}$

The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$

The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$

The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{160}$

The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$

The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$

The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{160}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C is correct.

**8) Answer (B)**

Alloy X has zinc and copper in ratio 6:9

If 40 grams is taken then weights of zinc and copper are:

Zinc = $\frac{6}{15}$ * 40 = 16 grams

Copper = $\frac{9}{15}$ * 40 = 24 grams

Alloy Y has zinc and copper in ratio 7:11

If 60 grams is taken then weights of zinc and copper are:

Zinc = $\frac{7}{18}$ * 60 = $\frac{70}{3}$ grams

Copper = 60 – $\frac{70}{3}$ = $\frac{110}{3}$ grams

When mixed together, the weights of Zinc and Copper are:

Zinc = 16 + $\frac{70}{3}$ = $\frac{118}{3}$

Copper = 24 + $\frac{110}{3}$ = $\frac{182}{3}$

Ratio = 118/3:182/3 = 59:91

**9) Answer (A)**

Expression : $S=\frac{\alpha\times\omega}{\tau+\rho\times\omega}$

=> $\frac{1}{S} = \frac{\tau+\rho\times\omega}{\alpha\times\omega}$

=> $\frac{1}{S} = \frac{\tau}{\alpha \omega} + \frac{\rho}{\omega}$

Since, $\tau, \rho$ and $\alpha$ are constant,

=> $\frac{1}{S} = \frac{k_1}{\omega} + k_2$

Thus, $S \propto \omega$

$\therefore$ When $\omega$ increases, S increases.

**10) Answer (C)**

Let the volume of the solution with 30 % acid content lie between $v_1$ and $v_2$, where we get a 20% acid solution for $v_1$

For $v_2$, we get a 25 % acid solution as the resultant mixture.

=> $15 \% (200) + 30 \% (v_1) = 20 \% (200 + v_1)$

=> $30 + 0.3 v_1 = 40 + 0.2 v_1$

=> $0.1 v_1 = 10$ => $v_1 = 10 \times 10 = 100$ litres

Similarly, $15 \% (200) + 30 \% (v_2) = 25 \% (200 + v_2)$

=> $30 + 0.3 v_2 = 50 + 0.25 v_2$

=> $0.05 v_2 = 20$ => $v_2 = 20 \times 20 = 400$ litres

$\therefore$ For the acid content in the resultant mixture to lie between 20 % and 25 %, the volume of the 30 % concentration acid solution must lie between 100 litres and 400 litres.

**11) Answer (B)**

Let x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $\frac{83x+76y+85z}{x+y+z}$

Substitute the relations in above equation we get,

$\frac{83x+76y+85z}{x+y+z}$ = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

**12) Answer (C)**

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.

So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k

So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k

50% of the maximum marks is 6.67k

So, the number of papers in which the student scored more than 50% is 4

**13) Answer (B)**

50 ml of potency 1 solution is equivalent to 1 tablet, 50 ml of potency 2 solution i equivalent to 2 tablets and so on.

Hence, 10 ml of potency 1 solution is equivalent to

= $\frac{10}{50} = \frac{1}{5}$

Similarly, 15 ml of potency 2 solution corresponds to = $\frac{15}{50} \times 2 = \frac{3}{5}$

and 30 ml of potency 4 solution corresponds to = $\frac{30}{50} \times 4 = \frac{12}{5}$

$\therefore$ Required dosage

= $\frac{1}{5} + \frac{3}{5} + \frac{12}{5}$

= $\frac{15}{5} = 3.2$ tablets

**14) Answer (A)**

Composition of alloy P

Silver:Copper:Aluminium = 45:0:55

Composition of alloy Q

Silver:Copper:Aluminium = 30:35:35

They are mixed in ratio of 1: 4.5

Let us consider alloy P is taken 200 grams and alloy Q is taken 900 grams.

Then for alloy P :-

Silver:Copper:Aluminium = 90:0:110

For alloy Q:

Silver:Copper:Aluminium = 270:315:315

Total weight of P and Q combined is 1100 grams.

When P and Q are mixed, the new combined ratio of

Silver:Copper:Aluminium = 360:315:425

Percentage of Silver in mixture = $\frac{360}{1100}$ x 100 $\cong$ 33%

Percentage of Copper in mixture = $\frac{315}{1100}$ x 100 $\cong$ 29%

**15) Answer (C)**

The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘A’ is 3 :4.Therefore, we can say that 14 kg of alloy ‘A’ will contain $\dfrac{3}{7} 14$ = 6 kg of ‘metal 1’ and $\dfrac{4}{7} 14$ = 8 kg of ‘metal 2’.

The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘B’ is 5 :8.Therefore, we can say that 26 kg of alloy ‘B’ will contain $\dfrac{5}{13} 26$ = 10 kg of ‘metal 1’ and $\dfrac{8}{13} 26$ = 16 kg of ‘metal 2’.

Hence, total weight of ‘metal 1’ in the new alloy = 6 + 10 = 16 kg

Total weight of ‘metal 2’ in the new alloy = 8 + 16 = 24 kg

Therefore, the ratio of ‘metal 1’ and ‘metal 2’ in the new alloy. = 16 : 24 = 2 :3. Hence, option C is the correct answer.

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