Ratio and Proportion is one of the most important topics in the Arithmetic section of the CAT Quants. It is an easy topic and so one must not avoid this topic. Every year a few questions are asked on Ratio and Proportion. You can check out these Ratio and Proportion based questions from **CAT Previous year’s papers**. Practice a good number of questions on CAT **Ratio and Proportion** questions so that you don’t miss out on the questions from this topic. In this article, we will look into some important Ratio and Proportion Questions for CAT Quants. These are a good source for practice; If you want to practice these questions, you can download this CAT Ratio and Proportion Questions PDF below, which is completely Free.

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**Question 1:Â **If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

a)Â 3.5

b)Â 2.5

c)Â 3

d)Â 4

**1)Â AnswerÂ (C)**

**Solution:**

Let the alloy contain x Kg silver and y kg copper

Now when mixed with 3Kg Pure silver

we getÂ $\frac{\left(x+3\right)}{x+y+3}=\frac{9}{10}$

we get 10x+30 =9x+9y+27

9y-x=3Â Â Â (1)

Now as per condition 2

silver in 2nd alloy = 2(0.9) =1.8

so we get$\frac{\left(x+1.8\right)}{x+y+2}=\frac{21}{25}$

we get 21y-4x =3 Â (2)

solving (1) and (2) we get y= 0.6 and x =2.4

so x+y = 3

**Question 2:Â **A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

**2)Â Answer:Â 34**

**Solution:**

Let price of smallest cup be 2x and medium be 5x and large be y

Now by conditionÂ 1

we getÂ $2x\ \times\ \ 5x\ \times\ y\ =800$

we getÂ $x^2y\ =80$Â Â Â (1)

Now as per second condition ;

$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$ Â Â Â Â (2)

Now dividing (2) and (1)

we get $\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$

we get $10x^2+42x+36\ =\ 40x^2$

we get $\ 30x^2-42x-36=0$

$5x^2-7x-6=0$

we get x=2

So 2x=4 and 5x=10

Now substituting in (1) we get y =20

Now therefore sum = 4+10+20 =34

**Question 3:Â **A person buys tea of three different qualities at â‚¹ 800, â‚¹ 500, and â‚¹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at â‚¹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

a)Â 653

b)Â 688

c)Â 692

d)Â 675

**3)Â AnswerÂ (B)**

**Solution:**

Considering the three kinds of tea are A, B, and C.

The price of kind A = Rs 800 per kg.

The price of kind B = Rs 500 per kg.

The price of kind C = Rs 300 per kg.

They were mixed in the ratio of 2 : 3: 5.

1/6 of the total mixture is sold for Rs 700 per kg.

Assuming the ratio of mixture to A =Â 12kg, B =Â 18kg, C =30 kg.

The total cost price is 800*12+500*18+300*30 = Rs 27600.

Selling 1/6 which is 10kg for Rs 700/kg the revenue earned is Rs 7000.

In order to have an overall profit of 50 percent on Rs 27600.

Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.

Hence he must sell the remaining 50 kg mixture for Rs 41400 – Rs 7000 = 34400.

Hence the price per kg is Rs 34400/50 = Rs 688

**Question 4:Â **From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

**4)Â Answer:Â 45**

**Solution:**

Let initial volume be V, final be F for milk.

The formula is given by :Â $F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$ n is the number of times the milk is drawn and replaced.

so we get $F=\ V\left(1-\frac{K}{V}\right)^{^2}$

here K =9

we get

$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{^2}$

we getÂ $1-\frac{9}{V}=\ \frac{4}{5}or\ -\frac{4}{5}$

If consideringÂ $1-\frac{9}{V}=-\frac{4}{5}$

V =5, but this is not possible because 9 liters is drawn every time.

Hence :Â $1-\frac{9}{V}=\frac{4}{5},\ V\ =\ 45\ liters$

**Question 5:Â **The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is

a)Â 3:2

b)Â 11:7

c)Â 11:3

d)Â 7:3

**5)Â AnswerÂ (C)**

**Solution:**

Let the amounts Neeta, Geeta, and Sita earn in a day be n, g, and s respectively.

Then, it has been given that:

n+g=6s -i

s+n=2g -ii

ii-i, we get: s-g = 2g-6s

7s = 3g.

Let g be 7a. Then s earns 3a.

Then n earns 6s-g = 18a-7a = 11a.

Thus, the ratio is 11a:3a = 11:3

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**Question 6:Â **A sum of money is split among Amal, Sunil and Mita so that the ratio of the sharesÂ of Amal and Sunil is 3:2, while the ratio of the shares of Sunil and Mita is 4:5. If theÂ difference between the largest and the smallest of these three shares is Rs.400,Â then Sunilâ€™s share, in rupees, is

**6)Â Answer:Â 800**

**Solution:**

Let the amount of money with Amal and Sunil be 6x and 4x. Now the amount of money with Mita be 5x. Difference between the largest and smallest amount is â‚¹400 i.e. 6x-4x=400 or 2x=400 or x=200 . Amount of money with Sunil is 200(4)=â‚¹800

**Question 7:Â **Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volumeÂ of the mixture is then doubled by adding solution A such that the resulting mixtureÂ has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol inÂ solution B is

a)Â 90%

b)Â 94%

c)Â 92%

d)Â 89%

**7)Â AnswerÂ (C)**

**Solution:**

Initially let’s consider A and B as one component

The volume of the mixture is doubled byÂ adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations ,Â $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol inÂ A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations ,Â $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

**Question 8:Â **An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a)Â 48

b)Â 84

c)Â 70

d)Â 96

**8)Â AnswerÂ (B)**

**Solution:**

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

**Question 9:Â **A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

**9)Â Answer:Â 8**

**Solution:**

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer.

**Question 10:Â **Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?

a)Â 6350

b)Â 6000

c)Â 7000

d)Â 7250

**10)Â AnswerÂ (D)**

**Solution:**

Assuming the investment ofÂ Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina’s interest income exceeds Amala by 20x-18x=2x=250Â => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

**Question 11:Â **In an examination, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali’s score exceeded Rama’s score by

a)Â 26

b)Â 32

c)Â 35

d)Â 24

**11)Â AnswerÂ (B)**

**Solution:**

Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given thatÂ Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali

r=$\ \frac{\ m+a}{12}$ â€”â€”â€”-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3

Let a+6 = 11x => a= 11x-6

m+6=10x => m=10x-6

r+ 6 =3x => r = 3x-6

Substituting these values in equation (1), we get

3x-6=$\ \frac{\ 21x-12}{12}$

12(3x-6) = 21x-12

x=4

Anjali’s score exceeds Rama’s score by (a-r)=8x=32

**Question 12:Â **The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Rameshâ€™s salary increased by 25% during 2010-2015, then the percentage increase in Rajeshâ€™s salary during this period is closest to

a)Â 10

b)Â 7

c)Â 9

d)Â 8

**12)Â AnswerÂ (B)**

**Solution:**

LetÂ the salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x, 7x respectively

Let the salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y, 3y respectively

It is given thatÂ Rameshâ€™s salary increased by 25% during 2010-2015,3y = 1.25*6x

y=2.5x

Percentage increase in Rajesh’s salary = 7.5-7/7=0.07

=7%

**Question 13:Â **The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a)Â 3 : 10

b)Â 1 : 3

c)Â 1 : 4

d)Â 2 : 5

**13)Â AnswerÂ (B)**

**Solution:**

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given thatÂ three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$Â $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

**Question 14:Â **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)Â 251 : 163

b)Â 239 : 161

c)Â 220 : 149

d)Â 229 : 141

**14)Â AnswerÂ (B)**

**Solution:**

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 thatÂ 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,Â A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

**Question 15:Â **A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

a)Â 30%

b)Â 40%

c)Â 50%

d)Â 60%

**15)Â AnswerÂ (C)**

**Solution:**

Let the volume of the first and the second solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture

= (20 + 300S)

Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final solution

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = $\frac{1}{2}$ = 50%

Hence, 50 is the correct answer.

**Question 16:Â **The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimalâ€™s new score to that of his original score is

a)Â 4 : 3

b)Â 8 : 5

c)Â 5 : 4

d)Â 3 : 2

**16)Â AnswerÂ (A)**

**Solution:**

Let the score of Amal and Bimal be 11k and 14k

Let the scores be increased by x

So, after increment, Amal’s score =Â 11k + x and Bimal’s score = 14k + x

According to the question,

$\dfrac{\text{11k + x}}{\text{14k + x}}$ = $\dfrac{47}{56}$

On solving, we get x = $\dfrac{42}{9}$k

Ratio of Bimal’s new score to his original score

=Â $\dfrac{\text{14k + x}}{\text{14k}}$

=$\dfrac{14k +\frac{42k}{9}}{14k}$

=$\dfrac{\text{168k}}{\text{14*9k}}$

=$\dfrac{4}{3}$

Hence, option A is the correct answer.

**Question 17:Â **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)Â $\frac{1}{5}$

b)Â $\frac{6}{19}$

c)Â $\frac{1}{4}$

d)Â $\frac{7}{33}$

**17)Â AnswerÂ (D)**

**Solution:**

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 +Â 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.

**Question 18:Â **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)Â 17 : 25

b)Â 18 : 25

c)Â 19 : 24

d)Â 21 : 25

**18)Â AnswerÂ (C)**

**Solution:**

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 =Â $\frac{3a+2b}{5}Â $

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

$3.3a+2.2b=200$ ——–(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 =Â $\frac{2a+3b}{5}$

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

$2.1a+3.15b=200$Â ——(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.

**Question 19:Â **Consider three mixtures â€” 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)Â The same amount of water and liquid B

b)Â The same amount of liquids B and C

c)Â More water than liquid B

d)Â More water than liquid A

**19)Â AnswerÂ (C)**

**Solution:**

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 $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$

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 $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C Â is correct.

**Question 20:Â **If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

a)Â 201

b)Â 205

c)Â 207

d)Â 210

**20)Â AnswerÂ (C)**

**Solution:**

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2

=> a = 3x, b = 4x , c = 2x

=> a + b + c = 9x

=> a + b + c is a multiple of 9.

From the given options only, option C is a multiple of 9

**Question 21:Â **Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

**21)Â AnswerÂ (B)**

**Solution:**

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4

Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$

The total volume of water is $\frac{2X}{9}+\frac{4}{13}$

They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$

Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

**Question 22:Â **A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

a)Â 1 : 1

b)Â 8 : 7

c)Â 4 : 3

d)Â 6 : 5

**22)Â AnswerÂ (A)**

**Solution:**

The ratio of L, S, J for popcorn = 7 : 17 : 16

Let them be 7$x$, 17$x$ and 16$x$

The ratio of L, S, J for chips = 6 : 15 : 14

Let them 6$y$, 15$y$ and 14$y$

Given, 40$x$ = 35$y$, $x = \frac{7y}{8}$

Jumbo popcor = 16$x$ = 16 *Â $\frac{7y}{8}$= 14$y$

Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1

**Instructions**

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

**Question 23:Â **Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?

a)Â 1.3

b)Â 1

c)Â 0.6

d)Â 2.3

**23)Â AnswerÂ (A)**

**Solution:**

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

**Question 24:Â **What is the maximum amount of sucrose (to the nearest gram) that can be added to one-gram of saccharin such that the final mixture obtained is atleast 100 times as sweet as glucose?

a)Â 7

b)Â 8

c)Â 9

d)Â 100

**24)Â AnswerÂ (B)**

**Solution:**

For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.

1 gm of saccharin = 675

Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.

So, (675 + N) / (N+1) = 74

=> 675 + N = 74N + 74

=> 601 = 73N => N = 8.23

When N=9, sweetness will be S = (675+9)/10 = 684/10 = 68.4

When N=8, sweetness will be S = (675+8)/9 = 683/9 = 75.8

So, option b) is the correct answer.

**Question 25:Â **There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

a)Â A > B

b)Â A < B

c)Â A = B

d)Â Cannot be determined

**25)Â AnswerÂ (C)**

**Solution:**

Let the volume of the cup be V.

Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V

Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.

So, in each cup, the amount of water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V} $

Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.