CMAT Average, Ratio, and Proportions Questions

The ratio in which Reema distributed chocolates = $$\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{5} : \dfrac{1}{8}$$

Now, $$LCM(2,3,5,8) = 120$$

So, the ratio in which Reema distributed chocolates = $$\dfrac{120}{2}:\dfrac{120}{3}:\dfrac{120}{5}:\dfrac{120}{8}=60:40:24:15$$

Let's say the common variable be $$x$$

So, the chocolates distributed by Reema = $$60x, 40x, 24x, 15x$$

So, total number of chocolates distributed = $$60x+40x+24x+15x$$ = $$139x$$

Now for minimum number of chocolates distributed, $$x$$ should be $$1$$.

So, minimum number of chocolates = $$139\times\ 1=139$$

Let's say amount of fertilizer alpha taken be $$a$$ units.

Amount of fertilizer beta taken be $$b$$ units.

Now, ratio of K in alpha = 10% and in final mixture, the ratio is 6%

So,$$\dfrac{10}{100}\times\ a+0=\dfrac{6}{100}\times\ \left(a+b\right)$$

or, $$0.1a=0.06a+0.06b$$

or, $$0.1a-0.06a=0.06b$$

or, $$0.04a=0.06b$$

or, $$\dfrac{a}{b}=\dfrac{6}{4}=\dfrac{3}{2}$$

So, let say $$a=3$$ and $$b=2$$

Let say percentage of P in Beta be $$p%$$

So, $$3\cdot\dfrac{70}{100}+2\cdot\dfrac{p}{100}=5\cdot\dfrac{68}{100}$$

or, $$210+2p=340$$

or, $$p=\dfrac{340-210}{2}=\dfrac{130}{2}=65\%$$

So, percentage of P in Beta = $$65\%$$

So, percentage of N in beta = $$\left(100-65\right)\%=35\%$$

So, correct answer is option C.

Given,

A.) $$4a=5b$$

or, $$\dfrac{a}{b}=\dfrac{5}{4}=\dfrac{45}{36}$$

$$7b=9c$$

or, $$\dfrac{b}{c}=\dfrac{9}{7}=\dfrac{36}{28}$$

So, $$a:b:c=45:36:28$$

So, option (II) is the correct answer for (A).

B.) $$A=\dfrac{1}{4}B$$ and $$B=\dfrac{1}{2}C$$

or, $$\dfrac{A}{B}=\dfrac{1}{4}$$ and $$\dfrac{B}{C}=\dfrac{1}{2}=\dfrac{4}{8}$$

So, $$A:B:C$$ is equal to $$1:4:8$$

So, option (III) is the correct answer for (B).

C.) $$P:Q:R=2:3:4$$

So, $$P:Q=2:3$$, $$Q:R=3:4$$, $$R:P=4:2$$

So, $$\dfrac{P}{Q}:\dfrac{Q}{R}:\dfrac{R}{P}=\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{2}=8:9:24$$

So, option (IV) is the correct answer for (C).

So, option (I) has to be the correct answer for (D).

So, option D is the correct answer.

Let say income of Biren and Deepak be ₹ $$x$$ and ₹ $$2x$$

So, income of Anand = ₹ $$(140+x)$$

Income of Chandu = ₹ $$(80+2x)$$

Now, the ratio of Anand's & Chandu's income is 2 : 3

So, $$\dfrac{140+x}{80+2x}=\dfrac{2}{3}$$

or, $$420+3x=160+4x$$

or, $$420-160=4x-3x$$

or, $$x=260$$

So, incomes of Anand, Biren, Chandu and Deepak = $$\left(140+x\right),x,\left(80+2x\right),2x$$

Putting $$x=260$$,

So, incomes of Anand, Biren, Chandu and Deepak = ₹$$400$$, ₹$$260$$, ₹$$600$$, ₹$$520$$

So, option C is the correct answer.

In bottle 1 (10 l), syrup and water is in the ratio 2 : 3

So, amount of syrup = 4l and amount of water = 6l.

In bottle 2 (21 l), syrup and water is in the ratio 3 : 4

So, amount of syrup = 9l and amount of water = 12l.

Let, say $$12x$$ l of solution is taken from bottle 3.

So, amount of syrup = $$7x$$ l and amount of water = $$5x$$ l.

So, total amount of syrup = $$4+9+7x=13+7x$$

Total amount of water = $$6+12+5x=18+5x$$ l.

Now, final ratio of syrup and water is 1 : 1

So, $$13+7x=18+5x$$

or, $$7x-5x=18-13$$

or, $$2x=5$$

or, $$x=\dfrac{5}{2}$$

So, amount from third bottle = $$12x=12\times\ \dfrac{5}{2}=30$$ litres.

A) $$16\frac{2}{3}$$ % = 0.1667
B) $$\frac{2}{15}$$ = 0.1333
C) 0.17 = 0.17
D) 0.25 % of 64 = 0.16

Hence, 

B < D < A < C

$$\dfrac{a}{100}\times a+\dfrac{b}{100}\times b\ =\ \dfrac{2}{100}\times ab$$

$$a^2+b^2=2ab$$

$$\left(a-b\right)^2=0$$

a = b

Thus, a is 100% of b.

The total volume of the mixture is 160 L and 70% of the milk and 30% of the water is removed. With this, 55% of the vessel is vacated, so 45% of the original mixture remains, which is 72 L.

Let the amount of Milk in the original mixture be $$x$$ liters so the amount of water in the vessel would be $$160-x$$ liters. 

Milk removed = 70% of $$x$$ = $$0.7x$$

Water removed = 30% of $$160-x$$ = 0.3(160−x).

Remaining milk: x − 0.7x = 0.3x

Remaining water: 160 − x − 0.3(160−x) = 0.7(160−x)

The remaining mixture is 72 liters. 

0.3x + 0.7(160−x) = 72

Upon solving, we get, x = 100

Therefore, the amount of Milk is 100 liters and the amount of water = 160 - 100 = 60 liters.

Given the ratio now is 46 : 41 : 34.

Assume the constant of the ratio to be $$x$$

So, now the coins they have are $$46x$$, $$41x$$ and $$34x$$ respectively.

Before selling/donating/losing the number of coins will be $$\left(46x+30\right)$$, $$\left(41x+30\right)$$ and $$\left(34x+25\right)$$ respectively.

We know $$46x+30+41x+30+34x+25=2505$$

$$121x=2420$$

$$x=20$$

So, the number of coins Purab received will be $$41x+30=41\times20+30=850$$

Hence, the answer is 850

Given that currently container has 6 parts of water and 10 parts of milk.

Assume the volume of water to be $$6x$$ and that of milk to be $$10x$$

So, the volume of the container is $$16x$$.

Assume we take out $$y$$ litre of mixture and replace it with water

So, the resultant mixture will have 

$$\left(6x-\dfrac{6x}{16x}\times y+y\right)$$ water and 

$$\left(10x-\dfrac{10x}{16x}\times y\right)$$ milk

Given the volume of water is half of the container i.e. $$8x$$

So, $$\left(10x-\dfrac{10x}{16x}\times y\right)=8x$$

$$10x-\dfrac{5}{8}y=8x$$

or, $$y=\dfrac{16x}{5}$$

In terms of container volume

$$\dfrac{y}{16x}=\dfrac{\dfrac{16x}{5}}{16x}=\dfrac{1}{5}$$

Hence, the answer is Option C

The Duplicate Ratio of 2 : 7 is the compound of it by itself i.e. 4 : 49

The compound ratio of 2 : 7, 5 : 3 and 4 : 7 is 2*5*4 : 7*3*7 i.e. 40 : 147

The ratio same as 2 : 7 is 4 : 14

The ratio same as 5 : 6 is 25 : 30

Hence, the answer is Option B.

A: If $$\frac{1}{x} : \frac{1}{y} : \frac{1}{z}$$ = 2 : 3 : 5, then x : y : z = 15 : 10 : 6

$$x:y:z\ =\ \frac{1}{2}:\frac{1}{3}:\frac{1}{5}=\frac{15}{30}:\frac{10}{30}:\frac{6}{30}$$

Hence, x : y : z = 15 : 10 : 6

Correct

B: If 4p = 6q = 9r, then p : q : r = 9 : 6 : 4

Let 4p = 6q = 9r = k

Therefore, p = k/4

q = k/6

r = k/9

Hence, p : q : r = $$\frac{k}{4}:\frac{k}{6}:\frac{k}{9}=\frac{9k}{36}:\frac{6k}{36}:\frac{4k}{36}=9:6:4$$

Correct

C: If 2A = 3B = 4C, then A : B : C = 3 : 4 : 6

Let 2A = 3B = 4C = k

then A = k/2

B = k/3

C = k/4

So, A:B:C = $$\frac{k}{2}:\frac{k}{3}:\frac{k}{4}=\frac{6k}{12}:\frac{4k}{12}:\frac{3k}{12}=6:4:3$$

Incorrect

D: If P : Q : R = 2 : 3 : 4, then $$\frac{P}{Q} : \frac{Q}{R} : \frac{R}{P}$$ = 9 : 8 : 24

Let P = 2k so Q and R will be 3k and 4k

SO, $$\frac{P}{Q}:\frac{Q}{R}:\frac{R}{P}=\frac{2k}{3k}:\frac{3k}{4k}:\frac{4k}{2k}=\frac{2}{3}:\frac{3}{4}:2=8:9:24$$

Incorrect

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M =  Girls in M + Boys in M

Girls in M =  $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C =  $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E -  Girls in C = Boys in C - Girls in E

Boys in E  +  Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.  

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

Total number of Girls = 51 + 23 + 28 = 102

The correct answer is option C.

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M = Girls in M + Boys in M

Girls in M = $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C = $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E - Girls in C = Boys in C - Girls in E

Boys in E + Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

The percentage number of girls who opted for Mechanical to the number of students who opted for Electronics is,

$$\dfrac{Girls\ in\ M}{E}\ \times\ 100\ =\ \dfrac{51}{60}\ \times\ 100\ =85\%\ $$

The correct answer is option C.

Let C be Computer Science, M be Mechanical, and E be Electronics

Total number of students = C + M + E = 240 - (1)

It is also given that C + E = M - (2)

M = Girls in M + Boys in M

Girls in M = $$\dfrac{42.5}{100}M$$

Boys in M = $$\dfrac{57.5}{100}M$$

C = Girls in C + Boys in C

Girls in C = $$\dfrac{1}{3}\times\ \dfrac{57.5}{100}M\ =\ \dfrac{57.5}{300}M$$

Boys in E - Girls in C = Boys in C - Girls in E

Boys in E + Boys in C = Girls in C + Boys in C

E = C - (3)

Girls in E = 28.

From (1), (2) and (3) we get

C + M + E = C + 2C + C = 240

4C = 240

C = E = 60

M = 120

Girls in M = $$\dfrac{42.5}{100}M$$ = $$\dfrac{42.5}{100}\ \times\ 120\ =\ 51$$

Girls in C = $$\dfrac{57.5}{300}\ \times\ 120\ =\ 23$$

Girls in E = 28

Boys in M = 120 - Girls in M = 120 - 51 = 69

E = 60

Ratio = $$\dfrac{69\ -\ 60}{60}\ \times\ 100\ =\ 15\%$$ 

The correct answer is option A.