# CRE 1 - Ratio | Ratio, Proportion and Partnership

**CRE 1 - Ratio | Ratio, Proportion and Partnership**

Anand divided 70 sweets among his daughters Rohini and Sushma in the ratio 4 : 3. How many sweets did Rohini get?

- A.
50

- B.
40

- C.
30

- D.
45

Answer: Option B

**Explanation** :

Since, the ratio of sweets that Rohini and Sushma get is 4 : 3, let the number of sweets received by Rohini is 4x and Sushma is 3x.

According to question, 4x + 3x = 70

⇒ 7x = 70

⇒ x = 10

∴ Sweets received by Rohini = 4x = 4 × 10 = 40.

Alternately,

Number of sweets that Rohini got = $\frac{4}{4+3}$ × 70 = 40

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If x : y = 4 : 3, find 5x : 7y.

- A.
15/28

- B.
3/4

- C.
20/21

- D.
20/37

Answer: Option C

**Explanation** :

$\frac{x}{y}=\frac{4}{3}$

$\frac{5x}{7y}=\frac{5}{7}\left(\frac{4}{3}\right)=\frac{20}{21}$

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

Ratio of two numbers is 4 : 5 and their sum is 144. Find the smaller of the two numbers.

- A.
60

- B.
55

- C.
64

- D.
58

Answer: Option C

**Explanation** :

Let the numbers be 4x and 5x.

4x + 5x = 144

x = 16

Smaller of the two numbers = 4x = 64.

**Alternately, **

Smaller number = 4/9 × 144 = 64.

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The greatest ratio out of 2 : 3, 5 : 4, 3 : 2 and 4 : 5 is?

- A.
4 : 5

- B.
3 : 2

- C.
5 : 4

- D.
2 : 3

Answer: Option B

**Explanation** :

Here, only 5/4 and 3/2 are greater than 1. So, let’s consider only these two ratios. Other 2 ratios are smaller than 1.

To find the greatest ratio, we must bring each of the ratio to common denominator.

Thus, $\frac{5}{4},\frac{3}{2}$ ≡$\frac{5}{4},\frac{3}{2}\times \frac{2}{2}$ ≡ $\frac{5}{4},\frac{6}{4}$

Hence, ratio 3 : 2 is the greatest ratio.

Hence, option (b).

**Alternately,**

If a/b > 1, ⇒ $\frac{a+x}{b+x}<\frac{a}{b}$

Hence, $\frac{5}{4}$ = $\frac{3+2}{2+2}$ < $\frac{3}{2}$.

**Alternately,**

Assume $\frac{5}{4}>\frac{3}{2}$

⇒ 5 × 2 > 3 × 4

⇒ 10 > 12, which is not true.

Hence our assumption was wrong.

∴ 3/2 is the greatest ratio.

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The smallest ratio out of 1 : 2, 2 : 1, 1 : 3 and 3 : 1 is?

- A.
3 : 1

- B.
1 : 3

- C.
2 : 1

- D.
1 : 2

Answer: Option B

**Explanation** :

Here, only 1/2 and 1/3 are less than 1. So, let’s consider only these two ratios. Other 2 ratios are greater than 1.

We have to find the smallest ratio here 1/2,1/3

Since numerator is same, the ratio with greater denominator will be smallest.

∴ 1/3 is the smallest of the ratios given.

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The ratio of two numbers is 3 : 5 and their sum is 72. Find the larger of the two numbers.

- A.
50

- B.
45

- C.
42

- D.
36

Answer: Option B

**Explanation** :

Let the numbers be 3k and 5k, where k is a constant.

Sum of 3k and 5k is 8k.

Now 8k = 72

⇒ k = 9

∴ The greater number is 5 × 6 i.e., 45.

**Alternately,**

Larger of the two numbers = 5/8 × 72 = 45.

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The present ages of two persons are in the ratio 15 : 17. Twenty-four years ago the ratio of their ages was 9 : 11. Find the present age of the older person.

- A.
64 years

- B.
72 years

- C.
56 years

- D.
68 years

Answer: Option D

**Explanation** :

Let their present ages be 15x and 17x years respectively.

Ratio of their ages 24 years ago:

$\frac{15x-24}{17x-24}=\frac{9}{11}$

⇒ 165x – 264 = 153x - 216

⇒ 12x = 48

⇒ x = 4

∴ The age of the older person is 17 × 4 i.e., 68 years

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The ratio of the number of students in 3 sections A, B and C is 3 : 7 : 8. If there are a total of 360 students in these sections, find the number of students in section A.

- A.
58

- B.
50

- C.
55

- D.
60

Answer: Option D

**Explanation** :

Let the number of students in A, B and C be 3x, 7x and 8x respectively.

3x + 7x + 8x = 360

⇒ x = 20

⇒ 3x = 60

Hence, 60.

**Alternately,**

Number of students in section A = $\frac{3}{3+7+8}$ × 360 = 60

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The present ages of Anand and Ashish are in the ratio 4 : 5. 10 years hence, the ratio of their ages will be 5 : 6. Find the present age of Anand. (in years)

- A.
40

- B.
35

- C.
50

- D.
45

Answer: Option A

**Explanation** :

Let the present ages of Anand and Ashish be 4x years and 5x years respectively.

The ratio of their ages 10 years hence = $\frac{4x+10}{5x+10}=\frac{5}{6}$

⇒ 24x + 60 = 25x + 50

⇒ 10 = x

Therefore, Anand’s present age is 4 × 10 = 40 years.

Hence, option (a).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

Find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second numbers added to the difference of the third and the second numbers is 42.

- A.
24, 16, 32

- B.
12, 8, 16

- C.
18, 12, 48

- D.
18, 12, 24

Answer: Option D

**Explanation** :

Let the numbers be p, q and r.

Given that p, q and r are in the ratio 3 : 2 : 4.

p : q : r = 3 : 2 : 4

Let p = 3x, q = 2x and r = 4x

Given (p + q) + (r – q) = 42

⇒ p + q + r – q = 42 ⇒ p + r = 42

⇒ 3x + 4x = 42

⇒ 7x = 42

⇒ x = 6

p, q, r are 3x, 2x, 4x.

∴ p, q, r are 18, 12, 24

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

156 is divided into two parts such that seven times the first part and five times the second part are in the ratio 5 : 2. Find the first part.

- A.
90

- B.
95

- C.
100

- D.
105

Answer: Option C

**Explanation** :

Let the numbers be x and 156 – x

$\frac{7x}{5(156-x)}=\frac{5}{2}$

⇒ 14x = 25 × 156 - 25x

⇒ 39x = 25 × 156

⇒ x = 25 × 4 = 100

∴ The first part is 100.

Hence, 100.

Alternately,

Let the numbers be a and b.

Hence, a + b = 156, and

$\frac{7a}{5b}=\frac{5}{2}$

⇒ $\frac{a}{b}=\frac{25}{14}$

⇒ a = 25/39 × 156 = 100.

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If a : b = 3 : 7, find the value of (4a + b) : (3a + 5b).

- A.
12 : 41

- B.
19 : 32

- C.
12 : 47

- D.
19 : 44

Answer: Option D

**Explanation** :

Let a = 3k, b = 7k

∴ $\frac{4a+b}{3a+5b}=\frac{4\times 3k+7k}{3\times 3k+5\times 7k}$ = $\frac{19k}{44k}=\frac{19}{44}$

Hence, option (d).

Alternately,

We have to find the value of $\frac{4a+b}{3a+5b}$

Dividing both numerator and denominator with b, we get

$\frac{4\left({\displaystyle \frac{a}{b}}\right)+1}{3\left({\displaystyle \frac{a}{b}}\right)+5}=\frac{4\left({\displaystyle \frac{3}{7}}\right)+1}{3\left({\displaystyle \frac{3}{7}}\right)+5}$ = $\frac{12+7}{9+35}=\frac{19}{44}$

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

In a group of 60 students, which of the following can’t be the ratio of males and females?

- A.
2 : 3

- B.
1 : 5

- C.
2 : 7

- D.
2 : 1

Answer: Option C

**Explanation** :

Let the ratio of males and females be m : f.

We know the number of males will be $\frac{m}{m+f}\times 60$

Consider option (c), m = 2 and f = 7. This cannot be the ratio because 60 is not divisible by (2 + 7) i.e. 9.

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If a : b = 3 : 2 and b : c = 7 : 5, then find a : b : c.

- A.
10 : 15 : 21

- B.
21 : 14 : 10

- C.
9 : 12 : 14

- D.
12 : 7 : 18

Answer: Option B

**Explanation** :

a/b = 3/2 = 21/14 (Multiplying and dividing by 7 to make the value of b same in both ratios)

b/c = 7/5 = 14/10 (Multiplying and dividing by 2 to make the value of b same in both ratios)

a : b : c = 21 : 14 : 10

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If A is 1/3^{rd} of B and B is 1/2 of C, then A : B : C is?

- A.
1 : 3 :6

- B.
2 : 3 : 6

- C.
3 : 2 : 6

- D.
None of these

Answer: Option A

**Explanation** :

A = B/3, B = C/2, C = C, then,

A : B : C = $\frac{B}{3}:\frac{C}{2}:\frac{C}{1}=\frac{C}{3\times 2}:\frac{C}{2}:\frac{C}{1}$ = $\frac{1}{6}:\frac{1}{2}:\frac{1}{1}$ = $\frac{6}{6}:\frac{6}{2}:\frac{6}{1}$ = 1 : 3 : 6

Hence, option (a).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If A : B = 3 : 2, B : C = 4 : 3, C : D = 5 : 4, then A : D is?

- A.
5 : 2

- B.
30 : 20

- C.
15 : 12

- D.
None of these

Answer: Option A

**Explanation** :

A/B = 3/2, B/C = 4/3, C/D = 5/4

Here, B = 2A/3 and B = 4C/3

⇒ 2A/3 = 4C/3 ⇒ C = A/2

∴ C/D = A/2D = 5/4

⇒ A/D = (5 × 2)/4 = 5/2 or 5 : 2

Hence, option (a).

**Alternately,**

A : B : C : D = 3 × 4 × 5 : 2 × 4 × 5 : 2 × 3 × 5 : 2 × 3 × 4

∴ A : B : C : D = 60 : 40 : 30 : 24

∴ A : D = 60 : 24 = 5 : 2

Hence, option (a).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The ratio of two numbers is 3 : 2 and their difference is 225, the smaller number is:

- A.
90

- B.
675

- C.
135

- D.
450

Answer: Option D

**Explanation** :

Let the two numbers be x and y whose ratio is 3 : 2, That is

x/y = 3/2 and x – y = 225

Let x = 3a and y = 2a.

∴ 3a – 2a = 225 ⇒ a = 225

⇒ x = 3 × 225 = 675 and y = 2 × a = 450.

Hence, smaller number of the ratio is 450.

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

What should be subtracted from the numbers in the ratio 9 : 16 so as to get 1 : 2?

- A.
8

- B.
14

- C.
2

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Two numbers are in the ratio of 9 : 16. Let the numbers be 9x and 16x.

Now, let a be subtracted to get new ratio as 1 : 2.

∴ $\frac{9x-a}{16x-a}=\frac{1}{2}$

⇒ 18x - 2a = 16x - a

⇒ 2x = a

Since we do not know the value of x, we cannot find out the value of a.

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If A : B = 2 : 3, B : C = 3 : 4 and C : D = 4 : 5, then, A : B : C : D is?

- A.
2 : 3 : 4 : 5

- B.
5 : 2 : 3 : 7

- C.
3 : 4 : 7 : 8

- D.
None of these

Answer: Option A

**Explanation** :

A/B = 2/3, B/C = 3/4, C/D = 4/5

⇒ If A = 2, B = 3, C = 4 and D = 5

∴ A : B : C : D = 2 : 3 : 4 : 5

Hence, option (a).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If $\frac{p+q}{2p+q}=\frac{3}{4}$, then find $\frac{3p+q}{2p+q}$.

- A.
4/5

- B.
5/4

- C.
6/7

- D.
3/4

Answer: Option B

**Explanation** :

Given that, $\frac{p+q}{2p+q}=\frac{3}{4}$

⇒ 4(p + q) = 3(2p + q)

⇒ 4p + 4q = 6p + 3q ⇒ q = 2p

Now, $\frac{3p+q}{2p+q}=\frac{3p+2p}{2p+2p}=\frac{5p}{4p}=\frac{5}{4}$

Hence, option (b).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If p + q + r = 180 and p = q/2 and q = r/3, find r.

- A.
10

- B.
50

- C.
100

- D.
120

Answer: Option D

**Explanation** :

p = $\frac{q}{2}$ ⇒ $\frac{p}{q}=\frac{1}{2}$

q = $\frac{r}{3}$ ⇒ $\frac{q}{r}=\frac{1}{3}=\frac{2}{6}$

So, p : q : r = 1 : 2 : 6

So, r = $\frac{6}{1+2+3}$ × 180 = 120.

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

If p : q = 3 : 1, find $\frac{p-2q}{3p-{q}^{2}}$.

- A.
2/7

- B.
1/8

- C.
3/7

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Given that p : q = 3 : 1

p/q = 3/1

p = 3q

$\frac{p-2q}{3p-{q}^{2}}=\frac{3q-2q}{3\left(3q\right)-{q}^{2}}$ (∵ p = 3q)

= $\frac{q}{9q-{q}^{2}}=\frac{1}{9-q}$

Since we don’t know the value of ‘q’, the answer cannot be determined.

Hence, option (d).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

At a party, there are a total of 56 people. If ‘m’ men join the party, the ratio of the number of ladies to that of men will change from 4 : 3 to 4 : 5. Find m.

- A.
8

- B.
12

- C.
16

- D.
15

Answer: Option C

**Explanation** :

Number of ladies at the party do not change.

Initial number of ladies = (4/7) × 56 = 32.

Initial number of men = 56 - 32 - 24.

After x men join,

$\frac{ladies}{men}=\frac{4}{5}$

the number of men would be 5/4 × (Number of ladies) = 5/4 × 32 = 40

∴ the number of men joining (m) = 40 – 24 = 16.

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

Rs. 1500 is divided among, A, B and C such that A receives 1/3^{rd} as much as B and C together. A’s share is?

- A.
Rs. 600

- B.
Rs. 525

- C.
Rs. 375

- D.
Rs. 0

Answer: Option C

**Explanation** :

Money received by A, B and C be a, b, c respectively then

a + b + c = 1500 …(i)

also, a = (b + c)/3 ⇒ b + c = 3a …(ii)

Then from (i) and (ii)

a + 3a = 1500

⇒ 4a = 1500 ⇒ a = 375

∴ A’s share is Rs. 375.

Hence, option (c).

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**CRE 1 - Ratio | Ratio, Proportion and Partnership**

The monthly salaries of A and B are in the ratio 2 : 3. The monthly expenditures of X and Y are in the ratio 3 : 4. Find the ratio of the monthly savings of X and Y.

- A.
5 : 3

- B.
3 : 5

- C.
4 : 7

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Let the monthly salaries of A and B be Rs. 2x and Rs. 3x respectively.

Let the monthly expenditures of A and B be Rs. 3y and Rs. 4y respectively.

Ratio of the monthly savings of A and B = $\frac{2x-3y}{3x-4y}=\frac{2\left({\displaystyle \frac{x}{y}}\right)-3}{3\left({\displaystyle \frac{x}{y}}\right)-4}$.

As x/y is unknown, the ratio cannot be found

Hence, option (d).

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