CRE 1 - Ratio

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).

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

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

Hence, option (c).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

A = 2B, 3B = 4C ⇒ C = 3B/4

Then, A : B : C = 2B : B : 3B/4 = 2 ∶ 1 ∶ 3/4 = $2\times 4:1\times 4:\frac{3\times 4}{4}$ = 8 ∶ 4 ∶ 3.

Hence, option (b).

The combined ratio:

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

= 16 : 24 : 18 : 45

Hence, option (a).

A/B = 2/3 = 4/6 and B/C = 6/5

∴ If A = 4, then B = 6 and C = 5.

⇒ A : B : C = 4 : 6 : 5.

∴ A’s share = $\frac{4}{4+6+5}$ × 750 = 200

Hence, option (c).

Remember speed = distance/time, since distance covered is the same, say d and if t₁, t₂, t₃ be the time taken by the three cars, then,

$\frac{d}{t_1}:\frac{d}{t_2}:\frac{d}{t_3}=4 : 3 : 2$ or $\frac{1}{t_1}:\frac{1}{t_2}:\frac{1}{t_3}=4 : 3 : 2$

To find t₁ : t₂ : t₃, we must find inverse ratio of 4 : 3 : 2, i.e. 1/4 : 1/3 : 1/2 = $\frac{1\times 12}{4}:\frac{1\times 12}{3}:\frac{1\times 12}{2}$ = 3 ∶ 4 ∶ 6

Hence, option (c).