Top

# Ratio & Proportion

1.

In a mixture $$60 litres,$$ the ratio of milk and water$$2 : 1.$$ If this ratio is to be $$1 : 2,$$ then the quanity of water to be further added is:

Quantity of milk $$= (60 \times \frac{2}{3}) littres = 40 littres.$$

Quantity of water in it $$= (60 - 40) littres = 20 littres.$$

New ratio $$= 1:2$$

Let quantity of water to be added further be $$x$$ $$littres.$$

Then, milk : water $$= (\frac{40}{20 + x})$$

Now, $$(\frac{40}{20 + x}) = \frac{1}{2}$$

$$\Rightarrow 20 + x = 80$$

$$\Rightarrow x = 60.$$

$$\therefore$$ Quantity of water to be added $$= 60$$  $$littres.$$

Enter details here

2.

In a bag, there are coins of $$25 p, 10 p$$ and $$5 p$$ in the ratio of $$1 : 2 : 3.$$ If there is $$Rs. 30$$ in all, how many $$5 p$$ coins are there?

Let the number of $$25 p, 10 p$$ and $$5 p$$ coins be $$x, 2x, 3x$$ respectively.

Then, sum of there value $$= Rs. (\frac{25x}{100} + \frac{10 \times 2x}{100} + \frac{5 \times 3x}{100}) = Rs. \frac{60x}{100}$$

$$\therefore \frac{60x}{100} = 30 \Leftrightarrow x = \frac{300 \times 100}{60} = 50.$$

Heance, the number of $$5 p$$ coins $$= (3 \times 50) = 150.$$

Enter details here

3.

If $$40$$% of a number is equal to two-third of another number, what is the ratio of first number to the second number?

Let $$40$$% of $$A = \frac{2}{3} B$$

Then, $$\frac{40A}{100} = \frac{2B}{3}$$

$$\Rightarrow \frac{2A}{5} = \frac{2B}{3}$$

$$\Rightarrow \frac{A}{B} = (\frac{2}{3} \times \frac{5}{2}) = \frac{5}{3}$$

$$\therefore A : B = 5 : 3.$$

Enter details here

4.

If $$0.75 : x :: 5 : 8,$$ then $$x$$ is equal to:

$$(x \times 5) = (0.75 \times 8 )$$

$$\Rightarrow x = (\frac{6}{5}) = 1.20$$

Enter details here

5.

$$A$$ and $$B$$ together have $$RS. 1210.$$ If $$\frac{4}{15}$$ of $$A,s$$ amount is equal to $$\frac{2}{5}$$ of $$B's$$ amount, how much amount does $$B$$ have?

$$\frac{4}{12}A = \frac{2}{5}B$$

$$\Rightarrow A = (\frac{2}{5} \times \frac{15}{4})B$$

$$\Rightarrow A = \frac{3}{2} B$$

$$\Rightarrow \frac{A}{B} = \frac{3}{2}$$

$$\Rightarrow A : B = 3 : 2.$$

$$\therefore B's$$ share $$= RS. (1210 \times \frac{2}{5}) = Rs. 484$$

Enter details here

6.

Two number are in the ratio $$3 : 5.$$ If $$9$$ is subtracted from each, the new numbers are in the ratio $$12 : 23.$$ The smaller number is:

Let the numbers be $$3x$$ and $$5x.$$

Then, $$\frac{3x - 9}{5x - 9} = \frac{12}{23}$$

$$\Rightarrow 23(3x - 9) = 12(5x - 9)$$

$$\Rightarrow 9x = 99$$

$$\Rightarrow x = 11.$$

$$\therefore$$ The smaller number $$= (3 \times 11) = 33.$$

Enter details here

7.

The fourth proportional to $$5, 8, 15$$ is:

Let the fourth proportional to $$5, 8 , 15$$ be $$x.$$

Then, $$5 : 8 : 15 : x$$

$$\Rightarrow 5x = (8 \times 15)$$

$$x = \frac{(8 \times 15)}{5} = 24.$$

Enter details here

8.

Two numbers are respectively $$20$$% and $$50$$% more than a $$3rd$$ number. The ratio of the two numbers is:

Let the $$3rd$$ number be $$x.$$

Then, first number $$= 120$$% of $$x = \frac{120x}{100} = \frac{6x}{5}$$

Second number $$= 150$$% of $$x = \frac{150x}{100} = \frac{3x}{2}$$

$$\therefore$$ Ratio of first two numbers $$= (\frac{6x}{5} : \frac{3x}{2}) = 12x : 15x = 4:5$$

Enter details here

9.

The ratio of the number of boys and girls in a college is $$7 : 8.$$ If the percentage increase in the number of boys and girls be $$20$$% and $$10$$% respectively, what will be the new ratio?

Let the number of boyes and girls in the collage be $$7x$$ and $$8x$$ respectively. Their increased number is $$(120$$% of $$7x)$$ and $$(110$$% of $$8x)$$.

$$\Rightarrow (\frac{120}{100} \times 7x)$$ and $$(\frac{110}{100} \times 8x)$$

$$\Rightarrow \frac{42x}{5}$$ and $$\frac{44x}{5}$$

$$\therefore$$ The required ratio $$= (\frac{42x}{5} : \frac{44x}{5}) = 21 :22.$$

Enter details here

10.

Seats for Mathematics, Physics and Biology in a school are in the ratio $$5 : 7 : 8.$$ There is a proposal to increase these seats by $$40$$%, $$50$$% and $$75$$% respectively. What will be the ratio of increased seats?

Let the number of seats for Mathematics, Physics and Biology be $$5x, 7x$$ and $$8x$$

Number of increased seats are $$(140$$% of $$5x),$$ $$(150$$% of $$7x)$$ and $$(175$$% of $$8x).$$

$$\Rightarrow (\frac{140}{100} \times 5x), (\frac{150}{100} \times 7x)$$ and $$(\frac{175}{100} \times 8x)$$

$$\Rightarrow 7x, \frac{21x}{2}$$ and $$14x.$$

$$\therefore$$ The required ratio $$= 7x : \frac{21x}{2} : 14x$$

$$\Rightarrow 14x : 21x : 28x$$

$$\Rightarrow 2 : 3 : 4.$$

Enter details here