Problems on ratios

You now know what a ratio is and how to share an amount in a ratio. In this les­son we use those ideas on real prob­lems: salaries, books in a library, stu­dents in a school, met­als in an alloy, and prices in a shop. Most of them are solved with the same small trick. Write the unknown amounts as $2k$, $3k$ and so on, then use the fact the ques­tion gives you.

Exam­ple 1: Salaries#

The salaries of Ravi and Sumit are in the ratio $2 : 3$. The salary of each is increased by ₹4000. Now the ratio becomes $40 : 57$. What is Sum­it's salary now?

Let Rav­i's old salary be $2k$ and Sum­it's old salary be $3k$. After the increase they are $2k + 4000$ and $3k + 4000$.

$$\displaystyle \frac{2k + 4000}{3k + 4000} = \frac{40}{57}$$

Cross mul­ti­ply:

$$\begin{aligned} 57(2k + 4000) &= 40(3k + 4000) \\ 114k + 57 \times 4000 &= 120k + 40 \times 4000 \\ 57 \times 4000 - 40 \times 4000 &= 120k - 114k \\ 4000 \times 17 &= 6k \\ 68000 &= 6k \end{aligned}$$

We want Sum­it's salary, which is $3k$. Half of $6k$ is $3k$, so $3k = 34000$.

Sum­it's old salary was ₹34,000. His salary now is $34000 + 4000 = 38000$, that is ₹38,000.

Com­par­ing ratios#

To see which of two ratios is big­ger, write them as frac­tions with the same denom­i­na­tor. This is just like com­par­ing frac­tions.

Exam­ple 2: Ratios with mixed num­bers#

If $\displaystyle 1\frac{3}{4} : 2\frac{1}{2} : 1\frac{1}{6}$ is $A : B : C$, which is true?

(a) $A = B = C$ (b) $B > A > C$ (c) $C > A > B$ (d) $B > C > A$

First change the mixed num­bers into improper frac­tions:

$$\displaystyle A : B : C = \frac{7}{4} : \frac{5}{2} : \frac{7}{6}$$

The L.C.M. of 4, 2 and 6 is 12. Write each frac­tion with denom­i­na­tor 12:

$$\displaystyle \frac{21}{12} : \frac{30}{12} : \frac{14}{12} = 21 : 30 : 14$$

Now it is easy to see that 30 is the biggest and 14 is the small­est. So $B > A > C$. The answer is (b).

A ratio equal to another ratio#

Exam­ple 3: Books in a library#

In a school library, the ratio of comic books to story books is the same as the ratio of story books to Hindi books. There are 540 story books and 300 Hindi books. Find the num­ber of comic books.

The ratio of story books to Hindi books is

$$\displaystyle \frac{540}{300} = \frac{9}{5}$$

The ratio of comic books to story books is the same:

$$\displaystyle \frac{\text{comic books}}{540} = \frac{9}{5}$$

So the num­ber of comic books is $\displaystyle 540 \times \frac{9}{5} = 108 \times 9 = 972$.

When the num­bers change#

Exam­ple 4: New stu­dents join#

A school has 600 stu­dents. The ratio of boys to girls is $13 : 11$. What will the new ratio be if 25 more girls join?

Total parts $= 13 + 11 = 24$, and $600 \div 24 = 25$.

Num­ber of boys $= 13 \times 25 = 325$.

Num­ber of girls $= 600 - 325 = 275$.

After 25 girls join, there are $275 + 25 = 300$ girls. The boys stay at 325.

$$\displaystyle \text{new ratio} = \frac{325}{300} = \frac{13}{12}$$

The new ratio of boys to girls is $13 : 12$.

Mix­ing two things#

Exam­ple 5: An alloy of gold and cop­per#

Gold is 19 times as heavy as water. Cop­per is 9 times as heavy as water. In what ratio should they be mixed to get an alloy that is 15 times as heavy as water?

"19 times as heavy as water" means that a piece of gold weighs 19 times as much as the same amount of water. So mea­sure the met­als in equal units of size, and let one unit of water weigh $w$. Then one unit of gold weighs $19w$ and one unit of cop­per weighs $9w$.

Take 1 unit of gold and $x$ units of cop­per. Together they make $(1 + x)$ units of alloy. The ratio we want is $1 : x$.

The alloy should weigh $15w$ for every unit, so

$$19w + x(9w) = (1 + x) \times 15w$$

Divide every term by $w$:

$$\displaystyle \begin{aligned} 19 + 9x &= 15 + 15x \\ 19 - 15 &= 15x - 9x \\ 4 &= 6x \\ x &= \frac{2}{3} \end{aligned}$$

So the ratio is $\displaystyle 1 : \frac{2}{3}$. Mul­ti­ply both terms by 3 to clear the frac­tion: $3 : 2$.

Check: 3 units of gold and 2 units of cop­per weigh as much as $3 \times 19 + 2 \times 9 = 57 + 18 = 75$ units of water. They make 5 units of alloy, and $75 \div 5 = 15$. So the alloy is 15 times as heavy as water. Cor­rect.

Find­ing the miss­ing part first#

Exam­ple 6: Cricket matches#

A school played 22 cricket matches and won 14 of them. What is the ratio of matches played to matches lost?

Matches lost $= 22 - 14 = 8$.

$$\displaystyle \frac{22}{8} = \frac{11}{4}$$

The ratio is $11 : 4$.

Exam­ple 7: Bulbs in a box#

A box has 40 bulbs, and 25 of them are defec­tive. If the ratio of defec­tive bulbs to good bulbs is $x : y$, find $x + y$.

Good bulbs $= 40 - 25 = 15$.

$$\displaystyle \frac{25}{15} = \frac{5}{3}$$

So $x : y = 5 : 3$ and $x + y = 5 + 3 = 8$.

Ratios with shapes#

Exam­ple 8: A square field and a rec­tan­gu­lar field#

The ratio of the area of a square field of side 14 m to the area of a rec­tan­gu­lar field is $4 : 3$. The breadth of the rec­tan­gu­lar field is 7 m. Find the ratio of the perime­ter of the square field to that of the rec­tan­gu­lar field.

Area of a square $= a \times a$. Area of a rec­tan­gle $= l \times b$.

$$\displaystyle \frac{14 \times 14}{l \times 7} = \frac{4}{3}$$

The left side is $\displaystyle \frac{196}{7l} = \frac{28}{l}$. So $\displaystyle \frac{28}{l} = \frac{4}{3}$, which gives $4l = 84$ and $l = 21$ m.

Perime­ter of the square $= 4 \times 14 = 56$ m. Perime­ter of the rec­tan­gle $= 2(21 + 7) = 56$ m.

The ratio of the perime­ters is $56 : 56 = 1 : 1$.

Ratios of prices#

Exam­ple 9: Cof­fee and tea#

Cof­fee costs ₹28 for 100 g. Tea costs ₹560 per kg. Find the ratio of the price of cof­fee to the price of tea.

Com­pare the prices for the same weight. 1 kg = 1000 g, which is 10 lots of 100 g. So cof­fee costs $28 \times 10 = 280$ rupees per kg.

$$\displaystyle \frac{280}{560} = \frac{1}{2}$$

The ratio is $1 : 2$.

Exam­ple 10: Mix­ing two milk mix­tures#

Two equal ves­sels A and B hold mix­tures of milk and water. In A the ratio of milk to water is $4 : 5$. In B it is $5 : 1$. In what ratio should we take the mix­tures from A and B to get a new mix­ture with milk and water in the ratio $5 : 4$?

Look at the part of each mix­ture that is milk.

In A, milk is 4 parts out of 9, so milk is $\displaystyle \frac{4}{9}$ of the mix­ture.

In B, milk is 5 parts out of 6, so milk is $\displaystyle \frac{5}{6}$ of the mix­ture.

In the new mix­ture, milk should be 5 parts out of 9, which is $\displaystyle \frac{5}{9}$.

Take $x$ litres from A and $y$ litres from B. The milk in them must be $\displaystyle \frac{5}{9}$ of $(x + y)$:

$$\displaystyle \frac{4}{9}x + \frac{5}{6}y = \frac{5}{9}(x + y)$$

Mul­ti­ply every term by 18 to clear the frac­tions:

$$\begin{aligned} 8x + 15y &= 10x + 10y \\ 15y - 10y &= 10x - 8x \\ 5y &= 2x \end{aligned}$$

So $\displaystyle \frac{x}{y} = \frac{5}{2}$. Take the mix­tures from A and B in the ratio $5 : 2$.

Check: 5 litres of A has $\displaystyle \frac{20}{9}$ litres of milk. 2 litres of B has $\displaystyle \frac{5}{3} = \frac{15}{9}$ litres of milk. Together that is $\displaystyle \frac{35}{9}$ litres of milk in 7 litres, and $\displaystyle \frac{35}{9} \div 7 = \frac{5}{9}$. Cor­rect.

Prac­tice#

1. The ages of Ravi and Sita are in the ratio $4 : 5$. After 6 years the ratio will be $6 : 7$. Find their present ages.
2. Arrange the ratios $2 : 3$, $4 : 5$ and $3 : 4$ from biggest to small­est.
3. The ratio of red pens to blue pens is the same as the ratio of blue pens to green pens. There are 60 blue pens and 45 green pens. How many red pens are there?
4. A school has 960 stu­dents and the ratio of boys to girls is $7 : 5$. If 40 more boys join, what is the new ratio?
5. A team played 30 matches and won 18. Find the ratio of matches won to matches lost.
6. A bas­ket has 60 man­goes and 12 of them are rot­ten. If the ratio of good man­goes to rot­ten man­goes is $x : y$, find $x + y$.
7. Rice costs ₹45 per kg. Wheat flour costs ₹6 for 200 g. Find the ratio of the price of rice to the price of wheat flour.
8. Metal A is 12 times as heavy as water and metal B is 4 times as heavy as water. In what ratio should they be mixed to get an alloy 10 times as heavy as water?
9. A square field has side 12 m. The ratio of its area to the area of a rec­tan­gu­lar field is $3 : 2$. The breadth of the rec­tan­gle is 8 m. Find the ratio of the perime­ter of the square to the perime­ter of the rec­tan­gle.

Answers#