4. Common Types

Some common type of questions on Linear Equations asked in the entrance tests are explained below.

4.1 Age-based questions

Example 6

$4$ years ago, Ram was $6$ times as old as his son. $5$ years from now, Ram will be thrice his son's age. What is the present age of Ram?

Solution

Let the present age of Ram and his son be $r$ and $s$ respectively. The following $2$ statements from the question can be used to form $2$ linear equations.

Statement $\bm{1}$: $4$ years back Ram was $6$ times as old as his son.
⇒ $ (r - 4) = 6(s - 4)$
⇒ $ r - 6s = -20$ -----($1$)

Statement $\bm{2}$: 5 years from now, Ram will be thrice his son's age.
⇒ $ r + 5 = 3(s + 5)$
⇒ $ r - 3s = 10$ -----($2$)

Eq($2$) $-$ Eq($1$) ⇒ $ -3s +6s = 10 + 20$
⇒ $ \bm{s = 10}$

Substituting $s = 10$ in Eq($1$), we get $\bm{r = 40}$

Answer: $40$ years

4.2 Digits-based questions

Example 7

When the digits of a $2$-digit number are reversed, its value reduces by $45$. If the product of the digits is $24$, then what is the original number?

Solution

Let the $2$-digit number be $10a + b$. When the digits are reversed, the number becomes $10b + a$.

$10a + b - (10b + a) = 45$
⇒ $ 9a - 9b = 45$
$a - b = 5$
⇒ $ a = b + 5$ -----($1$)

It is given that $ab = 24$
⇒ $ b(b + 5) = 24$ [Substituting $a = b + 5$ from Eq ($1$)]
⇒ $ b(b + 5) = 3(3 + 5)$
$\therefore \bm{b = 3}$

Substituting $b = 3$ in Eq($1$) ⇒ $ \bm{a = 8}$

$\therefore$ Original number $= 10a + b = 10 \times 8 + 3 = 83$

Answer: $83$

4.3 Fraction-based questions

Example 8

The sum of the numerator and denominator of a fraction is $6$. When $3$ is added to each of the numerator and the denominator, the fraction's value becomes $\dfrac{1}{2}$. What is the value of the initial fraction in decimal form?

Solution

Let the numerator and denominator of the initial fraction be $a$ and $b$ respectively.

Statement $\bm{1}$: The sum of the numerator and denominator of a fraction is $6$. $a + b = 6$ -----($1$)

Statement $\bm{2}$: When $3$ is added to each of the numerator and the denominator, the fraction's value is $\dfrac{1}{2}$

$\dfrac{a + 3}{b + 3} = \dfrac{1}{2}$ ⇒ $2a + 6 = b + 3$

⇒ $ 2a - b = -3$ -----($2$)

Eq($1$) + Eq($2$) ⇒ $ 3a = 3$
⇒ $ \bm{a = 1}$
$\therefore \bm{b = 5}$ [Substituting in Eq ($1$)]

Value of initial fraction in decimal form $= \dfrac{a}{b} = \dfrac{1}{5} = 0.2$

Answer: $0.2$

Example 9

When the numerator and denominator of a certain fraction are reduced by $2$ and $1$ respectively, then the fraction becomes $\dfrac{2}{5}$. However, if the numerator and denominator of the same fraction were increased by $1$ and $4$ respectively, the fraction would have become $\dfrac{3}{7}$. What is the sum of the numerator and denominator of the original fraction? ($1$) $20$ ($2$) $25$ ($3$) $40$ ($4$) $45$

Solution

Let the numerator and denominator of the initial fraction be $a$ and $b$ respectively

. Statement $\bm{1}$: When the numerator and denominator of a certain fraction are reduced by $2$ and $1$ respectively, then the fraction becomes $\dfrac{2}{5}$.
$\dfrac{a - 2}{b - 1} = \dfrac{2}{5}$ ⇒ $ 5a - 2b = 8$ -----($1$)

Statement $\bm{2}$: if the numerator and denominator of the same fraction were increased by $1$ and $4$ respectively, the fraction would have become $\dfrac{3}{7}$.

$\dfrac{a + 1}{b + 4} = \dfrac{3}{7}$ ⇒ $ 7a - 3b = 5$ -----($2$)

$3 \times $ Eq($1$) $-$ 2 $\times$ Eq($2$) ⇒ $ 15a - 14a = 24 - 10$
⇒ $ \bm{a = 14}$
Substituting $a = 14$ in Eq($1$), $\bm{b = 31}$

The total of the numerator and the denominator $= 14 + 31 = \bm{45}$

Answer: ($4$) $45$