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CAT 2025 Lesson : Linear Equations - Age, Digits & Fraction Based Questions

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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

44 years ago, Ram was 66 times as old as his son. 55 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 rr and ss respectively. The following 22 statements from the question can be used to form 22 linear equations.

Statement 1\bm{1}: 44 years back Ram was 66 times as old as his son.
⇒ (r−4)=6(s−4) (r - 4) = 6(s - 4)
⇒ r−6s=−20 r - 6s = -20 -----(11)

Statement 2\bm{2}: 5 years from now, Ram will be thrice his son's age.
⇒ r+5=3(s+5) r + 5 = 3(s + 5)
⇒ r−3s=10 r - 3s = 10 -----(22)

Eq(22) −- Eq(11) ⇒ −3s+6s=10+20 -3s +6s = 10 + 20
⇒ s=10 \bm{s = 10}

Substituting s=10s = 10 in Eq(11), we get r=40\bm{r = 40}

Answer: 4040 years


4.2 Digits-based questions

Example 7

When the digits of a 22-digit number are reversed, its value reduces by 4545. If the product of the digits is 2424, then what is the original number?

Solution

Let the 22-digit number be 10a+b10a + b. When the digits are reversed, the number becomes 10b+a10b + a.

10a+b−(10b+a)=4510a + b - (10b + a) = 45
⇒ 9a−9b=45 9a - 9b = 45
a−b=5a - b = 5
⇒ a=b+5 a = b + 5 -----(11)

It is given that ab=24ab = 24
⇒ b(b+5)=24 b(b + 5) = 24                 [Substituting a=b+5a = b + 5 from Eq (11)]
⇒ b(b+5)=3(3+5) b(b + 5) = 3(3 + 5)
∴b=3\therefore \bm{b = 3}

Substituting b=3b = 3 in Eq(11) ⇒ a=8 \bm{a = 8}

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

Answer: 8383


4.3 Fraction-based questions

Example 8

The sum of the numerator and denominator of a fraction is 66. When 33 is added to each of the numerator and the denominator, the fraction's value becomes 12\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 aa and bb respectively.

Statement 1\bm{1}: The sum of the numerator and denominator of a fraction is 66. a+b=6a + b = 6 -----(11)

Statement 2\bm{2}: When 33 is added to each of the numerator and the denominator, the fraction's value is 12\dfrac{1}{2}

a+3b+3=12\dfrac{a + 3}{b + 3} = \dfrac{1}{2} ⇒ 2a+6=b+32a + 6 = b + 3

⇒ 2a−b=−3 2a - b = -3 -----(22)

Eq(11) + Eq(22) ⇒ 3a=3 3a = 3
⇒ a=1 \bm{a = 1}
∴b=5\therefore \bm{b = 5} [Substituting in Eq (11)]

Value of initial fraction in decimal form =ab=15=0.2= \dfrac{a}{b} = \dfrac{1}{5} = 0.2

Answer: 0.20.2


Example 9

When the numerator and denominator of a certain fraction are reduced by 22 and 11 respectively, then the fraction becomes 25\dfrac{2}{5}. However, if the numerator and denominator of the same fraction were increased by 11 and 44 respectively, the fraction would have become 37\dfrac{3}{7}. What is the sum of the numerator and denominator of the original fraction? (11) 2020            (22) 2525            (33) 4040            (44) 4545           

Solution

Let the numerator and denominator of the initial fraction be aa and bb respectively

. Statement 1\bm{1}: When the numerator and denominator of a certain fraction are reduced by 22 and 11 respectively, then the fraction becomes 25\dfrac{2}{5}.
a−2b−1=25\dfrac{a - 2}{b - 1} = \dfrac{2}{5} ⇒ 5a−2b=8 5a - 2b = 8 -----(11)

Statement 2\bm{2}: if the numerator and denominator of the same fraction were increased by 11 and 44 respectively, the fraction would have become 37\dfrac{3}{7}.

a+1b+4=37\dfrac{a + 1}{b + 4} = \dfrac{3}{7} ⇒ 7a−3b=5 7a - 3b = 5 -----(22)

3×3 \times Eq(11) −- 2 ×\times Eq(22) ⇒ 15a−14a=24−10 15a - 14a = 24 - 10
⇒ a=14 \bm{a = 14}
Substituting a=14a = 14 in Eq(11), b=31\bm{b = 31}

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

Answer: (44) 4545


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