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Linear Equations

Linear Equations

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Basics of Equations
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Linear Equations 1
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Linear Equations 2
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Linear Equations : Level 1
Linear Equations : Level 2
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ALL MODULES

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

444 years ago, Ram was 666 times as old as his son. 555 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 rrr and sss respectively. The following 222 statements from the question can be used to form 222 linear equations.

Statement 1\bm{1}1: 444 years back Ram was 666 times as old as his son.
⇒ (r−4)=6(s−4) (r - 4) = 6(s - 4)(r−4)=6(s−4)
⇒ r−6s=−20 r - 6s = -20r−6s=−20 -----(111)

Statement 2\bm{2}2: 5 years from now, Ram will be thrice his son's age.
⇒ r+5=3(s+5) r + 5 = 3(s + 5)r+5=3(s+5)
⇒ r−3s=10 r - 3s = 10r−3s=10 -----(222)

Eq(222) −-− Eq(111) ⇒ −3s+6s=10+20 -3s +6s = 10 + 20−3s+6s=10+20
⇒ s=10 \bm{s = 10}s=10

Substituting s=10s = 10s=10 in Eq(111), we get r=40\bm{r = 40}r=40

Answer: 404040 years


4.2 Digits-based questions

Example 7

When the digits of a 222-digit number are reversed, its value reduces by 454545. If the product of the digits is 242424, then what is the original number?

Solution

Let the 222-digit number be 10a+b10a + b10a+b. When the digits are reversed, the number becomes 10b+a10b + a10b+a.

10a+b−(10b+a)=4510a + b - (10b + a) = 4510a+b−(10b+a)=45
⇒ 9a−9b=45 9a - 9b = 459a−9b=45
a−b=5a - b = 5a−b=5
⇒ a=b+5 a = b + 5a=b+5 -----(111)

It is given that ab=24ab = 24ab=24
⇒ b(b+5)=24 b(b + 5) = 24b(b+5)=24                 [Substituting a=b+5a = b + 5a=b+5 from Eq (111)]
⇒ b(b+5)=3(3+5) b(b + 5) = 3(3 + 5)b(b+5)=3(3+5)
∴b=3\therefore \bm{b = 3}∴b=3

Substituting b=3b = 3b=3 in Eq(111) ⇒ a=8 \bm{a = 8}a=8

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

Answer: 838383


4.3 Fraction-based questions

Example 8

The sum of the numerator and denominator of a fraction is 666. When 333 is added to each of the numerator and the denominator, the fraction's value becomes 12\dfrac{1}{2}21​. What is the value of the initial fraction in decimal form?

Solution

Let the numerator and denominator of the initial fraction be aaa and bbb respectively.

Statement 1\bm{1}1: The sum of the numerator and denominator of a fraction is 666. a+b=6a + b = 6a+b=6 -----(111)

Statement 2\bm{2}2: When 333 is added to each of the numerator and the denominator, the fraction's value is 12\dfrac{1}{2}21​

a+3b+3=12\dfrac{a + 3}{b + 3} = \dfrac{1}{2}b+3a+3​=21​ ⇒ 2a+6=b+32a + 6 = b + 32a+6=b+3

⇒ 2a−b=−3 2a - b = -32a−b=−3 -----(222)

Eq(111) + Eq(222) ⇒ 3a=3 3a = 33a=3
⇒ a=1 \bm{a = 1}a=1
∴b=5\therefore \bm{b = 5}∴b=5 [Substituting in Eq (111)]

Value of initial fraction in decimal form =ab=15=0.2= \dfrac{a}{b} = \dfrac{1}{5} = 0.2=ba​=51​=0.2

Answer: 0.20.20.2


Example 9

When the numerator and denominator of a certain fraction are reduced by 222 and 111 respectively, then the fraction becomes 25\dfrac{2}{5}52​. However, if the numerator and denominator of the same fraction were increased by 111 and 444 respectively, the fraction would have become 37\dfrac{3}{7}73​. What is the sum of the numerator and denominator of the original fraction? (111) 202020            (222) 252525            (333) 404040            (444) 454545           

Solution

Let the numerator and denominator of the initial fraction be aaa and bbb respectively

. Statement 1\bm{1}1: When the numerator and denominator of a certain fraction are reduced by 222 and 111 respectively, then the fraction becomes 25\dfrac{2}{5}52​.
a−2b−1=25\dfrac{a - 2}{b - 1} = \dfrac{2}{5}b−1a−2​=52​ ⇒ 5a−2b=8 5a - 2b = 85a−2b=8 -----(111)

Statement 2\bm{2}2: if the numerator and denominator of the same fraction were increased by 111 and 444 respectively, the fraction would have become 37\dfrac{3}{7}73​.

a+1b+4=37\dfrac{a + 1}{b + 4} = \dfrac{3}{7}b+4a+1​=73​ ⇒ 7a−3b=5 7a - 3b = 57a−3b=5 -----(222)

3×3 \times 3× Eq(111) −-− 2 ×\times× Eq(222) ⇒ 15a−14a=24−10 15a - 14a = 24 - 1015a−14a=24−10
⇒ a=14 \bm{a = 14}a=14
Substituting a=14a = 14a=14 in Eq(111), b=31\bm{b = 31}b=31

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

Answer: (444) 454545


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