Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 14 Linear Equations in one Variable

Exercise 14(A)

Solve the following equations:

Question 1. 20 = 6 + 2x
Answer:
Given equation:
\( 20 = 6 + 2x \)
Move \( 6 \) to the left side by subtracting it:
\( 20 - 6 = 2x \)
\( 14 = 2x \)
Now, divide both sides by \( 2 \):
\( x = \frac{14}{2} \)
\( x = 7 \)
In simple words: We want to find the value of x. First, we move 6 to the other side of the equals sign and subtract it from 20. Then, we divide 14 by 2 to get 7.

Exam Tip: You can check your answer by putting \( x = 7 \) back into the original equation to see if both sides are equal.

Question 2. 15 + x = 5x + 3
Answer:
Given equation:
\( 15 + x = 5x + 3 \)
Group the like terms by shifting \( x \) to the right and \( 3 \) to the left:
\( 15 - 3 = 5x - x \)
\( 12 = 4x \)
Divide both sides of the equation by \( 4 \):
\( x = \frac{12}{4} \)
\( x = 3 \)
In simple words: First, put the letters on one side and the numbers on the other side. Then, divide 12 by 4 to get 3.

Exam Tip: When moving a term to the other side of the equal sign, always remember to change its sign from plus to minus or minus to plus.

Question 3. \( \frac{3x+2}{x-6} = -7 \)
Answer:
Given equation:
\( \frac{3x+2}{x-6} = -7 \)
Multiply both sides by \( (x - 6) \):
\( 3x + 2 = -7(x - 6) \)
Expand the bracket on the right side:
\( 3x + 2 = -7x + 42 \)
Bring all \( x \) terms to the left side and constant numbers to the right side:
\( 3x + 7x = 42 - 2 \)
\( 10x = 40 \)
Now, divide by \( 10 \):
\( x = \frac{40}{10} \)
\( x = 4 \)
In simple words: Multiply the bottom part of the fraction with the right side. Open the bracket carefully, then group the x terms together and solve for x.

Exam Tip: Pay close attention to signs when multiplying. Here, \( -7 \times -6 \) becomes positive \( 42 \).

Question 4. 3a - 4 = 2(4 - a)
Answer:
Given equation:
\( 3a - 4 = 2(4 - a) \)
Multiply the terms inside the bracket by \( 2 \):
\( 3a - 4 = 8 - 2a \)
Move \( -2a \) to the left side and \( -4 \) to the right side:
\( 3a + 2a = 8 + 4 \)
\( 5a = 12 \)
Divide both sides by \( 5 \):
\( a = \frac{12}{5} \)
\( a = 2.4 \)
In simple words: First, multiply the outer number with everything inside the bracket. Next, group the letters on one side and numbers on the other side to find the value of a.

Exam Tip: If the final answer is a fraction like \( \frac{12}{5} \), convert it to a decimal to show a complete answer.

Question 5. 3(b - 4) = 2(4 - b)
Answer:
Given equation:
\( 3(b - 4) = 2(4 - b) \)
Expand the brackets on both sides:
\( 3b - 12 = 8 - 2b \)
Collect the variables on one side and the numerical values on the other side:
\( 3b + 2b = 8 + 12 \)
\( 5b = 20 \)
Divide by \( 5 \):
\( b = \frac{20}{5} \)
\( b = 4 \)
In simple words: Expand both brackets first. Then, move the letters to the left side and the numbers to the right side to get the value of b.

Exam Tip: Always multiply the number outside the bracket with each term inside the bracket, not just the first one.

Question 6. \( \frac{x+2}{9} = \frac{x+4}{11} \)
Answer:
Given equation:
\( \frac{x+2}{9} = \frac{x+4}{11} \)
Cross-multiply to remove the denominators:
\( 11(x + 2) = 9(x + 4) \)
Expand both sides:
\( 11x + 22 = 9x + 36 \)
Group the \( x \) terms together on the left and the constant terms on the right:
\( 11x - 9x = 36 - 22 \)
\( 2x = 14 \)
Divide both sides by \( 2 \):
\( x = \frac{14}{2} \)
\( x = 7 \)
In simple words: Multiply the bottom number of one side with the top part of the other side. This gets rid of the fractions, so we can solve it easily.

Exam Tip: When cross-multiplying, remember to put brackets around binomials like \( (x + 2) \) so you multiply the entire group.

Question 7. \( \frac{x-8}{5} = \frac{x-12}{9} \)
Answer:
Given equation:
\( \frac{x-8}{5} = \frac{x-12}{9} \)
Perform cross-multiplication:
\( 9(x - 8) = 5(x - 12) \)
Multiply through the brackets:
\( 9x - 72 = 5x - 60 \)
Shift the variables to the left and constants to the right:
\( 9x - 5x = -60 + 72 \)
\( 4x = 12 \)
Divide both sides by \( 4 \):
\( x = \frac{12}{4} \)
\( x = 3 \)
In simple words: Cross-multiply to clear the fractions. Then, open the brackets and move the terms to solve for x.

Exam Tip: Be extra careful with signs. Here, moving \( -72 \) to the right side changes it to \( +72 \).

Question 8. 5(8x + 3) = 9(4x + 7)
Answer:
Given equation:
\( 5(8x + 3) = 9(4x + 7) \)
Expand the expressions on both sides:
\( 40x + 15 = 36x + 63 \)
Rearrange terms by moving all \( x \) terms to the left side and numbers to the right side:
\( 40x - 36x = 63 - 15 \)
\( 4x = 48 \)
Divide by \( 4 \):
\( x = \frac{48}{4} \)
\( x = 12 \)
In simple words: Multiply the numbers outside with the terms inside the brackets. Then put the x terms on the left and the numbers on the right.

Exam Tip: Double-check your basic multiplication tables, as simple errors like \( 9 \times 7 \) can lose easy marks.

Question 9. 3(x + 1) = 12 + 4(x - 1)
Answer:
Given equation:
\( 3(x + 1) = 12 + 4(x - 1) \)
Expand the brackets on both sides:
\( 3x + 3 = 12 + 4x - 4 \)
Simplify the numbers on the right side:
\( 3x + 3 = 4x + 8 \)
Bring \( 4x \) to the left side and \( 3 \) to the right side:
\( 3x - 4x = 8 - 3 \)
\( -x = 5 \)
Multiply both sides by \( -1 \):
\( x = -5 \)
In simple words: Expand the brackets first. Group the x terms on one side and the numbers on the other side, then find the value of x.

Exam Tip: If you get \( -x = 5 \), remember that this means \( x = -5 \). Always solve for positive \( x \).

Question 10. \( \frac{3x}{4} - \frac{1}{4}(x - 20) = \frac{x}{4} + 32 \)
Answer:
Given equation:
\( \frac{3x}{4} - \frac{1}{4}(x - 20) = \frac{x}{4} + 32 \)
Expand the bracket:
\( \frac{3x}{4} - \frac{x}{4} + 5 = \frac{x}{4} + 32 \)
Rearrange terms to bring all the fraction terms with \( x \) to the left side:
\( \frac{3x}{4} - \frac{x}{4} - \frac{x}{4} = 32 - 5 \)
Combine the fractions over a common denominator of \( 4 \):
\( \frac{3x - x - x}{4} = 27 \)
\( \frac{x}{4} = 27 \)
Multiply both sides by \( 4 \):
\( x = 27 \times 4 \)
\( x = 108 \)
In simple words: Expand the bracket carefully. Put all terms with x on the left and normal numbers on the right, then multiply by 4 to find x.

Exam Tip: Be careful when multiplying \( -\frac{1}{4} \) with \( -20 \). The negative signs multiply to give a positive \( +5 \).

Question 11. \( 3a - \frac{1}{5} = \frac{a}{5} + 5\frac{2}{5} \)
Answer:
Given equation:
\( 3a - \frac{1}{5} = \frac{a}{5} + 5\frac{2}{5} \)
Rearrange the terms by bringing the variables to the left and constants to the right:
\( 3a - \frac{a}{5} = 5\frac{2}{5} + \frac{1}{5} \)
Convert the mixed fraction to an improper fraction:
\( 5\frac{2}{5} = \frac{5 \times 5 + 2}{5} = \frac{27}{5} \)
So the equation becomes:
\( 3a - \frac{a}{5} = \frac{27}{5} + \frac{1}{5} \)
To clear the denominators, multiply each term of the equation by \( 5 \):
\( 3a \times 5 - \left(\frac{a}{5} \times 5\right) = \left(\frac{27}{5} \times 5\right) + \left(\frac{1}{5} \times 5\right) \)
\( 15a - a = 27 + 1 \)
\( 14a = 28 \)
Divide by \( 14 \):
\( a = \frac{28}{14} \)
\( a = 2 \)
In simple words: Convert the mixed fraction to a normal fraction. Multiply the whole equation by 5 to get rid of the denominators, then solve for a.

Exam Tip: Multiplying every term by the common denominator is a quick way to eliminate fractions in an equation.

Question 12. \( \frac{x}{3} - 2\frac{1}{2} = \frac{4x}{9} - \frac{2x}{3} \)
Answer:
Given equation:
\( \frac{x}{3} - 2\frac{1}{2} = \frac{4x}{9} - \frac{2x}{3} \)
Convert the mixed number to an improper fraction:
\( \frac{x}{3} - \frac{5}{2} = \frac{4x}{9} - \frac{2x}{3} \)
Find the Least Common Multiple (LCM) of the denominators \( 3, 2, 9, 3 \), which is \( 18 \).
Multiply each term of the equation by \( 18 \) to clear the fractions:
\( \left(\frac{x}{3} \times 18\right) - \left(\frac{5}{2} \times 18\right) = \left(\frac{4x}{9} \times 18\right) - \left(\frac{2x}{3} \times 18\right) \)
\( 6x - 45 = 8x - 12x \)
Simplify the right side:
\( 6x - 45 = -4x \)
Move \( -4x \) to the left side and \( -45 \) to the right side:
\( 6x + 4x = 45 \)
\( 10x = 45 \)
Divide by \( 10 \):
\( x = \frac{45}{10} \)
\( x = 4.5 \)
In simple words: Turn the mixed fraction into a standard fraction. Multiply all terms by the LCM of denominators, which is 18, then group the x terms and solve.

Exam Tip: Finding the correct LCM first is crucial. If you multiply by the wrong number, the fractions won't clear completely.

Question 13. \( \frac{4(y+2)}{5} = 7 + \frac{5y}{13} \)
Answer:
Given equation:
\( \frac{4(y+2)}{5} = 7 + \frac{5y}{13} \)
Simplify the numerator on the left side and combine the terms on the right side over a common denominator of \( 13 \):
\( \frac{4y + 8}{5} = \frac{7 \times 13 + 5y}{13} \)
\( \frac{4y + 8}{5} = \frac{91 + 5y}{13} \)
Cross-multiply to eliminate the denominators:
\( 13(4y + 8) = 5(91 + 5y) \)
Expand both sides:
\( 52y + 104 = 455 + 25y \)
Group the variables on the left side and constants on the right side:
\( 52y - 25y = 455 - 104 \)
\( 27y = 351 \)
Divide both sides by \( 27 \):
\( y = \frac{351}{27} \)
\( y = 13 \)
In simple words: Write both sides as single fractions. Then, cross-multiply, expand the brackets, and solve for y.

Exam Tip: Keep your expansion steps clean. Be careful when multiplying \( 13 \times 8 = 104 \) and \( 5 \times 91 = 455 \).

Question 14. \( \frac{a+5}{6} - \frac{a+1}{9} = \frac{a+3}{4} \)
Answer:
Given equation:
\( \frac{a+5}{6} - \frac{a+1}{9} = \frac{a+3}{4} \)
Find the LCM of denominators \( 6, 9, 4 \), which is \( 36 \).
Multiply every term in the equation by \( 36 \) to clear the fractions:
\( 36 \left(\frac{a+5}{6}\right) - 36 \left(\frac{a+1}{9}\right) = 36 \left(\frac{a+3}{4}\right) \)
\( 6(a + 5) - 4(a + 1) = 9(a + 3) \)
Expand the brackets:
\( 6a + 30 - 4a - 4 = 9a + 27 \)
Combine the like terms on the left side:
\( 2a + 26 = 9a + 27 \)
Move \( 9a \) to the left side and \( 26 \) to the right side:
\( 2a - 9a = 27 - 26 \)
\( -7a = 1 \)
Divide both sides by \( -7 \):
\( a = -\frac{1}{7} \)
In simple words: Multiply all terms by the LCM of denominators, which is 36. Open the brackets, combine terms, and find the value of a.

Exam Tip: Be careful with the negative sign outside the brackets when expanding: \( -4(a + 1) \) becomes \( -4a - 4 \).

Question 15. \( \frac{2x-13}{5} - \frac{x-3}{11} = \frac{x-9}{5} + 1 \)
Answer:
Given equation:
\( \frac{2x-13}{5} - \frac{x-3}{11} = \frac{x-9}{5} + 1 \)
Find the LCM of denominators \( 5, 11 \), which is \( 55 \).
Multiply each term of the equation by \( 55 \) to eliminate the fractions:
\( 55 \left(\frac{2x-13}{5}\right) - 55 \left(\frac{x-3}{11}\right) = 55 \left(\frac{x-9}{5}\right) + 55(1) \)
\( 11(2x - 13) - 5(x - 3) = 11(x - 9) + 55 \)
Expand all brackets:
\( 22x - 143 - 5x + 15 = 11x - 99 + 55 \)
Simplify both sides:
\( 17x - 128 = 11x - 44 \)
Bring all terms containing \( x \) to the left side and numbers to the right side:
\( 17x - 11x = -44 + 128 \)
\( 6x = 84 \)
Divide both sides by \( 6 \):
\( x = \frac{84}{6} \)
\( x = 14 \)
In simple words: Multiply each term by 55 to clear the fractions. Expand the brackets carefully, group the x terms, and solve.

Exam Tip: Remember to multiply the standalone constant \( 1 \) by \( 55 \) as well. Every single term on both sides must be multiplied.

Question 16. 6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1
Answer:
Given equation:
\( 6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1 \)
Expand the brackets on both sides:
\( 36x - 30 - 35x + 40 = 48 - 12x + 1 \)
Combine terms on both sides of the equation:
\( x + 10 = 49 - 12x \)
Move \( -12x \) to the left side and \( 10 \) to the right side:
\( x + 12x = 49 - 10 \)
\( 13x = 39 \)
Divide both sides by \( 13 \):
\( x = \frac{39}{13} \)
\( x = 3 \)
In simple words: Expand the brackets first. Simplify both sides, group the terms, and solve to get the value of x.

Exam Tip: Watch out for the negative sign when multiplying: \( -5 \times -8 \) is \( +40 \). Sign errors are common here.

Question 17. (x - 5)(x + 3) = (x - 7)(x + 4)
Answer:
Given equation:
\( (x - 5)(x + 3) = (x - 7)(x + 4) \)
Multiply the binomial expressions on both sides:
\( x(x + 3) - 5(x + 3) = x(x + 4) - 7(x + 4) \)
\( x^2 + 3x - 5x - 15 = x^2 + 4x - 7x - 28 \)
Simplify the terms on both sides:
\( x^2 - 2x - 15 = x^2 - 3x - 28 \)
Subtract \( x^2 \) from both sides:
\( -2x - 15 = -3x - 28 \)
Rearrange terms by shifting \( -3x \) to the left side and \( -15 \) to the right side:
\( 3x - 2x = -28 + 15 \)
\( x = -13 \)
In simple words: Multiply the brackets first. The \( x^2 \) terms on both sides will cancel each other out, leaving a simple equation to solve.

Exam Tip: When \( x^2 \) appears on both sides with the exact same coefficient, you can cancel them out directly to simplify your work.

Question 18. (x - 5)^2 - (x + 2)^2 = -2
Answer:
Given equation:
\( (x - 5)^2 - (x + 2)^2 = -2 \)
Use the algebraic identities \( (a-b)^2 = a^2 - 2ab + b^2 \) and \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( (x^2 - 10x + 25) - (x^2 + 4x + 4) = -2 \)
Expand and remove the brackets:
\( x^2 - 10x + 25 - x^2 - 4x - 4 = -2 \)
Simplify by canceling out \( x^2 \) and \( -x^2 \):
\( -14x + 21 = -2 \)
Move \( 21 \) to the right side:
\( -14x = -2 - 21 \)
\( -14x = -23 \)
Divide both sides by \( -14 \):
\( x = \frac{-23}{-14} \)
\( x = \frac{23}{14} \)
\( x = 1\frac{9}{14} \)
In simple words: Use algebra rules to square the brackets. Remove the brackets, cancel out the \( x^2 \) parts, and simplify to find x as a mixed fraction.

Exam Tip: Be careful with the negative sign in front of the second bracket. It changes the sign of every term inside it when expanded.

Question 19. (x - 1)(x + 6) - (x - 2)(x - 3) = 3
Answer:
Given equation:
\( (x - 1)(x + 6) - (x - 2)(x - 3) = 3 \)
Expand the products of binomials:
\( (x^2 + 6x - x - 6) - (x^2 - 3x - 2x + 6) = 3 \)
\( (x^2 + 5x - 6) - (x^2 - 5x + 6) = 3 \)
Remove brackets carefully by reversing signs of terms in the second group:
\( x^2 + 5x - 6 - x^2 + 5x - 6 = 3 \)
Cancel out \( x^2 \) and simplify the remaining terms:
\( 10x - 12 = 3 \)
Move \( -12 \) to the right side:
\( 10x = 3 + 12 \)
\( 10x = 15 \)
Divide by \( 10 \):
\( x = \frac{15}{10} \)
Simplify the fraction to its lowest terms:
\( x = \frac{3}{2} \)
\( x = 1\frac{1}{2} \)
In simple words: Expand both parts. Subtract the second part by changing its signs, cancel \( x^2 \), and then solve for x.

Exam Tip: Remember to simplify the final fraction \( \frac{15}{10} \) to its simplest form, \( \frac{3}{2} \), and convert it to a mixed fraction.

Question 20. \( \frac{3x}{x+6} - \frac{x}{x+5} = 2 \)
Answer:
Given equation:
\( \frac{3x}{x+6} - \frac{x}{x+5} = 2 \)
Take the LCM of the denominators, which is \( (x+6)(x+5) \), and write the left side as a single fraction:
\( \frac{3x(x+5) - x(x+6)}{(x+6)(x+5)} = 2 \)
Expand the numerator and the denominator:
\( \frac{3x^2 + 15x - (x^2 + 6x)}{x^2 + 5x + 6x + 30} = 2 \)
\( \frac{3x^2 + 15x - x^2 - 6x}{x^2 + 11x + 30} = 2 \)
\( \frac{2x^2 + 9x}{x^2 + 11x + 30} = 2 \)
Cross-multiply to remove the fraction:
\( 2x^2 + 9x = 2(x^2 + 11x + 30) \)
\( 2x^2 + 9x = 2x^2 + 22x + 60 \)
Subtract \( 2x^2 \) from both sides:
\( 9x = 22x + 60 \)
Subtract \( 22x \) from both sides:
\( 9x - 22x = 60 \)
\( -13x = 60 \)
Divide by \( -13 \):
\( x = -\frac{60}{13} \)
\( x = -4\frac{8}{13} \)
In simple words: Find a common denominator to subtract the two fractions. Cross-multiply, cancel out the \( x^2 \) terms, and find x as a mixed fraction.

Exam Tip: Always keep binomial denominators in brackets like \( (x + 6)(x + 5) \) to avoid mistakes during expansion.

Question 21. \( \frac{1}{x-1} + \frac{2}{x-2} = \frac{3}{x-3} \)
Answer:
Given equation:
\( \frac{1}{x-1} + \frac{2}{x-2} = \frac{3}{x-3} \)
Combine the fractions on the left side over a common denominator:
\( \frac{1(x-2) + 2(x-1)}{(x-1)(x-2)} = \frac{3}{x-3} \)
Expand and simplify the numerator and denominator:
\( \frac{x - 2 + 2x - 2}{x^2 - 2x - x + 2} = \frac{3}{x-3} \)
\( \frac{3x - 4}{x^2 - 3x + 2} = \frac{3}{x-3} \)
Cross-multiply both sides:
\( (3x - 4)(x - 3) = 3(x^2 - 3x + 2) \)
Expand the expressions:
\( 3x^2 - 9x - 4x + 12 = 3x^2 - 9x + 6 \)
\( 3x^2 - 13x + 12 = 3x^2 - 9x + 6 \)
Rearrange terms by moving the variable terms to the left side:
\( 3x^2 - 13x - 3x^2 + 9x = 6 - 12 \)
Cancel out \( 3x^2 \) and \( -3x^2 \):
\( -4x = -6 \)
Divide both sides by \( -4 \):
\( x = \frac{-6}{-4} \)
Simplify the fraction:
\( x = \frac{3}{2} \)
\( x = 1\frac{1}{2} \)
In simple words: Combine the left fractions first. Then cross-multiply both sides, cancel out the \( x^2 \) terms, and simplify the fraction to find x.

Exam Tip: Be meticulous when expanding \( (3x - 4)(x - 3) \). Ensure each of the four multiplied products has the correct sign.

Question 22. \( \frac{x-1}{7x-14} = \frac{x-3}{7x-26} \)
Answer:
Given equation:
\( \frac{x-1}{7x-14} = \frac{x-3}{7x-26} \)
Cross-multiply to clear the denominators:
\( (x - 1)(7x - 26) = (7x - 14)(x - 3) \)
Expand the binomials on both sides:
\( 7x^2 - 26x - 7x + 26 = 7x^2 - 21x - 14x + 42 \)
\( 7x^2 - 33x + 26 = 7x^2 - 35x + 42 \)
Subtract \( 7x^2 \) from both sides:
\( -33x + 26 = -35x + 42 \)
Bring \( -35x \) to the left side and \( 26 \) to the right side:
\( 35x - 33x = 42 - 26 \)
\( 2x = 16 \)
Divide both sides by \( 2 \):
\( x = \frac{16}{2} \)
\( x = 8 \)
In simple words: Cross-multiply to remove the fractions. Open the brackets, cancel the \( x^2 \) terms, and solve for x.

Exam Tip: If you see a common factor like 7 in the denominator of the left side, you could factor it out, but direct cross-multiplication is safer and less prone to conceptual mistakes.

Question 23. \( \frac{1}{x-1} - \frac{1}{x} = \frac{1}{x+3} - \frac{1}{x+4} \)
Answer:
Given equation:
\( \frac{1}{x-1} - \frac{1}{x} = \frac{1}{x+3} - \frac{1}{x+4} \)
Combine the fractions on each side over their respective common denominators:
\( \frac{x - (x - 1)}{x(x - 1)} = \frac{(x + 4) - (x + 3)}{(x + 3)(x + 4)} \)
\( \frac{x - x + 1}{x(x - 1)} = \frac{x + 4 - x - 3}{(x + 3)(x + 4)} \)
\( \frac{1}{x(x - 1)} = \frac{1}{(x + 3)(x + 4)} \)
Since the numerators are equal to \( 1 \), equate the denominators:
\( (x + 3)(x + 4) = x(x - 1) \)
Expand both sides:
\( x^2 + 4x + 3x + 12 = x^2 - x \)
\( x^2 + 7x + 12 = x^2 - x \)
Subtract \( x^2 \) from both sides:
\( 7x + 12 = -x \)
Move \( -x \) to the left side and \( 12 \) to the right side:
\( 7x + x = -12 \)
\( 8x = -12 \)
Divide both sides by \( 8 \):
\( x = -\frac{12}{8} \)
Simplify the fraction to its lowest terms:
\( x = -\frac{3}{2} \)
\( x = -1\frac{1}{2} \)
In simple words: Simplify the fractions on both sides first. Since the top numbers are both 1, the bottom parts must be equal. Solve that new equation to find x.

Exam Tip: If the numerators are identical, you can simplify the work by directly equating the denominators.

Question 24. Solve: \( \frac{2x}{3} - \frac{x-1}{6} + \frac{7x-1}{4} = 2\frac{1}{6} \)
Hence, find the value of 'a', if \( \frac{1}{a} + 5x = 8 \).
Answer:
Given equation:
\( \frac{2x}{3} - \frac{x-1}{6} + \frac{7x-1}{4} = 2\frac{1}{6} \)
Convert the mixed fraction to an improper fraction:
\( \frac{2x}{3} - \frac{x-1}{6} + \frac{7x-1}{4} = \frac{13}{6} \)
The LCM of the denominators \( 3, 6, 4 \) is \( 12 \). Multiply each term by \( 12 \) to clear the fractions:
\( 12\left(\frac{2x}{3}\right) - 12\left(\frac{x-1}{6}\right) + 12\left(\frac{7x-1}{4}\right) = 12\left(\frac{13}{6}\right) \)
\( 4(2x) - 2(x - 1) + 3(7x - 1) = 2(13) \)
\( 8x - 2x + 2 + 21x - 3 = 26 \)
Group and simplify the terms on the left side:
\( 27x - 1 = 26 \)
\( 27x = 26 + 1 \)
\( 27x = 27 \)
\( x = 1 \)

Now, substitute \( x = 1 \) into the second equation to find \( a \):
\( \frac{1}{a} + 5x = 8 \)
\( \frac{1}{a} + 5(1) = 8 \)
\( \frac{1}{a} + 5 = 8 \)
Subtract \( 5 \) from both sides:
\( \frac{1}{a} = 8 - 5 \)
\( \frac{1}{a} = 3 \)
\( 3a = 1 \)
\( a = \frac{1}{3} \)
So, \( x = 1 \) and \( a = \frac{1}{3} \).
In simple words: First solve the big equation to find x, which is 1. Put this 1 in place of x in the second equation to find that a equals 1/3.

Exam Tip: This is a two-step question. Make sure you don't stop after finding \( x \); you must complete the second part to find \( a \) for full marks.

Question 25. Solve: \( \frac{4-3x}{5} + \frac{7-x}{3} + 4\frac{1}{3} = 0 \)
Hence, find the value of 'p', if \( 3p - 2x + 1 = 0 \).
Answer:
Given equation:
\( \frac{4-3x}{5} + \frac{7-x}{3} + 4\frac{1}{3} = 0 \)
Convert the mixed number to an improper fraction:
\( \frac{4-3x}{5} + \frac{7-x}{3} + \frac{13}{3} = 0 \)
The LCM of denominators \( 5, 3, 3 \) is \( 15 \). Multiply each term by \( 15 \) to clear the fractions:
\( 15\left(\frac{4-3x}{5}\right) + 15\left(\frac{7-x}{3}\right) + 15\left(\frac{13}{3}\right) = 15(0) \)
\( 3(4 - 3x) + 5(7 - x) + 5(13) = 0 \)
Expand the brackets:
\( 12 - 9x + 35 - 5x + 65 = 0 \)
Group like terms together:
\( -9x - 5x + 12 + 35 + 65 = 0 \)
\( -14x + 112 = 0 \)
Move \( 112 \) to the right side:
\( -14x = -112 \)
Divide both sides by \( -14 \):
\( x = \frac{-112}{-14} \)
\( x = 8 \)

Now, substitute \( x = 8 \) into the equation \( 3p - 2x + 1 = 0 \):
\( 3p - 2(8) + 1 = 0 \)
\( 3p - 16 + 1 = 0 \)
\( 3p - 15 = 0 \)
Move \( -15 \) to the right side:
\( 3p = 15 \)
Divide by \( 3 \):
\( p = \frac{15}{3} \)
\( p = 5 \)
In simple words: First, multiply everything by 15 to clear the fractions and find x, which is 8. Put 8 in the second equation and solve to find that p is 5.

Exam Tip: Be careful with the multiplication when clearing fractions. Ensure that the right-hand side, even if it is \( 0 \), is multiplied correctly (it remains \( 0 \)).

Question 26. Solve: \( 0.25 + \frac{1.95}{x} = 0.9 \)
Answer:
Given equation:
\( 0.25 + \frac{1.95}{x} = 0.9 \)
Multiply every term in the equation by \( x \) to remove the variable from the denominator:
\( 0.25x + 1.95 = 0.9x \)
Rearrange terms to group \( x \) terms together on the right side:
\( 1.95 = 0.9x - 0.25x \)
\( 1.95 = 0.65x \)
Divide both sides by \( 0.65 \):
\( x = \frac{1.95}{0.65} \)
Since both decimals have two decimal places, we can simplify this as:
\( x = \frac{195}{65} \)
\( x = 3 \)
In simple words: Multiply the entire equation by x to get rid of the fraction. Then group the x terms together and solve to get x equals 3.

Exam Tip: To divide decimals like \( \frac{1.95}{0.65} \), multiply the numerator and denominator by 100 to convert them into whole numbers first.

Question 27. Solve: \( 5x - \left(4x + \frac{5x-4}{7}\right) = \frac{4x-14}{3} \)
Answer:
Given equation:
\( 5x - \left(4x + \frac{5x-4}{7}\right) = \frac{4x-14}{3} \)
Simplify the terms inside the bracket first by finding a common denominator of \( 7 \):
\( 4x + \frac{5x-4}{7} = \frac{28x + 5x - 4}{7} = \frac{33x - 4}{7} \)
Substitute this back into the equation:
\( 5x - \left(\frac{33x-4}{7}\right) = \frac{4x-14}{3} \)
Now, simplify the left side of the equation:
\( \frac{35x - (33x - 4)}{7} = \frac{4x-14}{3} \)
\( \frac{35x - 33x + 4}{7} = \frac{4x-14}{3} \)
\( \frac{2x + 4}{7} = \frac{4x-14}{3} \)
Cross-multiply to clear the denominators:
\( 3(2x + 4) = 7(4x - 14) \)
Expand both sides:
\( 6x + 12 = 28x - 98 \)
Rearrange the terms by moving the variables to the right side and constants to the left side:
\( 12 + 98 = 28x - 6x \)
\( 110 = 22x \)
Divide by \( 22 \):
\( x = \frac{110}{22} \)
\( x = 5 \)
In simple words: Simplify the fraction inside the bracket first. Combine the left side terms, then cross-multiply and solve to get x equals 5.

Exam Tip: Be very careful with the negative sign in front of the fraction. When subtracting \( \frac{33x-4}{7} \), the negative sign applies to both terms, turning \( -4 \) into \( +4 \).

Exercise 14(B)

Question 1. Fifteen less than 4 times a number is 9. Find the number.
Answer: Assume that the unknown value is \( x \). Four times this value can be written as \( 4x \). Subtracting fifteen from this yields \( 4x - 15 \). Based on the given condition:
\( 4x - 15 = 9 \)
\( \implies 4x = 9 + 15 \)
\( \implies 4x = 24 \)
\( \implies x = \frac{24}{4} \)
\( \implies x = 6 \)
Therefore, the value we are looking for is 6.
In simple words: Let the number be x. We write the sentence as an equation: 4 times x minus 15 equals 9. Solving this gives x as 6.

Exam Tip: Always write down what x represents first. Clearly show each step of solving the equation to secure full marks.

Question 2. If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Answer: Let us assume Megha's current age is \( x \) years. Multiplying her age by three gives \( 3x \) years. As per the problem, adding these two values together gives 60:
\( x + 3x = 60 \)
\( \implies 4x = 60 \)
\( \implies x = \frac{60}{4} \)
\( \implies x = 15 \)
Thus, Megha is 15 years old.
In simple words: We can write Megha's age as x. Adding three times her age (3x) to x gives 4x, which equals 60. Dividing 60 by 4 shows she is 15.

Exam Tip: Do not forget to write the final unit, which is "years", in your final answer statement.

Question 3. 28 is 12 less than 4 times a number. Find the number.
Answer: Let the unknown number be represented by \( x \). Four times this number is written as \( 4x \). Taking away twelve from four times the number gives \( 4x - 12 \). According to the given problem:
\( 4x - 12 = 28 \)
\( \implies 4x = 28 + 12 \)
\( \implies 4x = 40 \)
\( \implies x = 10 \)
So, the required value is 10.
In simple words: Think of the mystery number as x. The sentence says that 4x minus 12 is equal to 28. Adding 12 to 28 gives 40, and dividing by 4 gives us 10.

Exam Tip: Check your work by substituting the final value back into the original word problem to see if 4 times 10 minus 12 really is 28.

Question 4. Five less than 3 times a number is -20. Find the number.
Answer: Let the target number be designated as \( x \). Three times this number is \( 3x \). Subtracting five from three times the number gives \( 3x - 5 \). Based on the statement:
\( 3x - 5 = -20 \)
\( \implies 3x = -20 + 5 \)
\( \implies 3x = -15 \)
\( \implies x = -5 \)
Hence, the required number is -5.
In simple words: Let the number be x. The equation is 3x minus 5 equals -20. When we solve it, we get x equals -5.

Exam Tip: Be very careful with negative numbers when adding and subtracting. Adding 5 to -20 results in -15, not -25.

Question 5. Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she ?
Answer: Let Neetu's age be represented by \( x \) years. Three times her age is \( 3x \) years. Adding fifteen to three times her age is written as \( 3x + 15 \) years. Four times her age is \( 4x \) years. According to the question:
\( 4x = 3x + 15 \)
\( \implies 4x - 3x = 15 \)
\( \implies x = 15 \)
Therefore, Neetu is 15 years old.
In simple words: If Neetu's age is x, then 3 times her age plus 15 is the same as 4 times her age. This gives the simple equation 4x minus 3x equals 15, which means x is 15.

Exam Tip: Always group the terms with variables on one side of the equation and constant numbers on the other side to solve quickly.

Question 6. A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.
Answer: Let the unknown number be \( x \). Reducing this number by thirty gives \( x - 30 \). Subtracting three times this number from fourteen gives \( 14 - 3x \). According to the condition:
\( x - 30 = 14 - 3x \)
\( \implies x + 3x = 14 + 30 \)
\( \implies 4x = 44 \)
\( \implies x = 11 \)
Thus, the required number is 11.
In simple words: Let the number be x. We set up the equation where x minus 30 equals 14 minus 3x. Adding the x terms together and numbers together gives 4x equals 44, so x is 11.

Exam Tip: Remember that "decreased by" means subtraction. Make sure you subtract in the correct order: "14 decreased by 3x" is written as \( 14 - 3x \), not \( 3x - 14 \).

Question 7. A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Answer: Let B's monthly salary be Rs. \( x \). Since A earns four times as much as B, A's salary is Rs. \( 4x \). Their total combined monthly earnings are Rs. 3,750:
\( x + 4x = 3750 \)
\( \implies 5x = 3750 \)
\( \implies x = \frac{3750}{5} \)
\( \implies x = 750 \)
Now, let's calculate A's earnings:
\( 4x = 4 \times 750 = 3000 \)
Therefore, A's monthly salary is Rs. 3,000 and B's monthly salary is Rs. 750.
In simple words: If B earns x, then A earns 4x. Together they earn 5x, which is Rs. 3,750. Dividing by 5 gives B's salary as Rs. 750, and multiplying by 4 gives A's salary as Rs. 3,000.

Exam Tip: Always write down the final salaries of both individuals clearly at the end of your solution, and don't forget to include the unit "Rs.".

Question 8. Separate 178 into two parts so that the first part is 8 less than twice the second part.
Answer: Let the second part be denoted as \( x \). Since the first part is eight less than twice the second part, we can write the first part as \( 2x - 8 \). The sum of these two parts is 178:
\( x + (2x - 8) = 178 \)
\( \implies 3x - 8 = 178 \)
\( \implies 3x = 178 + 8 \)
\( \implies 3x = 186 \)
\( \implies x = \frac{186}{3} \)
\( \implies x = 62 \)
Now, we can find the first part:
\( 2x - 8 = 2(62) - 8 = 124 - 8 = 116 \)
Thus, the two parts are 116 and 62.
In simple words: If we call the second part x, the first part is 2x minus 8. Adding them together gives 178. Solving this equation tells us the parts are 116 and 62.

Exam Tip: Verify your final parts by adding them together (116 + 62 = 178) to make sure they equal the original total.

Question 9. Six more than one-fourth of a number is two-fifth of the number. Find the number.
Answer: Let the required number be represented by \( x \). One-fourth of this number is \( \frac{x}{4} \). Two-fifths of the number is \( \frac{2x}{5} \). As stated in the problem:
\( \frac{2x}{5} = \frac{x}{4} + 6 \)
\( \implies \frac{2x}{5} - \frac{x}{4} = 6 \)
To eliminate the denominators, multiply every term by 20, which is the LCM of 5 and 4:
\( 20 \left( \frac{2x}{5} \right) - 20 \left( \frac{x}{4} \right) = 20(6) \)
\( \implies 8x - 5x = 120 \)
\( \implies 3x = 120 \)
\( \implies x = \frac{120}{3} \)
\( \implies x = 40 \)
Hence, the required number is 40.
In simple words: Let our mystery number be x. One-fourth of it plus 6 is equal to two-fifths of it. By finding a common denominator and solving, we find the number is 40.

Exam Tip: Multiplying the entire equation by the Least Common Multiple (LCM) of the denominators is a great way to clear fractions quickly.

Question 10. The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Answer: Let the width of the rectangle be \( x \) cm. Since the length is twice the width, it can be written as \( 2x \) cm. The formula for the perimeter of a rectangle is \( 2(\text{Length} + \text{Width}) \):
\( \text{Perimeter} = 2(2x + x) = 2(3x) = 6x \) cm.
Given that the perimeter is 54 cm:
\( 6x = 54 \)
\( \implies x = \frac{54}{6} \)
\( \implies x = 9 \)
Now, find the length:
\( \text{Length} = 2x = 2(9) = 18 \) cm.
Therefore, the length of the rectangle is 18 cm.
In simple words: If the width is x, the length is 2x. Adding all four sides gives a perimeter of 6x. Since the perimeter is 54, x is 9, which means the length is 18 cm.

Exam Tip: Read the final question carefully! Here, you are asked to find the length, not the width. Do not stop solving after finding \( x \).

Question 11. A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the original rectangle.
Answer: Let the width of the starting rectangle be \( x \) cm. Thus, the initial length is \( (2x - 5) \) cm. If we reduce the length by 5 cm, the new length becomes:
\( 2x - 5 - 5 = (2x - 10) \) cm.
Increasing the width by 2 cm gives the new width:
\( (x + 2) \) cm.
The formula for the new perimeter is:
\( 2[\text{New Length} + \text{New Width}] = 2[(2x - 10) + (x + 2)] = 2[3x - 8] = (6x - 16) \) cm.
We are given that this new perimeter is 74 cm:
\( 6x - 16 = 74 \)
\( \implies 6x = 74 + 16 \)
\( \implies 6x = 90 \)
\( \implies x = 15 \)
Thus, the width of the original rectangle is 15 cm. The length of the original rectangle is:
\( 2(15) - 5 = 30 - 5 = 25 \) cm.
Therefore, the original length is 25 cm and the original width is 15 cm.
In simple words: Let the original width be x, so the length is 2x minus 5. We adjust the sides and find that the new perimeter is 6x minus 16. Setting this equal to 74 shows us that x is 15, and the original length is 25 cm.

Exam Tip: Clearly state the dimensions of the original rectangle in your final line so the examiner can find your answer easily.

Question 12. The sum of three consecutive odd numbers is 57. Find the numbers.
Answer: Let the three sequential odd integers be \( x \), \( x + 2 \), and \( x + 4 \). Their sum is given as 57:
\( x + (x + 2) + (x + 4) = 57 \)
\( \implies 3x + 6 = 57 \)
\( \implies 3x = 57 - 6 \)
\( \implies 3x = 51 \)
\( \implies x = \frac{51}{3} \)
\( \implies x = 17 \)
The three numbers are:
\( x = 17 \)
\( x + 2 = 19 \)
\( x + 4 = 21 \)
Thus, the consecutive odd numbers are 17, 19, and 21.
In simple words: Odd numbers always have a gap of 2 between them. If the first is x, the next ones are x plus 2 and x plus 4. Adding them together gives 3x plus 6. Solving this tells us the numbers are 17, 19, and 21.

Exam Tip: Consecutive odd or consecutive even numbers both increase by 2 each time, so always set them up as \( x \), \( x+2 \), and \( x+4 \).

Question 13. A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Answer: Let the son's current age be \( x \) years. Consequently, the father's current age is \( 3x \) years. After 12 years:
The son's age will be \( (x + 12) \) years.
The father's age will be \( (3x + 12) \) years.
According to the given condition:
\( 3x + 12 = 2(x + 12) \)
\( \implies 3x + 12 = 2x + 24 \)
\( \implies 3x - 2x = 24 - 12 \)
\( \implies x = 12 \)
Now, calculating the father's current age:
\( 3x = 3(12) = 36 \)
So, the son is currently 12 years old and the father is 36 years old.
In simple words: If the son is x, the dad is 3x. In 12 years, we add 12 to both their ages. Since the dad will be twice as old as the son then, we write the equation and solve to find they are 12 and 36 years old.

Exam Tip: Remember to add 12 to BOTH the father's and the son's ages for the future condition, not just one of them.

Question 14. A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Answer: Currently, the father is 42 years old and the son is 12 years old. Suppose that after \( x \) years, the son's age becomes half of his father's age. In \( x \) years:
The father's age will be \( 42 + x \) years.
The son's age will be \( 12 + x \) years.
Based on the problem statement:
\( 12 + x = \frac{42 + x}{2} \)
\( \implies 2(12 + x) = 42 + x \)
\( \implies 24 + 2x = 42 + x \)
\( \implies 2x - x = 42 - 24 \)
\( \implies x = 18 \)
Thus, it will take 18 years for the son's age to be half of his father's age.
In simple words: Let the number of years be x. In x years, the son is 12 + x and the dad is 42 + x. Setting the son's age to half the dad's age and solving gives us 18 years.

Exam Tip: Always specify what the variable \( x \) stands for, such as "Let \( x \) be the number of years".

Question 15. A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Answer: The entire distance of the journey is 136 km. Let the distance traveled at the speed of 18 km/hr be \( x \) km. So, the distance traveled at the speed of 15 km/hr must be \( (136 - x) \) km. We know that \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). Time taken at 18 km/hr is \( \frac{x}{18} \) hours. Time taken at 15 km/hr is \( \frac{136 - x}{15} \) hours. The total duration of the trip is 8 hours:
\( \frac{x}{18} + \frac{136 - x}{15} = 8 \)
Multiplying each term by 90, which is the LCM of 18 and 15, we get:
\( 90 \left( \frac{x}{18} \right) + 90 \left( \frac{136 - x}{15} \right) = 90(8) \)
\( \implies 5x + 6(136 - x) = 720 \)
\( \implies 5x + 816 - 6x = 720 \)
\( \implies -x = 720 - 816 \)
\( \implies -x = -96 \)
\( \implies x = 96 \)
Therefore, the portion of the journey covered at 18 km/hr is 96 km.
In simple words: Let the distance traveled at 18 km/hr be x. The remaining distance is 136 minus x. We write an equation for the total time (which is 8 hours) using the formula time equals distance divided by speed, and solve for x.

Exam Tip: Using the LCM to clear fractions is extremely useful in speed-distance-time problems. Make sure to multiply every single term of the equation by the LCM.

Question 16. The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Answer: Let the smaller number be \( x \). Since the difference between the two numbers is 3, the larger number is \( x + 3 \). We are given that the difference between their squares is 69:
\( (x + 3)^2 - x^2 = 69 \)
Expanding the squared term:
\( x^2 + 6x + 9 - x^2 = 69 \)
\( \implies 6x + 9 = 69 \)
\( \implies 6x = 69 - 9 \)
\( \implies 6x = 60 \)
\( \implies x = \frac{60}{6} \)
\( \implies x = 10 \)
Now, let's find the larger number:
\( x + 3 = 10 + 3 = 13 \)
Therefore, the two numbers are 10 and 13.
In simple words: Let one number be x and the other be x plus 3. Their squares subtracted from each other equal 69. Solving this equation shows us the numbers are 10 and 13.

Exam Tip: When expanding \( (x+3)^2 \), remember to use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \) so you do not miss the middle term.

Question 17. Two consecutive natural numbers are such that one-fourth of the smaller exceeds one fifth of the greater by 1. Find the numbers.
Answer: Let the two successive natural numbers be \( x \) and \( x + 1 \). One-fourth of the smaller number is written as \( \frac{x}{4} \). One-fifth of the larger number is written as \( \frac{x + 1}{5} \). According to the problem, the first value is 1 greater than the second:
\( \frac{x}{4} = \frac{x + 1}{5} + 1 \)
\( \implies \frac{x}{4} - \frac{x + 1}{5} = 1 \)
Now, let us take the LCM of 4 and 5, which is 20:
\( \frac{5x - 4(x + 1)}{20} = 1 \)
\( \implies \frac{5x - 4x - 4}{20} = 1 \)
\( \implies \frac{x - 4}{20} = 1 \)
By cross-multiplying:
\( x - 4 = 20 \)
\( \implies x = 20 + 4 \)
\( \implies x = 24 \)
The larger number is:
\( x + 1 = 24 + 1 = 25 \)
Hence, the two consecutive numbers are 24 and 25.
In simple words: Let our two sequential numbers be x and x plus 1. One-fourth of x is 1 more than one-fifth of x plus 1. Solving this equation gives us the numbers 24 and 25.

Exam Tip: Always use parentheses when subtracting a grouped term like \( 4(x+1) \) to avoid mistakes with the negative sign.

Question 18. Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Answer: Let the three sequential whole numbers be \( x \), \( x + 1 \), and \( x + 2 \). As per the conditions given, when these are divided by 5, 3, and 4 respectively, the sum of their quotients is 40:
\( \frac{x}{5} + \frac{x + 1}{3} + \frac{x + 2}{4} = 40 \)
Multiplying each term by the LCM of 5, 3, and 4, which is 60:
\( 60 \left( \frac{x}{5} \right) + 60 \left( \frac{x + 1}{3} \right) + 60 \left( \frac{x + 2}{4} \right) = 60(40) \)
\( \implies 12x + 20(x + 1) + 15(x + 2) = 2400 \)
\( \implies 12x + 20x + 20 + 15x + 30 = 2400 \)
Group the \( x \) terms together and the constant numbers together:
\( 47x + 50 = 2400 \)
\( \implies 47x = 2400 - 50 \)
\( \implies 47x = 2350 \)
\( \implies x = \frac{2350}{47} \)
\( \implies x = 50 \)
The three numbers are:
\( x = 50 \)
\( x + 1 = 51 \)
\( x + 2 = 52 \)
Therefore, the three consecutive whole numbers are 50, 51, and 52.
In simple words: Let our three numbers be x, x plus 1, and x plus 2. We divide them by 5, 3, and 4, and add them to get 40. After solving, we find the three numbers are 50, 51, and 52.

Exam Tip: Ensure you calculate the LCM of 5, 3, and 4 correctly (which is 60) before clearing the denominators.

Question 19. If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Answer: Let the number to be added to each of the given numbers be \( x \). The four resulting values are \( 5 + x \), \( 11 + x \), \( 15 + x \), and \( 31 + x \). Since these numbers are in proportion, we can set up the ratio equation:
\( \frac{5 + x}{11 + x} = \frac{15 + x}{31 + x} \)
Now, cross-multiply the terms:
\( (5 + x)(31 + x) = (15 + x)(11 + x) \)
\( \implies 155 + 5x + 31x + x^2 = 165 + 11x + 15x + x^2 \)
Simplify both sides by combining like terms:
\( 155 + 36x + x^2 = 165 + 26x + x^2 \)
Subtract \( x^2 \) from both sides:
\( 155 + 36x = 165 + 26x \)
Rearrange the equation to solve for \( x \):
\( 36x - 26x = 165 - 155 \)
\( \implies 10x = 10 \)
\( \implies x = 1 \)
Thus, the number that must be added is 1.
In simple words: If we add x to each number, the new numbers form a proportion, which means the ratio of the first two equals the ratio of the last two. Cross-multiplying and solving the equation shows that x equals 1.

Exam Tip: Since \( x^2 \) appears on both sides of the equation with the same positive sign, they cancel each other out directly, simplifying the quadratic equation into a basic linear one.

Question 20. The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Answer: Suppose the son is currently \( x \) years of age. This means his father is currently \( 2x \) years of age. After 8 years pass, the son's age will be \( x + 8 \) years, while the father's age will be \( 2x + 8 \) years. As stated in the problem:
\( \frac{2x+8}{x+8} = \frac{7}{4} \)
\( \implies 4(2x + 8) = 7(x + 8) \)
\( \implies 8x + 32 = 7x + 56 \)
\( \implies 8x - 7x = 56 - 32 \)
\( \implies x = 24 \) So, the current age of the son is \( 24 \) years. The father's age is \( 2 \times 24 = 48 \) years. Thus, the father is \( 48 \) years old and the son is \( 24 \) years old.
In simple words: We can represent their ages today using \( x \). Then we find how old they will be in 8 years, set up a fraction, and solve for \( x \).

Exam Tip: Clearly define the variables for current ages before adding years to both of them. Remember to multiply both terms of the ratio when cross-multiplying.

Exercise 14(C)

Question 1. Solve:
(i) \( \frac { 1 }{ 3 }x - 6 = \frac { 5 }{ 2 } \)
(ii) \( \frac { 2x }{ 3 } - \frac { 3x }{ 8 } = \frac { 7 }{ 12 } \)
(iii) \( (x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0 \)
(iv) \( \frac { 1 }{ 10 } - \frac { 7 }{ x } = 35 \)
(v) \( 13(x - 4) - 3(x - 9) - 5(x + 4) = 0 \)
(vi) \( x + 7 - \frac { 8x }{ 3 } = \frac { 17x }{ 6 } - \frac { 5x }{ 8 } \)
(vii) \( \frac { 3x - 2 }{ 4 } - \frac { 2x + 3 }{ 3 } = \frac { 2 }{ 3 } - x \)
(viii) \( \frac { x + 2 }{ 6 } - \left(\frac { 11 - x }{ 3 } - \frac { 1 }{ 4 }\right) = \frac { 3x - 4 }{ 12 } \)
(ix) \( \frac { 2 }{ 5x } - \frac { 5 }{ 3x } = \frac { 1 }{ 15 } \)
(x) \( \frac { x + 2 }{ 3 } - \frac { x + 1 }{ 5 } = \frac { x - 3 }{ 4 } - 1 \)
(xi) \( \frac { 3x - 2 }{ 3 } + \frac { 2x + 3 }{ 2 } = x + \frac { 7 }{ 6 } \)
(xii) \( x - \frac { x - 1 }{ 2 } = 1 - \frac { x - 2 }{ 3 } \)
(xiii) \( \frac { 9x + 7 }{ 2 } - \left(x - \frac { x - 2 }{ 7 }\right) = 36 \)
(xiv) \( \frac { 6x + 1 }{ 2 } + 1 = \frac { 7x - 3 }{ 3 } \)
Answer:
(i) \( \frac{1}{3}x - 6 = \frac{5}{2} \)
Add \( 6 \) to both sides:
\( \implies \frac{1}{3}x = \frac{5}{2} + 6 \)
Take the common denominator on the right side:
\( \implies \frac{1}{3}x = \frac{5}{2} + \frac{12}{2} \)
\( \implies \frac{1}{3}x = \frac{17}{2} \)
Multiply both sides by \( 3 \) to solve for \( x \):
\( \implies x = \frac{17 \times 3}{2} \)
\( \implies x = \frac{51}{2} \)
Convert to a mixed fraction:
\( \implies x = 25\frac{1}{2} \)

(ii) \( \frac{2x}{3} - \frac{3x}{8} = \frac{7}{12} \)
First, find the Least Common Multiple (LCM) of the denominators \( 3 \) and \( 8 \), which is \( 24 \).
Rewrite each term with denominator \( 24 \):
\( \implies \frac{2x \times 8}{3 \times 8} - \frac{3x \times 3}{8 \times 3} = \frac{7}{12} \)
\( \implies \frac{16x}{24} - \frac{9x}{24} = \frac{7}{12} \)
\( \implies \frac{16x - 9x}{24} = \frac{7}{12} \)
\( \implies \frac{7x}{24} = \frac{7}{12} \)
Solve for \( x \):
\( \implies x = \frac{7 \times 24}{12 \times 7} \)
\( \implies x = 2 \)

(iii) \( (x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) = 0 \)
Expand each of the algebraic parts:
\( \implies [x^2 + 5x + 6] + [x^2 - 5x + 6] - [2x^2 + 2x] = 0 \)
Combine all similar terms:
\( \implies x^2 + x^2 - 2x^2 + 5x - 5x - 2x + 6 + 6 = 0 \)
\( \implies -2x + 12 = 0 \)
Subtract \( 12 \) from both sides:
\( \implies -2x = -12 \)
Divide by \( -2 \):
\( \implies x = 6 \)
Now, let us verify our answer by substituting \( x = 6 \) back into the left-hand side (L.H.S.):
\( \text{L.H.S.} = (6 + 2)(6 + 3) + (6 - 3)(6 - 2) - 2 \times 6(6 + 1) \)
\( \implies \text{L.H.S.} = (8)(9) + (3)(4) - 12(7) \)
\( \implies \text{L.H.S.} = 72 + 12 - 84 \)
\( \implies \text{L.H.S.} = 84 - 84 = 0 = \text{R.H.S.} \)
Hence, verified.

(iv) \( \frac{1}{10} - \frac{7}{x} = 35 \)
Move \( \frac{1}{10} \) to the right-hand side:
\( \implies -\frac{7}{x} = 35 - \frac{1}{10} \)
Find a common denominator on the right side:
\( \implies -\frac{7}{x} = \frac{350}{10} - \frac{1}{10} \)
\( \implies -\frac{7}{x} = \frac{349}{10} \)
Take the reciprocal of both sides:
\( \implies -\frac{x}{7} = \frac{10}{349} \)
Multiply both sides by \( 7 \) to solve for \( x \):
\( \implies -x = \frac{70}{349} \)
\( \implies x = -\frac{70}{349} \)

(v) \( 13(x - 4) - 3(x - 9) - 5(x + 4) = 0 \)
Expand the brackets:
\( \implies 13x - 52 - 3x + 27 - 5x - 20 = 0 \)
Collect like terms:
\( \implies (13x - 3x - 5x) + (-52 + 27 - 20) = 0 \)
\( \implies 5x - 45 = 0 \)
\( \implies 5x = 45 \)
Divide both sides by \( 5 \):
\( \implies x = 9 \)
Verification:
Substitute \( x = 9 \) into the left-hand side (L.H.S.):
\( \text{L.H.S.} = 13(9 - 4) - 3(9 - 9) - 5(9 + 4) \)
\( \implies \text{L.H.S.} = 13(5) - 3(0) - 5(13) \)
\( \implies \text{L.H.S.} = 65 - 0 - 65 = 0 = \text{R.H.S.} \)
Hence, verified.

(vi) \( x + 7 - \frac{8x}{3} = \frac{17x}{6} - \frac{5x}{8} \)
Combine the terms on each side of the equation. On the left, use common denominator \( 3 \):
\( \frac{3(x + 7) - 8x}{3} = \frac{3x + 21 - 8x}{3} = \frac{-5x + 21}{3} \)
On the right, the LCM of \( 6 \) and \( 8 \) is \( 24 \):
\( \frac{17x \times 4}{24} - \frac{5x \times 3}{24} = \frac{68x - 15x}{24} = \frac{53x}{24} \)
So, we have:
\( \implies \frac{-5x + 21}{3} = \frac{53x}{24} \)
Cross-multiply to clear the fractions:
\( \implies 24(-5x + 21) = 3(53x) \)
\( \implies -120x + 504 = 159x \)
Rearrange the terms with \( x \) on one side:
\( \implies 159x + 120x = 504 \)
\( \implies 279x = 504 \)
\( \implies x = \frac{504}{279} \)
Simplify the fraction by dividing the numerator and denominator by their common factor \( 9 \):
\( \implies x = \frac{56}{31} \)
Convert to a mixed fraction:
\( \implies x = 1\frac{25}{31} \)

(vii) \( \frac{3x - 2}{4} - \frac{2x + 3}{3} = \frac{2}{3} - x \)
For the left side, the LCM of \( 4 \) and \( 3 \) is \( 12 \). For the right side, write \( x \) as \( \frac{x}{1} \) and use denominator \( 3 \):
\( \implies \frac{3(3x - 2) - 4(2x + 3)}{12} = \frac{2 - 3x}{3} \)
Expand the numerators:
\( \implies \frac{9x - 6 - 8x - 12}{12} = \frac{2 - 3x}{3} \)
\( \implies \frac{x - 18}{12} = \frac{2 - 3x}{3} \)
Cross-multiply:
\( \implies 3(x - 18) = 12(2 - 3x) \)
\( \implies 3x - 54 = 24 - 36x \)
Group the like terms:
\( \implies 3x + 36x = 24 + 54 \)
\( \implies 39x = 78 \)
\( \implies x = \frac{78}{39} \)
\( \implies x = 2 \)

(viii) \( \frac{x + 2}{6} - \left(\frac{11 - x}{3} - \frac{1}{4}\right) = \frac{3x - 4}{12} \)
Simplify the expression inside the parentheses first, using LCM \( 12 \) for \( 3 \) and \( 4 \):
\( \implies \frac{x + 2}{6} - \left(\frac{4(11 - x) - 3}{12}\right) = \frac{3x - 4}{12} \)
\( \implies \frac{x + 2}{6} - \left(\frac{44 - 4x - 3}{12}\right) = \frac{3x - 4}{12} \)
\( \implies \frac{x + 2}{6} - \frac{41 - 4x}{12} = \frac{3x - 4}{12} \)
Now combine the terms on the left-hand side, using LCM \( 12 \) of \( 6 \) and \( 12 \):
\( \implies \frac{2(x + 2) - (41 - 4x)}{12} = \frac{3x - 4}{12} \)
\( \implies \frac{2x + 4 - 41 + 4x}{12} = \frac{3x - 4}{12} \)
\( \implies \frac{6x - 37}{12} = \frac{3x - 4}{12} \)
Multiply both sides by \( 12 \):
\( \implies 6x - 37 = 3x - 4 \)
Bring variables to one side and numbers to the other:
\( \implies 6x - 3x = 37 - 4 \)
\( \implies 3x = 33 \)
\( \implies x = 11 \)

(ix) \( \frac{2}{5x} - \frac{5}{3x} = \frac{1}{15} \)
Find a common denominator of \( 15x \) for the left-hand side:
\( \implies \frac{2 \times 3}{15x} - \frac{5 \times 5}{15x} = \frac{1}{15} \)
\( \implies \frac{6 - 25}{15x} = \frac{1}{15} \)
\( \implies \frac{-19}{15x} = \frac{1}{15} \)
Multiply both sides by \( 15 \):
\( \implies \frac{-19}{x} = 1 \)
\( \implies x = -19 \)

(x) \( \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 \)
On the left side, use LCM of \( 3 \) and \( 5 \) which is \( 15 \). On the right side, write \( 1 \) as \( \frac{4}{4} \):
\( \implies \frac{5(x + 2) - 3(x + 1)}{15} = \frac{x - 3 - 4}{4} \)
Simplify the expressions:
\( \implies \frac{5x + 10 - 3x - 3}{15} = \frac{x - 7}{4} \)
\( \implies \frac{2x + 7}{15} = \frac{x - 7}{4} \)
Cross-multiply to solve:
\( \implies 4(2x + 7) = 15(x - 7) \)
\( \implies 8x + 28 = 15x - 105 \)
Collect the terms of \( x \) on one side:
\( \implies 8x - 15x = -105 - 28 \)
\( \implies -7x = -133 \)
\( \implies x = \frac{-133}{-7} \)
\( \implies x = 19 \)

(xi) \( \frac{3x - 2}{3} + \frac{2x + 3}{2} = x + \frac{7}{6} \)
On the left side, the LCM of \( 3 \) and \( 2 \) is \( 6 \). Rewrite the right side with denominator \( 6 \):
\( \implies \frac{2(3x - 2) + 3(2x + 3)}{6} = \frac{6x + 7}{6} \)
Expand and simplify the numerator on the left side:
\( \implies \frac{6x - 4 + 6x + 9}{6} = \frac{6x + 7}{6} \)
\( \implies \frac{12x + 5}{6} = \frac{6x + 7}{6} \)
Multiply both sides by \( 6 \):
\( \implies 12x + 5 = 6x + 7 \)
Rearrange the terms:
\( \implies 12x - 6x = 7 - 5 \)
\( \implies 6x = 2 \)
\( \implies x = \frac{2}{6} \)
\( \implies x = \frac{1}{3} \)

(xii) \( x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} \)
Find a common denominator of \( 2 \) on the left side, and \( 3 \) on the right side:
\( \implies \frac{2x - (x - 1)}{2} = \frac{3 - (x - 2)}{3} \)
Simplify the numerators:
\( \implies \frac{2x - x + 1}{2} = \frac{3 - x + 2}{3} \)
\( \implies \frac{x + 1}{2} = \frac{5 - x}{3} \)
Cross-multiply to solve:
\( \implies 3(x + 1) = 2(5 - x) \)
\( \implies 3x + 3 = 10 - 2x \)
Collect the variables and constant terms:
\( \implies 3x + 2x = 10 - 3 \)
\( \implies 5x = 7 \)
\( \implies x = \frac{7}{5} \)

(xiii) \( \frac{9x + 7}{2} - \left(x - \frac{x - 2}{7}\right) = 36 \)
Simplify inside the parentheses first:
\( \implies \frac{9x + 7}{2} - \left(\frac{7x - (x - 2)}{7}\right) = 36 \)
\( \implies \frac{9x + 7}{2} - \left(\frac{6x + 2}{7}\right) = 36 \)
Now find a common denominator of \( 14 \):
\( \implies \frac{7(9x + 7) - 2(6x + 2)}{14} = 36 \)
\( \implies \frac{63x + 49 - 12x - 4}{14} = 36 \)
\( \implies \frac{51x + 45}{14} = 36 \)
Multiply both sides by \( 14 \):
\( \implies 51x + 45 = 504 \)
Subtract \( 45 \) from both sides:
\( \implies 51x = 459 \)
Divide by \( 51 \):
\( \implies x = 9 \)

(xiv) \( \frac{6x + 1}{2} + 1 = \frac{7x - 3}{3} \)
Rewrite the left side with a common denominator of \( 2 \):
\( \implies \frac{(6x + 1) + 2}{2} = \frac{7x - 3}{3} \)
\( \implies \frac{6x + 3}{2} = \frac{7x - 3}{3} \)
Cross-multiply:
\( \implies 3(6x + 3) = 2(7x - 3) \)
\( \implies 18x + 9 = 14x - 6 \)
Collect terms with \( x \) on the left and numbers on the right:
\( \implies 18x - 14x = -6 - 9 \)
\( \implies 4x = -15 \)
\( \implies x = -\frac{15}{4} \)
In simple words: To solve these equations, first simplify the fractions on both sides. Find a common denominator to combine terms, cross-multiply to remove the fractions, and then solve for the unknown variable.

Exam Tip: Be very careful with minus signs in front of brackets - they change the sign of every term inside. Always double-check your calculations by substituting the answer back into the original equation.

Question 2. After 12 years, I shall be 3 times as old as I was 4 years ago. Find my present age.
Answer: Suppose my current age is \( x \) years. Four years ago, my age was \( x - 4 \) years. After twelve years, my age will be \( x + 12 \) years. Using the information in the problem:
\( x + 12 = 3(x - 4) \)
\( \implies x + 12 = 3x - 12 \)
Rearrange the terms:
\( \implies 3x - x = 12 + 12 \)
\( \implies 2x = 24 \)
\( \implies x = 12 \) Therefore, my current age is \( 12 \) years.
In simple words: If we write down our age as \( x \), we can set up an equation comparing our age in 12 years with 3 times our age from 4 years ago. Solving it gives the answer.

Exam Tip: Make sure to subtract years for "ago" and add years for "after" or "hence" before setting up your equation.

Question 3. A man sold an article for Rs. 396 and gained 10% on it. Find the cost price of the article.
Answer: Let the cost price of the article be Rs. \( x \). The selling price (S.P.) is Rs. \( 396 \) and the gain percentage is \( 10\% \). We know the relation:
\( \text{S.P.} = \text{C.P.} \times \frac{100 + \text{Gain}\%}{100} \)
\( \implies 396 = x \times \frac{100 + 10}{100} \)
\( \implies 396 = \frac{110}{100}x \)
Now, solve for \( x \):
\( \implies x = \frac{396 \times 100}{110} \)
\( \implies x = 360 \) Hence, the cost price of the article is Rs. \( 360 \).
In simple words: The selling price of Rs. 396 includes a 10% profit. We use the formula to find the original price before this profit was added.

Exam Tip: Always state the formula relating selling price, cost price, and gain percentage clearly to secure full method marks.

Question 4. The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Answer: Let the two numbers be \( x \) and \( y \). From the first condition, their sum is \( 4500 \):
\( x + y = 4500 \quad \text{--- (i)} \)
From the second condition, \( 10\% \) of the first number equals \( 12.5\% \) of the second number:
\( \implies \frac{10}{100}x = \frac{12.5}{100}y \)
\( \implies 10x = 12.5y \)
\( \implies x = \frac{12.5}{10}y \quad \text{--- (ii)} \)
Substitute this value of \( x \) into equation (i):
\( \implies \frac{12.5}{10}y + y = 4500 \)
Multiply the entire equation by \( 10 \) to clear the fraction:
\( \implies 12.5y + 10y = 45000 \)
\( \implies 22.5y = 45000 \)
\( \implies y = \frac{45000}{22.5} = 2000 \)
Now, substitute \( y = 2000 \) back into equation (ii) to find \( x \):
\( \implies x = \frac{12.5}{10} \times 2000 \)
\( \implies x = 2500 \) Thus, the two required numbers are \( 2500 \) and \( 2000 \).
In simple words: We can name the two numbers as \( x \) and \( y \). Use the two clues given in the question to write two equations and solve them to find both numbers.

Exam Tip: When working with decimals like 22.5 in a fraction, multiplying the numerator and denominator by 10 makes the division much easier and prevents mistakes.

Question 5. The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \). According to the problem, their sum is \( 405 \):
\( x + y = 405 \quad \text{--- (i)} \)
The ratio of these numbers is \( 8 : 7 \), which gives:
\( \frac{x}{y} = \frac{8}{7} \)
\( \implies x = \frac{8}{7}y \quad \text{--- (ii)} \)
Substitute this value of \( x \) into equation (i):
\( \implies \frac{8}{7}y + y = 405 \)
Multiply the entire equation by \( 7 \):
\( \implies 8y + 7y = 405 \times 7 \)
\( \implies 15y = 2835 \)
\( \implies y = \frac{2835}{15} = 189 \)
Now, substitute \( y = 189 \) into equation (ii):
\( \implies x = \frac{8}{7} \times 189 \)
\( \implies x = 8 \times 27 = 216 \) Thus, the required numbers are \( 216 \) and \( 189 \).
In simple words: We can represent the two numbers in terms of a ratio. When we add them up, they equal 405. Solving this equation helps us find the two numbers.

Exam Tip: Alternatively, you can assume the numbers are \( 8a \) and \( 7a \). Then, \( 8a + 7a = 405 \implies 15a = 405 \implies a = 27 \). The numbers are \( 8 \times 27 = 216 \) and \( 7 \times 27 = 189 \). This is often a much quicker way to solve ratio problems!

Question 6. The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Answer: The current ages of A and B have a ratio of \( 7 : 5 \). Let the current age of A be \( 7x \) years, and that of B be \( 5x \) years. After ten years:
Age of A will be \( 7x + 10 \) years.
Age of B will be \( 5x + 10 \) years.
According to the problem, their age ratio in ten years will be \( 9 : 7 \):
\( \frac{7x + 10}{5x + 10} = \frac{9}{7} \)
By cross-multiplying, we get:
\( \implies 7(7x + 10) = 9(5x + 10) \)
\( \implies 49x + 70 = 45x + 90 \)
Bring terms with \( x \) to the left side and constant terms to the right side:
\( \implies 49x - 45x = 90 - 70 \)
\( \implies 4x = 20 \)
\( \implies x = 5 \)
Now, calculate their current ages:
Age of A = \( 7 \times 5 = 35 \) years.
Age of B = \( 5 \times 5 = 25 \) years.
Therefore, the current age of A is \( 35 \) years, and the current age of B is \( 25 \) years.
In simple words: We can write the current ages of A and B as \( 7x \) and \( 5x \). In ten years, we add 10 to both, set up their new ratio, and solve for \( x \).

Exam Tip: When a ratio is given, always use a common multiplier like \( x \) to write down the actual values. Remember to add 10 to both ages since time passes for both people equally!

Question 7. Find the number whose double is 45 greater than its half.
Answer: Let the unknown number be represented by \( x \). The double of this number is \( 2x \), and half of this number is \( \frac{x}{2} \). According to the given condition, the double is \( 45 \) more than its half:
\( 2x - \frac{x}{2} = 45 \)
\( \implies \frac{4x - x}{2} = 45 \)
\( \implies \frac{3x}{2} = 45 \)
Solve for \( x \):
\( \implies 3x = 90 \)
\( \implies x = 30 \) Hence, the required number is \( 30 \).
In simple words: We write the number as \( x \). Double of the number is \( 2x \) and half is \( \frac{x}{2} \). Subtracting half from the double gives 45, which helps us solve for \( x \).

Exam Tip: Always translate verbal phrases into algebraic expressions carefully. "Double" means multiplying by 2, and "half" means dividing by 2.

Question 8. The difference between the squares of two consecutive numbers is 31. Find the numbers.
Answer: Let the first number be \( x \). Since the numbers are consecutive, the second number is \( x + 1 \). As per the problem, the difference between their squares is \( 31 \):
\( (x + 1)^2 - x^2 = 31 \)
\( \implies (x^2 + 2x + 1) - x^2 = 31 \)
Combine the terms:
\( \implies 2x + 1 = 31 \)
\( \implies 2x = 30 \)
\( \implies x = 15 \) Thus, the first number is \( 15 \), and the next consecutive number is \( 15 + 1 = 16 \). Therefore, the two numbers are \( 15 \) and \( 16 \).
In simple words: Consecutive numbers are numbers that come one after another, like 15 and 16. Subtracting the square of the smaller number from the square of the larger number gives 31.

Exam Tip: Remember the algebraic identity \( (a + b)^2 = a^2 + 2ab + b^2 \) when expanding \( (x + 1)^2 \). It simplifies the equation rapidly as \( x^2 \) cancels out.

Question 9. Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Answer: Let the required number be \( x \). Five times the number is \( 5x \), and twice the number is \( 2x \). According to the given condition:
\( 5x - 5 = 2x + 4 \)
Rearrange the terms:
\( \implies 5x - 2x = 4 + 5 \)
\( \implies 3x = 9 \)
\( \implies x = 3 \) Thus, the required number is \( 3 \).
In simple words: We can represent the number with \( x \). We then write an equation following the steps in the question and solve for \( x \) to find the number.

Exam Tip: Group all terms with the variable on one side and constant numbers on the other side before simplifying.

Question 10. The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes \( \frac{4}{5} \). Find the original fraction.
Answer: Let the denominator of the original fraction be \( x \). Since the numerator is \( 5 \) less than the denominator, the numerator is \( x - 5 \). So, the original fraction is \( \frac{x - 5}{x} \). If we add \( 3 \) to both the numerator and the denominator, the new fraction is:
\( \frac{(x - 5) + 3}{x + 3} = \frac{x - 2}{x + 3} \)
According to the given condition, this new fraction is equal to \( \frac{4}{5} \):
\( \frac{x - 2}{x + 3} = \frac{4}{5} \)
By cross-multiplying:
\( \implies 5(x - 2) = 4(x + 3) \)
\( \implies 5x - 10 = 4x + 12 \)
Rearrange the terms:
\( \implies 5x - 4x = 12 + 10 \)
\( \implies x = 22 \)
Now, find the original fraction:
Numerator = \( 22 - 5 = 17 \).
Denominator = \( 22 \).
Therefore, the original fraction is \( \frac{17}{22} \).
In simple words: Let the denominator be \( x \) and the numerator be \( x - 5 \). If we add 3 to both and set it equal to \( \frac{4}{5} \), we can cross-multiply to find that \( x \) is 22. This gives us the fraction \( \frac{17}{22} \).

Exam Tip: When a question asks for a fraction, always write your final answer as a fraction \( \frac{\text{numerator}}{\text{denominator}} \) rather than just giving the value of \( x \).