CBSE Class 7 Mathematics Linear Equations In One Variable MCQs

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LINEAR EQUATION IN ONE VARIABLE

INTRODUCTION

Equation : A statement of equality which contains one or more unknown quantity or variable (literals) is called an equation. For example -

Ex. : \( 3x + 7 = 12 \), \( \frac{5}{2}x - 9 = 1 \), \( x^2 + 1 = 5 \) and \( \frac{x}{3} + 5 = \frac{x}{2} - 3 \) are equations in one variable \( x \).

Ex. : \( 2x + 3y = 15 \), \( 7x - \frac{y}{3} = 3 \) are equations in two variables \( x \) and \( y \).

Linear Equation : An equation involving only linear polynomials is called a linear equation.

Ex. : \( 3x - 2 = 7 \), \( \frac{3}{2}x + 9 = \frac{1}{2} \), \( \frac{y}{3} + \frac{y - 2}{4} = 5 \) are linear equations in one variable, because the highest power of the variable in each equation is one whereas the equations \( 3x^2 - 2x + 1 = 0 \), \( y^2 - 1 = 8 \) are not linear equations, because the highest power of the variable in each equation is not one.

Solution of a Linear Equation :

Solution : A value of the variable which when substituted for the variable in an equation, makes L.H.S. = R.H.S. is said to satisfy the equation and is called a solution or a root of the equation.

Rules for Solving Linear Equations in One Variable :

• Rule-1 Same quantity (number) can be added to both sides of an equation without changing the equality.
• Rule-2 Same quantity can be subtracted from both sides of an equation without changing the equality.
• Rule-3 Both sides of an equation may be multiplied by the same non-zero number without changing the equality.
• Rule-4 Both sides of an equation may be divided by the same non-zero number without changing the equality.

Solving Equations having Variable Terms on One Side and Number(s) on the Other Side :

EXAMPLES

Question. Solve the equation : \( \frac{x}{5} + 11 = \frac{1}{15} \) and check the result.
Answer: We have,
\( \frac{x}{5} + 11 = \frac{1}{15} \)
\( \Rightarrow \frac{x}{5} + 11 - 11 = \frac{1}{15} - 11 \) [Subtracting 11 from both sides]
\( \Rightarrow \frac{x}{5} = \frac{1}{15} - 11 \)
\( \Rightarrow \frac{x}{5} = \frac{1 - 165}{15} \)
\( \Rightarrow \frac{x}{5} = -\frac{164}{15} \)
\( \Rightarrow 5 \times \frac{x}{5} = 5 \times -\frac{164}{15} \)
\( \Rightarrow x = -\frac{164}{3} \)
Thus, \( x = -\frac{164}{3} \) is the solution of the given equation.

Question. Solve : \( \frac{1}{3}x - \frac{5}{2} = 6 \)
Answer: We have,
\( \frac{1}{3}x - \frac{5}{2} = 6 \)
\( \Rightarrow \frac{1}{3}x - \frac{5}{2} + \frac{5}{2} = 6 + \frac{5}{2} \) [Adding \( \frac{5}{2} \) on both sides]
\( \Rightarrow \frac{1}{3}x = 6 + \frac{5}{2} \)
\( \Rightarrow \frac{1}{3}x = \frac{12 + 5}{2} \)
\( \Rightarrow \frac{1}{3}x = \frac{17}{2} \)
\( \Rightarrow 3 \times \frac{1}{3}x = 3 \times \frac{17}{2} \) [Multiplying both sides by 3]
\( \Rightarrow x = \frac{51}{2} \)
Thus, \( x = \frac{51}{2} \) is the solution of the given equation.

Question. Solve : \( \frac{x}{2} + \frac{x}{3} - \frac{x}{4} = 7 \)
Answer: We have, \( \frac{x}{2} + \frac{x}{3} - \frac{x}{4} = 7 \)
LCM of denominators 2, 3, 4 on L.H.S. is 12.
Multiplying both sides by 12, we get
\( 6x + 4x - 3x = 7 \times 12 \)
\( \Rightarrow 7x = 7 \times 12 \)
\( \Rightarrow 7x = 84 \)
\( \Rightarrow \frac{7x}{7} = \frac{84}{7} \) [Dividing both sides by 7]
\( \Rightarrow x = 12 \)

Question. Solve : \( \frac{y - 1}{3} - \frac{y - 2}{4} = 1 \)
Answer: We have, \( \frac{y - 1}{3} - \frac{y - 2}{4} = 1 \)
LCM of denominators 3 and 4 on L.H.S. is 12.
Multiplying both sides by 12, we get
\( 12 \times (\frac{y - 1}{3}) - 12 \times (\frac{y - 2}{4}) = 12 \times 1 \)
\( \Rightarrow 4(y - 1) - 3(y - 2) = 12 \)
\( \Rightarrow 4y - 4 - 3y + 6 = 12 \Rightarrow 4y - 3y - 4 + 6 = 12 \)
\( \Rightarrow y + 2 = 12 \)
\( \Rightarrow y + 2 - 2 = 12 - 2 \) [Subtracting 2 from both sides]
\( \Rightarrow y = 10 \)
Thus, \( y = 10 \) is the solution of the given equation.

Transposition Method for Solving Linear Equations in One Variable

The transposition method involves the following steps:

• Step-I Obtain the linear equation.
• Step-II Identify the variable (unknown quantity) and constants(numerals).
• Step-III Simplify the L.H.S. and R.H.S. to their simplest forms by removing brackets.
• Step-IV Transpose all terms containing variable on L.H.S. and constant terms on R.H.S. Note that the sign of the terms will change in shifting them from L.H.S. to R.H.S. and vice-versa.
• Step-V Simplify L.H.S. and R.H.S. in the simplest form so that each side contains just one term.
• Step-VI Solve the equation obtained in step V by dividing both sides by the coefficient of the variable on L.H.S.

EXAMPLES

Question. Solve : \( \frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4} \)
Answer: We have, \( \frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4} \)
The denominators on two sides are 2, 5, 3 and 4. Their LCM is 60. Multiplying both sides of the given equation by 60, we get
\( 60 \times (\frac{x}{2} - \frac{1}{5}) = 60 (\frac{x}{3} + \frac{1}{4}) \)
\( \Rightarrow 60 \times \frac{x}{2} - 60 \times \frac{1}{5} = 60 \times \frac{x}{3} + 60 \times \frac{1}{4} \)
\( \Rightarrow 30x - 12 = 20x + 15 \)
\( \Rightarrow 30x - 20x = 15 + 12 \) [On transposing 20x to LHS and –12 to RHS]
\( \Rightarrow 10x = 27 \)
\( \Rightarrow x = \frac{27}{10} \)
Hence, \( x = \frac{27}{10} \) is the solution of the given equation.

Question. Solve : \( \frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t \)
Answer: We have, \( \frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t \)
The denominators on two sides are 4, 3 and 3. Their LCM is 12.
Multiplying both sides of the given equation by 12, we get
\( 12 (\frac{3t - 2}{4}) - 12 (\frac{2t + 3}{3}) = 12 (\frac{2}{3} - t) \)
\( \Rightarrow 3(3t - 2) - 4(2t + 3) = 12 (\frac{2}{3} - t) \)
\( \Rightarrow 9t - 6 - 8t - 12 = 12 \times \frac{2}{3} - 12t \)
\( \Rightarrow 9t - 6 - 8t - 12 = 8 - 12t \)
\( \Rightarrow t - 18 = 8 - 12t \)
\( \Rightarrow t + 12t = 8 + 18 \) [Transposing –12t to LHS and – 18 to RHS]
\( \Rightarrow 13t = 26 \)
\( \Rightarrow t = \frac{26}{13} \) [Dividing both sides by 13]
\( \Rightarrow t = 2 \)

Question. Solve : \( \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3x - 4}{12} \)
Answer: We have, \( \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3x - 4}{12} \)
The denominators on two sides of the given equation are 6, 3, 4 and 12. Their LCM is 24.
Multiplying both sides of the given equation by 24, we get
\( 24 (\frac{x + 2}{6}) - 24 (\frac{11 - x}{3} - \frac{1}{4}) = 24 (\frac{3x - 4}{12}) \)
\( \Rightarrow 4(x + 2) - 24 (\frac{11 - x}{3}) + 24 \times \frac{1}{4} = 2(3x - 4) \)
\( = 2(3x - 4) \)
\( \Rightarrow 4(x + 2) - 8(11 - x) + 6 = 2(3x - 4) \)
\( \Rightarrow 4x + 8 - 88 + 8x + 6 = 6x - 8 \)
\( \Rightarrow 12x - 74 = 6x - 8 \)
\( \Rightarrow 12x - 6x = 74 - 8 \) [Transposing 6x to LHS and – 74 to RHS]
\( \Rightarrow 6x = 66 \)
\( \Rightarrow x = \frac{66}{6} \) [Dividing both sides by 6]
\( \Rightarrow x = 11 \)

Question. Solve : \( x - \frac{2x + 8}{3} = \frac{1}{4} (x - \frac{2 - x}{6}) - 3 \)
Answer: We have, \( x - \frac{2x + 8}{3} = \frac{1}{4} (x - \frac{2 - x}{6}) - 3 \)
\( \Rightarrow x - \frac{2x + 8}{3} = \frac{x}{4} - \frac{2 - x}{24} - 3 \)
The denominators on the two sides of this equation are 3, 4 and 24. Their LCM is 24.
Multiplying both sides of this equation by 24, we get
\( 24x - 24 (\frac{2x + 8}{3}) = 24 \times \frac{x}{4} - 24 (\frac{2 - x}{24}) - 3 \times 24 \)
\( \Rightarrow 24x - 8(2x + 8) = 6x - (2 - x) - 72 \)
\( \Rightarrow 24x - 16x - 64 = 6x - 2 + x - 72 \)
\( \Rightarrow 8x - 64 = 7x - 74 \)
\( \Rightarrow 8x - 7x = 64 - 74 \) [Transposing 7x to LHS and – 64 to RHS]
\( \Rightarrow x = - 10 \)
Thus, \( x = - 10 \) is the solution of the given equation.

Question. Solve : \( 0.16(5x - 2) = 0.4x + 7 \)
Answer: We have, \( 0.16(5x - 2) = 0.4x + 7 \)
\( \Rightarrow 0.8x - 0.32 = 0.4x + 7 \) [Expanding the bracket on LHS]
\( \Rightarrow 0.8x - 0.4x = 0.32 + 7 \) [Transposing 0.4x to LHS and –0.32 to RHS]
\( \Rightarrow 0.4x = 7.32 \)
\( \Rightarrow \frac{0.4x}{0.4} = \frac{7.32}{0.4} \)
\( \Rightarrow x = \frac{732}{40} \Rightarrow x = \frac{183}{10} = 18.3 \)
Hence, \( x = 18.3 \) is the solution of the given equation.

Question. Solve : \( \frac{2}{5x} - \frac{5}{3x} = \frac{1}{15} \)
Answer: We have, \( \frac{2}{5x} - \frac{5}{3x} = \frac{1}{15} \)
Multiplying both sides by 15x, the LCM of 5x and 3x, we get
\( 15x \times \frac{2}{5x} - 15x \times \frac{5}{3x} = 15x \times \frac{1}{15} \)
\( \Rightarrow 6 - 25 = x \Rightarrow -19 = x \Rightarrow x = -19 \)
Hence, \( x = -19 \) is the solution of the given equation.

Question. Solve : \( \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 \)
Answer: Multiplying both sides by 60 i.e. the LCM of 3, 5, and 4, we get
\( 20(x + 2) - 12(x + 1) = 15(x - 3) - 1 \times 60 \)
\( \Rightarrow 20x + 40 - 12x + 12 = 15x - 45 - 60 \)
\( \Rightarrow 8x + 28 = 15x - 105 \)
\( \Rightarrow 8x - 15x = -105 - 28 \)
\( \Rightarrow - 7x = - 133 \)
\( \Rightarrow \frac{-7x}{-7} = \frac{-133}{-7} \) [Dividing both sides by –7]
\( \Rightarrow x = \frac{133}{7} = 19 \)
Thus, \( x = 19 \) is the solution of the given equation.

Cross-Multiplication Method for Solving Equations of the form :

\[ \frac{ax + b}{cx + d} = \frac{m}{n} \Rightarrow n(ax + b) = m (cx + d) \]

EXAMPLES

Question. Solve : \( \frac{2x + 1}{3x - 2} = \frac{9}{10} \)
Answer: We have, \( \frac{2x + 1}{3x - 2} = \frac{9}{10} \)
\( \Rightarrow 10 \times (2x + 1) = 9 \times (3x - 2) \) [By cross-multiplication]
\( \Rightarrow 20x + 10 = 27x - 18 \)
\( \Rightarrow 20x - 27x = - 18 - 10 \) [Using transposition]
\( \Rightarrow - 7x = - 28 \)
\( \Rightarrow \frac{-7x}{-7} = \frac{-28}{-7} \) [Dividing both sides by –7]
\( \Rightarrow x = 4 \)
Hence, \( x = 4 \) is the solution of the given equation.

Question. Solve : \( \frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8 \)
Answer: We have, \( \frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8 \)
\( \Rightarrow \frac{34 - 17x - 5x - 60}{1 - 7x} = \frac{8}{1} \)
\( \Rightarrow \frac{-22x - 26}{1 - 7x} = \frac{8}{1} \)
\( \Rightarrow 1 \times (- 22x - 26) = 8 \times (1 - 7x) \) [By cross-multiplication]
\( \Rightarrow - 22x - 26 = 8 - 56x \)
\( \Rightarrow - 22x + 56x = 8 + 26 \)
\( \Rightarrow 34x = 34 \)
\( \Rightarrow \frac{34x}{34} = \frac{34}{34} \)
Hence, \( x = 1 \) is the solution of the given equation.

Question. Solve : \( \frac{x + b}{a - b} = \frac{x - b}{a + b} \)
Answer: We have, \( \frac{x + b}{a - b} = \frac{x - b}{a + b} \)
\( \Rightarrow (x + b) \times (a + b) = (x - b) \times (a - b) \) [By cross-multiplication]
\( \Rightarrow x (a + b) + b(a + b) = x(a - b) - b(a - b) \)
\( \Rightarrow ax + bx + ba + b^2 = ax - bx - ba + b^2 \)
\( \Rightarrow ax + bx - ax + bx = - bx + b^2 - ba - b^2 \) (Wait, simplifying terms):
\( \Rightarrow bx + ba = -bx - ba \)
\( \Rightarrow 2bx = - 2ba \)
\( \Rightarrow \frac{2bx}{2b} = - \frac{2ab}{2b} \)
\( \Rightarrow x = - a \)
Hence, \( x = - a \) is the solution of the given equation.

Question. Solve : \( \frac{1}{x + 1} + \frac{1}{x + 2} = \frac{2}{x + 10} \)
Answer: We have, \( \frac{1}{x + 1} + \frac{1}{x + 2} = \frac{2}{x + 10} \)
Multiplying both sides by \( (x + 1)(x + 2)(x + 10) \) i.e., the LCM of \( x + 1 \), \( x + 2 \) and \( x + 10 \), we get
\( \frac{(x + 1)(x + 2)(x + 10)}{x + 1} + \frac{(x + 1)(x + 2)(x + 10)}{x + 2} = \frac{2(x + 1)(x + 2)(x + 10)}{x + 10} \)
\( \Rightarrow (x + 2)(x + 10) + (x + 1)(x + 10) = 2(x + 1)(x + 2) \)
\( \Rightarrow x^2 + 2x + 10x + 20 + x^2 + 10x + x + 10 = 2(x^2 + x + 2x + 2) \)
\( \Rightarrow 2x^2 + 23x + 30 = 2(x^2 + 3x + 2) \)
\( \Rightarrow 2x^2 + 23x + 30 = 2x^2 + 6x + 4 \)
\( \Rightarrow 2x^2 + 23x - 2x^2 - 6x = 4 - 30 \)
\( \Rightarrow 17x = - 26 \)
\( \Rightarrow x = - \frac{26}{17} \)
Hence, \( x = - \frac{26}{17} \) is the solution of the given equation.

Question. Ex.7 Solve : \( \frac{6x^2 + 13x - 4}{2x + 5} = \frac{12x^2 + 5x - 2}{4x + 3} \)
Answer: We have, \( \frac{6x^2 + 13x - 4}{2x + 5} = \frac{12x^2 + 5x - 2}{4x + 3} \)
\( \Rightarrow (6x^2 + 13x - 4)(4x + 3) = (12x^2 + 5x - 2)(2x + 5) \) [By cross-multiplication]
\( \Rightarrow (6x^2 + 13x - 4) \times 4x + (6x^2 + 13x - 4) \times 3 = (12x^2 + 5x - 2) \times 2x + (12x^2 + 5x - 2) \times 5 \)
\( \Rightarrow 24x^3 + 52x^2 - 16x + 18x^2 + 39x - 12 = 24x^3 + 10x^2 - 4x + 60x^2 + 25x - 10 \)
\( \Rightarrow 24x^3 + 70x^2 + 23x - 12 = 24x^3 + 70x^2 + 21x - 10 \)
\( \Rightarrow 24x^3 + 70x^2 + 23x - 24x^3 - 70x^2 - 21x = - 10 + 12 \)
\( \Rightarrow 2x = 2 \)
\( \Rightarrow x = 1 \)
Hence, \( x = 1 \) is the solution of the given equation.

Applications of Linear Equations to Practical Problems

The following steps should be followed to solve a word problem:

• Step-I Read the problem carefully and note what is given and what is required.
• Step-II Denote the unknown quantity by some letters, say x, y, z, etc.
• Step-III Translate the statements of the problem into mathematical statements.
• Step-IV Using the condition(s) given in the problem, form the equation.
• Step-V Solve the equation for the unknown.
• Step-VI Check whether the solution satisfies the equation.

EXAMPLES

Question. A number is such that it is as much greater than 84 as it is less than 108. Find it.
Answer: Let the number be \( x \). Then, the number is greater than 84 by \( x - 84 \) and it is less than 108 by \( 108 - x \). [Given]
\( \therefore x - 84 = 108 - x \)
\( \Rightarrow x + x = 108 + 84 \)
\( \Rightarrow 2x = 192 \)
\( \Rightarrow \frac{2x}{2} = \frac{192}{2} \)
\( \Rightarrow x = 92 \)
Hence, the number is 96.

Question. A number consists of two digits whose sum is 8. If 18 is added to the number its digits are reversed. Find the number.
Answer: Let one’s digit be \( x \).
Since the sum of the digits is 8. Therefore, ten’s digit = \( 8 - x \).
\( \therefore \) Number = \( 10 \times (8 - x) + x = 80 - 10x + x = 80 - 9x \) ... (i)
Now, Number obtained by reversing the digit = \( 10 \times x + (8 - x) = 10x + 8 - x = 9x + 8 \).
It is given that if 18 is added to the number its digits are reversed.
\( \therefore \) Number + 18 = Number obtained by reversing the digits
\( \Rightarrow 80 - 9x + 18 = 9x + 8 \)
\( \Rightarrow 98 - 9x = 9x + 8 \)
\( \Rightarrow 98 - 8 = 9x + 9x \)
\( \Rightarrow 90 = 18x \)
\( \Rightarrow \frac{18x}{18} = \frac{90}{18} \)
\( \Rightarrow x = 5 \)
Putting the value of \( x \) in (i), we get
Number = \( 80 - 9 \times 5 = 80 - 45 = 35 \)

Question. Divide 34 into two parts in such a way that \( \frac{4}{7} \text{th} \) of one part is equal to \( \frac{2}{5} \text{th} \) of the other.
Answer: Let one part be \( x \). Then, other part is \( (34 - x) \). It is given that
\( (\frac{4}{7})^{\text{th}} \text{ of one part} = (\frac{2}{5})^{\text{th}} \text{ of the other part} \)
\( \Rightarrow \frac{4}{7}x = \frac{2}{5}(34 - x) \)
\( \Rightarrow 20x = 14(34 - x) \) [Multiplying both sides by 35, the LCM of 7 and 5]
\( \Rightarrow 20x = 14 \times 34 - 14x \)
\( \Rightarrow 20x + 14x = 14 \times 34 \)
\( \Rightarrow 34x = 14 \times 34 \)
\( \Rightarrow \frac{34x}{34} = \frac{14 \times 34}{34} \) [Dividing both sides by 34]
\( \Rightarrow x = 14 \)
Hence, the two parts are 14 and \( 34 - 14 = 20 \)

Question. The numerator of a fraction is 4 less that the denominator. If 1 is added to both its numerator and denominator, it becomes 1/2. Find the fraction.
Answer: Let the denominator of the fraction be \( x \). Then, Numerator of the fraction = \( x - 4 \).
\( \therefore \text{Fraction} = \frac{x - 4}{x} \) ...(i)
If 1 is added to both its numerator and denominator, the fraction becomes \( \frac{1}{2} \).
\( \therefore \frac{x - 4 + 1}{x + 1} = \frac{1}{2} \)
\( \Rightarrow \frac{x - 3}{x + 1} = \frac{1}{2} \)
\( \Rightarrow 2(x - 3) = x + 1 \) [Using cross-multiplication]
\( \Rightarrow 2x - 6 = x + 1 \)
\( \Rightarrow 2x - x = 6 + 1 \)
\( \Rightarrow x = 7 \)
Putting \( x = 7 \) in (i), we get
Fraction = \( \frac{7 - 4}{7} = \frac{3}{7} \).
Hence, the given fraction is \( \frac{3}{7} \).

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LINEAR EQUATION IN ONE VARIABLE

Question. Arvind has Piggy bank. It is full of one-rupee and fifty-paise coins. It contains 3 times as many fifty paise coins as one rupee coins. The total amount of the money in the bank is Rs 35. How many coins of each kind are there in the bank ?
Answer: Let there be \( x \) one rupee coins in the bank.
Then,
Number of 50-paise coins \( = 3x \)
\( \therefore \) Value of \( x \) one rupee coins \( = \text{Rs } x \)
Value of \( 3x \) fifty-paise coins \( = 50 \times 3x \text{ paise} \)
\( = 150 x = \text{paise} = \text{Rs } \frac{150}{100} x = \text{Rs } \frac{3x}{2} \)
\( \therefore \) Total value of all the coins \( = \text{Rs } (x + \frac{3x}{2}) \)
But, the total amount of the money in the bank is given as Rs 35.
\( \therefore x + \frac{3x}{2} = 35 \)
\( \Rightarrow 2x + 3x = 70 \) [Multiplying both sides by 2]
\( \Rightarrow 5x = 70 \Rightarrow \frac{5x}{5} = \frac{70}{5} \Rightarrow x = 14 \)
\( \therefore \) Number of one rupee coins \( = 14 \), Number of 50 paise coins \( = 3x = 3 \times 14 = 42 \).

Question. Kanwar is three years older than Anima. Six years ago, Kanwar’s age was four times Anima’s age. Find the ages of Kanwar and Anima.
Answer: Let Anima’s age be \( x \) years. Then, Kanwar’s age is \( (x + 3) \) years.
Six years ago, Anima’s age was \( (x - 6) \) years
It is given that six years ago Kanwar’s age was four times Anima’s age.
\( \therefore x - 3 = 4(x - 6) \)
\( \Rightarrow x - 3 = 4x - 24 \)
\( \Rightarrow x - 4x = - 24 + 3 \)
\( \Rightarrow - 3x = - 21 \)
\( \Rightarrow \frac{-3x}{-3} = \frac{-21}{-3} \)
\( \Rightarrow x = 7 \)
Hence, Anima’s age \( = 7 \) years
Kanwar’s age \( = (x + 3) \) years
\( = (7 + 3) \) years \( = 10 \) years.

Question. Hamid has three boxes of different fruits. Box A weighs \( 2 \frac{1}{2} \) kg more than Box B and Box C weighs \( 10 \frac{1}{4} \) kg more than Box B. The total weight of the boxes is \( 48 \frac{3}{4} \) kg. How many kg does Box A weigh ?
Answer: Suppose the box B weights \( x \) kg.
Since box A weighs \( 2 \frac{1}{2} \) kg more than box B and C weighs \( 10 \frac{1}{4} \) kg more than box B.
\( \therefore \) Weight of box A \( = (x + 2 \frac{1}{2}) \text{ kg} \)
\( = (x + \frac{5}{2}) \text{ kg} \) ... (i)
Weight of box C \( = (x + 10 \frac{1}{4}) \text{ kg} \)
\( = (x + \frac{41}{4}) \text{ kg} \)
\( \therefore \) Total weight of all the boxes
\( = (x + \frac{5}{2} + x + x + \frac{41}{4}) \text{ kg} \)
But, the total weight of the boxes is given as
\( 48 \frac{3}{4} \text{ kg} = \frac{195}{4} \text{ kg} \)
\( \therefore x + \frac{5}{2} + x + x + \frac{41}{4} = \frac{195}{4} \)
\( \Rightarrow 4x + 10 + 4x + 4x + 41 = 195 \) [Multiplying both sides by 4]
\( \Rightarrow 12x + 51 = 195 \)
\( \Rightarrow 12x = 195 - 51 \)
\( \Rightarrow 12x = 144 \)
\( \Rightarrow \frac{12x}{12} = \frac{144}{12} \)
\( \Rightarrow x = 12 \)
Putting \( x = 12 \) in (i), we get
Weight of box A \( = (12 + \frac{5}{2}) \text{ kg} = 14 \frac{1}{2} \text{ kg} \).

Question. The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other, find the numbers.
Answer: Let the first number be \( x \). Then,
Second number \( = 2490 - x \).
[\( \therefore \) Sum of the numbers is given to be 2490]
Now,
6.5% of the first number \( = \frac{6.5}{100} \times x = \frac{65x}{1000} \)
and,
8.5% of the second number \( = \frac{8.5}{100} \times (2490 - x) \)
\( = \frac{85}{1000} (2490 - x) \)
It is given that 6.5% of the first number is equal to 8.5% of the other.
\( \therefore \frac{65x}{1000} = \frac{85}{1000} (2490 - x) \)
\( \Rightarrow 65x = 85 (2490 - x) \) [Multiplying both sides by 1000]
\( \Rightarrow 65x = 2490 \times 85 - 85x \)
\( \Rightarrow 65x + 85x = 2490 \times 85 \)
\( \Rightarrow 150x = 2490 \times 85 \)
\( \Rightarrow x = \frac{2490 \times 85}{150} \)
\( \Rightarrow x = 1411 \)
\( \therefore \) First number \( = 1411 \), Second number \( = 2490 - 1411 = 1079 \)

Question. The sum of two numbers is 45 and their ratio is 7 : 8. Find the numbers.
Answer: Let one of the numbers be \( x \). Since the sum of the two numbers is 45. Therefore, the other number will be \( 45 - x \).
It is given that the ratio of the numbers is 7 : 8.
\( \therefore \frac{x}{45 - x} = \frac{7}{8} \)
\( \Rightarrow 8 \times x = 7 \times (45 - x) \) [By cross-multiplication]
\( \Rightarrow 8x = 315 - 7x \)
\( \Rightarrow 8x + 7x = 315 \)
\( \Rightarrow 15x = 315 \)
\( \Rightarrow x = \frac{315}{15} = 21 \)
Thus, one number is 21 and, Other number \( = 45 - x = 45 - 21 = 24 \)

Question. An altitude of a triangle is five-thirds the length of its corresponding base. If the altitude were increased by 4 cm and the base be decreased by 2 cm, the area of the triangle would remain the same. Find the base and the altitude of the triangle.
Answer: Let the length of the base of the triangle be \( x \) cm. Then,
Altitude \( = (\frac{5}{3} \times x) \text{ cm} = \frac{5x}{3} \text{ cm} \)
\( \therefore \text{Area} = \frac{1}{2} (\text{Base} \times \text{Altitude}) \text{ cm}^2 \)
\( = \frac{1}{2} \times x \times \frac{5x}{3} \text{ cm}^2 = \frac{5x^2}{6} \text{ cm}^2 \)
When the altitude is increased by 4 cm and the base is decreased by 2 cm, we have
New base \( = (x - 2) \text{ cm} \), New altitude \( = \frac{5x}{3} + 4 \text{ cm} \)
\( \therefore \text{Area of the new triangle} \)
\( = \frac{1}{2} (\text{Base} \times \text{Altitude}) \)
\( = \frac{1}{2} \{ (\frac{5x}{3} + 4) \times (x - 2) \} \text{ cm}^2 \)
\( = \frac{1}{2} \{ (x - 2) \times (\frac{5x}{3} + 4) \} \text{ cm}^2 \)
\( = \frac{1}{2} \{ \frac{5x}{3}(x - 2) + 4(x - 2) \} \text{ cm}^2 \)
\( = \frac{1}{2} \{ \frac{5x^2}{3} - \frac{10x}{3} + 4x - 8 \} \text{ cm}^2 \)
\( = \frac{1}{2} \{ \frac{5x^2}{3} - \frac{5x}{3} + 2x - 4 \} \text{ cm}^2 \) [Wait, let's re-verify the expansion from the text:]
\( = \frac{1}{2} (\frac{5x^2}{6} - \frac{5x}{3} + 2x - 4) \text{ cm}^2 \)
It is given that the area of the given triangle is same as the area of the new triangle.
\( \therefore \frac{5x^2}{6} = \frac{5x^2}{6} - \frac{5x}{3} + 2x - 4 \)
\( \Rightarrow \frac{5x^2}{6} - \frac{5x^2}{6} + \frac{5x}{3} - 2x = - 4 \)
\( \Rightarrow \frac{5x}{3} - 2x = - 4 \) [Multiplying both sides by 3]
\( \Rightarrow - x = - 12 \)
\( \Rightarrow x = 12\text{cm} \)
Hence, base of the triangle \( = 12 \text{ cm} \).
Altitude of the triangle \( = \frac{5}{3} \times 12 \text{ cm} = 20 \text{ cm} \)