GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.4

Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1. Amina thinks of a number and subtracts \( \frac{5}{2} \) from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Answer: Let the number be 'x'.
Subtracting \( \frac{5}{2} \), we get \( x - \frac{5}{2} \).
According to the condition, we have
\( 8 (x - \frac{5}{2}) = 3x \)
\( 8x - 20 = 3x \)
Transposing -20 to RHS, we have
\( 8x = 3x + 20 \)
Again transposing 3x to LHS, we have
\( 8x - 3x = 20 \)
\( 5x = 20 \)
\( x = \frac{20}{5} \)
\( x = 4 \)
Thus, the required number is 4.
In simple words: Amina picked a number, took away two and a half, then multiplied what was left by eight. This final number was three times her original number. The number she thought of was 4.

Exam Tip: Carefully translate each phrase into an algebraic expression and set up the equation, ensuring correct order of operations.

Question 2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Answer: Let the positive number be \( 5x \).
The other number is \( x \).
On adding 21 to both numbers, we get \( (5x + 21) \) and \( (x + 21) \).
According to the condition, we have
\( 2 (x + 21) = 5x + 21 \)
\( 2x + 42 = 5x + 21 \)
Transposing 42 to RHS, we have
\( 2x = 5x + 21 - 42 \)
Transposing 5x to LHS, we have
\( 2x - 5x = -21 \)
\( -3x = -21 \)
Dividing both sides by (-3), we have
\( \frac{-3x}{-3} = \frac{-21}{-3} \)
\( x = 7 \)
And \( 5x = 5 \times 7 = 35 \).
Thus, the required numbers are 7 and 35.
In simple words: We have two numbers; one is five times the other. If we add 21 to both, one number becomes double the other. The two numbers are 7 and 35.

Exam Tip: Define the variables clearly for the original numbers, then express the new numbers after the given operation, and finally form the equation.

Question 3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Answer: Let the units digit be \( x \).
The tens digit will be \( (9-x) \).
The original number is \( 10(9 - x) + x \)
\( = 90 - 10x + x \)
\( = 90 - 9x \)
On interchanging the digits, the new number becomes
\( 10x + (9 - x) \)
\( = 10x + 9 - x \)
\( = 9x + 9 \)
According to the condition, we have
[New number] = [Original number] + 27
\( 9x + 9 = 90 - 9x + 27 \)
\( 9x + 9 = 117 - 9x \)
Transposing 9 to RHS:
\( 9x = 117 - 9 - 9x \)
Transposing (-9x) to LHS:
\( 9x + 9x = 108 \)
\( 18x = 108 \)
Dividing both sides by 18, we have
\( x = \frac{108}{18} \)
\( x = 6 \)
So, the units digit is 6.
The tens digit is \( 9 - x = 9 - 6 = 3 \).
The Original number = \( 90 - (9 \times 6) \)
\( = 90 - 54 = 36 \)
In simple words: We had a two-digit number where the digits add up to 9. When we swapped the digits, the new number was 27 larger than the first. The original number was 36.

Exam Tip: Remember that a two-digit number with tens digit 'a' and units digit 'b' is expressed as 10a + b. When digits are reversed, it becomes 10b + a.

Question 4. One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Answer: Let the digit at the unit place be \( x \).
The digit at the tens place will be \( 3x \).
The number is \( 10(3x) + x \)
\( = 30x + x = 31x \)
On interchanging the digits, the new number will be
\( 10x + 3x = 13x \)
According to the condition, we have
\( 31x + 13x = 88 \)
\( 44x = 88 \)
Dividing both sides by 44, we have
\( \frac{44x}{44} = \frac{88}{44} \)
\( x = 2 \)
The original number is \( 31 \times 2 = 62 \).
In simple words: In a two-digit number, one digit is three times the other. When you swap the digits and add the new number to the old one, the total is 88. The original number was 62.

Exam Tip: Always clearly define which digit is 'x' and which is '3x' (unit or tens place) to correctly form the original and new numbers.

Question 5. Shobo's mother's present age is six times Shobo's present age. Shobo's age five years from now will be one-third of his mother's present age. What are their present ages?
Answer: Let Shobo's present age be \( x \) years.
Mother's present age is \( 6x \) years.
After 5 years:
Shobo's age = \( (x + 5) \) years
Mother's age = \( (6x + 5) \) years
According to the condition, we have
(Mother's present age) = (Shobo's age after 5 years)
i.e., \( \frac{1}{3} (6x) = x + 5 \)
\( \frac{1}{3} \times 6x = x + 5 \)
\( 2x = x + 5 \)
Transposing \( x \) to LHS:
\( 2x - x = 5 \)
\( x = 5 \)
So, Shobo's present age = 5 years.
Mother's present age = \( 6 \times 5 = 30 \) years.
In simple words: Shobo's mother is six times older than him now. In five years, Shobo's age will be one-third of what his mother is now. Shobo is 5 years old, and his mother is 30 years old.

Exam Tip: When dealing with age problems, clearly define present ages, then express ages at different points in time (e.g., "5 years from now") before setting up the equation.

Question 6. There is a rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs. 100 per metre it will cost the village panchayat Rs. 75000 to fence the plot. What are the dimensions of the plot?
Answer: Let the length be \( 11x \) metres and breadth be \( 4x \) metres.
Perimeter = \( 2(\text{Length} + \text{Breadth}) \)
\( = 2(11x + 4x) \)
\( = 2 \times 15x = 30x \)
Cost of fencing = \( \text{Rs. } 100 \times 30x = \text{Rs. } 3000x \)
But the total cost of fencing is Rs. 75000.
\( 3000x = 75000 \)
\( x = \frac{75000}{3000} \)
\( x = 25 \)
Length = \( 11 \times 25 = 275 \) metres
Breadth = \( 4 \times 25 = 100 \) metres
In simple words: The school plot is shaped like a rectangle, with its length 11 times some unit and its width 4 times the same unit. Fencing it costs Rs. 100 per meter, and the total cost is Rs. 75,000. The length of the plot is 275 meters, and the breadth is 100 meters.

Exam Tip: For ratio problems, use a common multiplier 'x' to represent the unknown parts. Remember that fencing refers to the perimeter of the plot.

Question 7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs. 750 per metre and trouser material that costs him Rs. 90 per metre. For every 3 metres of the shirt material, he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs. 36,600. How much trouser material did he buy?
Answer: Let the length of cloth for trousers be \( 2x \) metres.
The length of cloth for shirts is \( 3x \) metres.
Cost of trouser's cloth = \( 2x \times \text{Rs. } 90 = \text{Rs. } 180x \)
Cost of shirt's cloth = \( 3x \times \text{Rs. } 50 = \text{Rs. } 150x \)
S.P. of trouser's cloth at 10% profit = \( \frac{110}{100} \times 180x = \text{Rs. } 198x \)
S.P. of shirt's cloth at 12% profit = \( \frac{112}{100} \times 150x = \text{Rs. } 168x \)
Total S.P = \( \text{Rs. } 198x + \text{Rs. } 168x = \text{Rs. } 366x \)
But the total S.P is Rs. 36,600.
\( 366x = 36,600 \)
Dividing both sides by 366:
\( x = \frac{36600}{366} \)
\( x = 100 \)
So, \( 2x = 2 \times 100 = 200 \).
Thus, he bought 200 metres of cloth for trousers.
In simple words: Hasan purchases fabric for uniforms. For every 3 meters of shirt fabric (costing Rs. 750/meter in question, but Rs. 50/meter as per solution's numbers), he buys 2 meters of trouser fabric (costing Rs. 90/meter). He sells them for a profit, making a total of Rs. 36,600. He bought 200 meters of trouser material.

Exam Tip: Be careful with profit calculations; 10% profit means multiplying by 110/100. Set up expressions for total cost and total selling price based on the 'x' multiplier.

Question 8. Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the total number of deer in the herd.
Answer: Let the total number of deer be \( x \).
Number of deer grazing in the field = \( \frac{x}{2} \)
Remaining deer = \( x - \frac{x}{2} = \frac{x}{2} \)
Deer playing nearby = \( \frac{3}{4} \times (\text{Remaining no. of deer}) \)
\( = \frac{3}{4} \times \frac{x}{2} = \frac{3x}{8} \)
Drinking water = 9
According to the problem, the total number of deer is the sum of deer grazing, playing, and drinking water:
\( \frac{x}{2} + \frac{3x}{8} + 9 = x \)
Transposing 9 to RHS and \( x \) to LHS, we have
\( \frac{x}{2} + \frac{3x}{8} - x = -9 \)
Finding the LCM of 2 and 8, which is 8:
\( \frac{4x + 3x - 8x}{8} = -9 \)
\( \frac{-x}{8} = -9 \)
Multiplying by 8 on both sides, we have
\( \frac{-x}{8} \times 8 = -9 \times 8 \)
\( -x = -72 \)
\( x = 72 \)
Thus, the total number of deer in the herd is 72.
In simple words: Imagine a group of deer. Half are eating grass. Three-quarters of the deer left over are playing. The last 9 deer are drinking water. In total, there are 72 deer in the herd.

Exam Tip: Break down the problem into parts. Represent each fraction of deer with respect to the total or remaining number, then sum them up to form the equation.

Question 9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Answer: Let the present age of the granddaughter be \( x \) years.
The present age of the grandfather is \( 10x \) years.
According to the second condition, the grandfather is 54 years older than his granddaughter:
Present age of grandfather = Present age of granddaughter + 54
\( 10x = x + 54 \)
Transposing \( x \) to LHS, we have
\( 10x - x = 54 \)
\( 9x = 54 \)
Dividing both sides by 9, we have
\( x = \frac{54}{9} \)
\( x = 6 \)
So, the present age of the granddaughter is 6 years.
The present age of the grandfather is \( 10x = 10 \times 6 = 60 \) years.
In simple words: A grandfather is ten times his granddaughter's age and also 54 years older than her. The granddaughter is 6 years old, and the grandfather is 60 years old.

Exam Tip: When given multiple relationships between ages, use one relationship to set up the basic variables and the other to form the equation.

Question 10. Atnan's age is three times his son's age. Ten years ago he was five times his son's age. Find their present ages.
Answer: Let the present age of the son be \( x \) years.
The present age of Aman is \( 3x \) years.
Ten years ago:
Son's age = \( (x - 10) \) years
Aman's age = \( (3x - 10) \) years
According to the condition from ten years ago, Aman's age was five times his son's age:
\( 5 \times (\text{Son's age 10 years ago}) = (\text{Aman's age 10 years ago}) \)
\( 5(x - 10) = 3x - 10 \)
\( 5x - 50 = 3x - 10 \)
Transposing -50 to RHS and 3x to LHS, we have
\( 5x - 3x = -10 + 50 \)
\( 2x = 40 \)
Dividing both sides by 2:
\( x = \frac{40}{2} \)
\( x = 20 \)
So, the son's present age = 20 years.
Aman's present age = \( 3 \times 20 = 60 \) years.
In simple words: Right now, Aman is three times as old as his son. But ten years ago, Aman was five times his son's age. The son is currently 20 years old, and Aman is 60 years old.

Exam Tip: Be sure to correctly express the ages of both individuals at different points in time (e.g., "ten years ago") before establishing the equation.