GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.2

Question 1. If you subtract \( \frac { 1 }{ 2 } \) from a number and multiply the result by \( \frac { 1 }{ 2 } \) you get \( \frac { 1 }{ 8 } \). What is the number?
Answer: Let the required number be \( x \).
According to the given condition, we have:
\( (x - \frac { 1 }{ 2 }) \times \frac { 1 }{ 2 } = \frac { 1 }{ 8 } \)
Multiply both sides by 2:
\( x - \frac { 1 }{ 2 } = \frac { 1 }{ 8 } \times 2 \)
\( x - \frac { 1 }{ 2 } = \frac { 1 }{ 4 } \)
Add \( \frac { 1 }{ 2 } \) to both sides:
\( x - \frac { 1 }{ 2 } + \frac { 1 }{ 2 } = \frac { 1 }{ 4 } + \frac { 1 }{ 2 } \)
\( x = \frac { 1 + 2 }{ 4 } \)
\( x = \frac { 3 }{ 4 } \)
Therefore, the number needed is \( \frac { 3 }{ 4 } \).
In simple words: We assumed the unknown number as \( x \). Then, we set up an equation based on the problem's description. By performing inverse operations like multiplying and adding on both sides, we found the value of \( x \), which is the number we were looking for.

Exam Tip: Always define your unknown variable clearly at the beginning. Remember to perform the same operation on both sides of the equation to maintain balance.

Question 2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Answer: The perimeter of the rectangular pool is 154 m.
Let the breadth of the pool be \( x \) meters.
So, the length will be \( 2(\text{breadth}) + 2 \) meters, which is \( (2x + 2) \) meters.
The formula for the perimeter of a rectangle is \( 2 \times (\text{length} + \text{breadth}) \).
Using the given information:
\( 2[(2x + 2) + x] = 154 \)
Divide both sides by 2:
\( (2x + 2) + x = \frac { 154 }{ 2 } \)
\( 3x + 2 = 77 \)
Move 2 to the right side (transpose 2 to RHS):
\( 3x = 77 - 2 \)
\( 3x = 75 \)
Divide both sides by 3:
\( x = \frac { 75 }{ 3 } \)
\( x = 25 \)
Therefore, the breadth of the pool is 25 m.
The length is \( 2(25) + 2 = 50 + 2 = 52 \) m.
In simple words: We set the breadth as \( x \) and the length as \( 2x + 2 \). Using the perimeter formula, we made an equation. Solving this equation gave us the breadth, and from that, we calculated the length of the pool.

Exam Tip: Always write down the formulas for geometric shapes. Clearly define your variables before setting up the equation.

Question 3. The base of an isosceles triangle is \( \frac { 4 }{ 3 } \) cm. The perimeter of the triangle is \( 4\frac { 2 }{ 15 } \) cm. What is the length of either of the remaining equal sides?
Answer: The base of the isosceles triangle is \( \frac { 4 }{ 3 } \) cm.
Let the length of either of the equal sides be \( x \) cm.
The perimeter of a triangle is the sum of its three sides.
So, Perimeter \( = \frac { 4 }{ 3 } + x + x = \frac { 4 }{ 3 } + 2x \)
We are given that the perimeter of the triangle is \( 4\frac { 2 }{ 15 } \) cm.
Convert the mixed fraction to an improper fraction: \( 4\frac { 2 }{ 15 } = \frac { (4 \times 15) + 2 }{ 15 } = \frac { 60 + 2 }{ 15 } = \frac { 62 }{ 15 } \) cm.
Now, set up the equation:
\( \frac { 4 }{ 3 } + 2x = \frac { 62 }{ 15 } \)
Move \( \frac { 4 }{ 3 } \) to the right side (transpose \( \frac { 4 }{ 3 } \) to RHS):
\( 2x = \frac { 62 }{ 15 } - \frac { 4 }{ 3 } \)
To subtract the fractions, find a common denominator, which is 15.
\( 2x = \frac { 62 }{ 15 } - \frac { 4 \times 5 }{ 3 \times 5 } \)
\( 2x = \frac { 62 }{ 15 } - \frac { 20 }{ 15 } \)
\( 2x = \frac { 62 - 20 }{ 15 } \)
\( 2x = \frac { 42 }{ 15 } \)
Divide both sides by 2:
\( x = \frac { 42 }{ 15 } \times \frac { 1 }{ 2 } \)
\( x = \frac { 21 }{ 15 } \)
Simplify the fraction by dividing both numerator and denominator by 3:
\( x = \frac { 7 }{ 5 } \)
Convert the improper fraction to a mixed fraction:
\( x = 1\frac { 2 }{ 5 } \) cm.
Therefore, the length of each of the equal sides is \( 1\frac { 2 }{ 5 } \) cm.
In simple words: We used the triangle's perimeter and the length of its base to figure out the length of its two equal sides. We set up an equation, then solved it by moving terms and finding a common bottom number for the fractions.

Exam Tip: Remember that an isosceles triangle has two equal sides. Always convert mixed fractions to improper fractions before performing addition or subtraction in equations.

Question 4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer: Let the smaller number be \( x \).
Then the other number is \( x + 15 \).
According to the given condition, the sum of the two numbers is 95:
\( x + (x + 15) = 95 \)
Combine like terms:
\( 2x + 15 = 95 \)
Subtract 15 from both sides (transpose 15 to RHS):
\( 2x = 95 - 15 \)
\( 2x = 80 \)
Divide both sides by 2:
\( x = \frac { 80 }{ 2 } \)
\( x = 40 \)
So, the smaller number is 40.
The other number is \( 40 + 15 = 55 \).
The two numbers are 40 and 55.
In simple words: We called the smaller number \( x \). Since the other number is 15 more, we called it \( x + 15 \). We then added them together to equal 95 and solved the equation to find both numbers.

Exam Tip: When dealing with "one exceeds the other by X," represent the numbers as \( x \) and \( x+X \). Always check your answer by summing the two numbers to ensure it matches the given total.

Question 5. Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Answer: Let the two numbers be \( 5x \) and \( 3x \), as they are in the ratio 5:3.
According to the condition, they differ by 18, which means their difference is 18:
\( 5x - 3x = 18 \)
Combine like terms:
\( 2x = 18 \)
Divide both sides by 2:
\( x = \frac { 18 }{ 2 } \)
\( x = 9 \)
Now, find the two numbers using the value of \( x \):
First number: \( 5x = 5 \times 9 = 45 \)
Second number: \( 3x = 3 \times 9 = 27 \)
The required numbers are 45 and 27.
In simple words: We represented the numbers as \( 5x \) and \( 3x \) because of their ratio. Since their difference is 18, we wrote an equation \( 5x - 3x = 18 \). Solving for \( x \) allowed us to find the actual values of both numbers.

Exam Tip: When numbers are in a ratio, represent them as multiples of a variable (e.g., 5x and 3x). This helps in setting up and solving the equation correctly.

Question 6. Three consecutive integers add up to 51. What are these integers?
Answer: Let the three consecutive integers be \( x \), \( x + 1 \) and \( x + 2 \).
According to the condition, their sum is 51:
\( x + (x + 1) + (x + 2) = 51 \)
Combine like terms:
\( x + x + 1 + x + 2 = 51 \)
\( 3x + 3 = 51 \)
Move 3 to the right side (transpose 3 to RHS):
\( 3x = 51 - 3 \)
\( 3x = 48 \)
Divide both sides by 3:
\( x = \frac { 48 }{ 3 } \)
\( x = 16 \)
Now, find the other two integers:
Second integer: \( x + 1 = 16 + 1 = 17 \)
Third integer: \( x + 2 = 16 + 2 = 18 \)
Thus, the required three consecutive numbers are 16, 17, and 18.
In simple words: We represented three numbers that follow each other as \( x \), \( x+1 \), and \( x+2 \). We added them up and set the sum to 51. After solving this equation, we found the first number, and then the next two.

Exam Tip: Consecutive integers always differ by 1. For problems involving sums, setting them up as \( x, x+1, x+2 \) simplifies the equation significantly.

Question 7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer: Let the three consecutive multiples of 8 be \( x \), \( x + 8 \) and \( x + 16 \).
According to the condition, their sum is 888:
\( x + (x + 8) + (x + 16) = 888 \)
Combine like terms:
\( x + x + 8 + x + 16 = 888 \)
\( 3x + 24 = 888 \)
Move 24 to the right side (transpose 24 to RHS):
\( 3x = 888 - 24 \)
\( 3x = 864 \)
Divide both sides by 3:
\( x = \frac { 864 }{ 3 } \)
\( x = 288 \)
Now, find the other two multiples:
Second multiple: \( x + 8 = 288 + 8 = 296 \)
Third multiple: \( x + 16 = 288 + 16 = 304 \)
Thus, the required multiples of 8 are 288, 296, and 304.
In simple words: We used \( x \), \( x+8 \), and \( x+16 \) to show three multiples of 8 that come one after another. We added them up to 888 and then worked through the equation to find the value of \( x \), which helped us find all three multiples.

Exam Tip: Consecutive multiples of a number \( N \) are represented as \( x, x+N, x+2N \), etc. Ensure your multiples are consistently increasing by the given number.

Question 8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Answer: Let the three consecutive integers be \( x \), \( (x + 1) \) and \( (x + 2) \).
According to the condition, when multiplied by 2, 3, and 4 respectively, their sum is 74:
\( 2x + 3(x + 1) + 4(x + 2) = 74 \)
Distribute the numbers:
\( 2x + 3x + 3 + 4x + 8 = 74 \)
Combine like terms:
\( (2x + 3x + 4x) + (3 + 8) = 74 \)
\( 9x + 11 = 74 \)
Move 11 to the right side (transpose 11 to RHS):
\( 9x = 74 - 11 \)
\( 9x = 63 \)
Divide both sides by 9:
\( x = \frac { 63 }{ 9 } \)
\( x = 7 \)
Now, find the other two integers:
Second integer: \( x + 1 = 7 + 1 = 8 \)
Third integer: \( x + 2 = 7 + 2 = 9 \)
The required integers are 7, 8, and 9.
In simple words: We picked three integers that follow each other: \( x \), \( x+1 \), and \( x+2 \). We then multiplied each by 2, 3, and 4 in that order. When we added these new numbers together, they equaled 74. We solved this equation to find the values of all three integers.

Exam Tip: Remember to apply the distributive property correctly when multiplying a number by an expression in parentheses, such as \( 3(x+1) \).

Question 9. The ages Of Rahul and Haroon are in the ratio 5: 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Answer: Since the ages of Rahul and Haroon are in the ratio 5:7,
Let their present ages be \( 5x \) years and \( 7x \) years, respectively.
Four years later:
Rahul's age will be \( (5x + 4) \) years.
Haroon's age will be \( (7x + 4) \) years.
According to the condition, the sum of their ages after four years will be 56 years:
\( (5x + 4) + (7x + 4) = 56 \)
Combine like terms:
\( 5x + 4 + 7x + 4 = 56 \)
\( 12x + 8 = 56 \)
Move 8 to the right side (transpose 8 to RHS):
\( 12x = 56 - 8 \)
\( 12x = 48 \)
Divide both sides by 12:
\( x = \frac { 48 }{ 12 } \)
\( x = 4 \)
Now, calculate their present ages:
Rahul's present age: \( 5x = 5 \times 4 = 20 \) years.
Haroon's present age: \( 7x = 7 \times 4 = 28 \) years.
In simple words: We set Rahul's age as \( 5x \) and Haroon's as \( 7x \). We then added 4 years to each age and made an equation where their new total age was 56. Solving this equation for \( x \) helped us find their current ages.

Exam Tip: For ratio problems, use a common multiplier \( x \). When dealing with future ages, remember to add the number of years to each person's current age before summing them up.

Question 10. The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Answer: The ratio of the number of boys to girls is 7:5.
Let the number of boys be \( 7x \) and the number of girls be \( 5x \).
According to the condition, the number of boys is 8 more than the number of girls:
\( 7x = 5x + 8 \)
Move \( 5x \) to the left side (transpose \( 5x \) to LHS):
\( 7x - 5x = 8 \)
\( 2x = 8 \)
Divide both sides by 2:
\( x = \frac { 8 }{ 2 } \)
\( x = 4 \)
Now, calculate the number of boys and girls:
Number of boys: \( 7x = 7 \times 4 = 28 \)
Number of girls: \( 5x = 5 \times 4 = 20 \)
Total class strength = Number of boys + Number of girls
Total class strength = \( 28 + 20 = 48 \) students.
In simple words: We used \( 7x \) for boys and \( 5x \) for girls based on their ratio. Since there are 8 more boys than girls, we set up the equation \( 7x = 5x + 8 \). Solving for \( x \) allowed us to find the actual number of boys and girls, and then we added them to get the total class size.

Exam Tip: Pay close attention to how "more than" is phrased in a word problem. It often indicates an addition or subtraction relationship between two quantities.

Question 11. Baichung's father is 26 years younger than Baichung's grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each?
Answer: Let Baichung's age be \( x \) years.
Baichung's father is 29 years older than Baichung, so his father's age is \( (x + 29) \) years.
Baichung's father is 26 years younger than Baichung's grandfather.
This means Baichung's grandfather is 26 years older than Baichung's father.
So, Baichung's grandfather's age is \( (x + 29 + 26) \) years, which simplifies to \( (x + 55) \) years.
According to the condition, the sum of their ages is 135 years:
\( x + (x + 29) + (x + 55) = 135 \)
Combine like terms:
\( x + x + 29 + x + 55 = 135 \)
\( 3x + 84 = 135 \)
Move 84 to the right side (transpose 84 to RHS):
\( 3x = 135 - 84 \)
\( 3x = 51 \)
Divide both sides by 3:
\( x = \frac { 51 }{ 3 } \)
\( x = 17 \)
Now, calculate each person's age:
Baichung's age = 17 years.
Baichung's father's age = \( 17 + 29 = 46 \) years.
Baichung's grandfather's age = \( 46 + 26 = 72 \) years.
In simple words: We started by calling Baichung's age \( x \). Then we figured out his father's age and his grandfather's age based on the given information, all in terms of \( x \). We added all three ages together, made an equation that equals 135, and solved it to find each person's exact age.

Exam Tip: When dealing with multiple related ages, define one age as \( x \) and express all other ages in terms of \( x \). Be careful with "younger than" and "older than" relationships.

Question 12. Fifteen years from now Ravi's age will be four times his present age. What is Ravi s present age?
Answer: Let Ravi's present age be \( x \) years.
Four times his present age would be \( 4x \) years.
Fifteen years from now, Ravi's age will be \( (x + 15) \) years.
According to the condition, fifteen years from now, his age will be four times his present age:
\( x + 15 = 4x \)
Move \( x \) to the right side (transpose \( x \) to RHS):
\( 15 = 4x - x \)
\( 15 = 3x \)
Divide both sides by 3:
\( x = \frac { 15 }{ 3 } \)
\( x = 5 \)
Therefore, Ravi's present age is 5 years.
In simple words: We said Ravi's age now is \( x \). In 15 years, he'll be \( x+15 \). The problem says this future age is four times his current age, so \( x+15 = 4x \). We solved this equation to find Ravi's age today.

Exam Tip: Be sure to distinguish between present age and future age. "From now" indicates addition, and "times" indicates multiplication.

Question 13. A rational number is such that when you multiply it by \( \frac {5}{ 2 } \) and add \( \frac { 2 }{ 3 } \) to the product, you get \( \frac { 7 }{ 12 } \). What is the number?
Answer: Let the required rational number be \( x \).
According to the condition, when you multiply it by \( \frac { 5 }{ 2 } \) and add \( \frac { 2 }{ 3 } \) to the product, you get \( -\frac { 7 }{ 12 } \).
So, we have the equation:
\( x \times \frac { 5 }{ 2 } + \frac { 2 }{ 3 } = -\frac { 7 }{ 12 } \)
\( \frac { 5x }{ 2 } + \frac { 2 }{ 3 } = -\frac { 7 }{ 12 } \)
Move \( \frac { 2 }{ 3 } \) to the right side (transpose \( \frac { 2 }{ 3 } \) to RHS):
\( \frac { 5x }{ 2 } = -\frac { 7 }{ 12 } - \frac { 2 }{ 3 } \)
To subtract the fractions, find a common denominator, which is 12.
\( \frac { 5x }{ 2 } = -\frac { 7 }{ 12 } - \frac { 2 \times 4 }{ 3 \times 4 } \)
\( \frac { 5x }{ 2 } = -\frac { 7 }{ 12 } - \frac { 8 }{ 12 } \)
\( \frac { 5x }{ 2 } = \frac { -7 - 8 }{ 12 } \)
\( \frac { 5x }{ 2 } = \frac { -15 }{ 12 } \)
Multiply both sides by \( \frac { 2 }{ 5 } \) to isolate \( x \):
\( x = \frac { -15 }{ 12 } \times \frac { 2 }{ 5 } \)
\( x = \frac { -3 \times 5 \times 2 }{ 12 \times 5 } \)
\( x = \frac { -3 \times 2 }{ 12 } \)
\( x = \frac { -6 }{ 12 } \)
Simplify the fraction:
\( x = -\frac { 1 }{ 2 } \)
The required rational number is \( -\frac { 1 }{ 2 } \).
In simple words: We called the unknown number \( x \). We followed the problem's steps to build an equation: \( \frac{5x}{2} + \frac{2}{3} = -\frac{7}{12} \). We then solved this equation by moving terms and combining fractions until we found the value of \( x \).

Exam Tip: Be careful with signs, especially when transposing terms or multiplying/dividing negative numbers. Always find a common denominator before adding or subtracting fractions.

Question 14. Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3: 5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?
Answer: Let the number of notes be:
Number of Rs 100 notes \( = 2x \)
Number of Rs 50 notes \( = 3x \)
Number of Rs 10 notes \( = 5x \)
Calculate the total value from each denomination:
Value from Rs 100 notes \( = 2x \times \text{Rs } 100 = \text{Rs } 200x \)
Value from Rs 50 notes \( = 3x \times \text{Rs } 50 = \text{Rs } 150x \)
Value from Rs 10 notes \( = 5x \times \text{Rs } 10 = \text{Rs } 50x \)
According to the condition, the total cash is Rs 4,00,000:
\( 200x + 150x + 50x = 400000 \)
Combine like terms:
\( 400x = 400000 \)
Divide both sides by 400:
\( x = \frac { 400000 }{ 400 } \)
\( x = 1000 \)
Now, find the number of notes of each denomination:
Number of Rs 100 notes \( = 2 \times 1000 = 2000 \)
Number of Rs 50 notes \( = 3 \times 1000 = 3000 \)
Number of Rs 10 notes \( = 5 \times 1000 = 5000 \)
Thus, Lakshmi has 2000 Rs 100 notes, 3000 Rs 50 notes, and 5000 Rs 10 notes.
In simple words: We used \( 2x \), \( 3x \), and \( 5x \) for the number of notes for each value. We then calculated the total money from each type of note. Adding these values together gave us the total cash, Rs 4,00,000. Solving this equation gave us \( x \), which helped us find out exactly how many notes of each kind Lakshmi has.

Exam Tip: When dealing with ratios of quantities and their total value, multiply each part of the ratio by \( x \) and then multiply by the respective value to form the total value equation.

Question 15. I have a total of Rs 300 in coins of denomination Rs 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer: Let the number of Rs 5 coins be \( x \).
The number of Rs 2 coins is 3 times the number of Rs 5 coins, so the number of Rs 2 coins \( = 3x \).
The total number of coins is 160.
The number of Rs 1 coins \( = \text{Total coins} - (\text{Number of Rs 2 coins} + \text{Number of Rs 5 coins}) \)
Number of Rs 1 coins \( = 160 - (3x + x) = 160 - 4x \).
Now, calculate the total value from each type of coin:
Value from Rs 5 coins \( = 5 \times x = \text{Rs } 5x \)
Value from Rs 2 coins \( = 2 \times 3x = \text{Rs } 6x \)
Value from Rs 1 coins \( = 1 \times (160 - 4x) = \text{Rs } (160 - 4x) \)
According to the condition, the total value of all coins is Rs 300:
\( 5x + 6x + (160 - 4x) = 300 \)
Combine like terms:
\( 5x + 6x + 160 - 4x = 300 \)
\( (5x + 6x - 4x) + 160 = 300 \)
\( 7x + 160 = 300 \)
Move 160 to the right side (transpose 160 to RHS):
\( 7x = 300 - 160 \)
\( 7x = 140 \)
Divide both sides by 7:
\( x = \frac { 140 }{ 7 } \)
\( x = 20 \)
Now, find the number of coins of each denomination:
Number of Rs 5 coins = 20
Number of Rs 2 coins \( = 3 \times 20 = 60 \)
Number of Rs 1 coins \( = 160 - (4 \times 20) = 160 - 80 = 80 \)
In simple words: We started by setting the number of Rs 5 coins as \( x \). Then, we found the number of Rs 2 coins and Rs 1 coins in terms of \( x \). We figured out the total money from each type of coin. Adding all these money amounts gave us the total of Rs 300. Solving the equation helped us find how many of each coin there are.

Exam Tip: This problem involves two constraints: total number of coins and total value. Make sure your variables correctly represent both these relationships. Always clearly define each type of coin in terms of the variable \( x \).

Question 16. The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.
Answer: Let the number of winners be \( x \).
The total number of participants is 63.
So, the number of participants who are not winners \( = (63 - x) \).
Calculate the total prize money given to winners:
Prize money for winners \( = x \times \text{Rs } 100 = \text{Rs } 100x \).
Calculate the total prize money given to non-winners:
Prize money for non-winners \( = (63 - x) \times \text{Rs } 25 = \text{Rs } (63 \times 25 - 25x) = \text{Rs } (1575 - 25x) \).
According to the condition, the total prize money distributed is Rs 3,000:
\( 100x + (1575 - 25x) = 3000 \)
Combine like terms:
\( 100x + 1575 - 25x = 3000 \)
\( 75x + 1575 = 3000 \)
Move 1575 to the right side (transpose 1575 to RHS):
\( 75x = 3000 - 1575 \)
\( 75x = 1425 \)
Divide both sides by 75:
\( x = \frac { 1425 }{ 75 } \)
\( x = 19 \)
Thus, the number of winners is 19.
In simple words: We let \( x \) be the number of winners. This means \( 63-x \) participants did not win. We then added the prize money for all winners (Rs \( 100x \)) and all non-winners (Rs \( 25(63-x) \)), and set the total to Rs 3,000. Solving this equation showed us that there were 19 winners.

Exam Tip: Clearly define variables for each category of participants (winners and non-winners). Remember to multiply the number of participants in each category by their respective prize money before summing them to the total prize amount.