Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 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?
Solution:
Let the required number be x.
By subtracting \(\frac {1}{2}\) from x, we get x-\(\frac {1}{2}\) and by multiplying this result by \(\frac {1}{2}\),
we get \(\frac {1}{2}\)(x – \(\frac {1}{2}\))
But, the result is \(\frac {1}{8}\)
\(\frac {1}{2}\)(x – \(\frac {1}{2}\)) = \(\frac {1}{8}\)
∴ \(\frac {1}{2}\)(x – \(\frac {1}{2}\)) × 2 = \(\frac {1}{8}\) × 2 (Multiplying both the sides by 2)
∴ x – \(\frac {1}{2}\) = \(\frac {1}{4}\)
∴ x = \(\frac{1}{4}+\frac{1}{2}\) (Transposing –\(\frac {1}{2}\) to RHS)
∴ x = \(\frac{1+2}{4}\) (LCM = 4)
∴ x = \(\frac {3}{4}\)
Thus, the required number = \(\frac {3}{4}\)
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?
Solution:
Perimeter of the pool = 154 m
Let breadth = x metres
Length is 2 m more than twice its breadth.
Length = 2 (breadth) + 2
= (2x + 2) metres
Perimeter of a rectangle = 2 (length + breadth)
2 (length + breadth) = Perimeter
∴ 2[(2x + 2) + x] = 154
∴ 2 [2x + 2 + x] = 154
∴ 2 (3x + 2) = 154
∴ \(\frac{2(3 x+2)}{2}=\frac{154}{2}\) (Dividing both the sides by 2)
∴ 3x + 2 = 77
∴ 3x = 77 – 2 (Transposing 2 to RHS)
∴ 3x = 75
∴ \(\frac{3 x}{3}=\frac{75}{3}\) (Dividing both the sides by 3) x — 25
Breadth = 25 m
Length = 2x + 2
= 2 (25) + 2
= 50 + 2
= 52 m
Thus, the length of the pool is 52 m and its breadth is 25 m.
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?
Solution:
Base of an isosceles triangle = \(\frac {4}{3}\) cm
Let the length of each of the equal sides = x cm
Perimeter of the triangle = \(\frac {4}{3}\) + x + x
= \(\frac {4}{3}\) + 2x
Perimeter of the triangle = 4\(\frac {2}{15}\) cm (Given)
∴ \(\frac {4}{3}\) + 2x = 4\(\frac {2}{15}\)
∴ \(\frac {4}{3}\) + 2x = \(\frac {62}{15}\)
∴ 2x = \(\frac{62}{15}-\frac{4}{3}\)(Transposing to RHS)
∴ 2x = \(\frac{62-20}{15}\) (LCM = 15)
∴ 2x = \(\frac {42}{15}\)
∴ \(\frac{2 x}{2}=\frac{42}{15} \times \frac{1}{2}\) (Dividing both the sides by 2)
∴ x = \(\frac {21}{15}\)
∴ x = \(\frac{7 \times 3}{5 \times 3}\)
∴ x = \(\frac {7}{5}\)
∴ x = 1\(\frac {2}{5}\)
Thus, the required length of either of the remaining equal sides is 1\(\frac {2}{5}\) cm.
Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number be x
∴ The greater number = x + 15
Their sum is 95.
∴ x + (x + 15) = 95
∴ x + x + 15 = 95
∴ 2x + 15 = 95
∴ 2x = 95 – 15 (Transposing 15 to RHS)
∴ 2x = 80
∴ \(\frac{2 x}{2}=\frac{80}{2}\) (Dividing both the sides by 2)
∴ x = 40
The smaller number = x = 40
The greater number = x + 15 = 40 + 15 = 55
Thus, 40 and 55 are the required numbers.
Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers ?
Solution:
Ratio of the two numbers = 5 : 3
Let the two numbers be 5x and 3x.
Difference = 18
∴ 5x – 3x = 18
∴ 2x = 18
∴ \(\frac{2 x}{2}=\frac{18}{2}\) (Dividing both the sides by 2)
∴ x = 9
∴ Greater number = 5x = 5 × 9 = 45
Smaller number = 3x = 3 × 9 = 27
Thus, 45 and 27 are the numbers.
Question 6.
Three consecutive integers add up to 51. What are these integers ?
Solution:
Let the consecutive integers be, x, x + 1 and x + 2.
Their sum is 51.
∴ x + (x + 1) + (x + 2) = 51
∴ x + x + 1 + x + 2 = 51
∴ 3x + 3 = 51
∴ 3x = 51 – 3 (Transposing 3 to RHS)
∴ 3x = 48
∴ \(\frac{3 x}{3}=\frac{48}{3}\) (Dividing both the sides by 3)
∴ x = 16
∴ First number = x = 16
Second number = x + 1 = 16 + 1 = 17
Third number = x + 2 = 16 + 2 = 18
Thus, the required consecutive integers are 16, 17 and 18.
Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three multiples of 8 be x, x + 8, and x + 8 + 8 = x + 16.
Their sum is 888.
∴ x + (x + 8) + (x + 16) = 888
∴ x + x + 8 + x + 16 = 888
∴ 3x + 24 = 888
∴ 3x = 888 – 24 (Transposing 24 to RHS)
∴ 3x = 864
∴ \(\frac{3 x}{3}=\frac{864}{3}\) (Dividing both the sides by 3)
∴ x = 288
∴ First number = x = 288
∴ Second number = x + 8 = 288 + 8 = 296
Third number = x + 16 = 288 + 16 = 304
Thus, the required three consecutives multiples of 8 are 288, 296 and 304.
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.
Solution:
Let the three consecutive integers be, x, (x + 1) and (x + 2).
According to the condition,
∴ 2 × (x) + 3 × (x + 1) + 4 × (x + 2) = 74
∴ 2x + 3x + 3 + 4x + 8 = 74
∴ 9x + 11 = 74
∴ 9x = 74 – 11 (Transposing 11 to RHS)
∴ 9x = 63
∴ \(\frac{9 x}{9}=\frac{63}{9}\) (Dividing both the sides by 9)
∴ x = 7
∴ First integer = x – 7
Second integer = x + 1 = 7 + 1 = 8
Third integer = x + 2 = 7 + 2 = 9
Thus, the required integers are 7, 8 and 9.
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?
Solution:
Ages of Rahul and Haroon are in the ratio of 5 : 7.
Let their present ages be 5x and 7x years.
∴ 4 years later,
the age of Rahul will be 5x + 4 years and the age of Haroon will be 7x + 4 years.
According to the condition,
(5x + 4) + (7x + 4) = 56
∴ 5x + 4 + 7x + 4 = 56
∴ 12x + 8 = 56
∴ 12x = 56 – 8 (Transposing 8 to RHS)
∴ 12x = 48
∴ \(\frac{12 x}{12}=\frac{48}{12}\) (dividing both the sides by 12)
∴ x = 4
Present age of Rahul = 5x = 5 × 4
= 20 years
Present age of Haroon = 7x = 7 × 4
= 28 years
Thus, present age of Rahul is 20 years and that of Haroon is 28 years.
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?
Solution:
Number of boys : Number of girls = 7 : 5
Let the number of boys be 7x, and the number of girls be 5x.
According to the condition
7x = 5x + 8
∴ 7x – 5x = 8 (Transposing 5x to LHS)
∴ 2x = 8
∴ \(\frac{2 x}{2}=\frac{8}{2}\) (Dividing both the sides by 2)
∴ x = 4
Number of boys = 7x = 7 × 4 = 28
Number of girls = 5x = 5 × 4 = 20
Total class strength = 28 + 20 = 48
Thus, total class strength is 48.
Question 11.
Bharat’s father is 26 years younger than Bharat’s grandfather and 29 years older than Bharat. The sum of the ages of all the three is 135 years. What is the age of each one of them ?
Solution:
Let the age of Bharat be x years,
His father’s age = (x + 29) years
His grandfather’s age = x + 29 + 26
= (x + 55) years
Sum of their ages is 135 years.
∴ x + (x + 29) + (x + 55) = 135
∴ x + x + 29 + x + 55 = 135
∴ 3x + 84 = 135
∴ 3x = 135 – 84 (Transposing 84 to RHS)
∴ 3x = 51
∴ \(\frac{3 x}{3}=\frac{51}{3}\) (Dividing both the sides by 3)
∴ x = 17
Bharat’s age = x = 17 years
His father’s age = x + 29
= 17 + 29
= 46 years
His grandfather’s age = x + 55
= 17 + 55
= 72 years
Thus, Bharat’s age is 17 years, his father’s age is 46 years and his grandfather’s age is 72 years.
Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age be x years.
4 times his present age be 4x years.
15 years from now his age be x + 15 years.
According to the condition, x + 15 = 4x
∴ 4x = x + 15 (Interchanging both the sides)
∴ 4x – x = 15 (Transposing x to LHS)
∴ 3x = 15
∴ \(\frac{3 x}{3}=\frac{15}{3}\) (Dividing both the sides by 3)
∴ x = 5
Thus, Ravi’s present age is 5 years.
Question 13.
A rational number is such that when you multiply it by \(\frac {5}{2}\) an\(\frac {2}{3}\) add g to the product, you get –\(\frac {7}{12}\). What is the number ?
Solution:
Let the required rational number be x.
We get \(\frac{5 x}{2}\) by multiplying x with \(\frac {5}{2}\)
By adding \(\frac {2}{3}\) to it we get \(\frac{5 x}{2}+\frac{2}{3}\)
But, the result is \(\frac {-7}{12}\)
∴ \(\frac{5 x}{2}+\frac{2}{3}=\frac{-7}{12}\)
\(\frac{5 x}{2}=-\frac{7}{12}-\frac{2}{3}\) (Transposing \(\frac {2}{3}\) to RHS)
∴ \(\frac{5 x}{2}=\frac{-7-8}{12}\) (LCM = 12)
∴ \(\frac{5 x}{2}=\frac{-15}{12}\)
∴ \(\frac{5 x}{2} \times \frac{2}{5}=\frac{-15}{12} \times \frac{2}{5}\) (Multiplying both the sides by \(\frac {2}{5}\))
∴ x = –\(\frac {1}{2}\)
Thus, the required rational number is –\(\frac {1}{2}\).
Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of
₹ 100 notes be 2x,
₹ 50 notes be 3x,
₹ 10 notes be 5x.
Value of ₹ 100 notes = 2x × 100 = ₹ 200x
Value of ₹ 50 notes = 3x × 50 = ₹ 150x
Value of ₹ 10 notes = 5x × 10 = ₹ 50x
According to the condition, value of
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00,000
∴ 200x + 150x + 50x = 4,00,000
∴ 400x = 400000
∴ \(\frac{400 x}{400}=\frac{400000}{400}\)
∴ x = 1000
Number of ₹ 100 notes = 2x
= 2 × 1000
= 2000
Number of ₹ 50 notes = 3x
= 3 × 1000
= 3000
Number of ₹ 10 notes = 5x
= 5 × 1000
– 5000
Thus, Lakshmi has 2000 notes of ₹ 100, 3000 notes of ₹ 50 and 5000 notes of ₹ 10.
Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x.
Then, the number of ₹ 2 coins = 3x
Total number of coins = 160
Number of ₹ 1 coins = 160 – 3x – x
= 160 – 4x
Now, value of
₹ 5 coins = ₹ 5 × x = ₹ 5x
₹ 2 coins = ₹ 2 × 3x = ₹ 6x
₹ 1 coins = ₹ 1 × (160 – 4x)
= ₹ (160 – 4x)
According to the condition,
5x + 6x + (160 – 4x) = 300
∴ 5x + 6x + 160 – 4x = 300
∴ 11x – 4x + 160 = 300
∴ 7x + 160 = 300
∴ 7x = 300 – 160 (Transposing 160 to RHS)
7x = 140
∴ \(\frac{7 x}{7}=\frac{140}{7}\) (Dividing both the sides by 7)
∴ x = 20
Number of
₹ 5 coins = x = 20
₹ 2 coins = 3x = 3 × 20 = 60
₹ 1 coins = 160 – 4x
= 160 – 4 × 20
= 160 – 80
= 80
Thus, I have 20 coins of ₹ 5, 60 coins of ₹ 2 and 80 coins of ₹ 1.
Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be x
∴ Number of participants who are not winners = (63 – x)
Prize money given to winners = x × ₹ 100 = ₹ 100x
Prize money given to non-winner
participants = ₹ 25 × (63 -x)
= ₹ 25 × 63 – ₹ 25x
= ₹ 1575 – ₹ 25x
According to the condition
100x + 1575 – 25x = 3000
∴ 75x + 1575 = 3000
∴ 75x = 3000 – 1575 (Transposing 1575 to RHS)
∴ 75x = 1425
\(\frac{75 x}{75}=\frac{1425}{75}\) (Dividing both the sides by 75)
∴ x = 19
Thus, the number of winners are 19.