CHAPTER 2: LINEAR EQUATIONS IN ONE VARIABLE
Exercise 2.1
Solve the following equations.
1. x – 2 = 7
Solution: We have,
x – 2 = 7
⇒ x = 7 + 2 [Transposing – 2 to R.H.S.]
⇒ x = 9
2. y + 3 = 10
Solution: We have,
3. 6 = z + 2
Solution: We have,
4.
Solution: We have,
5.
Solution: We have,
6.
Solution: We have,
7.
Solution: We have,
8.
Solution: We have,
9.
Solution: We have,
10.
Solution: We have,
14y – 8 = 13
⇒ 14y = 13 + 8
11.
Solution: We have,
17 + 6p = 9
⇒ 6p = 9 – 17 [Transposing 17 to RHS]
12.
Solution: We have,
Exercise 2.2
1. If you subtract from a number and multiply the result by, you get. What is the number?
Solution: Let the number be x.
A/Q,
⸫The required number is
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: Let the breadth of the pool be x.
⸫ The length of the pool = 2x+2
A/Q,
2(2x + 2 + x) = 154
⇒ 2(3x + 2) = 154
⇒ 6x + 4 =154
⇒ 6x = 154 – 4 [Transposing 4 to RHS]
⇒ 6x =150
⇒ x = 25
2x + 2 = 2× 25 + 2 = 52
⸫The breadth of the pool=25m
The length of the pool=52m
3. The base of an isosceles triangle is cm. The perimeter of the triangle is cm. What is the length of either of the remaining equal sides?
Solution: Let length of each equal sides be x.
A/Q,
⸫The length of each equal side iscm.
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 larger number = x + 15
A/Q,
⇒ 2x = 80
⸫
⸫The numbers are 40 and 55.
5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Solution: Let the numbers be 5x and 3x.
A/Q,
5x – 3x = 18
⇒ 2x = 18
⸫
⸫The numbers are 45 and 27.
6. Three consecutive integers add up to 51. What are these integers?
Solution: Let the numbers be x, x +1 and x + 2.
A/Q,
⸫ The numbers are 16, 17 and 18
7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution: Let the numbers be x, x+8 and x+16
A/Q,
⸫The numbers are 288, 296 and 304
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 numbers be x, x + 1 and x+2.
A/Q,
⇒ 9x = 74 – 11 ⇒ 9x = 63
⸫The numbers are 7, 8 and 9
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: Let present ages of Rahul and Haroon be 5x and 7x respectively.
⸫Four years later Rahul’s age=5x+4
⸫Four years later Haroon’s age=7x+4
A/Q,
⸫ Present age of Rahul=20 years and
Present age of Haroon=28 years
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: Let the numbers of boys and girls be 7x and 5x respectively.
⸫ Total class strength=7x+5x=12x
A/Q,
7x = 5x + 8
⇒ 7x – 5x = 8
⇒ 2x = 8
⸫ Total class strength=48 students
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 one of them?
Solution: Let Baichung’s age be x.
⸫Baichung’s father’s age=x+29
⸫Baichung’s grandfather’s age =x+29+26 =x+55
A/Q,
x + (x + 29) + (x + 55) =135
⸫Baichung’s age=17 years
Baichung’s father’s age= 46 years
Baichung’s grandfather’s age=72 years
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.
Fifteen years from now Ravi’s age= x+15 years
A/Q,
Ravi’s present age=5 years
13. A rational number is such that when you multiply it by and add to the product, you get. What is the number?
Solution: Let the number be x.
A/Q,
⸫The number=
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 numbers of ₹100, ₹50 and ₹10 notes be 2x, 3x and 5x.
A/Q,
⸫The numbers of ₹100, ₹50 and ₹10 notes are 2000, 3000 and 5000 respectively.
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.
⸫The number of ₹2 coins=3x
⸫The number of ₹1 coins=160 – (x+3x)
= 160 – 4x
A/Q,
⇒ 160 – 4x + 6x + 5x = 300
⇒ 160 + 7x = 300
The number of ₹1 coins=80
The number of ₹2 coins=60
The number of ₹5 coins=20
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 ₹3,000. Find the number of winners, if the total number of participants is 63.
Solution: Let the number of winners be x.
⸫The number of participants does not win=63–x
A/Q,
⸫The number of winners=19
Exercise 2.3
1.
Solution:
2.
Solution:
5t – 3 = 3t – 5
⇒ 5t – 3t = – 5 + 3 ⇒ 2t = – 2
3.
Solution:
4.
Solution:
5.
Solution:
6.
Solution:
8x + 4 = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
7.
Solution:
8.
Solution:
9.
Solution:
10.
Solution:
Exercise 2.4
1. Amina thinks of a number and subtracts 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?
Solution: Let the number be x.
A/Q,
⸫The number is 4
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?
Solution: Let the smaller number be x.
The larger number=5x
A/Q,
⸫The numbers are 7 and 35
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?
Solution: Let the ones place digit be x.
⸫The tens place digit = 9 – x
⸫The number=10(9 – x) + x = 90 –10x + x = 90 – 9x
After interchanging the digits, the new number = 10x + (9 – x) = 10x + 9 – x = 9x + 9
A/Q,
The required number = 36
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?
Solution: Let the ones place digit be x.
⸫The tens place digit = 3x
⸫The number=10×3x+x=31x
After interchanging the digits, the new number=10x+3x=13x
A/Q,
⸫The number=62
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?
Solution: Let Shobo’s present age be x.
⸫ Shobo’s mother’s present age=6x
A/Q,
⸫ Shobo’s present age= 5 years
⸫ Shobo’s mother’s present age= 30years
6. There is a narrow 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 ₹100 per metre it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?
Solution: Let the length and breadth of the plot be 11x and 4x.
Perimeter of the fence=2(11x+4x) = 2×15x =30x
A/Q,
⸫The length and breadth of the plot are 275m and 100m respectively
7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 3 meters 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 ₹36,600. How much trouser material did he buy?
Solution: Let the length of shirt materials be 3x m and length of trouser material be 2x m respectively.
Cost price of shirt materials=50×3x=₹150x
Cost price of trouser materials=90×2x=₹180x
⸫Profit on shirt materials= 12% of 150x
⸫Profit on trouser materials = 10% of 180x
⸫SP of shirt materials=150x+18x=168x
⸫SP of trouser materials=180x+18x=198x
A/Q,
⸫ The length of trouser material=200 m
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 number of deer in the herd.
Solution: Let the number of deer in the herd be x.
⸫Number of deer grazing=
Number of deer playing=
Number of deer drinking = 9
A/Q,
⸫The number of deer=72
9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution: Let the present age of granddaughter be x years
⸫The present age of grandfather = 10x
A/Q,
The present age of granddaughter is 6 years and the present age of grandfather is 60 years
10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution: Let Aman’s son’s age be x
⸫ Aman’s age=3x
A/Q,
⸫ Aman’s age is 60 years and Aman’s son’s age is 20 years
Exercise 2.5
1.
Solution:
2.
Solution:
3.
Solution:
4.
Solution:
⇒ 5x – 25 = 3x – 9
⇒ 5x – 3x = – 9 + 25
5.
Solution:
6.
Solution:
Simplify and solve the following linear equations.
7.
Solution:
8.
Solution:
9.
Solution:
10.
Solution:
Exercise 2.6
Solve the following equations.
1.
Solution:
2.
Solution:
3.
Solution:
4.
Solution:
5.
Solution:
6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Solution: Let the ages of Hari and Harry be 5x and 7x respectively
A/Q,
⸫ The ages of Hari and Harry are 20 years and 28 years respectively.
7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is. Find the rational number.
Solution: Let the numerator be x
⸫The denominator = x + 8
⸫The rational number=
A/Q,
⸫The rational number=
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