Karnataka Class 8 Math Solution Chapter Linear Equations in one variable by Experience Math Teacher. Karnataka Board Class 8 Math Part 1 PDF Chapter 3 Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4, Exercise 3.5, Exercise 3.6 Solution.
Karnataka Class 8 Math Solution Chapter Linear Equations in one variable
EXERCISE 3.1
Solve the following equations.
1.) x – 2 = 7
ANS:
Here, given that x – 2 = 7
We transfer – 2 to right-hand side (RHS)
X = 7 + 2
X = 9
2.) y + 3 = 10
ANS:
Here, given that y + 3 = 10
We transfer 3 to right-hand side (RHS)
Y = 10 – 3
Y = 7
3.) 6 = z + 2
ANS:
Here, given that 6 = z + 2
We transfer 2 to Left Hand Side (LHS)
6 – 2 = Z
Z = 4
4.) 3/7 + x = 17/7
ANS:
Here, given that 3/7 + x = 17/7
We transfer 3/7 to right-hand side (RHS)
x = 17/7 -3/7
x = 14/7
x = 2
5.) 6x = 12
ANS:
Here, given that 6x = 12
We transfer 6 to right-hand side (RHS)
X = 12 / 6
X = 2
6.) t/5 = 10
ANS:
Here, given that t/5 = 10
We transfer 5 to right-hand side (RHS)
t = 5 x 10
t = 50
(2x)/3 = 18
ANS:
Here, given that (2x)/3 = 18
We transfer 3 to right-hand side (RHS)
2x = 18 x 3
2x = 54
We transfer 2 to right-hand side (RHS)
X = 54 / 2
X = 27
8.) 1.6 = y/1.5
ANS:
Here, given that 1.6 = y/1.5
We transfer 1.5 to Left Hand Side (LHS)
1.6 x 1.5 = y
2.4 = y
I.e. y = 2.4
9.) 7x – 9 = 16
ANS:
Here, given that 7x – 9 = 16
We transfer -9 to right-hand side (RHS)
7x = 16 + 9
7x = 25
We transfer 7 to right-hand side (RHS)
X = 25 / 7
10). 14v – 8 = 13
ANS:
Here, given that 14v – 8 = 13
We transfer -8 to right-hand side (RHS)
14v = 13 + 8
14v = 21
We transfer 14 to right-hand side (RHS)
V = 21 / 14
V = 3 / 2
11.) 17 + 6p = 9
ANS:
Here, given that 17 + 6p = 9
We transfer 17 to right-hand side (RHS)
6p = 9 – 17
6p = – 8
P = – 8 / 6
P = -4 / 3
12.) X/3 + 1 = 7/15
ANS:
Here, given that X/3 + 1 = 7/15
We transfer 1 to right-hand side (RHS)
X/3 = 7/15 – 1
X/3 = – 8 / 15
We transfer 3 to right-hand side (RHS)
X = – 8 / 15 x 3
X = – 8 / 5
EXERCISE 3.2
1.) If you subtract 1/2 from a number and multiply the result by 1/2 you get 1/8 what is the number?
ANS:
Let the number be x.
Then,
1/ 2 is subtracted from a number i.e. x – (1/2)
Then,
Result is multiplied by 1/2 i.e. 1/2 [x – (1/2)]
Answer is 1 / 8
I.e. 1/2 [x – (1/2)] = 1 / 8
Multiplying with 2 on both side
We get,
[x – (1/2)] = 1/ 4
We transfer – (1/2) to right-hand side (RHS)
X = 1/ 4 + 1/2
We doing denominator same.
X = 3 / 4
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?
ANS:
Given that,
i) Perimeter of a rectangular swimming pool is 154m.
ii) Its length is 2m more than twice its breadth.
We form a linear equation by using the formula for the perimeter of a rectangle.
Let,
The breadth of swimming pool be x m.
Therefore, the length of the swimming pool will be (2x + 2) m
We know,
Perimeter of rectangular swimming pool: 2(Length + Breadth)
Therefore 2 x (X + 2x + 2) = 154m.
(X + 2x + 2) = 154 / 2
3x + 2 = 77
3x = 77 – 2
3x = 75
We transfer 3 to right-hand side (RHS)
x = 75 / 3
X = 25 m.
But, x is the breadth of swimming pool.
The breadth of swimming pool = 25 m.
The length of the swimming pool will be (2x + 2) m
2 x 25 + 2= 52 m.
The length of the swimming pool = 52 m.
The breadth of swimming pool = 25 m.
3.) The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 (2/15) cm. What is the length of either of the remaining equal sides?
ANS:
Given that,
The base of an isosceles triangle is 4/3 cm.
The perimeter of the triangle is 4 (2/15) cm.
We know,
Perimeter of a Triangle = Sum of the Lengths of all Three Sides
Let,
The length of any one equal side be x cm
We form equation.
Perimeter of a Triangle = x + x + 4/3 = 4 (2/15)
2x + 4/3 = 4 (2/15)
2x + 4/3 = 62 / 15
We transfer 4 / 3 to right-hand side (RHS)
2x = 62 / 15 – 4/ 3
2x = 42 / 15
We transfer 2 to right-hand side (RHS)
X = 42 / 15 x 2
X = 42 / 30
X = 7 / 5 cm.
But x is length of any one equal side.
The length of any one equal side is 7 / 5 cm.
4.) Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
ANS:
Given that,
Some of two numbers is 95.
One exceeds the other by 15.
Let, one of the numbers be x.
Then the other number is 95 – x
If we assume x is greater.
x – (95 – x) = 15
x – 95 + x = 15
2x – 95 = 15
We transfer -95 to right-hand side (RHS)
2x = 15 + 95
2x = 110
We transfer 2 to right-hand side (RHS)
X = 110 / 2
X = 55
But x is one of the numbers
1st number is 55.
2nd number is 95 – x = 95 – 55 = 40
5.) Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
ANS:
Given that,
Two number are in ratio 5:3.
They differ by 18.
Let, the number are 5x and 3x.
The two numbers differ by 18
5x – 3x = 18
2x = 18
We transfer 2 to right-hand side (RHS)
X = 18 / 2
X = 9
The number are 5x and 3x
1st number is 5x = 5 x 9 = 45
2nd number is 3x = 3 x 9 = 27
6.) Three consecutive integers add up to 51. What are these integers?
ANS:
Given that,
Three consecutive integers add up to 51.
Let the first integer be x.
Then next consecutive two integers are (x + 1) and (x + 2)
X + (x + 1) + (x + 2) = 51
3x + 3 = 51
We transfer 3 to right-hand side (RHS)
3x = 51 -3
3x = 48
We transfer 3 to right-hand side (RHS)
X = 48 / 3
X = 16
But x is the first integer.
1st integer is 16.
2nd is 17
3rd is 18.
7.) The sum of three consecutive multiples of 8 is 888. Find the multiples.
ANS:
Given that,
The sum of three consecutive multiples of 8 is 888.
Let, the first multiple be x
Then second consecutive multiple of 8 = x + 8
Third consecutive multiple of 8 = x + 16
Sum of three consecutive multiples is 888.
X + (x + 8) + (x + 16) = 888
3x + 24 = 888
We transfer 24 to right-hand side (RHS)
3x = 888 – 24
3x = 864
We transfer 3 to right-hand side (RHS)
X = 864 / 3
X = 288
But x is first multiple.
1st multiple 288.
2nd is 296
3rd is 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.
ANS:
Let the first integer be x.
Then next consecutive two integers are (x + 1) and (x + 2)
They are taken in increasing order and multiplied by 2,3 and 4 respectively.
X x 2 = 2x
(x + 1) 3 = 3x + 3
(x + 2) 4 = 4x + 8
They add up to 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
We transfer 11 to right-hand side (RHS)
9x = 74 – 11
9x = 63
We transfer 9 to right-hand side (RHS)
X = 63/9
X = 7
But x is first integer.
1st integer is 7.
2nd is 8
3rd is 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?
ANS:
Given that,
Ages of Rahul and Haroon are in ratio 5:7
Four years later, sum of their ages will be 56 years.
Let,
Present ages of Rahul and Haroon are 5x and 7x respectively.
Four years later, sum of their ages will be 56 years.
Four years later, age of Rahul = 5x + 4
Four years later, age of Haroon = 7x + 4
5x + 4 + 7x + 4 = 56
12x + 8 = 56
We transfer 8 to right-hand side (RHS)
12x = 56 – 8
12x = 48
X = 48 / 12
X = 4
Present ages of Rahul and Haroon are 5x and 7x respectively.
Present ages of Rahul 5x = 5 x 4 = 20 year
Present ages of Haroon 7x = 7 x 4 = 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?
ANS:
Given that,
Number of boys and girls in a class are in ratio 7:5
Number of boys is 8 more than the number of girls.
Number of boys in class = 7x
Number of girls in class = 5x
Number of boys is 8 more than number of girls.
7x = 5x + 8
We transfer 5x to Left hand side (LHS)
7x – 5x = 8
2x = 8
We transfer 2 to right-hand side (RHS)
X = 8/2
X = 4.
Number of boys in class = 7x
Number of boys in class = 7 x 4 = 28
Number of girls in class = 5x = 5 x 4 = 20
The total class strength is 28 + 20 = 48
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?
ANS:
Given that,
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung
Sum of the ages of all the three is 135 years.
Let the age of Baichung be x years.
Age of Baichung’s father = x + 29 years
Age of Baichung’s grandfather = age of Baichung father + 26 years
Age of Baichung’s grandfather = x + 29 + 26 years
Age of Baichung’s grandfather = x + 55 years
Sum of ages of all the three is 135 years
X + x + 29 + x + 55 = 135 years
3x + 84 = 135
We transfer 84 to right-hand side (RHS)
3x = 51
We transfer 3 to right-hand side (RHS)
X = 17.
But x is age of Baichung.
Age of Baichung is 17 years.
Age of Baichung’s father = x + 29 years = 17 + 29 years = 46 years
Age of Baichung’s grandfather = x + 55 years = 17 + 55 years = 72 years
12.) Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
ANS:
Let the present age of Ravi be x years.
15 years from now, Ravi’s age will be 4 times his present age
X + 15 = 4x
We transfer x to right-hand side (RHS)
15 = 3x
X = 5
But x is present age of Ravi.
Present age of Ravi is 5 years.
13.) A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get – 7/12 What is the number?
ANS:
Let the rational number be x
Multiply by 5 / 2 = 5x / 2
Add 2/ 3 in the product = 5x / 2 + 2/ 3
We get – 7/12.
5x / 2 + 2/ 3 = – 7/12
On solving we get,
X = -1/2
But x is rational number.
The rational number is -1/2.
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?
ANS:
Given that,
Cashier Lakshmi has currency notes of denominaton Rs. 100, Rs.50, and Rs.10.
Ratio of the number of notes is 2:3:5.
Total cash with Lakshmi is Rs. 4, 00,000
Number of notes are in ratio 2:3:5
Therefore number of notes is 2x, 3x and 5x respectively.
I.e. Rs. 100 x 2x = 200x
Rs.50 x 3x = 150x
Rs.10 x 5x = 50x
Total cash with Lakshmi is Rs. 4, 00,000
200x + 150x + 50x = Rs. 4, 00,000
400x = Rs. 4, 00,000
We transfer 400 to right-hand side (RHS)
X = 1000
Number of notes is 2x, 3x and 5x respectively.
Number of Rs.100 notes = 2 x 1000 = 2000
Number of Rs.50 notes = 3 x 1000 = 3000
Number of Rs.10 notes = 5 x 1000 = 5000
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?
ANS:
Given that,
Total amount = RS. 300
Coin denominations are Rs. 1, Rs.2 and Rs. 5
Number of Rs.2 coins are 3 times Rs. 5 coins.
Total coins = 160
Let the number of Rs 5 coins be x.
Then the number of Rs.2 coins is 3x.
Number of Rs.1 coin is = 160 – 4x
Now, amount of Rs 5 coins = 5x
Amount of Rs 2 coins = 2 x 3x = 6x
Amount of Rs 1 coins = 1 x (160 – 4x)
Given that
Total amount = RS. 300
5x + 6x + 160 – 4x = 300
7x + 160 = 300
We transfer 160 to right-hand side (RHS)
7x = 140
X = 20
But x is number of Rs 5 coins
Number of Rs 5 coins Is 20.
The number of Rs.2 coins is 3x. = 3 x 20 = 60
Number of Rs.1 coin is = 160 – 4x = 160 – 80 = 80
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.
ANS:
Given that,
Winner of essay competition gets a prize of Rs. 100
Participant who does not win gets a prize of Rs. 25
Total prize of money distributed is Rs. 3000
Total number of participants is 63.
Let, number of winners be x
Then, the number of participants who do not win = 63 – x
Total prize money = Rs. 3000
X x 100 + (63 – x) x 25 = Rs. 3000
100x + 1575 – 25x = Rs. 3000
75x + 1575 = Rs. 3000
We transfer 1575 to right-hand side (RHS)
75x = 1425
X = 19
But x is number of winners.
The number of winners is 19.
EXERCISE 3.3
Solve the following equations and check your results.
1.)3x = 2x + 18
Ans:
Given, 3x = 2x + 18
We transfer 2x to Left hand side (RHS).
3x – 2x = 18
X = 18
2.) 5t – 3 = 3t – 5
Ans:
Given, 5t – 3 = 3t – 5
By arranging,
5t – 3t = – 5 + 3
2t = -2
We transfer 2 to right-hand side (RHS)
t = -1
3.) 5x + 9 = 5 + 3x
Ans:
Given,5x + 9 = 5 + 3x
By arranging,
5x – 3x = 5 – 9
2x = – 4
We transfer 2 to right-hand side (RHS)
X = -4 / 2
X = -2
4.) 4z + 3 = 6 + 2z
Ans:
Given, 4z + 3 = 6 + 2z
By arranging,
4z – 2z = 6 – 3
2z = 3
We transfer 2 to right-hand side (RHS)
Z = 3 / 2
5.) 2x – 1 = 14 – x
Ans:
Given, 2x – 1 = 14 – x
By arranging,
2x + x = 14 + 1
3x = 15
We transfer 3 to right-hand side (RHS)
X = 15 / 3
X = 5
6.) 8x + 4 = 3(x – 1) + 7
Ans:
Given, 8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4
By arranging,
8x – 3x = 4 – 4
5x = 0
X = 0
7.) x = 4/5 x (x + 10)
Ans:
Given, x = 4/5 x (x + 10)
By arranging,
5x = 4 (x + 10)
5x = 4x + 40
By arranging,
X = 40
8.) (2x)/3 + 1 = (7x)/15 + 3
Ans:
Given, (2x)/3 + 1 = (7x)/15 + 3
By arranging,
(2x)/3 – (7x)/15 = 3 – 1
By doing same denominator.
10x – 7x / 15 = 2
3x / 15 = 2
3x = 30
X = 10
9.) 2y + 5/3 = 26/3 – y
Ans:
Given, 2y + 5/3 = 26/3 – y
By arranging,
2y + y = 26/ 3 – 5 / 3
3y = 21/3 = 7
We transfer 3 to right-hand side (RHS)
Y = 7 / 3
10). 3m = 5m – 8/5
Ans:
Given, 3m = 5m – 8/5
By arranging,
3m – 5m = – 8/5
-2m = – 8/5
We transfer -2 to right-hand side (RHS)
M = 4 / 5
EXERCISE 3.4
1.) Amina thinks of a number and subtracts 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?
Ans:
Given that,
5/2 is subtracted from a number
Then multiplied by 8
Result is 3 times the actual number
Let, the number be x.
According to the given question, we form equation.
8(x – 5/2) = 3x
8x – 20 = 3x
We transfer 3x to Left Hand Side (LHS)
8x – 3x = 20
5x = 20
We transfer 5 to Right Hand Side (RHS)
X = 20 / 5
X = 4
But x is the number.
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?
Ans:
Given that,
A positive number is 5 times another number
21 is added to both the numbers
Then one of the new numbers becomes twice the other new number
Let, the numbers be x and 5x.
According to the question, we form a equation.
21 + 5x = 2(x + 21)
21 + 5x = 2x + 42
We transfer 2x to Left Hand Side (LHS)
5x – 2x = 42 – 21
3x = 21
We transfer 3 to Right Hand Side (RHS)
x = 7
First number is x = 7
Second number is 5x = 5 x 7 = 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?
Ans:
Given that,
Sum of the digits of a two-digit number is 9
Interchanging the digits result in a new number greater than the original number by 27
Let, the digits at tens place and ones place be x and 9 – x respectively.
From Questions, original number = 10x + (9 – x) = 9x + 9
Now,
On interchanging the digits, the digits at ones place and tens place will be x and 9 – x respectively.
Therefore, new number after interchanging the digits:
= 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
According to the given question, we form Equation.
New number = Original number + 27
90 – 9x = 9x + 9 + 27
90 – 9x = 9x + 36
We transfer 9x to RHS and 36 to LHS
90 – 36 = 18x
54 = 18x
We transfer 18 to Left Hand Side.
Weget,
x = 3
But x is digits at tens place.
9 – x is digits at ones place.
9 – x = 6
The two digit number is 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?
Ans:
Given that,
One of the two digits of a two-digit number is three times the other digit.
Interchanging the digit of this two-digit number and adding the resulting number to the original number results in 88.
Let, the digits at tens place and ones place be x and 3x respectively.
Therefore, original number = 10x + 3x = 13x
On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.
Number after interchanging = 30x + x = 31x
According to the given question, we form Equation.
Original number + New number = 88
13x + 31x = 88
44x = 88
We transfer 44 on Right Hand Side (RHS)
x = 2
But x is digits at tens place 2
Now digits at ones place is 3x = 3 x 2 = 6
The number is 26.
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?
Ans:
Given that,
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.
Let, Shobo’s age be x years.
Therefore, his mother’s age will be 6x years.
According to the given question,
After 5years, Shobo’s age = Shobo’s mother’s present age / 3
Now, we form Equation.
x + 5 = (6x)/3
x + 5 = 2x
We transfer x to RHS
5 = 2x – x
5 = x
But x is Shobo’s age.
Shobo’s age is 5 years.
Shobo’s mother’s age is 6x = 6 x 5 = 30 years.
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?
Ans:
Given that,
The length and breadth of the plot are in the ratio 11:4
At the rate Rs. 100 per metre it will cost Rs. 75, 000 to fence the plot
Let, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.
We know,
Perimeter of the plot = 2(Length + Breadth)
= [2(11x + 4x)] m
Perimeter of the plot = = 30x m
Now, It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75,000.
100 x Perimeter = 75000
100 x 30x = 75000
3000x = 75000
We transfer 3000 to Right Hand Side (RHS)
x = 25
Now, Length is 11x m.
The Length of rectangular Plot
= (11 x 25) m
The Length of rectangular Plot = 275m
Now,
Breadth of rectangular Plot = 4x m
= (4 x 25) m
Breadth of rectangular Plot = 100m
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?
Ans:
Given that,
Shirt material that costs him Rs 50 per meter and trouser material that costs him Rs 90 per meter
For every 2 meters of the trouser material, 3 meters of the shirt material is purchased.
Selling of the materials takes place at 12% and 10% profit respectively
Total sale is Rs 36660.
Let 2x m of trouser material and 3x m. of shirt material be bought by him.
Per metre selling price of trouser material
= Rs (90 + (90 x 12)/100)
= Rs 100.80
Now,
Per metre selling price of shirt material
= Rs (50 + (50 x 10)/100)
= Rs. 55
Given that, total amount of selling = Rs 36660
We form Equation.
100.80x x (2x) + 55(3x) = 36660
201.6x + 165x = 36660
366.6x = 36660
We transfer 366.6 to Right Hand Side. (RHS)
X = 100
Now,
Trouser material is 2x m
= (2 x 100) m
Trouser material required = 200m
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.
Ans:
Given that,
Half of the herd of deer are grazing in the field
Three-fourths of the remaining are playing nearby
The rest 9 are drinking water from the pond.
Let the number of deer be x.
Number of deer grazing in the field = x / 2
Number of deer playing nearby = 3/4 x Number of remaining deer
= 3/4 x (x – x/2)
= 3/4 x X/2
= (3x)/8
Now, given that, Number of deer drinking water from the pond =9
We form Equation,
x – (x/2 + (3x)/8) = 9
On solving we get,
x = 72
But x is the total number of deer.
The total number of deer is 72.
9.) A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Ans:
Given that,
Grandfather is ten times older than his granddaughter also he is 54 years older than her.
Let the granddaughter’s age be x years.
Therefore, grandfather’s age will be 10x years.
According to the question, we form Equation.
Grandfather’s age = Granddaughter’s age + 54 years
10x = x + 54
We transfer x to Left Hand Side
10x – x = 54
9x = 54
We transfer 9 to Right Hand Side.
x = 6
But x is Granddaughter’s age.
Granddaughter’s age is 6 years.
Now,
Grandfather’s age = 10x years.
= (10 x 6) years
Grandfather’s age 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.
Ans:
Given that,
Aman’s age is three times his son’s age.
Ten years ago he was five times his son’s age.
Let, Aman’s son’s age be x years.
Therefore, Aman’s age will be 3x years.
Ten years ago, their age was (x – 10) years and (3x – 10) years respectively.
10 years ago, Aman’s age = 5 × Aman’s son’s age 10 years ago
We form Equation.
3x – 10 = 5(x – 10)
3x – 10 = 5x – 50
We transfer 3x to RHS and 50 to LHS
50 – 10 = 5x – 3x
40 = 2x
We transfer 2 to LHS
20 = x
But x is Aman’s son’s age
Aman’s son’s age is 20 years.
Now,
Aman’s age = 3x years
= (3 x 20) years
Aman’s age is 60 years.
EXERCISE 3.5
Solve the following linear equations.
1.) x/2 – 1/5 = x/3 + 1/4
Ans:
Given, x/2 – 1/5 = x/3 + 1/4
Here, we take LCM of the denominators.
I.e.2, 3, 4, and 5, which is 60.
We multiply both sides by 60, we get,
60(x/2 – 1/5) = 60(x/3 + 1/4)
30x – 12 = 20x + 15
We transfer 20x to LHS and -12 to RHS.
30x – 20x = 15 + 12
10x = 27
We transfer 10 to RHS.
x = 27/10
2.) n/2 – (3n)/4 + (5n)/6 = 21
Ans:
Given, n/2 – (3n)/4 + (5n)/6 = 21
Here, we take LCM of the denominators.
I.e.2, 4, and 6, which is 12.
We multiply both sides by 12, we get,
6n – 9n + 10n = 252
7n = 252
n = 252/7
n = 36
3.) x + 7 – (8x)/3 = 17/6 – (5x)/2
Ans:
Given, x + 7 – (8x)/3 = 17/6 – (5x)/2
Here, we take LCM of the denominators.
i.e. 2, 3, and 6, which is 6.
We multiply both sides by 6, we get,
6x + 42 – 16x = 17 – 15x
6x – 16x + 15x = 17 – 42
5x = – 25
x = – 25/5
x = – 5
4.) (x – 5)/3 = (x – 3)/5
Ans:
Given, (x – 5)/3 = (x – 3)/5
Here, we take LCM of the denominators.
3 and 5, which is 15.
We multiply both sides by 15, we get,
5(x – 5) = 3(x – 3)
5x – 25 = 3x – 9
5x – 3x = 25 – 9
2x = 16
x = 16/2
x = 8
5.) (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Ans:
Given, (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Here, we take LCM of the denominators.
3 and 4, which is 12.
We multiply both sides by 12, we get
3(3t – 2) – 4(2t + 3) = 8 – 12t
= 9t – 6 – 8t – 12 = 8 – 12t
9t – 8t + 12t = 8 + 6 + 12
13t = 26
t = 26/13
t = 2
6.) m – (m – 1)/2 = 1 – (m – 2)/3
Ans:
Given, m – (m – 1)/2 = 1 – (m – 2)/3
Here, we take LCM of the denominators.
2 and 3, which is 6.
We multiply both sides by 6, we get,
6m – 3(m – 1) = 6 – 2(m – 2)
6m – 3m + 3 = 6 – 2m + 4
6m – 3m + 2m = 6 + 4 – 3
5m = 7
m = 7/5
Simplify and solve the following linear equations.
7.) 3(t – 3) = 5(2t + 1)
Ans:
Given, 3(t – 3) = 5(2t + 1)
3t – 9 = 10t + 5
– 9 – 5 = 10t – 3t
– 14 = 7t
t = – 14/7
t = – 2
8.) 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Ans:
Given, 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
18y – 12 = 0
18y = 12
y = 12/18
y = 2/3
9.) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Ans:
Given, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
15z – 21 – 18z + 22 = 32z – 52 – 17
– 3z + 1 = 32z – 69
– 3z – 32z = – 69 – 1
– 35z = – 70
z = 70/35
z = 2
10.) 0.25(4f – 3) = 0.05(10f – 9)
Ans:
Given, 0.25(4f – 3) = 0.05(10f – 9)
0.25(4f – 3) = 0.05(10f – 9)
We put 0.25 as 1/4 and 0.05 as 1/20
1/4 x (4f – 3) = 1/20 x (10f – 9)
We multiply both sides by 20, we obtain
5(4f – 3) = 10f – 9
20f – 15 = 10f – 9
20f – 10f = – 9 + 15
10f = 6
f = 3/5
EXERCISE 3.6
Solve the following equations.
1.) (8x – 3)/ (3x) = 2
Ans:
Given, (8x – 3)/ (3x) = 2
We multiply both sides by 3x, we get
8x – 3 = 6x
8x – 6x = 3
2x = 3
x = 3/2
2.) (9x)/ (7 – 6x) = 15
Ans:
Given, (9x)/ (7 – 6x) = 15
We multiply both sides by 7 – 6x we get,
9x = 15(7 – 6x)
9x = 105 – 90x
9x + 90x = 105
99x = 105
x = 105/99
x = 35/33
3.) z/(z + 15) = 4/9
Ans:
Given, z/(z + 15) = 4/9
We multiply both sides by 9(z + 15) we get,
9z = 4(z + 15)
9z = 4z + 60
9z – 4z = 60
5z = 60
z = 12
4.) (3y + 4)/ (2 – 6y) = – 2/5
Ans:
Given, (3y + 4)/ (2 – 6y) = – 2/5
We multiply both sides by 5(2 – 6y), we get,
5(3y + 4) = – 2(2 – 6y)
15y + 20 = – 4 + 12 y
15y – 12y = – 4 – 20
3y = – 24
y = – 8
5.) (7y + 4)/(y + 2) = – 4/3
Ans:
Given, (7y + 4)/(y + 2) = – 4/3
We multiply both sides by 3(y + 2) we get,
3(7y + 4) = – 4(y + 2)
21y + 12 = – 4y – 8
21y + 4y = – 8 – 12
25y = – 20
y = – 4/5
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.
Ans:
Given that,
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.
Let the common ratio between their ages be x.
Therefore, Hari’s age will be 5x years and Harry’s age 7x years.
Four years later, Hari’s age will be (5x + 4) years and Harry’s age (7x + 4) years.
Now, we form equation.
(5x + 4)/ (7x + 4) = 3/4
By cross multiplication,
4(5x + 4) = 3(7x + 4)
20x + 16 = 21x + 12
16 – 12 = 21x – 20x
4 = x
Now, Hari’s age = 5x years
= (5 x 4) years
Hari’s age is 20 years.
Harry’s age = 7x years
= (7 x 4) years
Harry’s age is 28 years.
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 3/2 Find the rational number.
Ans:
Given that,
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 3 /2.
Let the numerator of the rational number be X.
Therefore, its denominator will be x + 8
The rational number will be x/(x + 8)
Now, we form equation.
(x + 17)/(x + 8 – 1) = 3/2
(x + 17)/(x + 7) = 3/2
By cross multiplication,
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
34 – 21 = 3x – 2x
13 = x
But x is Numerator of the Rational Number.
Numerator of the Rational Number is 13.
Now,
Denominator of the Rational Number = x+8
=13+8
Denominator of the Rational Number is 21.
Rational Number is 13 / 21.
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