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Equations In One Variable

Class 8th Mathematics (new) MHB Solution
Practice Set 12.1
  1. Each equation is followed by the values of the variable. Decide whether these values…
  2. Each equation is followed by the values of the variable. Decide whether these values…
  3. Each equation is followed by the values of the variable. Decide whether these values…
  4. Each equation is followed by the values of the variable. Decide whether these values…
  5. Solve the following equations17p – 2 = 49
  6. Solve the following equations2m + 7 = 9
  7. Solve the following equations3x + 12 = 2x – 4
  8. Solve the following equations5(x – 3) = 3(x + 2)
  9. Solve the following equations {9x}/{8} + 1 = 10
  10. Solve the following equations {y}/{7} + frac {y-4}/{3} = 2
  11. Solve the following equations 13x-5 = {3}/{2}
  12. Solve the following equations3(y + 8) = 10(y – 4) + 8
  13. Solve the following equations {x-9}/{x-5} = frac {5}/{7}
  14. Solve the following equations {y-4}/{3} + 3y = 4
  15. Solve the following equations { b + (b+1) + (b+2) }/{4} = 21
Practice Set 12.2
  1. Mother is 25 years older than her son. Find son’s age if after 8 years ratio of son’s…
  2. The denominator of a fraction is greater than its numerator by 12. If the numerator is…
  3. The ratio of weights of copper and zinc in brass is 13:7. Find the weight of zinc in a…
  4. Find three consecutive whole numbers whose sum is more than 45 but less than 54.…
  5. In a two-digit number, a digit at the ten’s place is twice the digit at unit’s place.…
  6. Some tickets of Rs.200 and some of Rs.100, of a drama in a theatre were sold. The…
  7. Of the three consecutive natural numbers, five times the smallest number is 9 more than…
  8. Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs.54. Then he sold…
  9. A Cricket player scored 180 runs in the first match and 257 runs in the second match.…
  10. Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half…
Practice Set 12.1
Question 1.

Each equation is followed by the values of the variable. Decide whether these values are the solutions to that equation.

x – 4 = 3, x = – 1, 7, – 7
Answer:

given x – 4 = 3


adding 4 on both sides


so x = 4 + 3 = 7


7 is the only solution of this given equation
Question 2.

Each equation is followed by the values of the variable. Decide whether these values are the solutions to that equation.

9m = 81, m = 3, 9, – 3
Answer:

Given 9m = 81


Dividing by 9 on both sides



So,


is the only solution of this given equation
Question 3.

Each equation is followed by the values of the variable. Decide whether these values are the solutions to that equation.

2 a + 4 = 0, a = 2, – 2, 1
Answer:

Given 2 a + 4 = 0


Adding – 4 on both sides


2a = – 4


a = – 2


– 2 is the only solution of this given equation
Question 4.

Each equation is followed by the values of the variable. Decide whether these values are the solutions to that equation.

3 – y = 4, y = – 1, 1, 2
Answer:

Given 3 – y = 4


Adding – 4 + y on both side


y = – 1


– 1 is the only solution of this given equation
Question 5.

Solve the following equations

17p – 2 = 49
Answer:

Given 17p – 2 = 49


Adding on both sides


17p = 51


Dividing by on both sides
Question 6.

Solve the following equations

2m + 7 = 9
Answer:

Given 2m + 7 = 9


Adding on both sides


2m = 2


Dividing by on both sides


m = 1
Question 7.

Solve the following equations

3x + 12 = 2x – 4
Answer:

Given 3x + 12 = 2x – 4


Adding – 12 – 2x on both sides


3x – 2x = – 12 – 4


X = – 16
Question 8.

Solve the following equations

5(x – 3) = 3(x + 2)
Answer:

Given 5(x – 3) = 3(x + 2)


Expanding the equation


5x – 15 = 3x + 6


Adding on both sides


5x – 3x = 15 + 6


2x = 21


x = 21/2
Question 9.

Solve the following equations
Answer:

Given




Multiplying on both sides
Question 10.

Solve the following equations
Answer:

Given


Taking LCM of 7 and 3, that is 21


Now multiplying both side of given equation by


3y + 7(y – 4) = 42


Expanding the given equation


10y – 28 = 42


Adding on both sides


10y = 70


y = 7
Question 11.

Solve the following equations
Answer:

Given


Multiplying by on both sides


26x – 10 = 3


Adding on both sides


26x = 13


Dividing by on both sides
Question 12.

Solve the following equations

3(y + 8) = 10(y – 4) + 8
Answer:

Given 3(y + 8) = 10(y – 4) + 8


expanding


3y + 24 = 10y – 40 + 8


Adding –3y on both sides


24 = 7y – 40 + 8


Adding 32 on both sides


7y = 56


Dividing by on both sides


y = 8
Question 13.

Solve the following equations
Answer:

Given


Multiplying 7(x – 5) on both sides


7(x – 9) = 5(x – 5)


expanding


7x – 63 = 5x – 25


Adding 63 – 5x on both sides


2x = 38


Dividing by 2 on both sides


x = 19
Question 14.

Solve the following equations
Answer:

Given


Multiplying by on both sides


y – 4 + 9y = 12


10y – 4 = 12


Adding 4 on both sides


10y = 16


Dividing by on both sides
Question 15.

Solve the following equations
Answer:

Given


Multiplying by on both sides


b + (b + 1) + (b + 2) = 84


Adding – 3 on both sides


3b + 3 = 84


Dividing both side by 3


3b = 81


b = 27
Practice Set 12.2
Question 1.

Mother is 25 years older than her son. Find son’s age if after 8 years ratio of son’s age to mother’s age will be 4/9
Answer:

let the age of son be x, so age of mother is x + 25

After 8 years, gather e of son is x + 8 and age of mother is x + 25 + 8


So according to give conditions


Now solving the equation


Multiplying both sides by



Multiplying both sides by 4(x + 25 + 8)


9(x + 8) = 4(x + 33)


Expanding the equation


9x + 72 = 4x + 132


Adding – 4x – 72 on both sides


5x = 60


Dividing by 10 on both sides


x = 12


So, age of son is 12 years
Question 2.

The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent with 1/2. Find the fraction.
Answer:

let the numerator be x, so dthe enominator is x + 12

New numerator is x – 2 , newthe denominator is x + 12 + 7


So according to given conditions


Multiplying both sides by



Multiplying by x + 19 on both sides


2(x – 2) = x + 19


2x – 4 = x + 19


Adding – x on both sides


x – 4 = 19


Adding 4 on both sides


x = 23


So, numerator is 23 and denominator is 12 + 23 = 35


Required fraction is



Question 3.

The ratio of weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Answer:

let the weight of zinc be

copper/zinc = copper/x = 13/7


copper = 13x/7


So according to given conditions


x + 13x/7 = 700


Multiplying both side by 7


7x + 13x = 4900


20x = 4900


Dividing both side by 20


x = 245


So, weight of zinc is 245g
Question 4.

Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Answer:

let the lowest number be , so other numbers are x + 1

And x + 2


According to given conditions 45 < x + (x + 1) + (x + 2) < 54


Solving this equation, we have 45 < 3x + 3 < 54


Adding – 3 on both sides


42 < 3x < 51


Dividing by 3 on both sides


14 < x < 17


So x = 15 or x = 16


So consecutive numbers are 15,16,17 or 16,17,18
Question 5.

In a two-digit number, a digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number
Answer:

let the number at digit place be

So, da igit at tens place is 2x.


Number is 10×2x + x = 21x


Now interchanging the digits, at digit place we have 2x and at tens place x.number is 10×x + 2x = 12x


According to the given condition’s 12x + 21x = 66


33x = 66


Dividing both side by 33


X = 2


Digit at tens place is 2 × 2 = 4


Number is 42
Question 6.

Some tickets of Rs.200 and some of Rs.100, of a drama in a theatre were sold. The number of tickets for Rs.200 sold was 20 more than the number of tickets for Rs.100. The total amount received by the theatre by the sale of tickets was Rs.37000. Find the number of Rs.100 tickets sold.
Answer:

let the number of Rs.100 ticket sold be x

Number of Rs.200 ticket sold is x + 20


According to given conditions 100x + 200(x + 20) = 37000


300x + 4000 = 37000


Adding - 4000 on both sides


300x = 33000


Dividing both side by 300


x = 110


Number of Rs.100 tickets sold are 110
Question 7.

Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Answer:

let the numbers be x,x + 1,x + 2 of which being the smallest. According to given conditions 5x = 9 + 4(x + 2)

Expanding


5x = 9 + 4x + 8


Adding - 4x on both sides


x = 17


So, numbers are 17,18,19
Question 8.

Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs.54. Then he sold the bicycle to Nikhil for Rs.1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Answer:

let the cost of the cycle for Raju be




So according to given conditions total cost = total selling price



Adding on both sides



Multiplying by on both side


100x + 8x = 108000


108x = 108000


Dividing by on both side


x = 1000


So, cost of cycle to Raju is Rs.1000
Question 9.

A Cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Answer:

let the runs required in third match be

So according to given conditions


Multiplying both side by


257 + 180 + x = 690


Subtracting from both sides


257 + x = 510


Subtracting 257 from both sides


x = 253


Therefore 253 runs are required.
Question 10.

Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6, then find Viru’s age.
Answer:

let the age of Viru be

Sudhir age = 3x + 5



According to given conditions,



Multiplying on both side



Multiplying 3(3x + 5) on both sides


2(4x + 5) = 3(3x + 5)


Expanding


8x + 10 = 9x + 15


Adding on both side


X + 15 = 10


Adding on both side


x = – 5


But since age cannot be negative so, x = 5


Hence age is 5 years.