RS Aggarwal Class 6 Math Ninth Chapter Linear Equation In One Variable Exercise 9C Solution

EXERCISE 9C

(1) If 9 is added to a certain number, the result is 36. Find the number.

Solution: Let the required number is x.

Then,   X+9=36

Or, x = 36 – 9

Or, x = 27

(2) If 11 is subtracted from 4 times a number is 89. Find the number.

Solution: Let the required number is x.

(3) Find a number which when multiplied by 5 is increased by 80.

Solution: Let the required number is x.

(4) The sum of three consecutive natural numbers is 114. Find the numbers.

Solution: Let the required numbers are x, x+1, x+2

So, the numbers are 37, 38 and 39.

(5) When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

Solution: Let the required number is x.

So, the required number is 13.

(6) If a number is tripled and the result is increased by 5, we get 50. Find the number.

Solution: Let the required number is x.

So, the required number is 15.

(7) Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

Solution: Let the required numbers are x and x+18

So, the numbers are 37 and (37+18)=55.

(8) Find out of two numbers is thrice the other. If their sum is 124, find the numbers.

Solution: Let the required numbers are x and 3x.

So, the numbers are 31 and (3×31)=93

(9) Find two numbers such that one of them is five times the other and their difference is 132.

Solution: Let the required numbers are x and 5x.

So, the numbers are 33 and (33 x 5)=165.

(10) The sum of two consecutive even numbers is 74. Find the numbers.

Solution: Let the required numbers are x and x+2.

So, the numbers are 36 and (36+2)=38.

(11) The sum of three consecutive odd numbers is 21. Find the numbers.

Solution: Let the required numbers are x, x+2 and x+4

So, the numbers are 5, (5+2)=7 and (5+4)=9

(12) Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

Solution: Let Ajay’s age is x years and Reena’s age is x+6

So, the present age of Ajay is 11 and Reena is (11+6)=17.

(13) Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.
Solution: Let the age of Vikash is x years. And Deepak is 2x years.
Then, 2x – x = 11
or, x = 11
So, Their present ages are 11 and (2 x 11) = 22 years.

(14) Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.
Solution: Let Rekha’s age is x years and Mrs goel’s age is x+27 years.
After 8 years the age of Rekha is x+8 and Mrs Goel is x+27+8 years.
Then, x + 27 + 8 = 2 (x+8)
or, x + 35 = 2x + 16
or, 2x – x = 35 – 16
or, x = 19
So, the age of Rekha is 19 years and Mrs Goel is (19+27)=46.

(15) A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.
Solution: Let the age of son is x and his father age is 4x.
After 16 years the age of son is x+16 and his father age is 4x+16.

So, the present age of son is 8 years and his father age is (4×8)=32 years.

(16) A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.
Solution: Let the age of son is x and the age of the man is 3x
Before 5 years their age are (x – 5) and (3x – 5)
Then, 3x – 5 = 4 (x – 5)
or, 3x – 5 = 4x – 20
or, 3x – 4x = – 20 + 5
or, -x = -15
or,x=15
So, their present age are 15 and (3×15) =45 years.

(17) After 16 years, Fatima will be three times as old as she is now. Find her present ages.

Solution: Let, the Fatima’s age is x years. After 16 years her age is (x+16).

So, the present age of Fatima is 8 years.

(18) After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

Solution: Let the age of Rahim is x years. After 32 years his age is (x+32) years. Before 8 years ago his age was

(x-8).

So, Rahim is 18 years old today.

(19) A bag contains 25-paisa and 50-paisa coins whose total value is Rs 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.
Solution: Let the amount of 50 paise is x and 25 paise is 4x.
50x + (25 × 4x)
=50x + 100x = 150x
And, Rs 30 = 3000 paise

So, the required number of 50 paise is 20 and 25 paise is (4×20)=80.

(20) Five times the price of a pen is Rs 17 more than three times its price. Find the price of the pen.

Solution: Let, the price of each pen is = x.

So, the price of the pen is 8.50 rupees.

(21) The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

Solution: Let the number of girls is x.

So, the number of girls is 119.

(22) The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, find its dimensions.

Solution: Let the breadth of the park is x m and length is 3x m.

So, The breadth is 21 m and length is (3×21)=63 m.

(23) The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

Solution: Let the breadth of the hall is x m and length is (x+5) m.

So, The breadth is 16 m and the length is (16+5)=21 m.

(24) A wire of length 86 cm is bent in the form of a rectangle such that length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

Solution: Let the breadth is x cm and the length is (x+7) cm.

So, the breadth of rectangle formed is 18 cm and the length is (18+7) = 25 cm.