RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 9- Linear Equations in One Variable
RS Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable Exercise 9C is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable
Exercise 9C
1. If 9 is added to a certain number, the result is 36. Find the number.
Solution:
Let the number be x
According to the question:
⇒ x + 9 = 36
⇒ x = 36 − 9
∴ x = 27
Thus, the required number is 27.
2. If 11 is subtracted from 4 times a number, the result is 89. Find the number.
Solution:
Let the required number be x.
According to the question:
4x −11 = 89
⇒ 4x = 89 + 11
⇒ 4x = 100
⇒ x = `frac\{cancel 100^25}{cancel 4}`
∴ x = 25
Thus, the required number is 25.
3. Find a number which when multiplied by 5 is increased by 80.
Solution:
Let the required number be x.
According to the question:
⇒ 5x = x + 80
⇒ 5x − x = 80
⇒ 4x = 80
⇒ x = `frac\{cancel 80^20}{cancel 4}`
∴ x = 20
Thus, the required number is 20.
4. The sum of three consecutive natural numbers is 114. Find the numbers.
Solution:
Let the three consecutive natural numbers be x, (x+1), (x+2).
According to the question:
x + (x + 1) + (x + 2) = 114
⇒ x + x + 1 + x + 2 = 114
⇒ 3x + 3 = 114
⇒ 3x = 114 − 3
⇒ 3x = 111
⇒ x = `frac\{cancel 111^37}{cancel 3}`
∴ x = 37
Required numbers are:
x = 37
⇒ x + 1 = 37 + 1 = 38
⇒ x + 2 = 37 + 2 = 39
Thus, the required 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 be x.
When Raju multiplies it with 17, the number becomes 17x.
According to the question :
17x + 4 = 225
⇒ 17x = 225 − 4
⇒ 17x = 221
⇒ x = `frac\{cancel 221^13}{cancel 17}`
∴ x = 13
Thus, 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 be x.
According to the question,
The number is tripled and 5 is added to it
∴ 3x + 5
⇒ 3x + 5 = 50
⇒ 3x = 50 − 5
⇒ 3x = 45
⇒ x = `frac\{cancel 45^15}{cancel 3}`
∴ x = 15
Thus, 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 one of the number be x.
∴ The other number = x + 18
According to the question:
x + x + 18 = 92
⇒ 2x + 18 = 92
⇒ 2x = 92 − 18
⇒ 2x = 74
⇒ x = `frac\{cancel 74^37}{cancel 2}`
∴ x = 37
Required numbers are:
One number = x = 37
Other number = x + 18
= 37 + 18
= 55
8. One out of two numbers is thrice the other. If their sum is 124. find the numbers.
Solution:
Let one of the number be x
∴ Second number = 3x
According to the question:
x + 3x = 124
⇒ 4x = 124
⇒ x = `frac\{cancel 124^31}{cancel 4}`
∴ x = 31
Thus, the required number is x = 31
and Second number = 3x = 3 × 31 = 93.
9. Find two numbers such that one of them is five times the other and their difference is 132.
Solution:
Let one of the number be x.
∴ Second number = 5x
According to the question:
5x − x = 132
⇒ 4x = 132
⇒ x = `frac\{cancel 132^33}{cancel 4}`
∴ x = 33
Thus, the required numbers is x = 33
and Other number = 5x = 5 × 33 = 165.
10. The sum of two consecutive even numbers is 74. Find the numbers.
Solution:
Let one of the even number be x.
Then, the other consecutive even number is (x + 2).
According to the question:
x + x + 2 = 74
⇒ 2x + 2 = 74
⇒ 2x = 74 − 2
⇒ 2x = 72
⇒ x = `frac\{cancel 72^36}{cancel 2}`
∴ x = 36
Thus, the required numbers are x = 36
and x + 2 = 38.
11. The sum of three consecutive odd numbers is 21. Find the numbers.
Solution:
Let the three consecutive odd numbers be x, (x + 2) and (x + 4).
According to the question,
x + (x + 2) + (x + 4) = 21
⇒ 3x + 6 = 21
⇒ 3 x = 21 – 6
⇒ 3 x = 15
⇒ x = `frac\{cancel 15^5}{cancel 3}`
∴ x = 5
The required numbers are 5,
and 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 the present age of Ajay be x years.
Then, the present age of Reena = (x + 6) years
According to the question,
x + (x + 6) = 28
⇒ 2x + 6 = 28
⇒ 2x = 28 – 6
⇒ 2 x = 22
⇒ x = `frac\{cancel 22^11}{cancel 2}`
∴ x = 11
Present age of Ajay = 11 years
and present age of Reena = 11 + 6 = 17 years.
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 present age of Vikas be x years.
Then, the present age of Deepak = 2x years
According to the question,
2x – x = 11
∴ x = 11
Present age of Vikas = 11 years
and present age of Deepak = 2 × 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 the present age of Rekha be x years
Then, the present age of Mrs. Goel = (x + 27) years
After 8 years,
Age of Rekha = (x + 8) years
Age of Mrs. Goel = (x + 27 + 8) years
= (x + 35) years
According to the question,
x + 35 = 2 (x + 8)
⇒ x + 35 = 2x + 16
⇒ x – 2x = 16 – 35
⇒ – x = – 19
∴ x = 19
Present age of Rekha = 19 years
and present age of Mr. Goel = 19 + 27 = 46 years.
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 present age of the son be x years.
Then, the present age of the man = 4x years
After 16 years,
Son’s age = (x + 16) years
Man’s age = (4 x + 16) years
According to the question,
4x + 16 = 2 (x + 16)
⇒ 4x + 16 = 2x + 32
⇒ 4x – 2x = 32 – 16
⇒ 2x = 16
⇒ x = `frac\{cancel 16^8}{cancel 2}`
∴ x = 8
Present age of the son = 8 years
and present age of the man = 8 × 4 = 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 present age of the son be x years.
Then, the present age of the man = 3x years
Five years ago,
the age of the son = (x – 5) years
the age of the man = (3x – 5) years
According to the question,
3x – 5 = 4(x – 5)
⇒ 3x – 5 = 4x – 20
⇒ 3x – 4x = – 20 + 5
⇒ – x = – 15
∴ x = 15
Present age of the son = 15 years
and present age of the man = 3 × 15 = 45 years.
17. After 16 years, Fatima will be three times as old as she is now. Find her present age.
Solution:
Let the present age of Fatima be x years.
After 16 years, age of Fatima be (x + 16) years.
According to the question,
x + 16 = 3 x
⇒ x – 3x = – 16
⇒ – 2 x = – 16
⇒ x = `frac\{– cancel 16}{– cancel 2}`
∴ x = 8
Present age of Fatima = 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 present age of Rahim be x years
Rahim’s age after 32 years = (x + 32) years
Rahim’s age 8 years ago = (x – 8) years
According to the question,
x + 32 = 5 (x – 8)
⇒ x + 32 = 5 x – 40
⇒ x – 5x = – 40 – 32
⇒ – 4 x = – 72
⇒ x = `frac\{cancel 72^18}{cancel 4}`
Present age of Rahim =18 years
19. A bag contains 25-paise and 50-paise coins whose total value is Rs. 30. If the number of 25-paise coins is four times that of 50-paise coins, find the number of each type of coins.
Solution:
Let the number of 50 paise coins be x.
Then, the number of 25 paise coins = 4x
Total value of 50 paise coins = 50 x paisa
and total value of 25 paise coins = 25 x 4 = 100 x paisa
But total value of both the coins = Rs. 30 (Given)
= 30 × 100 paise
= 3000 paisa
According to the question,
50 x + 100 x = 3000
⇒ 150 x = 3000
⇒ x = `frac\{cancel 3000^20}{cancel 150}`
∴ x = 20
Number of 50 paise coins = 20
and number of 25 paise coins = 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 the pen be x rupees.
According to the question,
5x = 3x + 17
⇒ 5x – 3 x = 17
⇒ 2x = 17
⇒ x = `frac\{17}{2}`
∴ x = 8.50
Price of the pen = `frac\{17}{2}` rupees
= Rs. 8.50.
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 in the school be x.
Then, the number of boys in the school = (x + 334)
According to the question,
x + (x + 334) = 572
⇒ 2x + 334 = 572
⇒ 2x = 572 – 334
⇒ 2x = 238
⇒ x = `frac\{cancel 238^119}{cancel 2}`
∴ x = 119
Number of girls in the school = 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 be x metres.
Then, the length of die park = 3x metres.
According to the question,
Perimeter of the park = 168 metres
⇒ 2 (x + 3 x) = 168
⇒ 2 × 4x = 168
⇒ 8x = 168
⇒ x = `frac\{cancel 168^21}{cancel 8}`
∴ x = 21
Breadth of the park = 21 metres
and Length of the park = 3 × 21 = 63 metres.
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 be x metres.
Then, the length of the hall = (x + 5) metres .
According to question,
Perimeter of the hall = 74 metres
⇒ 2 (x + x + 5) = 74
⇒ 2 (2x + 5) = 74
⇒ 4x + 10 = 74
⇒ 4x = 74 – 10
⇒ 4x = 64
⇒ x = `frac\{cancel 64^16}{cancel 4}`
∴ x = 16
Breadth of the hall = 16 metres
and length of the hall = 16 + 5 = 21 metres.
24. A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.
Solution:
Since a wire of length 86 cm is bent to form rectangle,
So, the perimeter of the rectangle = 86 cm.
Let the breadth of the rectangle = x cm
Then, the length of the rectangle = (x + 7) cm
2 (x + x + 7) = 86
⇒ 2 (2x + 7) = 86 .
⇒ 4x + 14 = 86
⇒ 4x = 86 – 14
⇒ 4x = 72
⇒ x = `frac\{cancel 72^18}{cancel 4}`
∴ x = 18
Breadth of the rectangle = 18 cm
Length of the rectangle = (18 + 7) = 25 cm.
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