Linear Equations in One Variable (Exercise 8.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download

 Page 1


 
 
 
 
 
 
 
Exercise 8.4        Page No: 8.26 
 
1. If 5 is subtracted from three times a number, the result is 16. Find the number. 
 
Solution: 
Let the required number be x.  
Then, given that 5 subtracted from 3 times x i.e. 3x – 5 
? 3x – 5 = 16 
Adding 5 to both sides, we get  
? 3x – 5 + 5 = 16 + 5 
? 3x = 21 
Dividing both sides by 3, we get 
? 3x/3 = 21/3 
? x = 7 
Thus, the required number is x = 7. 
 
2. Find the number which when multiplied by 7 is increased by 78. 
 
Solution: 
Let the required number be x.  
Given that, when multiplied by 7, it gives 7x, and x increases by 78. 
According to the question we can write as 7x = x + 78 
Transposing x to LHS, we get 
? 7x – x = 78 
? 6x = 78 
Dividing both sides by 6, we get 
? 6x/6 = 78/6 
? x =13 
Thus, the required number is x = 13. 
 
3. Find three consecutive natural numbers such that the sum of the first and second is 
15 more than the third. 
 
Solution: 
Let first number be x. 
According to the question second number is x + 1 and the third is x + 2 
Sum of first and second numbers = (x) + (x + 1). 
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Exercise 8.4        Page No: 8.26 
 
1. If 5 is subtracted from three times a number, the result is 16. Find the number. 
 
Solution: 
Let the required number be x.  
Then, given that 5 subtracted from 3 times x i.e. 3x – 5 
? 3x – 5 = 16 
Adding 5 to both sides, we get  
? 3x – 5 + 5 = 16 + 5 
? 3x = 21 
Dividing both sides by 3, we get 
? 3x/3 = 21/3 
? x = 7 
Thus, the required number is x = 7. 
 
2. Find the number which when multiplied by 7 is increased by 78. 
 
Solution: 
Let the required number be x.  
Given that, when multiplied by 7, it gives 7x, and x increases by 78. 
According to the question we can write as 7x = x + 78 
Transposing x to LHS, we get 
? 7x – x = 78 
? 6x = 78 
Dividing both sides by 6, we get 
? 6x/6 = 78/6 
? x =13 
Thus, the required number is x = 13. 
 
3. Find three consecutive natural numbers such that the sum of the first and second is 
15 more than the third. 
 
Solution: 
Let first number be x. 
According to the question second number is x + 1 and the third is x + 2 
Sum of first and second numbers = (x) + (x + 1). 
 
 
 
 
 
 
 
According to question: 
? (x) + (x + 1) = 15 + (x + 2) 
? 2x + 1 = 17 + x 
Transposing x to LHS and 1 to RHS, we get  
?2x – x = 17 – 1 
? x = 16 
So, first number = x = 16,  
Second number = x + 1 = 16 + 1 = 17  
And third number = x + 2 = 16 + 2 = 18 
Thus, the required consecutive natural numbers are 16, 17 and 18. 
 
4. The difference between two numbers is 7. Six times the smaller plus the larger is 77. 
Find the numbers. 
 
Solution: 
Let the smaller number be ‘x’.  
So, the larger number = x + 7. 
According to question: 
? 6x + (x + 7) = 77 
? 6x + x + 7 = 77 
On simplifying we get 
? 7x + 7 = 77 
Subtracting 7 from both sides, we get 
? 7x + 7 – 7 = 77 – 7 
? 7x = 70 
Dividing both sides by 7, we get 
? 7x/7 = 70/7 
? x = 10 
Thus, the smaller number = x = 10 
And the larger number = x + 7 = 10 + 7 = 17. 
The two required numbers are 10 and 17. 
 
5. A man says, “I am thinking of a number. When I divide it by 3 and then add 5, my 
answer is twice the number I thought of”. Find the number. 
 
Solution: 
Let the required number be x. 
Page 3


 
 
 
 
 
 
 
Exercise 8.4        Page No: 8.26 
 
1. If 5 is subtracted from three times a number, the result is 16. Find the number. 
 
Solution: 
Let the required number be x.  
Then, given that 5 subtracted from 3 times x i.e. 3x – 5 
? 3x – 5 = 16 
Adding 5 to both sides, we get  
? 3x – 5 + 5 = 16 + 5 
? 3x = 21 
Dividing both sides by 3, we get 
? 3x/3 = 21/3 
? x = 7 
Thus, the required number is x = 7. 
 
2. Find the number which when multiplied by 7 is increased by 78. 
 
Solution: 
Let the required number be x.  
Given that, when multiplied by 7, it gives 7x, and x increases by 78. 
According to the question we can write as 7x = x + 78 
Transposing x to LHS, we get 
? 7x – x = 78 
? 6x = 78 
Dividing both sides by 6, we get 
? 6x/6 = 78/6 
? x =13 
Thus, the required number is x = 13. 
 
3. Find three consecutive natural numbers such that the sum of the first and second is 
15 more than the third. 
 
Solution: 
Let first number be x. 
According to the question second number is x + 1 and the third is x + 2 
Sum of first and second numbers = (x) + (x + 1). 
 
 
 
 
 
 
 
According to question: 
? (x) + (x + 1) = 15 + (x + 2) 
? 2x + 1 = 17 + x 
Transposing x to LHS and 1 to RHS, we get  
?2x – x = 17 – 1 
? x = 16 
So, first number = x = 16,  
Second number = x + 1 = 16 + 1 = 17  
And third number = x + 2 = 16 + 2 = 18 
Thus, the required consecutive natural numbers are 16, 17 and 18. 
 
4. The difference between two numbers is 7. Six times the smaller plus the larger is 77. 
Find the numbers. 
 
Solution: 
Let the smaller number be ‘x’.  
So, the larger number = x + 7. 
According to question: 
? 6x + (x + 7) = 77 
? 6x + x + 7 = 77 
On simplifying we get 
? 7x + 7 = 77 
Subtracting 7 from both sides, we get 
? 7x + 7 – 7 = 77 – 7 
? 7x = 70 
Dividing both sides by 7, we get 
? 7x/7 = 70/7 
? x = 10 
Thus, the smaller number = x = 10 
And the larger number = x + 7 = 10 + 7 = 17. 
The two required numbers are 10 and 17. 
 
5. A man says, “I am thinking of a number. When I divide it by 3 and then add 5, my 
answer is twice the number I thought of”. Find the number. 
 
Solution: 
Let the required number be x. 
 
 
 
 
 
 
 
So, according to question: 
? x/3 + 5 = 2x 
Transposing x/3 to RHS, we get 
? 5 = 2x – (x/3) 
? 5 = (6x – x)/3 
? 5 = (5x/3) 
Multiplying both sides by 3 we get, 
? 5 × 3 = (5x/3) × 3 
? 15 = 5x 
Dividing both sides by 5 we get 
? 15/5 = 5x/5 
? 3 = x 
Thus the number thought of by the man is 3. 
 
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 question: 
? 3x + 5 = 50 
Subtracting 5 from both sides, we get 
? 3x + 5 – 5 = 50 – 5 
? 3x = 45 
Dividing both sides by 3, we get 
? 3x/3 = 45/3 
? x = 15 
Therefore, the required number is 15. 
  
7. Shikha is 3 years younger to her brother Ravish. If the sum of their ages 37 years, 
what are their present age? 
 
Solution: 
Let the present age of Shikha be x years. 
Therefore, the present age of Shikha’s brother Ravish = (x + 3) years. 
So, sum of their ages = x + (x+ 3) 
? x +(x + 3) = 37 
? 2x + 3 = 37 
Page 4


 
 
 
 
 
 
 
Exercise 8.4        Page No: 8.26 
 
1. If 5 is subtracted from three times a number, the result is 16. Find the number. 
 
Solution: 
Let the required number be x.  
Then, given that 5 subtracted from 3 times x i.e. 3x – 5 
? 3x – 5 = 16 
Adding 5 to both sides, we get  
? 3x – 5 + 5 = 16 + 5 
? 3x = 21 
Dividing both sides by 3, we get 
? 3x/3 = 21/3 
? x = 7 
Thus, the required number is x = 7. 
 
2. Find the number which when multiplied by 7 is increased by 78. 
 
Solution: 
Let the required number be x.  
Given that, when multiplied by 7, it gives 7x, and x increases by 78. 
According to the question we can write as 7x = x + 78 
Transposing x to LHS, we get 
? 7x – x = 78 
? 6x = 78 
Dividing both sides by 6, we get 
? 6x/6 = 78/6 
? x =13 
Thus, the required number is x = 13. 
 
3. Find three consecutive natural numbers such that the sum of the first and second is 
15 more than the third. 
 
Solution: 
Let first number be x. 
According to the question second number is x + 1 and the third is x + 2 
Sum of first and second numbers = (x) + (x + 1). 
 
 
 
 
 
 
 
According to question: 
? (x) + (x + 1) = 15 + (x + 2) 
? 2x + 1 = 17 + x 
Transposing x to LHS and 1 to RHS, we get  
?2x – x = 17 – 1 
? x = 16 
So, first number = x = 16,  
Second number = x + 1 = 16 + 1 = 17  
And third number = x + 2 = 16 + 2 = 18 
Thus, the required consecutive natural numbers are 16, 17 and 18. 
 
4. The difference between two numbers is 7. Six times the smaller plus the larger is 77. 
Find the numbers. 
 
Solution: 
Let the smaller number be ‘x’.  
So, the larger number = x + 7. 
According to question: 
? 6x + (x + 7) = 77 
? 6x + x + 7 = 77 
On simplifying we get 
? 7x + 7 = 77 
Subtracting 7 from both sides, we get 
? 7x + 7 – 7 = 77 – 7 
? 7x = 70 
Dividing both sides by 7, we get 
? 7x/7 = 70/7 
? x = 10 
Thus, the smaller number = x = 10 
And the larger number = x + 7 = 10 + 7 = 17. 
The two required numbers are 10 and 17. 
 
5. A man says, “I am thinking of a number. When I divide it by 3 and then add 5, my 
answer is twice the number I thought of”. Find the number. 
 
Solution: 
Let the required number be x. 
 
 
 
 
 
 
 
So, according to question: 
? x/3 + 5 = 2x 
Transposing x/3 to RHS, we get 
? 5 = 2x – (x/3) 
? 5 = (6x – x)/3 
? 5 = (5x/3) 
Multiplying both sides by 3 we get, 
? 5 × 3 = (5x/3) × 3 
? 15 = 5x 
Dividing both sides by 5 we get 
? 15/5 = 5x/5 
? 3 = x 
Thus the number thought of by the man is 3. 
 
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 question: 
? 3x + 5 = 50 
Subtracting 5 from both sides, we get 
? 3x + 5 – 5 = 50 – 5 
? 3x = 45 
Dividing both sides by 3, we get 
? 3x/3 = 45/3 
? x = 15 
Therefore, the required number is 15. 
  
7. Shikha is 3 years younger to her brother Ravish. If the sum of their ages 37 years, 
what are their present age? 
 
Solution: 
Let the present age of Shikha be x years. 
Therefore, the present age of Shikha’s brother Ravish = (x + 3) years. 
So, sum of their ages = x + (x+ 3) 
? x +(x + 3) = 37 
? 2x + 3 = 37 
 
 
 
 
 
 
 
Subtracting 3 from both sides, we get 
? 2x+ 3 – 3 = 37 – 3 
? 2x = 34 
Dividing both sides by 2, we get 
? 2x/2 = 34/2 
? x = 17 
Therefore, the present age of Shikha = 17 years,  
And the present age of Ravish = x + 3 = 17 + 3 = 20 years. 
 
8. Mrs Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as 
old as Nilu. Find their present ages? 
 
Solution: 
Let the present age of Nilu be x years 
Therefore the present age of Nilu’s mother = (x + 27) years 
So, after 8 years, 
Nilu’s age = (x + 8), and Mrs Jain’s age = (x + 27 + 8) = (x + 35) years 
? x + 35 = 2(x + 8) 
Expanding the brackets, we get 
? x + 35 = 2x + 16 
Transposing x to RHS and 16 to LHS, we get 
? 35 – 16 = 2x – x 
? x = 19 
So, the present age of Nilu = x = 19 years,  
And the present age of Nilu’s mother that is Mrs Jain = x+ 27 = 19 + 27 = 46 years. 
 
9. A man 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 = x years. 
Therefore, the present age of his father = 4x years. 
So, after 16 years, 
Son’s age = (x + 16) and father’s age = (4x + 16) years 
According to question: 
? 4x + 16 = 2(x + 16) 
? 4x + 16 = 2x + 32 
Page 5


 
 
 
 
 
 
 
Exercise 8.4        Page No: 8.26 
 
1. If 5 is subtracted from three times a number, the result is 16. Find the number. 
 
Solution: 
Let the required number be x.  
Then, given that 5 subtracted from 3 times x i.e. 3x – 5 
? 3x – 5 = 16 
Adding 5 to both sides, we get  
? 3x – 5 + 5 = 16 + 5 
? 3x = 21 
Dividing both sides by 3, we get 
? 3x/3 = 21/3 
? x = 7 
Thus, the required number is x = 7. 
 
2. Find the number which when multiplied by 7 is increased by 78. 
 
Solution: 
Let the required number be x.  
Given that, when multiplied by 7, it gives 7x, and x increases by 78. 
According to the question we can write as 7x = x + 78 
Transposing x to LHS, we get 
? 7x – x = 78 
? 6x = 78 
Dividing both sides by 6, we get 
? 6x/6 = 78/6 
? x =13 
Thus, the required number is x = 13. 
 
3. Find three consecutive natural numbers such that the sum of the first and second is 
15 more than the third. 
 
Solution: 
Let first number be x. 
According to the question second number is x + 1 and the third is x + 2 
Sum of first and second numbers = (x) + (x + 1). 
 
 
 
 
 
 
 
According to question: 
? (x) + (x + 1) = 15 + (x + 2) 
? 2x + 1 = 17 + x 
Transposing x to LHS and 1 to RHS, we get  
?2x – x = 17 – 1 
? x = 16 
So, first number = x = 16,  
Second number = x + 1 = 16 + 1 = 17  
And third number = x + 2 = 16 + 2 = 18 
Thus, the required consecutive natural numbers are 16, 17 and 18. 
 
4. The difference between two numbers is 7. Six times the smaller plus the larger is 77. 
Find the numbers. 
 
Solution: 
Let the smaller number be ‘x’.  
So, the larger number = x + 7. 
According to question: 
? 6x + (x + 7) = 77 
? 6x + x + 7 = 77 
On simplifying we get 
? 7x + 7 = 77 
Subtracting 7 from both sides, we get 
? 7x + 7 – 7 = 77 – 7 
? 7x = 70 
Dividing both sides by 7, we get 
? 7x/7 = 70/7 
? x = 10 
Thus, the smaller number = x = 10 
And the larger number = x + 7 = 10 + 7 = 17. 
The two required numbers are 10 and 17. 
 
5. A man says, “I am thinking of a number. When I divide it by 3 and then add 5, my 
answer is twice the number I thought of”. Find the number. 
 
Solution: 
Let the required number be x. 
 
 
 
 
 
 
 
So, according to question: 
? x/3 + 5 = 2x 
Transposing x/3 to RHS, we get 
? 5 = 2x – (x/3) 
? 5 = (6x – x)/3 
? 5 = (5x/3) 
Multiplying both sides by 3 we get, 
? 5 × 3 = (5x/3) × 3 
? 15 = 5x 
Dividing both sides by 5 we get 
? 15/5 = 5x/5 
? 3 = x 
Thus the number thought of by the man is 3. 
 
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 question: 
? 3x + 5 = 50 
Subtracting 5 from both sides, we get 
? 3x + 5 – 5 = 50 – 5 
? 3x = 45 
Dividing both sides by 3, we get 
? 3x/3 = 45/3 
? x = 15 
Therefore, the required number is 15. 
  
7. Shikha is 3 years younger to her brother Ravish. If the sum of their ages 37 years, 
what are their present age? 
 
Solution: 
Let the present age of Shikha be x years. 
Therefore, the present age of Shikha’s brother Ravish = (x + 3) years. 
So, sum of their ages = x + (x+ 3) 
? x +(x + 3) = 37 
? 2x + 3 = 37 
 
 
 
 
 
 
 
Subtracting 3 from both sides, we get 
? 2x+ 3 – 3 = 37 – 3 
? 2x = 34 
Dividing both sides by 2, we get 
? 2x/2 = 34/2 
? x = 17 
Therefore, the present age of Shikha = 17 years,  
And the present age of Ravish = x + 3 = 17 + 3 = 20 years. 
 
8. Mrs Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as 
old as Nilu. Find their present ages? 
 
Solution: 
Let the present age of Nilu be x years 
Therefore the present age of Nilu’s mother = (x + 27) years 
So, after 8 years, 
Nilu’s age = (x + 8), and Mrs Jain’s age = (x + 27 + 8) = (x + 35) years 
? x + 35 = 2(x + 8) 
Expanding the brackets, we get 
? x + 35 = 2x + 16 
Transposing x to RHS and 16 to LHS, we get 
? 35 – 16 = 2x – x 
? x = 19 
So, the present age of Nilu = x = 19 years,  
And the present age of Nilu’s mother that is Mrs Jain = x+ 27 = 19 + 27 = 46 years. 
 
9. A man 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 = x years. 
Therefore, the present age of his father = 4x years. 
So, after 16 years, 
Son’s age = (x + 16) and father’s age = (4x + 16) years 
According to question: 
? 4x + 16 = 2(x + 16) 
? 4x + 16 = 2x + 32 
 
 
 
 
 
 
 
Transposing 2x to LHS and 16 to RHS, we get 
? 4x – 2x = 32 – 16 
? 2x = 16 
Dividing both sides by 2, we get 
? 2x/2 = 16/2 
? x = 8 
So, the present age of the son = x = 8 years,  
And the present age of the father = 4x = 4(8) = 32 years. 
 
10. The difference in age between a girl and her younger sister is 4 years. The younger 
sister in turn is 4 years older than her brother. The sum of the ages of the younger 
sister and her brother is 16. How old are the three children? 
 
Solution: 
Let the age of the girl = x years. 
So, the age of her younger sister = (x – 4) years. 
Thus, the age of the brother = (x – 4 – 4) years = (x – 8) years. 
According to question: 
? (x – 4) + (x – 8) = 16 
? x + x – 4 – 8 = 16 
? 2x – 12 = 16 
Adding 12 to both sides, we get 
? 2x – 12 + 12 = 16 + 12 
? 2x = 28 
Dividing both sides by 2, we get 
? 2x/2 = 28/2 
? x = 14 
Thus, the age of the girl = x = 14 years, 
The age of the younger sister = x – 4 = 14 – 4 = 10 years,  
The age of the younger brother = x – 8 = 14 – 8 = 6 years. 
 
11. One day, during their vacation at beach resort, Shella found twice as many sea 
shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella 
found 16 sea shells. How many did each of them find? 
 
Solution: 
Let the number of sea shells found by sandy = x