# Class 8 RD Sharma Solutions- Chapter 9 Linear Equation In One Variable – Exercise 9.4 | Set 1

### Question 1. Four-fifth of a number is more than three-fourth of the number by 4. Find the number.

Solution:

Let us consider the number as ‘x’
Three-fourth of the number = 3x/4
Fourth-fifth of the number = 4x/5
4x/5 – 3x/4 = 4
Now we will take LCM of 5 and 4 is 20
(16x – 15x)/20 = 4
Now by doing cross-multiplying we get,
16x – 15x = 4(20)
x = 80
The number is 80.

### Question 2. The difference between the squares of two consecutive numbers is 31. Find the numbers.

Solution:

Let us assume that two consecutive numbers be x and (x – 1)
According to question
x2 – (x-1)2 = 31
As we know the formula (a-b)2 = a2 + b2 – 2ab
x2 – (x2 – 2x + 1) = 31
x2 – x2 + 2x – 1 = 31
2x – 1 = 31
2x = 31+1
2x = 32
x = 32/2 = 16
Two consecutive numbers are x and (x-1) : 16 and (16-1) = 15
The two consecutive numbers are 16 and 15.

### Question 3. Find a number whose double is 45 greater than its half.

Solution:

Let us assume that the number is ‘x’
2x – x/2 = 45
(4x-x)/2 = 45
Now we will do cross-multiplying,
3x = 90
x = 90/3 = 30
The number will be 30.

### Question 4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4 more than twice the number.

Solution:

Let us assume that number is ‘x’
Then, five times the number will be 5x (According to the question)
two times the number will be 2x
5x – 5 = 2x + 4
5x – 2x = 5 + 4
3x = 9
x = 9/3
x = 3

### Question 5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

Solution:

Let us assume that number is ‘x’
x/5 + 5 = x/4 – 5
x/5 – x/4 = -5 – 5
Now take LCM for 5 and 4 which is 20
(4x-5x)/20 = -10
Now we will do cross-multiplying,
4x – 5x = -10(20)
-x = -200
x = 200
The number is 200.

### Question 6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.

Solution:

Let us assume that one of the digit be ‘x’
The other digit is 9-x
The two digit number is 10(9-x) + x
The number obtained after interchanging the digits is 10x + (9-x) [According to question]
10(9-x) + x – 27 = 10x + (9-x)
By doing simplification,
90 – 10x + x – 27 = 10x + 9 – x
-10x + x – 10x + x = 9 – 90 + 27
-18x = -54
x = 54/18
= 9/3
= 3

The two digit number is 10(9-x) + x
By substituting the value of x we get,
10(9-x) + x
10(9 – 3) + 3
10(6) + 3
60+3 = 63
The number is 63.

### Question 7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Solution:

Let assume that one of the number be ‘x’
And the other number as 184 – x
According to question, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM of 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
By doing cross-multiplying,
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
10x = 720
x = 720/10 = 72
One of the number is 72
other number is 184 – x
= 184 – 72 = 112.

### Question 8. The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?

Solution:

Let us assume that denominator as x and numerator as (x-6)
As we know that formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
Now By doing cross-multiplying
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
x = 9
The denominator is x = 9,then numerator is (x-6) = (9-6) = 3
fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3

### Question 9. A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.

Solution:

Let us assume that number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount of 10Rs notes = 10 × x = 10x
Amount of 20Rs notes = 20 × (50 – x) = 1000 – 20x
The total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10 = 20
The number of 10Rs notes = 20
Number of 20Rs notes are 50 – 20 = 30

### Question 10. Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty- five paise coins as she has fifty- paise coins. How many coins of each kind does she have?

Solution:

Let assume that number of fifty paise coins be x
Number of twenty-five paise coins will be 2x
Amount of fifty paise coins = (50×x)/100 = 0.50x
Amount of twenty-five paise coins = (25×2x)/100 = 0.50x
The total amount is Rs 9
0.50x + 0.50x = 9
1x = 9
x = 9
The number of fifty paise coins is x = 9
Number of twenty-five paise coins 2x = 2×9 = 18

### Question 11. Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?

Solution:

Let assume that present age of Ashima be ‘x’ years
Let the present age of Sunita is 2x years
So , Ashima’s new age = (x – 6) years
Sunita’s new age = (2x + 4) years
(2x + 4) = 4 (x – 6)
2x + 4 = 4x – 24
2x – 4x = -24 – 4
-2x = -28
x = -28/-2 = 14
Age of Ashima is x years = 14 years
Age of Sunita is 2x years = 2(14) = 28 years
Now ,
Two years ago, age of Ashima is 14 – 2 = 12 years
Age of Sunita = 28 – 2 = 26 years.

### Question 12. The ages of Sonu and Monu are in the ratio 7:5 Ten years hence, the ratio of their ages will be 9:7 find their present ages.

Solution:

Let us assume that present age of Sonu be 7x years
Present age of Monu will be 5x years
Sonu’s age after 10 years will be = (7x + 10) years
Monu’s age after 10 years will be = (5x + 10) years
(7x + 10) / (5x + 10) = 9/7
By doing cross-multiplication,
7(7x + 10) = 9(5x + 10)
49x + 70 = 45x + 90
49x – 45x = 90 – 70
4x = 20
x = 20/4 = 5
Present age of Sonu will be 7x = 7(5) = 35years
Present age of Monu will be 5x = 5(5) = 25years

### Question 13. Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Solution:

Let us assume that age of son five years ago be x years
The age of man five years ago will be 7x years
After five years, son’s age will be x + 5 years
After five years father’s age will be 7x + 5 years
So, five years the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
4x = 20
x = 5
Present father’s age will be 7x + 5 = 7(5) + 5 = 35 + 5 = 40 years
Present son’s age will be x + 5 = 5 + 5 = 10 years

### Chapter 9 Linear Equation In One Variable – Exercise 9.4 | Set 2

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