Chapter 9: Linear Equation in One Variable Exercise – 9.4

Question: 1

Four-fifth of a number is more than three-fourths of the number by 4. Find the number.

Solution:

Let the number be x.
According to the question,

Thus, the required number is 80.

 

Question: 2

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

Solution:

Let the numbers be x and x + 1.

According to the question,

(x + 1)² – x² = 31

or x² + 2x + 1 – x²  = 31

or 2x = 31 – 1

or x = 30/2

or x = 15

Thus, the numbers are 15 and 16.

Question: 3

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

Solution:

Let the number be x.
According to the question,

 

Question: 4

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

Solution:

Let the number be x.
According to the question,

5x – 5 = 2x + 4

Or 5x – 2x = 4 + 5

Or 3x = 9

Or x = 9/3

Or x = 3

Thus, the number is 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 the number be x.
According to the question,

 

Question: 6

A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number.

Solution:

Let the units digit be x.

Since, Sum of two digits = 9

∴ Tens digit = (9 – x)

∴ Original number = 10 × (9 – x) + x

Reversed number = 10x + (9 – x)

According to the question,

10 × (9 – x) + x – 27 = 10x + (9 – x)

Or 90 – 10x + x – 27 = 10x + 9 – x

Or 9x + 9x = 90 – 27 – 9

Or 18x = 54

Or x = 54/18 = 3

The number = 10 × (9 – 3) + 3 = 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 the first part of 184 be x.

Therefore, the other part will be (184 – x).

According to the question,

10x – 552 = 168 [After cross multiplication]

Or 10x = 168 + 552

Or 10x = 720

Or x = 720/10 = 72

Thus, the parts of 184 are 72 and 112 (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 the denominator of the fraction be x.

Therefore, the numerator will be (x – 6).

 

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 the number of Rs 10 notes be x.

Therefore, the number of Rs 20 notes will be (50 – x).

Value of Rs 10 notes = 10x

Value of Rs 20 notes = 20 (50 – x)

According to the question,

10x + 20(50 – x) = 800

Or 10x + 1000 – 20x = 800

Or 10x = 1000 – 800

Or x = 200/10 = 20

Therefore, Number of Rs 10 notes = 20

Number of Rs 20 notes = (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 the number of 50 paise coins be x.

Therefore, the number of 25 paise coins will be 2x.

Value of 50 paise coins = Rs. 0.5x

Value of 25 paise coins = Rs. 0.25 × 2x

According to the question,

0.5x + 0.25 × 2x = 9

or x = 9

∴ Number of fifty paise coins = 9

Number of twenty five paise coins = 2 × 9 = 18

Total number of coins = 9 + 18 = 27.

 

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 the age of Ashima be x years.

Therefore, the age of Sunita will be 2x years.

According to the question,

4(x – 6) = 2x + 4

or 4x – 24 = 2x + 4

or 4x – 2x = 4 + 24

or 2x = 28

or x = 14

∴ Age of Ashima = 14 years.

Age of Sunita = 2 × 14 = 28 years.

 

Question: 12

The ages of Sonu and Monti are in the ratio 7: 5. Ten years hence, the ratio of their ages will be 9: 7. Find their present ages.

Solution:

It is given that the ratio of the ages of Sonu and Monu is 7: 5.

Let the present ages of Sonu and Monu be 7x and 5x years.

After ten years:

Age of Sonu = 7x + 10 years

Age of Monu = 5x + 10 years

According to the question,

Or 49x + 70 = 45x + 90

Or 49x – 45x = 90 – 70

Or 4x = 20

Or x = 5

∴ Present age of Sonu = 7 × 5 = 35 years.

Present age of Monu = 5 × 5 = 25 years.

 

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:

Five years ago:

Let the age of the son be x years.

Therefore, the age of the father will be 7x years.

∴ Present age of the son = (x + 5) years

Present age of the father = (7x + 5) years

After five years:

Age of the son = (x + 5 + 5) = (x + 10) years

Age of the father = (7x + 5 + 5) = (7x + 10) years

According to the question, 7x + 10 = 3 (x + 10)

or 7x – 3x = 30 – 10

or 4x = 20

or x = 5

∴ Present age of the son = (5 + 5) = 10 years.

Present age of the father = (7 × 5 + 5) = 40 years.

 

Question: 14

I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?

Solution:

Let the age of my son be x years.

Therefore, my age will be 5x years.

After 6 years:

Age of my son = (x + 6) years

My age = (5x + 6) years

According to the question,

5x + 6 = 3 (x + 6)

or 5x – 3x = 18 – 6

or 2x = 12

or x = 6

∴ Age of my son = 6 years.

My age = 5 × 6 = 30 years.

 

Question: 15

I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Solution:

Let the number of five – rupee notes be x.

Therefore, the number of ten- rupee notes will be (x +10).

Now,

Value of five – rupee notes = Rs. 5x

Value of ten – rupee notes = Rs. 10 (x + 10)

According to the question,

5x + 10 (x + 10) = 1000

or 15x = 1000 – 100

or x = 90015 = 60

∴ Number of five – rupee notes = 60.

Number of ten – rupee notes = 60 + 10 = 70.

 

Question: 16

At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two-fifths drank fruit juice and just three did not drink anything. How many guests were in all?

Solution:

Let the total number of guests be x.

Therefore, the number of guests, who drank colas, would be (1/4)x.

The number of guests, who drank squash, would be (1/3)x.

The number of guests, who drank fruit juice, would be (2/5)x.

The number of guests, who did not drink, would be 3.

According to the question,

or x = 180

Thus, total number of guests = 180.

Question: 17

There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

Solution:

Let the number of correctly answered questions be x.

Therefore, the number of unattempted or wrongly answered questions will be (180 – x).

According to the question,

4x – 1 (180 – 1) = 450

or 5x = 450 + 180

or x = 630/5 = 126

Thus, number of correctly answered questions = 126.

Number of unattempted or wrongly answered questions = 180 – 126 = 54.