RD Sharma Solutions Class 7 Linear Equations In One Variable Exercise 8.4

RD Sharma Class 7 Solutions Chapter 8 Ex 8.4 PDF Free Download

RD Sharma Solutions Class 7 Chapter 8 Exercise 8.4

Exercise 8.4

Q 1. If 5 is subtracted from three times a number, the result is 16. Find the number.

SOLUTION :

Let us consider that the required number be ‘y’. Then subtract 5 from 3y = 3y – 5.

=> 3y – 5 = 16

Add 5 to both sides, we get = 3y – 5 + 5 = 16 + 5

=> 3y = 21

Divide both sides by three, we get

\(\frac{3y}{3}\) = \(\frac{21}{3}\)

y =7

Therefore, the required no. is 7.

Q 2. Find the number which when multiplied by 7 is increased by 78.

SOLUTION :

Let us consider that the number required to be ‘y’. Let us multiply it by 7, it gives 7y, and y is increased by 78.

7y =y + 78

Transpose y to LHS, we will  get

7y – y= 78

6y = 78

Divide both sides by 6, we get

\(\frac{6y}{6}\) = \(\frac{78}{6}\)

y =13

Therefore, the required number is 13.

Q 3. Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.

SOLUTION :

Let us consider that the first number be ‘y’.

So, the 2nd number = y + 1 and the third number = y + 2.

=> Sum of 1st and 2nd numbers = (y) + (y + 1).

Acc to question:

(y) + (y + 1) = 15 + (y + 2)

=> 2y + 1 = 17 + y

Transposing x to LHS and 1 to RHS, we get

=> 2y – y = 17 – 1

=>y = 16

So, 1st number = y = 16, 2nd number = y + 1 = 16 + 1 = 17 and 3rd number = y + 2 = 16 + 2 = 18

Therefore, the required consecutive natural numbers are 16, 17 and 18.

Q 4. The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

SOLUTION :

Let us consider that the smaller number be ‘y’.

So, the larger number is = y + 7.

Acc to question:

6y + (y + 7) = 77

6y + y+ 7 = 77

7y + 7 = 77

Subtract 7 from both sides, we will get

7y + 7 – 7 = 77 – 7

7y = 70

Divide both sides by 7, we will get

\(\frac{7y}{7}\) = \(\frac{70}{7}\)

y = 10

Therefore, the smaller number = y = 10,

and the larger number = y + 7 = 10 + 7 = 17.

The 2 required no.s are 10 and 17.

Q 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 us consider that the number assumed by the man be ‘y’.

So, Acc to question:

\(\frac{y}{3}\) + 5 = 2y

Transpose \(\frac{y}{3}\) to RHS, we get

5 = 2y – \(\frac{y}{3}\)

5 = 6y – \(\frac{y}{3}\)

5 = \(\frac{5y}{3}\)

Multiply both sides by 3, we will get

5\(\times\)3 = \(\frac{5y}{3}\) \(\times\)3

15 = 5y

Divide both the sides by 5, we will get

\(\frac{15}{5}\)= \(\frac{5y}{5}\)

y = 3

Therefore, the number assumed by the man is 3.

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

SOLUTION :

Let us consider that the required number be ‘y’.

So, Acc. to question:

=> 3y + 5 = 50

Subtract 5 from both sides, we get

=> 3y + 5 – 5 = 50 – 5

=> 3y = 45

Divide both sides by 3, we get

\(\frac{3y}{3}\) = \(\frac{45}{3}\)

y = 15

Therefore, the required number is 15.

Q 7. Shikha is 3 years younger to her brother Ravish. If the sum of their ages is 37 years, what are their present ages?

SOLUTION :

Let us consider that the present age of Shikha = ‘y’ years.

So, the present age of Shikha’s brother Ravish = (y + 3) years.

So, sum of their ages = y + (y+ 3)

=> y +(y + 3 )= 37

=> 2y + 3 = 37

Subtract 3 from both sides, we get

=> 2y+ 3 – 3 = 37 – 3

=> 2y = 34

Divide both sides by 2, we get

=> \(\frac{2y}{2}\) = \(\frac{34}{2}\)

=>Therefore, y = 17

So, the present age of Shikha = 17 years, and the present age of Ravish = y + 3 = 17 + 3 = 20 years.

Q 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 us consider that the present age of Nilu = ‘y’ years.

Thus, the present age of Nilu’s mother, Mrs. Jain = (y + 27) years.

Therefore, after 8 years,

Nilu’s age = (y + 8), and Mrs. Jain’s age = (y + 27 + 8) = (y + 35) years

=> y + 35 = 2(y + 8)

Expanding the brackets, we get

=> y + 35 = 2y + 16

Transpose y to RHS and 16 to LHS, we get

=> 35 – 16 = 2y – y

=> y = 19 years

Therefore, the present age of Nilu = y = 19 years,

and the present age of Nilu’s mother = y+ 27 = 19 + 27 = 46 years.

Q 9. A man is 4 times as old as his son . After 16 years, he will be only twice as old as his son . Find the their present ages.

SOLUTION :

Let us consider that the present age of the son = ‘y’ years.

Therefore, the present age of his father = ‘4y’ years.

Thus, after 16 years,

Son’s age = (y + 16) and father’s age = (4y + 16) years

Acc.to question:

=> 4y + 16 = 2(y + 16)

=> 4y + 16 = 2y + 32

Transpose 2y to LHS and 16 to RHS, we get

=> 4y – 2y = 32 – 16

=> 2y= 16

Divide both sides by 2, we will get

\(\frac{2y}{2}\) = \(\frac{16}{2}\)

=>y = 8

So, the present age of the son = y = 8 years,

& the present age of the father = 4y = 4 (8) = 32 years.

Q 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 us consider that the girl’s = ‘y’ years.

So, the age of her younger sister is = (y – 4) years.

And so, the age of the brother = (y – 4 – 4) years i.e =  (y – 8) years.

According to question:

(y – 4) + (y – 8) = 16

y + y – 4 – 8 = 16

2y – 12 = 16

Add 12 to both sides, we get

2y – 12 + 12 = 16 + 12

2y = 28

Divideboth sides by 2, we get

\(\frac{2y}{2}\) = \(\frac{28}{2}\)

y = 14

Thus, the age of the girl = y= 14 years, the age of the younger sister = y – 4 = 14 – 4 = 10 years, and the age of the younger brother = y- 8 = 14 – 8 = 6 years.

Q 11. One day, during their vacation at a beach resort, Shelia found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Sheila found 16 sea shells. How many did each of them find?

SOLUTION :

Let us consider that the number of seashells found by Sandy = ‘y’.

So, the number of seashells found by Anita = (y + 5).

The number of seashells found by Shelia = 2 (y + 5 ).

According to the question,

=> y + 2(y + 5) = 16

=> y + 2y +10 = 16 = 3y + 10 = 16

Subtract 10 from both sides, we get

=> 3y + 10 – 10 = 16 – 10

=> 3y = 6

Divide both sides by 3, we get

=> \(\frac{3y}{6}\) = \(\frac{6}{3}\)

=> y = 2

Therefore, the number of sea shells found by Sandy = y = 2, the number of sea shells found by Anita = y + 5 = 2 + 5 = 7,

and the no. of sea shells found by Shelia = 2(y + 5) = 2(2 + 5) = 2(7 ) = 14.

Q 12. Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has 75 marbles more than Sandy. How many does each of them have?

SOLUTION :

Let us consider that the number of marbles with Pandy = ‘y’.

So, the no. of marbles with Andy = ‘2y’.

Thus, the number of marbles with Sandy = \(\frac{y}{2}\) + \(\frac{2y}{2}\) = \(\frac{3y}{2}\).

According to the question,

2y + 75 = \(\frac{3y}{2}\)

2y – \(\frac{3y}{2}\) = -75

\(\frac{4y-3y}{2}\) = -75

\(\frac{y}{2}\) = -75

y = -150

Since, no. of marbles cannot be negative.

Therefore,y = 150

Therefore, Pandy has 150 marbles, Andy has 2y = 2(150) = 300 marbles,

and Sandy has \(\frac{3y}{2}\) = 225 marbles.

Q 13. 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 us consider that the number of 50 paise coins = ‘y’.

So, the money value contribution of 50 paise coins = 0.5y.

The no. of 25 paise coins = ‘4y’.

The money value contribution of 25 paise coins = 0.25(4y) = y.

Acc to the question,

0.5y + y= 30

=> 1.5y = 30

Divide both sides by 1.5, we get

=> \(\frac{1.5y}{1.5}\)= \(\frac{30}{1.5}\)

=> y = 20

Therefore, the number of 50 paise coins = ‘y’ = 20, and the number of 25 paise coins = ‘4y’ = 4 (20) = 80.

Q 14. The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres, find the dimensions of the field.

SOLUTION :

Let us consider that the breadth of the rectangle = ‘y’ metres.

According to the question,

Length of the rectangle = ‘2y’ metres

Perimeter of a rectangle = 2 (length + breadth)

So, 2 (2y + y) = 228

=> 2 (3y) = 228

=> 6y = 228

Divide both sides by 6, we get

=> \(\frac{6y}{6}\) = \(\frac{228}{6}\)

=> y = 38

Therefore, the breadth of the rectangle = y= 38 metres,

& the length of the rectangle = 2y = 2(38) = 76 metres.

Q 15. There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.

SOLUTION :

Let us consider that the no. of 25-paise coins in the purse be ‘y’.

So, the value of money in the purse = 0.25y.

But 0.25y= 17.5.

Divide both sides by 0.25, we get

=> \(\frac{0.25y}{0.25}\) = \(\frac{17.5}{0.25}\)

=> y = 70

Therefore, the number of 25-paise coins in the purse = 70.

Q 16. In a hostel mess, 50 kg rice are consumed everyday. If each student gets 400 gm of rice per day , find the number of students who take meals in the hostel mess.

SOLUTION :

Let us consider that the number of students in the hostel be ‘y’.

The quantity of rice consumed by each student = 400 gm.

So the daily rice consumption in the hostel mess = 400(y).

But, the daily rice consumption = 50 kg = 50 \(\times\) 1000 = 50000 gm [since 1 kg = 1000 gm].

Acc. to the question,

400y = 50000

Divide both sides by 400, we get

=> \(\frac{400y}{400}\) = \(\frac{50000}{400}\)

=> y = 125

Therefore, 125 students have their meals in the hostel mess.

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