Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.

Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number, you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

(g) Anwar thinks of a number. If he takes away 7 from \(\frac { 5 }{ 2 }\) of the number, the result is 23.

Solution:

(a) Let the required number be x.

∴ Eight times of the number = 8x

According to the condition, we have

8x + 4 = 60

Transposing 4 to R.H.S., we have

8x = 60 – 4

or 8x = 56

Dividing both sides by 8, we have

\(\frac { 8x }{ 8 }\) = \(\frac { 56 }{ 8 }\)

or x = 7

∴ The required number = 7

(b) Let a number be x.

∴ One-fifth of the number = \(\frac { 1 }{ 5 }\)x

According to the condition, we have

\(\frac { 1 }{ 5 }\)x – 4 = 3 or \(\frac { 1 }{ 5 }\)x = 3 + 4

[Transposing – 4 to R.H.S.]

or \(\frac { x }{ 5 }\) = 7

Multiplying both sides by 5, we have

\(\frac { x }{ 5 }\) x 5 = 7 x 5

or x = 35

Thus, the required number = 35

(c) Let the number be x.

Then three-fourths of the number = \(\frac { 3 }{ 4 }\)x

According to the condition, we have

\(\frac { 3 }{ 4 }\)x + 3 = 21

or \(\frac { 3 }{ 4 }\)x = 21 – 3

[Transposing 3 to R.H.S.]

or \(\frac { 3 }{ 4 }\)x = 18

Multiplying both sides by \(\frac { 4 }{ 3 }\), we have

\(\frac { 3 }{ 4 }\)x × \(\frac { 4 }{ 3 }\) = 18 x \(\frac { 4 }{ 3 }\)

or x = 6 x 4

x = 24

Thus, the required number = 24

(d) Let the number be x.

∴ Twice the number = 2x

According to the condition, we have

2x – 11 = 15

or 2x = 15 + 11

[Transposing – 11 to R.H.S.]

2x = 26

Dividing both sides by 2, we get

\(\frac { 2x }{ 2 }\) = \(\frac { 26 }{ 2 }\)

∴ x = 13

Thus, the required number =13

(e) Let, Munna has x notebooks.

∴ Thrice the number of notebooks = 3x

Now, according to the condition, we have

50 – 3x = 8

Transposing 50 to R.H.S. we have

– 3x = 8 – 50

or -3x = – 42

Dividing both sides by (-3), we have

\(\frac { -3x }{ -3 }\) = \(\frac { -42 }{ -3 }\)

or x = 14

Thus, the required number of notebooks = 14

(f) Let the number be x.

Then, according to the condition, we have

Thus, the required number = 21

(g) Let the number be x.

∴ \(\frac { 5 }{ 2 }\) of the number = \(\frac { 5 }{ 2 }\)x

According to the condition, we have

\(\frac { 5 }{ 2 }\)x – 7 = 23

Transposing – 7 from L.H.S. to R.H.S., we have

\(\frac { 5 }{ 2 }\)x = 23 + 7 or \(\frac { 5 }{ 2 }\) = 30

Dividing both sides by \(\frac { 5 }{ 2 }\), we have

\(\frac { 5 }{ 2 }\)x ÷ \(\frac { 5 }{ 2 }\) = 30 ÷ \(\frac { 5 }{ 2 }\)

or \(\frac { 5 }{ 2 }\)x x \(\frac { 2 }{ 5 }\) = 30 x \(\frac { 2 }{ 5 }\)

or x = 12

Thus, the required number = 12

Question 2.

Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? {Remember, the sum of three angles of a triangle is 180°.)

(c) Sachin scored twice as many runs as Rahul. Together their runs fell two short of a double century. How many runs did each one score?

Solution:

(a) Let the lowest marks scored be x.

Then twice the lowest marks = 2x According to the condition, we have [Twice the lowest marks] + 7 = 87

∴ 2x + 7 = 87

or 2x = 87 – 7

[Transposing 7 from L.H.S. to R.H.S.]

or 2x =80

Dividing both sides by 2, we have

\(\frac { 2x }{ 2 }\) = \(\frac { 80 }{ 2 }\)

or x =40

∴ The lowest marks = 40

(b) Let the base angle be x°.

∴ The other base angle = x°

∵ Vertex angle = 40°

∴ Sum of the angles = x + x + 40°

= 2x + 40°

According to the condition, we have

2x + 40° =180°

or 2x = 180° – 40°

[Transposing 40° from L.H.S. to R.H.S.] or 2x = 140°

Dividing both sides by 2, we have

\(\frac { 2x }{ 2 }\) = \(\frac { 140° }{ 2 }\)

or x =70°

∴ The base angles of the triangle = 70° each.

(c) Let the number of runs scored by Rahul be x

∴ Sachin’s runs =2x

Sum of their runs = x + 2x = 3x

According to the condition, we have

Sum of the runs = [Double century] – 2

3x = 200 – 2

or 3x = 198

Dividing both sides by 3, we have

\(\frac { 3x }{ 3 }\) = \(\frac { 198 }{ 3 }\)

or x =66 = Rahul’s runs

and 2x =2 x 66 = 132 = Sachin’s runs

Thus, Rahul scored = 66 runs and Sachin scored = 132 runs.

Question 3.

Solve the following:

(i) Irfan says that he has 1 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmi s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruits trees planted if the number of non-fruit trees planted was 77?

Solution:

(i) Let the number of marbles with Parmit be x

5 times x = 5x

Number of marbles with Irfan = 37

According to the condition, we have

5x + 7 = 37

or 5x = 37 – 7

[Transposing 7 from L.H.S. to R.H.S.] or 5x = 30

Dividing both sides by 5, we have

\(\frac { 5x }{ 5 }\) = \(\frac { 30 }{ 5 }\)

or x = 6

∴ Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years

Three times Laxmi’s age = 3JC .

According to the condition, we have

[Three times Laxmi’s age] + 4 = Father’s age

3x + 4 = 49

or 3x = 49 – 4

[Transposing 4 from L.H.S. to R.H.S.]

or 3x = 45

Dividing both sides by 3, we have

\(\frac { 3x }{ 5 }\) = \(\frac { 45 }{ 3 }\)

or x = 15

∴ Laxmi’s age is 15 years.

(iii) Let the number of fruit trees be x

∴Three times fruit trees = 3x

Number of non-fruit trees = 2 + [three times the fruit trees]

According to the condition, we have

2 + 3x = 77

or 3x = 77 – 2

(Transposing 2 to R.H.S.)

3x = 75

or x = \(\frac { 75 }{ 3 }\) = 25

(Dividing both sides by 3)

∴ The number of fruit trees = 25

Question 4.

Solve the following riddle:

I am a number, Tell my identity!

Take me seven times over And add a fifty!

To reach a triple century You still need forty!

Solution:

Let the required number be x.

7 times x = 7x

Triple century = 3 x 100 = 300

According to the riddle, we have

[7 times the number] + 50 = [Triple century] – 40

∴ 7x + 50 = 300 – 40

or 7x + 50 = 260

or 7x = 260 – 50

[Transposing 50 from L.H.S. to R.H.S.]

or 7x = 210

Dividing both sides by 7, we have

\(\frac { 7x }{ 7 }\) = \(\frac { 210 }{ 7 }\)

or x = 30

∴ The required number is 30.