Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.1

Question 1.

Complete the last column of the table:

Solution:

Question 2.

Check whether the value given in the brackets is a solution to the given equation or not:

(a) n + 5 = 19 (n = 1)

(b) 7n + 5 = 19 (n = – 2)

(c) In + 5 = 19 (n = 2)

(d) 4p – 3 = 13 (p = 1)

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

Solution:

(a) n + 5 = 19 at n = 1

Putting n = 1 in n + 5, we have 1 + 5 = 6 ā 19 [ āµ RHS = 19]

ā“ n = 1 is not a solution to n + 5 = 19.

(b) 7n + 5 = 19 at n = – 2

Putting n = – 2 in In + 5, we have 7 x (- 2) + 5 = – 14 + 5 = – 9 ā 19

[ āµ RHS = 19]

ā“ n = – 2 is not a solution to In + 5 = 19.

(c) 7n + 5 = 19 at n = 2

Putting n = 2 in 7n + 5, we have:

7 x 2 + 5 = 14 + 5 = 19 = RHS

ā“ n = 2 is a solution to 7n + 5 = 19.

(d) 4p – 3 = 13 at p = 1

Putting p = 1 in 4p – 3, we have 4 x 1 – 3 = 4 – 3 = 1 ā 13

[ āµ RHS = 13]

ā“ p = 1 is not a solution to 4p – 3 = 13.

(e) 4p – 3 = 13 at p = – 4

Putting p = – 4 in 4p – 3, we have 4 x (- 4) – 3 = – 16 – 3 = – 19 ā 13

[ āµ RHS = 13]

ā“ p = – 4 is not a solution to 4p – 3 = 13.

(f) 4p – 3 = 13 at p = 0

Putting p = 0 in 4p – 3, we have

(4 x 0) – 3 = 0 – 3 = – 3 ā 13

[āµ RHS = 13]

ā“ p = 0 is not a solution to 4p – 3 = 13.

Question 3.

Solve the following equations by trial and error method:

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

Solution:

(i) 5p + 2 = 17

Value of 5p + 2

When p = 0, then (5 x 0) + 2 = 0 + 2 = 2 ā 17 [ āµ RHS = 17]

When p = 1, then (5 x 1) + 2 = 5 + 2 = 7 ā 17

When p = 2, then (5 x 2) + 2 = 10 + 2 = 12

But 12 ā 17

When p = 3, then (5 x 3) + 2 = 15 + 2 = 17

ā“ L.H.S. = R.H.S.

Thus, p = 3 is the solution to 5p + 2 = 17.

(ii) 3m – 14 = 4

If m = 0, then

L.H.S. = 3m – 14 = 3(0) – 14 = 0 – 14

= – 11 ā RHS [ āµ RHS = 4]

ā L.H.S. ā R.H.S.

If m = 1, then

L.H.S. = 3m – 14 = (3 x 1) – 14 = 3 – 14

= – 11 ā RHS [ āµ RHS = 4]

ā“ L.H.S. ā R.H.S.

If m = 2, then

L.H.S. = 3m – 14 = (3 x 2) – 14 = 6- 14 = – 8 ā RHS [āµ RHS = 4]

ā“ L.H.S. ā R.H.S.

If m = 3, then

L.H.S. = 3m – 14 = (3 x 3) – 14 = 9- 14 = – 5 ā RHS [āµ RHS = 4]

ā“ L.H.S. ā R.H.S.

If m = 4, then

L.H.S. = 3m – 14 = (3 x 4) – 14 =12- 14 = – 2 ā RHS [āµ RHS = 4]

ā“ L.H.S. ā R.H.S.

If m = 5, then

L.H.S. = 3m – 14 = (3 x 5)- 14 = 15 – 14 = 1 ā RHS [āµRHS = 4]

ā“ L.H.S. ā R.H.S.

If m = 6, then

L.H.S. = 3m – 14 = (3 x 6)- 14 =18- 14 = 4 = R.H.S.

ā“ L.H.S. = R.H.S.

Thus, m = 6 is a solution to 3m – 14 = 4.

Question 4.

Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) 2 subtracted from y is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three- fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(vii) One- fourth of a number x minus 4 gives 4.

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one- third of z, you get 30.

Solution:

(i) x + 4 = 9

(ii) y – 2 = 8

(iii) 10a = 70

(iv) \(\frac { b }{ 5 }\) = 6

(v) \(\frac { 3t }{ 4 }\)

(vi) 7m + 7 = 77

(vii) ( \(\frac { 1 }{ 4 }\) )x – 4 = 4

(viii) 6y – 6 = 60

(ix) \(\frac { z }{ 3 }\) + 3 = 30

Question 5.

Write the following equations in statement

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv) \(\frac { m }{ 5 }\)

(v) \(\frac { 3m }{ 5 }\) = 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii) \(\frac { p }{ 2 }\) +2 = 8

Solution:

(i) The sum of p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Twice a number m is 7.

(iv) One-fifth of a number m is 3.

(v) Three-fifth of a number m is 6.

(vi) 4 added to 3 times a number p is 25.

(vii) 2 subtracted from four times a number p is 18.

(viii) 2 added to half of a number p is 8.

Question 6.

Set up an equation in each of the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Permit has. Irfan has 37 marbles. (Take m to be the number of Parmitās marbles.)

(ii) Laxmiās father is 49 years old. He is 4 years older than three times Laxmiās age. (Take Laxmiās age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 1. The highest score is 87. (Take the lowest score to be l.)

(iv) In an isosceles triangle the vertex angle is twice either base angle. {Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees.)

Solution:

(i) Let Permit has āmā marbles.

ā“ 5 times of Parmitās marbles = 5m

Since, Irfanās marbles = [Parmitās marbles] + 7

ā“ Irfanās marbles = (5m + 7)

But Irfanās marbles = 37

ā“ 5m + 7 = 37

(ii) Let Laxmiās age = y years

ā“ 3 times Laxmiās age = 3y years

Age of Laxmiās father = [3 times Laxmiās age] + 4 years

= 3y + 4 years

But Laxmiās father is 49 years old ā“ 3y + 4 = 49

Let the lowest score (marks) = l

ā“ Twice the lowest marks = 2l

Since, highest marks = [twice the lowest marks] + 7

= 2l + 7

But the highest marks = 87

ā“ 2l + 7 = 87

(iv) Let the base angle be b degrees.

ā“ The base angles of an isosceles triangle are equal

ā“ The other base angle = b degrees

Since, the vertex angle = Twice either base angle

= 2b degrees

Also, the sum of three angles of a triangle = 180Ā°

ā“ b + b + 2b = 180Ā° or 4b = 180Ā°