Students can Download Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.1

Question 1.

Complete the last column of the table.

Solution:

Last column to be filled.

(i) No

(ii) No

(iii) Yes

(iv) No

(v) Yes

(vi) No

(vii) Yes

(viii) No

(ix) No

(x) No

(xi) Yes

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)

Solution:

1 + 5 ≠ 19

6 ≠ 19

∴ LHS ≠ RHS

∴ n = 1 is not a solution to the given equation n + 5 = 19.

b) 7n + 5 = 19(n = -2)

Solution:

7 × -2 + 5 = 19

-14 + 5 = 19

-9 = 19

LHS ≠ RHS

∴ n = -2 is not a solutions the given equation 7n + 5 = 19 (n = 2)

7 × 2 + 5 = 19

14 + 5 = 19

19 = 19

LHS = RHS

∴ n = 2 is a solution to the given equation 7n + 5 = 19

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

Solution:

7 × 2+5 = 19

14 + 5 = 19

19 = 19

LHS = RHS

∴ n = 2 is a solution to the given equation 7n + 5 = 19.

d) 4p – 3= 13 (p = 1)

4(1) – 3 = 13

4 – 3 = 13

1 ≠ 13

LHS ≠ RHS

∴ p = 1 is not a solution to the given equation 4p – 3 = 13.

e) 4p – 3 = 13 (p = -4)

4(-4) – 3 = 13

– 16 – 3 = 13

-19 ≠ 13

LHS ≠ RHS

∴ p = -4 is not a solution to the given equation 4p – 3 = 13.

f) 4p – 3 = 13 (p = 0)

4(0) – 3 = 13

0 – 3 = 13

– 3 ≠ 13

LHS ≠ RHS

∴ p = 0 is not a solution to the given equation 4p – 3 = 13.

Question 3.

Solve the following equations by trial and error method :

i) 5p + 2 = 17

Solution:

If p = 0 then 5(0) + 2 = 17, 0 + 2 ≠ 17

If p = 1 then 5(1) + 2 = 17, 5 + 2 ≠ 17

If p = 2 then 5(2) + 2 = 17, 10 + 2 ≠ 17

If p = 3 then 5(3) + 2 = 17, 15 + 2 ≠ 17

∴ p = 3 is the solution to the given equation 5p + 2 = 17.

ii) 3m – 14 = 4

If m = 0 then 3(0) – 14

0 – 14 = – 14 ≠ 4

3m – 14 = 4

If m = 1 then 3(1) – 14

3 – 14 = -11

If m = 2 then 3(2) – 14

6 – 14 = – 8

If m = 3 then 3(3) – 14 = -5

If m = 4 then 3(4) – 14 = -2

If m = 5 then 3(5) – 14 = 1

If m = 6 then 3(6) – 14 = 4

LHS = RHS

∴ m = 6 is the solution to the given equation 3m – 14 = 4.

Question 4.

Write equations for the following statements :

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

Solution:

x + 4 = 9

ii) 2 subtracted from y is 8.

Solution:

y – 2 = 8

iii) Ten times a is 70.

Solution:

10a = 70

iv) The number b divided by 5 gives 6.

Solution:

\(3 / 4 t=15\)

v) Three-fourth of t is 15.

Solution:

t = 15

vi) Seven times m plus 7 gets you 77.

Solution:

7m + 1 = 11

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

Solution:

\(\frac{1}{4}\) x – 4 = 4

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

Solution:

6y – 6 = 60

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

Solution:

\(\frac{1}{3}\)z + 3 = 30

Question 5.

Write the following equations in statement forms :

i) p + 4 = 15

Solution:

The sum of P and 4 is equal to 15 .

ii) m – 7 = 3

Solution:

Seven subtracted from m is equal to 3.

iii) 2m = 1

Solution:

Two times a number m is equal to 7.

iv) \(\frac{\mathrm{m}}{5}\) = 3

Solution:

One – fifth of ‘m’ number is equal to 3 3m

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

Solution:

Three – fifth of ‘m’ number is equal to 6

vi) 3p + 4 = 25

Solution:

Three times a number ‘P’ is added to 4 is equal to 25.

vii) 4p – 2 = 18

Solution:

Two is subtracted from 4 times ‘p’ is equal to 18.

viii)

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

Solution:

Two is added to the half the number of p is equal to 8.

Question 6.

Set up an equation in the following cases:

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

Solution:

Let the number of marbles Parmit had be ‘m’

∴ Number of marbles Irfan has = Five times of parmits marble 5m.

To this 7 marbles more means = 5m + 7

∴ 5m + 7 = 37.

ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to bey years)

Solution:

Let the age of Lakshmi be ‘y’ years

Three times of age = 3y

More than 4 years = 3y + 4

∴ Lakshmi’s father age be = 3y + 4

∴ 3y + 4 = 49

iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)

Solution:

Let the lowest score be ‘l’

Twice the lowest score be = 2l

Plus seven = 2l + 1

∴ The highest score = 2l + 7

∴ 2l + 7 = 87

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

Solution:

Let the base angle be b°

Vertex angle = Twice the base angle

∴ Vertex angle = 2b

Sum of the angles of an isosceles triangle is =180°

∴ 2b + b + b = 180° = 4b = 180°