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## Karnataka State Syllabus 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)

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:

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:

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)

Solution:

Two is added to half 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 is 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, 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°