# Ncert Solutions For Class 7 Maths Ex 4.1

## Ncert Solutions For Class 7 Maths Chapter 4 Ex 4.1

Q-1) Complete the last column of the table.

 Sr. No. Equation Value Say, whether the equation is satisfied. (Yes/No) 1 a + 4 = 0 a = 4 – 2 a + 5 = 0 a = 0 – 3 a + 4 = 0 a = – 4 – 4 a – 8 = 1 a = 8 – 5 a – 8 = 1 a = 9 – 6 5a = 25 a = 0 – 7 5a = 25 a = 5 – 8 5a = 25 a = – 5 – 9 x/3 = 3 x = – 9 – 10 x/3 = 3 x = 0 – 11 x/3 = 3 x = 9 –

Ans.)

1. a + 4 = 0, a = 4

L.H.S = a + 4

By putting, a = 4;

L.H.S=4+4=8R.H.S$L.H.S = 4 + 4 = 8 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

2. a + 5 = 0, a = 0

L.H.S = a + 5

By putting, a = 0;

L.H.S=0+5=5R.H.S$L.H.S = 0 + 5 = 5 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

3. a + 4 = 0, a = – 4

L.H.S = a + 4

By putting, a = – 4;

L.H.S = – 4 + 4 = 0 = R.H.S

Therefore, “Yes”, the equation is satisfied.

4. a – 8 = 1, a = 8

L.H.S = a – 8

By putting, a = 8;

L.H.S=88=0R.H.S$L.H.S = 8 – 8 = 0 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

5. a – 8 = 1, a = 9

L.H.S = a – 8

By putting, a = 8;

L.H.S = 9 – 8 = 1 = R.H.S

Therefore, “Yes”, the equation is satisfied.

6. 5a = 25, a = 0,

L.H.S = 5a

By putting, a = 0;

L.H.S=50=0R.H.S$L.H.S = 5*0 =0 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

7. 5a = 25, a = 5,

L.H.S = 5a

By putting, a = 5;

L.H.S = 5*5 = 25 = R.H.S

Therefore, “Yes”, the equation is satisfied.

8. 5a = 25, a = – 5,

L.H.S = 5a

By putting, a = – 5;

L.H.S=5(5)=25R.H.S$L.H.S = 5*(-5) =-25 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

9. x/3 = 3, x = – 9,

L.H.S = x/3,

By putting, x = – 9;

L.H.S=9/3=3R.H.S$L.H.S = -9/3 = -3 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

10. x/3 = 3, x = 0,

L.H.S = x/3,

By putting, x = 0;

L.H.S=0/3=0R.H.S$L.H.S = 0/3 = 0 \neq R.H.S$

Therefore, “No”, the equation is not satisfied.

11. x/3 = 3, x = 9,

L.H.S = x/3,

By putting, x = 9;

L.H.S = 9/3 = 3 = R.H.S

Therefore, “Yes”, the equation is satisfied.

Q-2) Verify whether the value in the brackets is the solution of the given equation.

1. x + 5 = 18 ( x = 1)
2. 7x + 5 = 17 ( x = – 2)
3. 7x + 5 = 19 ( x = 2)
4. 4y – 3 = 11 (y = 1)
5. 4y – 3 = 12 ( y = – 4)
6. 4y – 3 = 14 ( y = 0)

Ans.)

1) Here, x + 5 is  H.S, 18 is R.H.S and x = 1  ( given data)

L.H.S = x + 5,

By putting, x = 1,

L.H.S=1+5=6R.H.S$L.H.S = 1 + 5 = 6 \neq R.H.S$

As, L.H.SR.H.S$L.H.S \neq R.H.S$ , so x = 1 is not a solution.

2) Here, 7x + 5 is  H.S, 17 is R.H.S and x = – 2  ( given data)

L.H.S = 7x + 5,

By putting, x = – 2,

L.H.S=7(2)+5=9L.H.S$L.H.S = 7(-2) + 5 = -9 \neq L.H.S$

As, L.H.SR.H.S$L.H.S \neq R.H.S$ , so x = – 2 is not a solution.

3) Here, 7x + 5 is  H.S, 19 is R.H.S and x = 2  ( given data)

L.H.S = 7x + 5,

By putting, x = 2,

L.H.S = 7(2) + 5 = 19 = R.H.S

As, L.H.S = R.H.S, so x = 2 is a solution.

4) Here, 4y – 3 is  H.S, 11 is R.H.S and y = 1  ( given data)

L.H.S = 4y – 3,

By putting, y = 1,

L.H.S=4(1)3=1L.H.S$L.H.S = 4(1) – 3 = 1 \neq L.H.S$

As, L.H.SR.H.S$L.H.S \neq R.H.S$, so y = 1 is not a solution.

5) Here, 4y – 3 is  H.S, 12 is R.H.S and y = – 4  ( given data)

L.H.S = 4y – 3,

By putting, x = – 4,

L.H.S=4(4)3=19R.H.S$L.H.S = 4(-4) – 3 = -19 \neq R.H.S$

As, L.H.SR.H.S$L.H.S \neq R.H.S$, so y = – 4 is not a solution.

6) Here, 4y – 3 is  H.S, 14 is R.H.S and y = 0  ( given data)

L.H.S = 4y – 3,

By putting, x = 0,

L.H.S=4(0)3=3R.H.S$L.H.S = 4(0) – 3 = -3 \neq R.H.S$

As, L.H.SR.H.S$L.H.S \neq R.H.S$, so y = 0 is not a solution.

Q-3) Solve the equation given below by trial and error method;

1. 5m + 1 = 11
2. 3p – 15 = 6

Ans.)

1) 5m + 1 = L.H.S

By putting, m = 0,

5(0)+1=111$5(0) + 1 = 1 \neq 11$

By putting, m = 1,

5(1)+1=211$5(1) + 1 = 2 \neq 11$

By putting, m = 2,

5(2) + 1 = 11 = R.H.S

2) 3p – 15 = L.H.S

By putting, p = 5,

3(5)15=06$3(5) – 15 = 0 \neq 6$

By putting, p = 6,

3(6)15=36$3(6) – 15 = 3 \neq 6$

By putting, p = 7,

3(7) – 15 = 6 = R.H.S

Q-4) Make the questions of the statements given below

1. The sum of numbers and 5 is 10.
2. 6 subtracted from is 9.
3. Fifteen times xis 45.
4. The number divided by 8 gives 2.
5. One-third of is 9.
6. Ten times plus 5 gives you 49.
7. Two-third of a number minus 5 gives 7.
8. If you take away 7 from 7 times w, you get 56.
9. If you add 4 to One-fourth of v, you get 35.

Ans.)

1. + 5 = 10
2. q– 6 = 9
3. 15x= 45
4. m/8 = 2
5. 1/3n= 9
6. 10+ 5 = 49
7. 2/3h– 5 = 7
8. 7w– 7 = 56
9. 1/4v+ 4 = 35

Q-5) Write the equations given below in the statement form;

1. x + 6 = 8
2. y – 9 = 10
3. 4k = 14
4. 1/3x = 9
5. 1/5p = 8
6. 1/2x + 6 = 9
7. 4t + 3 = 27
8. 3t – 4 = 25

Ans.)

1. The sum of and 6 is 8.
2. 9 subtracted from is 10.
3. Four times is 14.
4. One-third of is 9.
5. One-fifth of is 8.
6. When 6 is added to the half of a number x, it gives 9.
7. Four times of a number t, when added to 3, gives 27.
8. When 4 is subtracted from three times of number t, gives 25.