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

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

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

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

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

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

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

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

As, L.H.SR.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

As, L.H.SR.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

As, L.H.SR.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

As, L.H.SR.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

As, L.H.SR.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

By putting, m = 1,

5(1)+1=211

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

By putting, p = 6,

3(6)15=36

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.