These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions

NCERT In-text Question Page No. 78

Question 1.

The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.

Answer:

Value of y | Value of expression (10y – 20) |

0 | 10 × 0 – 20 = -20 |

1 | 10 × 1 – = – 10 |

2 | 10 × 2 – 20 = 0 |

3 | 10 × 3 – 20 = 10 |

4 | 10 × 4 – 20 = 20 |

5 | 10 × 5 – 20 = 30 |

6 | 10 × 6 – 20 = 40 |

7 | 10 × 7 – 20 = 50 |

Thus the condition 10y – 20 is true for y = 7

NCERT In-text Question Page No. 80

Question 1.

Write atleast one other form for each equation (ii), (iii) and (iv)

Equation | Other form of the equation |

(ii) 5p = 20 | (a) 5p – 4 = 16 |

(iii) 3n + 7 = 1 | (iii) 4n + 9 = 45 |

(iv) \(\frac{\mathrm{m}}{\mathrm{s}}\) | (iv) \(\frac{\mathrm{m}}{\mathrm{s}}\) |

Statement of other form of the equation:

(a) Taking away 4 from five times p gives 16.

(b) Add 9 to four times n to get 45.

(c) Add 8 to one-third of m to get 17.

NCERT In-text Question Page No. 88

Question 1.

Start with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5.

Answer:

(i) x = 5

Multiplying both sides by 2, we have

2x = 10

Adding 6 to both sides, we have

or 2x + 6 = 10 + 6

or 2x + 6 = 16 is an equation.

(ii) x = 5

Dividing both sides by 3, we have

\(\frac{x}{3}=\frac{5}{3}\)

Subtracting 2 from both sides, we have

\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5}{3}\) – 2

or

\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5-6}{3}=\frac{-1}{3}\)

Thus,

\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\) is an equation.

Solution to I

2x + 6 = 16

Subtracting 6 from both sides, we have

2x + 6- 6 = 16 – 6

or 2x = 10

Dividing both sides by 2, we have

\(\frac{2 \mathrm{x}}{2}=\frac{10}{2}\) or x = 5

Solution to II

\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\)

Adding 2 to both sides, we have

\(\frac{\mathrm{x}}{3}\) – 2 + 2 = \(\frac{-1}{3}\) + 2 or \(\frac{x}{3}=\frac{5}{3}\)

Multiplying both sides by 3, we have

\(\frac{\mathrm{x}}{3}\) × 3 = \(\frac{5}{3}\) × 3 or x = 5

NCERT In-text Question Page No. 88

Question 1.

Try to make two number puzzles, one with the solution 11 and another with 100.

Answer:

I. A puzzle having the solution as 11:

Think of a number. Multiply bit by 3 and add 7. Tell me the sum.

If the sum is 40, then the number is 11.

II. A puzzle having the solution as 100:

Think of a number. Divide it by 4 and add 5. Tell me what you get.

If you get 30, then the number is 100.

Note:

Instead of making the same operation on both sides, we can move a number from one side to another by changing its sign from (+) to (-) and (-) to (+). This is called ‘transposing a number’. Thus, transposing a number is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed.

NCERT In-text Question Page No. 90

Question 1.

(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?

(ii) What is that number one-third of which added to 5 gives 8?

Answer:

(i) Let the number be x.

∴ Multiply the number by 6, we have 6x.

Now, according to the condition, we have

6x = 7 + 5 = 12

Dividing both sides by 6, we have

\(\frac{6 x}{6}=\frac{12}{6}\)

or x = 2

∴ The required number = 2

(ii) Let the required number be x.

∵ One-third of the number = \(\frac{1}{3}\)x

∴ According to the condition, we have

5 + \(\frac{1}{3}\)x = 8

Transposing 5 from L.H.S. to R.H.S., we have

\(\frac{1}{3}\)x = 8 – 5 = 3

Multiplying both sides by 3, we have

3 × \(\frac{1}{3}\)x = 3 × 3

or x = 9

Thus, the required number = 9.

NCERT In-text Question Page No. 90

Question 1.

There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of he smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?

Answer:

Let the number of mangoes contained in the smaller box be x.

∴ Number of mangoes in 8 smaller boxes = 8x

Now, according to the condition, we have

8x + 4 = 100

Transposing 4 to R.H.S., we have

8x = 100 – 4 or 8x = 96

Dividing both sides by 8, we have

\(\frac{8 x}{8}=\frac{96}{8}\)

or x = 12

Thus, the number of mangoes in the smaller box = 12.