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.