Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations InText Questions and Answers.
Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations InText Questions
Try These (Page No. 78):
Question 1.
The value of the expression (lOy – 20) depends on the value of y. Verify this by giving 5 different values to y and finding the value of (lOy – 20) for each. From the different values of (lOy – 20) you obtain, do you see a solution to 10y – 20 = 50 ? If there
is no solution, try giving more values of y and find whether the condition 10y – 20 = 50 is met.
Solution:
The value of an expression depends upon the value of the variable from which the expression is formed.
10y – 20 = 50?
L.H.S. = 10y – 20
If y = 0, 10y – -20 = 10 × 0 – 20 = 0 – 20 = -20
If y = 1, 10y – -20 = 10 × 1 – 20 = 10 – 20 = -10
If y = 2, 10y – -20 = 10 × 2 – 20 = 20 – 20 = 0
If y = 3, 10y – -20 = 10 × 3 – 20 = 30 – 20 = 10
If y = 4, 10y – -20 = 10 × 4 – 20 = 40 – 20 = 20
If y = 5, 10y – -20 = 10 × 5 – 20 = 50 – 20 = 30
If y = 6, 10y – -20 = 10 × 6 – 20 = 60 – 20 = 40
If y = 7, 10y – -20 = 10 × 7 – 20 = 70 – 20 = 50
if, y = 7, 10y – 20 = 50
Try These (Page No. 80):
Question 1.
Write at least one other from each equation.
(i) 5P = 20, (ii) 3n + 7 = 1, (iii) \(\frac{m}{5}\) – 2 = 6
Solution:
(i) (a) five times a number p is 20. (b) multiple of p is 20.
(ii) (a) Add 7 to three times n to get 1. (b)
Three times n plus 7 gets you 1.
(iii) (a) You get 6, when you subtract 2 from one fifth of a number m.
(b) One fifth of a number m mumus 2 leaves 6.
Try These (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.
Solution:
x = 5
Multiply both sides by 2, we get,
2x = 10
Multiply both sides by 2, we get
4x = 20
Check: If x = 5, 2 x 5 = 10
10 = 10
If x = 5, 4 x 5 = 20
20 = 20
Try These (Page No. 88):
Question 1.
Try to make two number puzzles, one with the sulution 11 and another with 100.
Solution:
(i) If x = 11
Multiply both sides by 2.2x = 22
Add 4 to both sides 2x + 4 = 26
(ii) If y = 100
Multiply both sides by 2.3y = 300
Subtract 100 from
both sides 3y -100 = 200
Try These (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 ?
Solution:
(i) Let x be the number.
hence, 6x – 5 =7
or, Qx = 7 + 5 = 12
or, x = \(\frac{12}{6}\) = 2
Hence number be 2.
(ii)Let y be the number.
hence, \(\frac{y}{3}\) + 5 = 8
\(\frac{y+15}{3}\)
or, y + 15 = 24
∴ y = 24 – 15 = 9
Hence number be 9.
Try These (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 the smaller type. Each larger box contains 100 mangoes. Find the number of managoes contained in the smaller box ?
Solution:
Let the number of mangoes contained in the smaller box be x.
According to questions, 8x + 4 = 100
or, 8x = 100 – 4
or, 8x = 96
or, x = \(\)\frac{96}{8}\(\) = 12
Hence, 8 mangoes contained in the smaller box.