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

CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

Class 7 Maths NCERT Solutions Chapter 4 Simple Equations InText Questions

Try These (Page 78)

Question 1.5
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.
Solution:
Let us find the value of the expression (10 y – 20) for different values of y.
NCERT Solutions for Class 7 maths Integers chapter 3 img 12
Clearly, the value of (10y – 20) depends on the value of y.
But, none of the values is 50. Thus, y = 1 or y = 2 or y = 3 or y = 4 or y = 5 is not the solution of 10y – 20 = 50.
Let us try by giving more values to y and find whether the condition 10y – 20 = 50 is met.
When, y = 6, then l0y – 20 = 10 x 6 – 20 = 60 – 20 = 40 ≠ 50. So, y = 6 is not its solution.
When, y = 7, then l0y – 20 = 10 x 7 – 20 = 70 – 20 = 50. So, y = 7 is its solution.

Try These (Page 80)

Question 1.
Write at least one other form for each equation (ii), (iii), and (iv).
Solution:
Other forms are as under:
(i) p multiplied by 5 gives 20
(ii) 7 added to thrice of n gives 1
(iii) one-fifth of m less 2 gives 6.

Try These (Page 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:
For 1 st equation :
Starting with x = 5
Multiplying both sides by 5, 5x, = 25
Adding 5 to both sides, 5x + 5 = 30
Let us solve it,
NCERT Solutions for Class 7 maths Integers chapter 3 img 13
For IInd equation :
Starting with x = 5
Multiplying both sides by 3, 3x = 15
Subtracting 3 from both sides, 3x – 3 = 12
Let us solve it,
NCERT Solutions for Class 7 maths Integers chapter 3 img 14

Try These (Page 88)

Question 1.
Try to make two number puzzles, one with the solution 11 and another with 100.
Solution:
The first puzzle with solution 11: Think of a number, multiply it by 3 and add 2 to the product. The result obtained is 35. Tell me the number. The second puzzle with solution 100: Think of a number, divide it by 10 and subtract 5 from the quotient. The result obtained is 5. Tell me the number.

Try These (Page 90)

(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell me what the number is?
Solution:
Let the number be x
Then, according to the question,
NCERT Solutions for Class 7 maths Integers chapter 3 img 15

(ii) What is that number one-third of which added to 5 gives 8?
Solution:
Let the number by y
Then, according to the question
NCERT Solutions for Class 7 maths Integers chapter 3 img 16

Try These (Page 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 mangoes contained in the smaller box?
Solution:
Let the number of mangoes in each smaller box be x.
Then, the number of mangoes in 8 such boxes = 8x.
Therefore, number of mangoes in each larger box = 8x + 4
NCERT Solutions for Class 7 maths Integers chapter 3 img 17
Thus, the number of mangoes in the smaller box is 12.

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