GSEB Class 7 Maths Solutions Chapter 4 Simple Equations InText Questions

 

Try These (Page 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: We can check the value of the expression \( (10y - 20) \) for different values of \( y \) as shown in the table below.

From the table, we can observe that the condition \( 10y - 20 = 50 \) is satisfied when \( y = 7 \). Thus, \( y=7 \) is the solution.
In simple words: We tested different values for \( y \) in the expression \( 10y - 20 \). We found that when \( y \) is 7, the expression equals 50, which solves the equation.

Exam Tip: When asked to verify an equation by giving values, always create a table for clarity and extend the values until the condition is met, showing the step-by-step evaluation.

 

Try These (Page 80)

Question 1. Write at least one other form for each equation (ii), (iii) and (iv) of Example 2 on Page 80.
Answer: Here are other forms for the given equations:

(a) Taking away 4 from five times \( p \) gives 16.
(b) Adding 9 to four times \( n \) will give 45.
(c) Adding 8 to one-third of \( m \) will result in 17.
In simple words: We changed each equation slightly to make a new one that still has the same solution. For \( 5p=20 \), we created \( 5p-4=16 \). For \( 3n+7=1 \), we wrote \( 4n+9=45 \). For \( \frac{m}{5}-2=6 \), we came up with \( \frac{m}{3}+8=17 \).

Exam Tip: To find "other forms" of an equation, you can perform the same arithmetic operations (addition, subtraction, multiplication, division) on both sides of the original equation without changing its solution.

 

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 \( x = 5 \).
Answer: We will start with \( x = 5 \) and create two different equations.
I. Starting with \( x = 5 \)
Multiplying both sides by 2, we get:
\( 2x = 10 \)
Adding 6 to both sides, we have:
\( 2x + 6 = 10 + 6 \)
\( 2x + 6 = 16 \)
So, \( 2x + 6 = 16 \) is one equation.
Solution to I:
\( 2x + 6 = 16 \)
Subtracting 6 from both sides, we get:
\( 2x + 6 - 6 = 16 - 6 \)
\( 2x = 10 \)
Dividing both sides by 2, we have:
\( \frac { 2x }{ 2 } = \frac { 10 }{ 2 } \)
\( x = 5 \)
II. Starting with \( x = 5 \)
Dividing both sides by 3, we get:
\( \frac { x }{ 3 } = \frac { 5 }{ 3 } \)
Subtracting 2 from both sides, we have:
\( \frac { x }{ 3 } - 2 = \frac { 5 }{ 3 } - 2 \)
\( \frac { x }{ 3 } - 2 = \frac { 5-6 }{ 3 } \)
\( \frac { x }{ 3 } - 2 = \frac { -1 }{ 3 } \)
Thus, \( \frac { x }{ 3 } - 2 = \frac { -1 }{ 3 } \) is another equation.
Solution to II:
\( \frac { x }{ 3 } - 2 = \frac { -1 }{ 3 } \)
Adding 2 to both sides, we have:
\( \frac { x }{ 3 } - 2 + 2 = \frac { -1 }{ 3 } + 2 \)
\( \frac { x }{ 3 } = \frac { -1 + 6 }{ 3 } \)
\( \frac { x }{ 3 } = \frac { 5 }{ 3 } \)
Multiplying both sides by 3, we have:
\( \frac { x }{ 3 } \times 3 = \frac { 5 }{ 3 } \times 3 \)
\( x = 5 \)
In simple words: We began with \( x = 5 \) and did the same math steps to both sides to make two different equations. When we solved these new equations, both gave us \( x = 5 \) again. This shows that performing identical operations on both sides keeps the solution the same.

Exam Tip: When creating equations, remember that any operation performed on one side must also be performed on the other side to maintain equality and ensure the original solution remains valid.

 

Try These (Page 88)

Question 1. Try to make two number puzzles, one with the solution 11 and another with 100.
Answer: Here are two number puzzles with the specified solutions:
I. A puzzle having the solution as 11:
"Think of a number. Multiply it 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."
In simple words: We created two simple riddles. The first puzzle's answer is 11, and the second one's answer is 100.

Exam Tip: When creating number puzzles, work backward from the desired solution. Choose a simple sequence of operations, and then determine the final result to pose the question.

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.

 

Try These (Page 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 unknown number be \( x \).
Multiplying the number by 6 gives \( 6x \).
According to the given condition, we have:
\( 6x - 5 = 7 \)
Transposing 5 from the Left Hand Side (L.H.S.) to the Right Hand Side (R.H.S.), we get:
\( 6x = 7 + 5 \)
\( 6x = 12 \)
Dividing both sides by 6, we have:
\( \frac { 6x }{ 6 } = \frac { 12 }{ 6 } \)
\( x = 2 \)
Therefore, the required number is 2.
(ii) Let the unknown number be \( x \).
One-third of the number is \( \frac { 1 }{ 3 }x \).
According to the condition, we have:
\( 5 + \frac { 1 }{ 3 }x = 8 \)
Transposing 5 from the L.H.S. to the R.H.S., we get:
\( \frac { 1 }{ 3 }x = 8 - 5 \)
\( \frac { 1 }{ 3 }x = 3 \)
Multiplying both sides by 3, we have:
\( 3 \times \frac { 1 }{ 3 }x = 3 \times 3 \)
\( x = 9 \)
Thus, the required number is 9.
In simple words: For the first problem, the number is 2. For the second problem, the number is 9. We found these by setting up equations based on the puzzle's clues and then solving them.

Exam Tip: For word problems, always define your unknown variable clearly (e.g., "Let the number be x"). Then, translate each part of the sentence into a mathematical operation to form the equation correctly.

 

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?
Answer: Let the number of mangoes contained in a smaller box be \( x \).
Then, the number of mangoes in 8 smaller boxes would be \( 8x \).
According to the condition, a larger box contains 4 more mangoes than \( 8x \), so a larger box has \( 8x + 4 \) mangoes.
We are also told that each larger box contains 100 mangoes.
So, we can set up the equation:
\( 8x + 4 = 100 \)
Transposing 4 to the R.H.S., we get:
\( 8x = 100 - 4 \)
\( 8x = 96 \)
Dividing both sides by 8, we have:
\( \frac { 8x }{ 8 } = \frac { 96 }{ 8 } \)
\( x = 12 \)
Thus, the number of mangoes in the smaller box is 12.
In simple words: We set up an equation where \( x \) stands for mangoes in a small box. Since a big box has 100 mangoes and also 4 more than 8 small boxes, we found that each small box has 12 mangoes.

Exam Tip: When solving word problems, carefully break down each sentence to define variables and form the equation. Pay attention to keywords like "more than," "less than," "times," and "half" to correctly translate them into mathematical operations.