GSEB Class 6 Maths Solutions Chapter 11 Algebra Exercise 11.5

Gujarat Board Textbook Solutions Class 6 Maths Chapter 11 Algebra Ex 11.5

 

Question 1. State which of the following are equations (with a variable). Give a reason for your answer. Identify the variable from the equations with a variable.
(a) \( 17 = x + 7 \)
(b) \( (t - 7) > 5 \)
(c) \( \frac { 4 }{ 2 } = 2 \)
(d) \( (7 \times 3) - 19 = 8 \)
(e) \( 5 \times 4 - 8 = 2x \)
(f) \( x - 2 = 0 \)
(g) \( 2m < 30 \)
(h) \( 2n + 1 = 11 \)
(i) \( 7 = (11 \times 5) - (12 \times 4) \)
(j) \( 7 = (11 \times 2) + p \)
(k) \( 20 = 5y \)
(l) \( \frac { 3q }{ 2 } < 5 \)
(m) \( z + 12 > 24 \)
(n) \( 20 - (10 - 5) = 3 \times 5 \)
(o) \( 7 - x = 5 \)
Answer:
(a) This is an equation. It has a variable \( x \).
(b) This is not an equation. It does not contain an equality sign \( (=) \).
(c) This is not an equation. It does not contain any variable.
(d) This is not an equation. It does not contain any variable.
(e) This is an equation. It has a variable \( x \).
(f) This is an equation. It has a variable \( x \).
(g) This is not an equation. It does not contain an equality sign \( (=) \).
(h) This is an equation. It has a variable \( n \).
(i) This is not an equation. It does not contain any variable.
(j) This is an equation. It has a variable \( p \).
(k) This is an equation. It has a variable \( y \).
(l) This is not an equation. It does not contain an equality sign \( (=) \).
(m) This is not an equation. It does not contain an equality sign \( (=) \).
(n) This is not an equation. It does not contain any variable.
(o) This is an equation. It has a variable \( x \).
In simple words: An equation needs an equals sign and at least one letter that stands for a changing value. If it is an equation, we also need to say what that letter is.

Exam Tip: Remember that an "equation" always contains an equals sign (=), while an "expression" does not. A "variable" is typically represented by a letter that can stand for different numbers.

 

Question 2. Complete the entries in the 4th column of the table.
Answer:

In simple words: To see if an equation is satisfied, you just put the given number in place of the letter. If both sides of the equals sign then match, it is satisfied; otherwise, it is not.

Exam Tip: When testing if an equation is satisfied, be careful with negative numbers and fractions. Always perform the arithmetic accurately on both sides of the equation.

 

Question 3. Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) \( 5m = 60 \) (10, 5, 12, 15)
(b) \( n + 12 = 20 \) (12, 8, 20, 0)
(c) \( p - 5 = 5 \) (0, 10, 5, -5)
(d) \( \frac { q }{ 2 } = 7 \) (7, 2, 10, 14)
(e) \( r - 4 = 0 \) (4, -4, 8, 0)
(f) \( x + 4 = 2 \) (-2, 0, 2, 4)
Answer:
(a) \( 5m = 60 \):
For \( m = 10 \), we have: LHS \( = 5 \times 10 = 50 \) and RHS \( = 60 \). Since, LHS \( \neq \) RHS, \( m = 10 \) is not a solution.
For \( m = 5 \), we have: LHS \( = 5 \times 5 = 25 \) and RHS \( = 60 \). Since, LHS \( \neq \) RHS, \( m = 5 \) is not a solution.
For \( m = 12 \), we have: LHS \( = 5 \times 12 = 60 \) and RHS \( = 60 \). Since, LHS \( = \) RHS, \( m = 12 \) is a solution.
For \( m = 15 \), we have: LHS \( = 5 \times 15 = 75 \) and RHS \( = 60 \). Since, LHS \( \neq \) RHS, \( m = 15 \) is not a solution.

(b) \( n + 12 = 20 \):
For \( n = 12 \), LHS \( = 12 + 12 = 24 \) and RHS \( = 20 \). Since, LHS \( \neq \) RHS, \( n = 12 \) is not a solution.
For \( n = 8 \), we have: LHS \( = 8 + 12 = 20 \) and RHS \( = 20 \). Since, LHS \( = \) RHS, \( n = 8 \) is a solution.
For \( n = 20 \), we have: LHS \( = 20 + 12 = 32 \) and RHS \( = 20 \). Since, LHS \( \neq \) RHS, \( n = 20 \) is not a solution.
For \( n = 0 \), we have: LHS \( = 0 + 12 = 12 \) and RHS \( = 20 \). Since, LHS \( \neq \) RHS, \( n = 0 \) is not a solution.

(c) \( p - 5 = 5 \):
For \( p = 0 \), LHS \( = 0 - 5 = -5 \) and RHS \( = 5 \). Since, LHS \( \neq \) RHS, \( p = 0 \) is not a solution.
For \( p = 10 \), LHS \( = 10 - 5 = 5 \) and RHS \( = 5 \). Since, LHS \( = \) RHS, \( p = 10 \) is a solution.
For \( p = 5 \), LHS \( = 5 - 5 = 0 \) and RHS \( = 5 \). Since, LHS \( \neq \) RHS, \( p = 5 \) is not a solution.
For \( p = -5 \), LHS \( = -5 - 5 = -10 \) and RHS \( = 5 \). Since, LHS \( \neq \) RHS, \( p = -5 \) is not a solution.

(d) \( \frac { q }{ 2 } = 7 \):
For \( q = 7 \), LHS \( = \frac { 7 }{ 2 } \) and RHS \( = 7 \). Since, LHS \( \neq \) RHS, \( q = 7 \) is not a solution.
For \( q = 2 \), LHS \( = \frac { 2 }{ 2 } = 1 \) and RHS \( = 7 \). Since, LHS \( \neq \) RHS, \( q = 2 \) is not a solution.
For \( q = 10 \), LHS \( = \frac { 10 }{ 2 } = 5 \) and RHS \( = 7 \). Since, LHS \( \neq \) RHS, \( q = 10 \) is not a solution.
For \( q = 14 \), LHS \( = \frac { 14 }{ 2 } = 7 \) and RHS \( = 7 \). Since, LHS \( = \) RHS, \( q = 14 \) is a solution.

(e) \( r - 4 = 0 \):
For \( r = 4 \), LHS \( = 4 - 4 = 0 \) and RHS \( = 0 \). Since, LHS \( = \) RHS, \( r = 4 \) is a solution.
For \( r = -4 \), LHS \( = -4 - 4 = -8 \) and RHS \( = 0 \). Since, LHS \( \neq \) RHS, \( r = -4 \) is not a solution.
For \( r = 8 \), LHS \( = 8 - 4 = 4 \) and RHS \( = 0 \). Since, LHS \( \neq \) RHS, \( r = 8 \) is not a solution.
For \( r = 0 \), LHS \( = 0 - 4 = -4 \) and RHS \( = 0 \). Since, LHS \( \neq \) RHS, \( r = 0 \) is not a solution.

(f) \( x + 4 = 2 \):
For \( x = -2 \), LHS \( = -2 + 4 = 2 \) and RHS \( = 2 \). Since, LHS \( = \) RHS, \( x = -2 \) is a solution.
For \( x = 0 \), LHS \( = 0 + 4 = 4 \) and RHS \( = 2 \). Since, LHS \( \neq \) RHS, \( x = 0 \) is not a solution.
For \( x = 2 \), LHS \( = 2 + 4 = 6 \) and RHS \( = 2 \). Since, LHS \( \neq \) RHS, \( x = 2 \) is not a solution.
For \( x = 4 \), LHS \( = 4 + 4 = 8 \) and RHS \( = 2 \). Since, LHS \( \neq \) RHS, \( x = 4 \) is not a solution.
In simple words: To find the answer, you need to test each number from the list by putting it into the equation. The number that makes both sides of the equation equal is the correct solution. All other numbers will not make the equation true.

Exam Tip: Systematically test each given value. Write down the Left Hand Side (LHS) and Right Hand Side (RHS) for each substitution to clearly show whether they are equal or not.

 

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation \( m + 10 = 16 \)
(b) Complete the table and by inspection of the table, find the solution to the equation \( 5t = 35 \).
(c) Complete the table and find the solution to the equation \( \frac { z }{ 3 } = 4 \) using the table.
(d) Complete the table and find the solution to the equation \( m - 7 = 3 \).
Answer:
(a) By inspection, we have:

For \( m = 6 \), \( m + 10 = 16 \). Thus, \( m = 6 \) is the solution to \( m + 10 = 16 \).

(b) By inspection, we have:

For \( t = 7 \), \( 5t = 35 \). Thus, \( t = 7 \) is the solution to \( 5t = 35 \).

(c) By inspection, we have:

For \( z = 12 \), \( \frac { z }{ 3 } = 4 \). Thus, \( z = 12 \) is the solution to \( \frac { z }{ 3 } = 4 \).

(d) By inspection, we have:

For \( m = 10 \), \( m - 7 = 3 \). Thus, \( m = 10 \) is the solution to \( m - 7 = 3 \).
In simple words: To complete the table, simply put each given 'm' (or 't' or 'z') value into the equation and write down the result. The number that makes the equation true is the answer.

Exam Tip: For problems involving tables and inspection, carefully substitute each value into the equation. Double-check your arithmetic to avoid errors when filling in the table.

 

Question 5. Solve the following riddles, you may yourself construct such riddles.
(i) Go round a square
Counting every corner Thrice and no more!
Add the count to me
To get exactly thirty-four!
(ii) For each day of the week Make an account from me If you make no mistake You will get twenty-three!
(iii) I am a special number Take away from me a six! A whole cricket team You will still be able to fix!
(iv) Tell me who I am I shall give a pretty clue! You will get me back If you take me out of twenty-two!
Answer:
(i) Suppose I am 'x'.
Since there are 4 corners on a square and each corner is counted three times,
\( 4 \times 3 = 12 \)
According to the condition,
\( x + 12 = 34 \)
By inspection, we have
\( 22 + 12 = 34 \)
\( x = 22 \). Thus, I am 22.

(ii) There are 7 days in a week and let I am 'x'.
Accounting from JC for 7, the sum = 23
i.e. \( x + 7 = 23 \)
By inspection, we have \( 16 + 7 = 23 \)
\( x = 16 \). Thus, I am 16.

(iii) Let the special number be x and there are 11 members in a cricket team.
\( x - 6 = 11 \)
By inspection, we have \( 17 - 6 = 11 \),
\( x = 17 \). Thus, I am 17.

(iv) Suppose I am 'x'.
According to the problem,
\( 22 - x = x \)
By inspection, we have: \( 22 - 11 = 11 \)
\( x = 11 \). Thus, I am 11.
In simple words: For each riddle, turn the words into a math problem where 'x' is the unknown number. Then, solve the simple equation to find 'x', which is the answer to the riddle.

Exam Tip: For riddles, break down each line into a mathematical operation. Identify the unknown as a variable (like 'x') and form an equation. Solve the equation to find the answer.