Chapter 11 Algebra Ex 11.5
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Question 2.
Complete the entries in the third column of the table.
S. No. | Equation | Value of variable | Equations satisfied Yes /No |
(a) | 10y = 80 | y = 10 | |
(b) | 10y = 80 | y = 8 | |
(c) | 10y = 80 | y = 5 | |
(d) | 4l = 20 | l = 20 | |
(e) | 4l = 20 | l = 80 | |
(f) | 4l = 20 | l = 5 | |
(g) | b + 5 = 9 | b = 5 | |
(h) | b + 5 = 9 | b = 9 | |
(i) | b + 5 = 9 | b = 4 | |
(J) | h – 8 = 5 | h = 13 | |
(k) | h – 8 = 5 | h = 8 | |
(l) | h – 8 = 5 | h = 0 | |
(m) | P + 3 = 1 | p = 3 | |
(n) | p + 3 = 1 | p = 1 | |
(o) | p + 3 = 1 | p = 0 | |
(P) | p + 3 = 1 | p = -1 | |
(q) | p + 3 = 1 | p = -2 |
Solution:
S. No. | Equation | Value of variable | Equations satisfied Yes /No |
(a) | 10y = 80 | y = 10 | No |
(b) | 10y = 80 | y = 8 | Yes |
(c) | 10y = 80 | y = 5 | No |
(d) | 4l = 20 | l = 20 | No |
(e) | 4l = 20 | l = 80 | No |
(f) | 4l = 20 | l = 5 | Yes |
(g) | b + 5 = 9 | b = 5 | No |
(h) | b + 5 = 9 | b = 9 | No |
(i) | b + 5 = 9 | b = 4 | Yes |
(J) | h – 8 = 5 | h = 13 | Yes |
(k) | h – 8 = 5 | h = 8 | No |
(l) | h – 8 = 5 | h = 0 | No |
(m) | P + 3 = 1 | p = 3 | No |
(n) | p + 3 = 1 | p = 1 | No |
(o) | p + 3 = 1 | p = 0 | No |
(P) | p + 3 = 1 | p = -1 | No |
(q) | p + 3 = 1 | p = -2 | Yes |
![](https://rajboardexam.in/wp-content/uploads/2022/06/image-870.png)
(b) n + 12 = 20 (12, 8, 20, 0)
For n = 12, LHS = 12 + 12 = 24, RHS = 20
Here, LHS ≠ RHS
∴ n = 12 is not the solution of the equation
For n = 8, LHS = 8 + 12 = 20, RHS = 20
Here, LHS = RHS
∴ n = 8 is the solution of the equation
For n = 20, LHS = 20 + 12 = 32, RHS = 20
Here, LHS ≠ RHS
∴ n = 20 is not the solution of the equation
For n = 0, LHS = 0 + 12 – 12, RHS = 20
Here, LHS ≠ RHS
∴ n= 0 is not the solution of the equation
(c) p – 5 = 5 (0, 10, 5, -5)
For p = 0, LHS = 0 – 5 = -5, RHS = 5
Here, LHS ≠ RHS
∴ p = 0 is not the solution of the equation
For p = 10, LHS = 10 – 5 = 5, RHS = 5
Here, LHS = RHS
∴ p = 10 is the solution of the equation
For p = 5, LHS = 5-5-0, RHS = 5
Here LHS ≠ RHS
∴ p = 5 is not the solution of the equation
For p = 5, LHS = 5 – 5 = 0, RHS = 5
Here, LHS ≠ RHS
∴ p = -5 is not the solution of the equation
![](http://104.199.181.145/wp-content/uploads/2022/04/word-image-33.png)
(e) r – 4 = 0 (4, -4, 8, 0)
For r = 4, LHS = 4 – 4 = 0, RHS = 0
Here, LHS = RHS
∴ r = 4 is the solution of the equation
For r = -4, LHS = -4 – 4 = -8, RHS = 0
Here, LHS ≠ RHS
∴ r = -4 is not the solution of the equation
For r = 8, LHS = 8 – 4 = 4, RHS = 0
Here, LHS ≠ RHS
For r = 8 is not the solution of the equation
For r = 0, LHS = 0 – 4 = – 4, RHS = 0
Here, LHS ≠ RHS
∴ r = 0 is not the solution of the equation
(f) x + 4 = 2 (-2, 0, 2, 4)
For x = -2, LHS = -2 + 4 = 2, RHS = 2
Here, LHS – RHS
∴ x = -2 is the solution of the equation
For x = 0, LHS = 0 + 4 – 4, RHS = 2
Here, LHS ≠ RHS
∴ x = 0 is not the solution of the equation
For x = – 2, LHS = -2 + 4 – 6, RHS = 2
Here, LHS ≠ RHS
∴ x = 2 is not the solution of the equation
For r = 4, LHS = 4 + 4 = 8, RHS = 2
Here, LHS ≠ RHS
∴ x = 4 is not the solution of the equation
Question 4.
(a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 6
(b) Complete the table and by inspection of the table find the solution to the equation 51 – 35
(c) Complete the table and find the solution of the equation g = 4 using the table.
(d) Complete the table and find the solution to the equation m – 7 = 3
Solution:
(a) By inspections, we have
So, m – 6 is the solution of the equation.
(b) Given that 5t = 35
So, t = 7 is the solution of the equation.
![](https://rajboardexam.in/wp-content/uploads/2022/06/image-871.png)
So, z = 12 is the solution of the equation.
(d) Given that m – 7 = 3
So, m = 10 is the solution of the equation.
Question 5.
Solve the following riddles, you may yourself construct such riddles. Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!
![](https://rajboardexam.in/wp-content/uploads/2022/06/image-872.png)
(ii) For each day of the week
Make an upcount from me
If you make no mistake
you will get twenty three!
![](https://rajboardexam.in/wp-content/uploads/2022/06/image-873.png)
(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 you a pretty clue!
you will get me back
If you take me out of twenty two!
Solution:
(i) According to the condition,
I + 12 = 34 or x + 12 = 34
∴ By inspection, we have
22 + 12 = 34
So, I am 22.
(ii) Let I am ‘x’.
We know that there are 7 days in a week.
∴ upcounting from x for 7, the sum = 23
By inspections, we have
16 + 7 = 23
∴ x = 16
Thus I am 16.
(iii) Let the special number be x and there are 11 players in cricket team.
∴ Special Number -6 = 11
∴ x – 6 = 11
By inspection, we get
17 – 6 = 11
∴ x = 17
Thus I am 17.
(iv) Suppose I am ‘x’.
∴ 22 – I = I
or 22 – x = x
By inspection, we have
22 – 11 = 11
∴ x = 11
Thus I am 11.
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