Exercise 8.1 | Class 8 | SCERT Solution | General Mathematics

1. A watch was bought for Rs. 250 and sold it for Rs. 260. Find the profit and profit per cent.
Solution:
Given, CP = Rs. 250 and SP = Rs. 260
We have,
Profit = SP-CP = Rs. 260 - Rs. 250 = Rs. 10
We also have,
Profit per cent $\ = \frac{profit}{CP}$ ×100%
$\ = \frac{10}{250}$ ×100% = 4 %


2. Buying a pen at Rs. 60, at what price should it be sold to get a profit of 15% ?
Solution:
Given, CP = Rs. 60
SP = ?
Also given, percentage of profit, p = 15 %

$\ ⇨ \frac{SP-CC}{CP}×100 = 15$

$\ ⇨ \frac{SP-60}{60}×100 = 15$

$\ ⇨ (SP-60)×100 = 15×60$

$\ ⇨ SP - 60 = \frac{900}{100}$

$\ ⇨ SP = 9+60$
$\ ⇨ SP = 69$
⸫ The required SP should be Rs. 69.

Aternative Mathod
Given, CP = Rs. 60
percentage of profit, p = 15 %

We have,
$\ SP = \left(\frac{100+p}{100}\right)×CP$

$\ ⇨ SP = \left(\frac{100+15}{100}\right)× 60$

$\ ⇨ SP = \left(\frac{115}{100}\right)× 60$

$\ ⇨ SP = 1.15× 60$
$\ ⇨ SP = 69$
⸫ The required SP should be Rs. 69.


3. Ramen sold a mobile for Rs. 13500 at a loss of 20%. Find the CP of the mobile.
Solution:
Given, SP = Rs. 13500
loss percentage, l = 20%
We have,
$\ CP = \left(\frac{100}{100-20}\right)×13500$

$\ ⇨ CP = \left(\frac{100}{80}\right)×13500$

$\ ⇨ CP = \left(\frac{100}{8}\right)×1350$

$\ ⇨ CP = 25×675$
$\ ⇨ CP = 16875$ (Rs.)


4. If the SP of 10 pens is equal to the CP of 8 pens, calculate the loss or gain percentage.
Solution:
Let the SP of the one pen be $\ x$
⸫ the SP of the 10 pens be $\ 10x$ ...(1)
A/Q, CP of 8 pens $\ = 10x$ [using (1)]

⸫ CP of one pen $\ = \frac{10x}{8}$

⸪ $\ \frac{10x}{8} > x$ , i.e., CP>SP

⸫ There will be a loss.

Loss $\ = \frac{10x}{8}-x = \frac{10x-8x}{8} = \frac{2x}{8}= \frac{x}{4}$

⸫ Percentage of loss $\ = \frac{Loss~ in~ one~ pen}{CP~ of~ one~ pen}$×100%

$\ = \frac{\frac{x}{4}}{\frac{10x}{8}}$×100%

$\ = \frac{x}{4}×\frac{8}{10x}$×100%

$\ = 20$%


5. A cycle bought for Rs. 5000 is again sold at a profit of 12%. Find the SP of the cycle.
Solution:
Given, CP = Rs. 5000
percentage of profit, p = 12 %

We have,
$\ SP = \left(\frac{100+p}{100}\right)×CP$

$\ ⇨ SP = \left(\frac{100+12}{100}\right)× 5000$

$\ ⇨ SP = \left(\frac{112}{100}\right)× 5000$

$\ ⇨ SP = 112× 50$
$\ ⇨ SP = 5600$ (Rs.)


6. Kamal bought a water filter for Rs. 4500 and sold it at Rs. 4230. Calculate the loss per cent.
Solution:
Given, CP = Rs. 4500 and SP = Rs. 4230
We have,
Loss per cent $\ = \frac{CP-SP}{CP}$ ×100%

$\ = \frac{4500-4230}{4500}$ ×100%

$\ = \frac{270}{45}$%

$\ = 6$%


7. A shopkeeper sold a watch for Rs. 785 at a loss of 5%. What is the CP of the watch?
Solution:
Given, SP = Rs. 785
loss percentage, l = 5%
We have,
$\ CP = \left(\frac{100}{100-5}\right)×785$

$\ ⇨ CP = \left(\frac{100}{95}\right)×785$

$\ ⇨ CP = \left(\frac{20}{19}\right)×785$

$\ ⇨ CP = 1.052×785$
$\ ⇨ CP = 826.32$ (Rs.)


8. If the selling price of 10 items is equal to the cost price of 11 items of the same types, find the profit or loss percentage.
Solution:
Let the SP of the one item be $\ x$
⸫ the SP of the 10 items be $\ 10x$ ...(1)
A/Q, CP of 11 items $\ = 10x$ [using (1)]

⸫ CP of one item $\ = \frac{10x}{11}$

⸪ $\ x > \frac{10x}{11}$ , i.e., SP>CP

⸫ There will be a profit.

Profit $\ = x - \frac{10x}{11}= \frac{11x-10x}{11} = \frac{x}{11}$

⸫ Percentage of loss $\ = \frac{\frac{x}{11}}{\frac{10x}{11}}$×100%

$\ = \frac{x}{11}×\frac{11}{10x}$×100%

$\ = 10$%


9. A man bought two cars for Rs. 99000 each. He sold one of them at a profit of 10% and one at a loss of 10%. Calculate the profit or loss percentage in the whole transaction.
Solution:
As the percentage of profit and loss for both the cars are same which is 10%.
So, there will be a loss by $\ \frac{10^2}{100}$% $\ = \frac{100}{100}$% = 1 %