Fractions in Disguise Class 8 Long Question Answer

Fractions in Disguise Class 8 Long Question Answer

Fractions in Disguise Class 8 Long Question Answer

Question 1.

A shopkeeper bought two TV sets at ₹ 10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.

Solution:

For the first TV, we have :

CP = ₹ 10000 and profit = 10%.

∴ SP = ₹ [latex]\frac{110}{100}[/latex] × 10000

= ₹ (110 × 100) = ₹ 11000.

For the second TV, we have :

CP = ₹ 10000 and loss = 10%.

∴ SP = ₹ [latex]\frac{90}{100}[/latex] × 10000

= ₹ (90 × 100)

= ₹ 9000.

Total cost paid in buying two TVs

= ₹ 10000 + ₹ 10000

= ₹ 20000.

and total SP = ₹ (11000 + 9000)

= ₹ 20000.

Thus, SP = CP.

Hence, he gains neither profit nor loss.

Question 2.

The cost of an article is ₹ 15500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Solution:

The effective cost price of the article is equal to the price at which it was bought plus the repair cost.

∴ CP of the given article

= ₹ (15500 + 450)

= ₹ 15950.

Profit = 15%.

∴ Profit = 15% of ₹ 15950

= ₹ ([latex]\frac{15}{100}[/latex] × 15950)

= ₹ 2392.50

∴ SP = CP + Profit

= ₹ (15950 + 2392.50)

= ₹ 18342.50.

Question 3.

₹ 6,050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 years.

Solution:

We have : P = Principal = ₹ 6050, R = Rate of interest per annum = 6.5 and T = Time = 3 years.

∴ Simple interest (S.I.)

= [latex]\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}[/latex] = ₹ [latex]\left(\frac{6050 \times 6.5 \times 3}{100}\right)[/latex]

= ₹ 1179.75.

Now, Amount = Principal + S.I.

= ₹ (6050 + 1179.75)

= ₹ 7229.75.