Exercise: 7-E
Competency Focused Questions
Q1: Which of the following does mean a 100% profit?
Step 1: Understand what 100% profit means:
100% profit means profit = cost price
So, selling price = cost price + profit = 2 × cost price
Step 2: Check each option:
Option a: C.P. = ₹100, S.P. = ₹50 → Loss of ₹50 → Not 100% profit ❌
Option b: C.P. = ₹100, S.P. = ₹100 → No gain or loss ❌
Option c: C.P. = ₹50, S.P. = ₹100 → Gain = ₹50 = 100% of ₹50 ✅
Option d: C.P. = ₹50, S.P. = ₹150 → Gain = ₹100 → Gain% = \(\frac{100}{50} \times 100 = 200\%\) ❌
Answer: c. Buying at ₹50 and selling at ₹100.
Q2: A shopkeeper buys a ball and a bat for ₹200 and ₹500 respectively. For selling, he marks the bat at ₹525. What price should he ask for the ball, if he wants to make a profit of 5% on the whole transaction?
Step 1: Total cost price of ball and bat = ₹200 + ₹500 = ₹700
Step 2: Required profit = 5% of ₹700 = ₹\(\frac{5}{100} \times 700 = 35\)
Step 3: Required selling price = ₹700 + ₹35 = ₹735
Step 4: Selling price of bat is fixed at ₹525
So, required selling price of ball = ₹735 – ₹525 = ₹210
Answer: c. ₹210
Q3: A shopkeeper bought an item and plans to sells it in two ways:
i. The shopkeeper gives a discount of x% on the marked price and adds x% tax on the discounted price of the item.
ii. The shopkeeper hikes the marked price by x% and then gives x% discount on the hiked price of the item.
Which option is better for the shopkeeper?
Step 1: Assume the Marked Price (M.P.) = ₹100 for simplicity.
Case i:
Discount = x% ⇒ Discounted Price = ₹100 × \(\left(1 – \frac{x}{100}\right)\)
Tax = x% on discounted price ⇒ Final price = ₹100 × \(\left(1 – \frac{x}{100}\right) \times \left(1 + \frac{x}{100}\right)\)
This is of the form: \((1 – \frac{x^2}{10000})\)
So, Final price = ₹100 × \(\left(1 – \frac{x^2}{10000}\right)\)
Case ii:
Hike = x% ⇒ New Price = ₹100 × \(\left(1 + \frac{x}{100}\right)\)
Discount = x% on new price ⇒ Final price = ₹100 × \(\left(1 + \frac{x}{100}\right) \times \left(1 – \frac{x}{100}\right)\)
Again, this is of the form: \((1 – \frac{x^2}{10000})\)
So, Final price = ₹100 × \(\left(1 – \frac{x^2}{10000}\right)\)
Step 2: Compare both cases:
Both final amounts = ₹100 × \(\left(1 – \frac{x^2}{10000}\right)\)
Answer: c. both will give the same revenue
Q4: Amazon offers you 10% cashback (discount) on an article of ₹500 and adds 15% delivery charges on the discounted amount of the article. How much do you need to pay?
Step 1: Marked Price (M.P.) = ₹500
Step 2: Cashback (discount) = 10% of ₹500 = ₹50
Discounted Price = ₹500 – ₹50 = ₹450
Step 3: Delivery charges = 15% of ₹450 = ₹\(\frac{15}{100} \times 450 = ₹67.50\)
Step 4: Total amount to be paid = ₹450 + ₹67.50 = ₹517.50
Now match with the options:
Only option (d) matches:
₹500 – 10% of ₹500 + 15% of ₹450 = ₹500 – ₹50 + ₹67.50 = ₹517.50
Answer: d. ₹500 – 10% of ₹500 + 15% of ₹450
Q5: A shopkeeper sells an item for 25% discount and still makes 25% profit. The discount he needs to offer to achieve a profit of 36% is:
Step 1: Let the Cost Price (C.P.) = ₹100
He gives 25% discount and still makes 25% profit.
So, Selling Price (S.P.) = ₹100 + 25% of 100 = ₹125
Let the Marked Price (M.P.) be x.
Since 25% discount is given:
S.P. = M.P. × (1 – 25/100) = x × 0.75
So,
x × 0.75 = 125
⇒ x = \(\frac{125}{0.75} = ₹166.67\)
Step 2: Now for 36% profit, new S.P. = ₹100 + 36% of 100 = ₹136
Let new discount be D%. Then:
136 = M.P. × (1 – D/100) = ₹166.67 × (1 – D/100)
Divide both sides by 166.67:
\[
\frac{136}{166.67} = 1 – \frac{D}{100} \\
0.816 = 1 – \frac{D}{100} \\
\Rightarrow \frac{D}{100} = 0.184 \\
\Rightarrow D = 18.4\%
\]Answer: c. 18.4%
Q6: A teacher buys a book of cost price ₹100. As she goes to school next day, her friend asks her to sells that to her. As a friend, she the book at 20% loss to her. And then her friend sells that book at 20% gain to other colleague. The final selling price of the book is:
Step 1: Teacher’s Cost Price = ₹100
Loss = 20%
Selling Price to friend = ₹100 – 20% of ₹100
= ₹100 – ₹20 = ₹80
Step 2: Friend’s Cost Price = ₹80
Gain = 20%
Selling Price to colleague = ₹80 + 20% of ₹80
= ₹80 + ₹16 = ₹96
Answer: c. ₹96
Q7: A dealer allows a discount of 10% on the marked price. How much above the cost price must he mark his goods to make a profit of 8%?
Step 1: Let the Cost Price (C.P.) = ₹100
Profit = 8% ⇒ Selling Price (S.P.) = ₹100 + 8% of ₹100 = ₹108
Step 2: Let the Marked Price (M.P.) = ₹x
Given: 10% discount on M.P. ⇒ S.P. = x – 10% of x = x × (1 – 10/100) = x × 0.9
So, x × 0.9 = 108
⇒ x = 108 ÷ 0.9 = ₹120
Step 3: Percentage above C.P.:
\[
\text{Markup \%} = \frac{120 – 100}{100} \times 100 = 20\%
\]Answer: d. 20%
Q8: A dishonest seller sells fruits with a 960 g weight instead of a I kg weight. If he sold the fruits at the cost price, his gain is:
Step 1: Let the Cost Price (C.P.) of 1 kg of fruit = ₹100
Since the seller gives only 960 g but charges for 1000 g (1 kg), he actually sells 960 g for ₹100.
Step 2: Cost Price of 960 g = \( \frac{960}{1000} \times ₹100 = ₹96 \)
Selling Price of 960 g = ₹100 (he charges for full 1 kg)
Step 3: Gain = S.P. – C.P. = ₹100 – ₹96 = ₹4
\[
\text{Gain \%} = \frac{4}{96} \times 100 = \frac{100}{24} = 4\frac{1}{6}\%
\]Answer: d. \(4\frac{1}{6}\)%
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