A food stall owner sold some burgers in such a way that selling price

1.	A food stall owner sold some burgers in such a way that selling price of 100 burgers is the same as the cost price of 105 burgers. So the percentage of gain or loss is:1. 20%2. 7%3. 10%4. 5%
Answer» Correct Answer - Option 4 : 5% Given: SP of 100 burgers = CP of 105 burgers Formula used: P = S.P. – C.P. L = C.P. – S.P. P% = (P/C.P.) × 100 L% = (L/C.P.) × 100 Where, P → Profit L → Loss SP → Selling price CP → Cost price Calculations: Let the CP of 1 burger be Rs.y and its SP be Rs.x. So the SP of 100 burgers = 100x And the CP of 105 burgers = 105y According to the question, 100x = 105y ⇒ x ∶ y = 21 ∶ 20 Now let x and y be 21k and 20k respectively. Since x > y So P = 21k – 20k = k P% = (P/CP) × 100 ⇒ P% = (k/20k) × 100 = 5% ∴ The required Profit% is 5%.

A food stall owner sold some burgers in such a way that selling price of 100 burgers is the same as the cost price of 105 burgers. So the percentage of gain or loss is:1. 20%2. 7%3. 10%4. 5%

Answer» Correct Answer - Option 4 : 5%

Given:

SP of 100 burgers = CP of 105 burgers

Formula used:

P = S.P. – C.P.

L = C.P. – S.P.

P% = (P/C.P.) × 100

L% = (L/C.P.) × 100

Where,

P → Profit

L → Loss

SP → Selling price

CP → Cost price

Calculations:

Let the CP of 1 burger be Rs.y and its SP be Rs.x.

So the SP of 100 burgers = 100x

And the CP of 105 burgers = 105y

According to the question,

100x = 105y

⇒ x ∶ y = 21 ∶ 20

Now let x and y be 21k and 20k respectively.

Since x > y

So P = 21k – 20k = k

P% = (P/CP) × 100

⇒ P% = (k/20k) × 100 = 5%

∴ The required Profit% is 5%.

Discussion

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