A shopkeeper bought an article for Rs.1000 and marked its price as Rs.

Question

A shopkeeper bought an article for Rs.1000 and marked its price as Rs.2160. He gave three successive discounts of a%, b% and c%, where a + b + c = 50. If he made a profit of x% finally, then what is the difference between maximum value of x and minimum value of x ?

A. 10
B. 17
C. 24
D. 30
E. 33.33

A. 10
B. 17
C. 24
D. 30
E. 33.33

KarishmaB · Accepted Answer

First consider that a single discount of 50% is greater than two discounts of 25% each because after the first discount of 25%, the next 25% will be applied on a smaller amount.

So maximum discount will be when two discounts are 0 and one discount is the entire 50%.
Sale price = 2160 * (1/2) = 1080
Profit = (80/1000) * 100 = 8%

To get the maximum profit, discount must be least. So we break it down into smaller units.
So if sum of all discounts is 50%, each discount is 50/3% = 1/6
Hence, the sale price = 2160 *(5/6)*(5/6)*(5/6) = 1250
Profit = 250/1000 * 100 = 25%

Diff between max and min value of x is 25% - 8% = 17%

WiziusCareers1 · Answer

Final selling price is 2160 * (100 - a)/100 * (100 - b)/100 * (100 - c)/100

For the minimum selling price, take a = b = 0 and c = 50
=> Minimum selling price = 1080
=> Profit percentage, x = 8%

For the maximum selling price, take a = b = c = 50/3
=> Maximum selling price = 1250.3
=> Profit percentage = 25%

The difference between the two x values = 25 - 8 = 17
The answer is (B).

exc4libur · Answer

maximum value: when successive discounts are equal, (a,b,c)=50/3%
minimum value: largest possible successive discount, (a,b,c)=50,0,0%

original net%=sale/cost-1=2160/1000-1=2.16-1=1.16=116%
min. net%=2160(1-50/100)/1000-1=1.08-1=0.08=8%
max. net%=2160(1-50/3/100)^3/1000-1=216(5/6)^3/100-1…
max. net%=216(125/216)/100-1=1.25-1=0.25=25%
max-min net%=25-8=17%

Ans (B)

ScottTargetTestPrep · Answer

Recall that the order in which we apply the discount doesn’t matter (for example, a discount of 10% followed by a discount of 20% is same as a discount of 20% followed by a discount of 10%). Also, let’s assume that any values of a, b, and c can be 0 and they can also be decimals or fractions.

The minimum discounted price of the article after 3 discounts occurs when a, b, and c are as far apart as possible. That is, if we let a = 50, b = 0, and c = 0, then the minimum discounted price is 2160 x 0.5 x 1 x 1 = 1080.

The maximum discounted price of the article after 3 discounts occurs when a, b, and c are as close to each other as possible. That is, if we let a = b = c = 50/3 = 16⅔ (notice that 16⅔% = 1/6), then the maximum discounted price is 2160 x 5/6 x 5/6 x 5/6 = 2160 x 125/216 = 1250.

In the first case (minimum discounted price), the profit is 80, so the profit margin is 8%. In other words, the minimum value of x is 8. In the latter case (maximum discounted price), the profit is 250, so the profit margin is 25%. In other words, the maximum value of x is 25.

So the difference between the maximum and minimum values of x is 25 - 8 = 17.

Answer: B

riri3026 · Answer

To get the maximum profit, discount must be least. So we break it down into smaller units.
So if sum of all discounts is 50%, each discount is 50/3% = 1/6
Hence, the sale price  = 2160 *(5/6)*(5/6)*(5/6) = 1250

@ KarishmaB , can you please help me understand how did you reach to the conclusion of the bolded.
Thank you!

MS26 · Answer

SP = MP (1-d%) as the three discount are give then we can write as SP = MP (1-d%)*(1-d%)*(1-d%) , SP= 2160 *(1-50/300)(1-50/300)(1-50/300)=1250 Profit = 1250-1000 /1000 = 25% (max profit)

A shopkeeper bought an article for Rs.1000 and marked its price as Rs.

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A shopkeeper bought an article for Rs.1000 and marked its price as Rs.

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