During a sale of 20% on everything in a store, a kid is successful in

Use the standard formula for Mark-up, Discount and Profit which is (1 + m/100)*(1 - d/100) = (1 + p/100)

Note that there are 2 discounts - the 20% which is the discount on everything and another which the kid gets by taking 20 candies for the discounted price of 14. This is another 6/20 = 30/100 = 30% discount.

(1 + m/100)(1 - 20/100)(1 - 30/100) = (1 + 12/100)

(1 + m/100) = (112/100) * (5/4) * (10/7)

m = 100

Answer (A)

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/09 ... questions/
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Hi Akhil, I can share the way I solved it..

Let's say Marked price = M

So, there's a discount of 20% on M So, new S.P. = 80% of M
Now, the child convinces the owner to sell 20 candies for the price of 14 candies .
Let's say each candy after discount is 1$. So, S.P. of 20 candies = 20$. THe child bought it for 14 $ So, he got a discount of 6/20 *100 = 30%

So, the latest S.P. = 70% of 80% of M = 0.7* 0.8 M

Now, we are given that the shopkeeper still makes a profit of 12%. So we have ,

0.7 * 0.8 * M = 1.12 C.P

So, we get, M= 2 C.P.

i.e. Marked price was kept 100% above C.P.


Hope it is clear.

1st Way
Candy Marked Up price be x
so during sale candy price = 0.8x
Now,
20 candies for the price of 14 candies
so
0.8x * 14 = 20 * y (Y is money he paid for 20 candies)
y = 0.56x

still store makes 12% profit
SO,
0.56x = 1.12z (z = OG price)
x = 2Z

100% markup

2nd Way
0.8 of Marked up price
20 candies for 14 candies so each candy marked down to 0.7 (14/20) of discounted price
Total down price 0.8*0.7
Now Put options,
let OG price be a
option 1
100% markup
so marked up price 2a
2a * 0.8 *0.7 = 1.12a
12% profit
Over.

i picked numbers for this one..as it seemed faster for me...
suppose 1 candy -> 100$.
discount 20% => 80$ (still pretty expensive candy :D )
now, the seller sold 20 candies for the price of 14 discounted ones -> 14*80 -> 28x40 -> 56x20 -> 112x10 -> 1120 $
now, 1120/20 = 112/2 -> 56 $ - price sold for 1 candy.
56 is the cost + 12% profit.
12% = 3/25
56 = 28x/25
28x = 56x25
28X = 28x50
x=50.
so the real price of the candy is 50$.
the regular price of the candy is 100$.

the markup is:
(100-50)/50 * 100% => 100%

Let’s denote the cost of each candy by c and the mark-up percentage by p. Then, the regular sale price of the candy is (1+p/100)c.

Since the regular sale price was discounted by 20%, the discounted sale price is 0.8(1+p/100)c.

Since the kid paid the discounted price of 14 candies, he paid 14(0.8)(1+p/100)c.

Since the kid received 20 candies, he paid (1/20)14(0.8)(1+p/100)c per candy.

The store made a profit of 12% on this sale, which means if we subtract the cost of a candy from (1/20)14(0.8)(1+p/100)c, we will find 12% of the cost of a candy. In other words, we have the following equation:

(1/20)14(0.8)(1+p/100)c - c = 0.12c

Let’s simplify this equation:

0.56(1 + p/100)c = 1.12c

0.56(1 + p/100) = 1.12

1 + p/100 = 1.12/0.56

1+ p/100 = 2

p/100 = 1

p = 100

Answer: A
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I chose smart numbers to solve:

1) 1 candy = $1 therefore 20 candies = $20

2) The kid was able to negotiate 20 candies for the price of 14 therefore 20 candies = $14

3) The store is giving 20% off all products so therefore the kid gets 20% off $14 = $11.2 (for 20 candies)

4) The store still makes a 12% profit on selling 20 candies for $11.2 which makes the cost of candies ($11.2/1.12) $10.

5) The 20 candies cost the store $10 however they originally sell the 20 candies for $20, so there's a 100% markup.

For each candy let:
C.P. be x, M.P. be y , and S.P. = \(\frac{80}{100}\)* y

Discounted S.P. price of 14 candies = \(\frac{80}{100}\)* y *14
But the kid actually got 20 candies.
The 20 candies cost (C.P.)the shopkeeper = 20x

But the shopkeeper still made a profit of 12% on C.P. so , \(\frac{(S.P. - C.P)}{C.P.}\) = \(\frac{(\frac{80}{100} *y*14)-20x}{20x}= \frac{12}{100}\) , After simplification: 5600y=11200x or y = 2x , so M.P. is 100% greater than C.P.
Answer A
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