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

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Manager
Joined: 11 Aug 2011
Posts: 165
Location: United States
Concentration: Economics, Finance
GMAT Date: 10-16-2013
GPA: 3
WE: Analyst (Computer Software)
During a sale of 20% on everything in a store, a kid is successful in  [#permalink]

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02 Oct 2014, 23:52
2
9
00:00

Difficulty:

85% (hard)

Question Stats:

58% (02:56) correct 42% (03:17) wrong based on 146 sessions

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During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy?

(A) 100%
(B) 80%
(C) 75%
(D) 66+2/3%
(E) 55%

This question can be solved in multiple ways and it would be good to discuss what are the different ways in which to solve this problem.

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Joined: 03 Jul 2012
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GMAT 1: 710 Q50 V36
GPA: 3.9
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Re: During a sale of 20% on everything in a store, a kid is successful in  [#permalink]

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03 Oct 2014, 00:06
2
akhil911 wrote:
During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy?

(A) 100%
(B) 80%
(C) 75%
(D) 66+2/3%
(E) 55%

This question can be solved in multiple ways and it would be good to discuss what are the different ways in which to solve this problem.

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 $So, he got a discount of 6/20 *100 = 30% So, the latest S.P. = 70% of 80% of M = 0.7* 0.8 M Now, we are given that the shopkeeper still makes a profit of 12%. So we have , 0.7 * 0.8 * M = 1.12 C.P So, we get, M= 2 C.P. i.e. Marked price was kept 100% above C.P. Hope it is clear. Manager Joined: 13 Sep 2014 Posts: 87 WE: Engineering (Consulting) Re: During a sale of 20% on everything in a store, a kid is successful in [#permalink] ### Show Tags 26 Jul 2015, 20:13 1st Way Candy Marked Up price be x so during sale candy price = 0.8x Now, 20 candies for the price of 14 candies so 0.8x * 14 = 20 * y (Y is money he paid for 20 candies) y = 0.56x still store makes 12% profit SO, 0.56x = 1.12z (z = OG price) x = 2Z 100% markup 2nd Way 0.8 of Marked up price 20 candies for 14 candies so each candy marked down to 0.7 (14/20) of discounted price Total down price 0.8*0.7 Now Put options, let OG price be a option 1 100% markup so marked up price 2a 2a * 0.8 *0.7 = 1.12a 12% profit Over. Board of Directors Joined: 17 Jul 2014 Posts: 2620 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Re: During a sale of 20% on everything in a store, a kid is successful in [#permalink] ### Show Tags 22 Feb 2016, 19:20 i picked numbers for this one..as it seemed faster for me... suppose 1 candy -> 100$.
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% Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8673 Location: Pune, India Re: During a sale of 20% on everything in a store, a kid is successful in [#permalink] ### Show Tags 22 Feb 2016, 21:27 2 1 akhil911 wrote: During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy? (A) 100% (B) 80% (C) 75% (D) 66+2/3% (E) 55% This question can be solved in multiple ways and it would be good to discuss what are the different ways in which to solve this problem. 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: http://www.veritasprep.com/blog/2014/09 ... questions/ _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4294 Location: United States (CA) Re: During a sale of 20% on everything in a store, a kid is successful in [#permalink] ### Show Tags 01 Jun 2017, 09:12 akhil911 wrote: During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy? (A) 100% (B) 80% (C) 75% (D) 66+2/3% (E) 55% 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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 08 May 2011 Posts: 21 Location: Canada Re: During a sale of 20% on everything in a store, a kid is successful in [#permalink] ### Show Tags 02 Jun 2017, 12:39 2 akhil911 wrote: During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy? (A) 100% (B) 80% (C) 75% (D) 66+2/3% (E) 55% This question can be solved in multiple ways and it would be good to discuss what are the different ways in which to solve this problem. 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.
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Re: During a sale of 20% on everything in a store, a kid is successful in  [#permalink]

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14 Dec 2017, 05:51
akhil911 wrote:
During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy?

(A) 100%
(B) 80%
(C) 75%
(D) 66+2/3%
(E) 55%

This question can be solved in multiple ways and it would be good to discuss what are the different ways in which to solve this problem.

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.
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Re: During a sale of 20% on everything in a store, a kid is successful in &nbs [#permalink] 14 Dec 2017, 05:51
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