Harry bought 3 items and got 20% discount off regular price

Question

Harry bought 3 items and got 20% discount off regular price of most expensive item and 10% discount off regular price of the other two. Was the total amount of the 3 discounts greater than 15% of the sum of the regular prices of the 3 items?

(1) Regular price of most expensive item was $50 and regular price of next most expensive was $20
(2) Regular price of least expensive item was $15

IanStewart · Accepted Answer

You can avoid algebra completely here. The question describes a weighted average. If you buy two things, and one is discounted by 20%, the other by 10%, then the overall discount will be 15% if the two original prices are equal. If the 20% item is more expensive, before the discount, than the 10% one, the discount will be closer to 20% (so will be greater than 15%).

So here we can think of the two cheaper items as a single purchase by combining their prices. From Statement 1, the most expensive item cost $50, and the two cheaper items cost at most $40, so the overall discount was closer to 20% than to 10%, and was thus greater than 15%. Statement 2 is not sufficient alone, so the answer is A.

ulm · Answer

Let "a" the most exp, "b" the second expensive, "c" the third
we need to know whether  0.2a+0.1b+0.1c>0.15(a+b+c) 
after easy calculation we have:  a-b-c>0  ?

from 1)
 a=50, b=20 
 50-20-c>0 ?
c should be 19 or smaller (consider we use only integers)
so,  50-20-19>0  yes
sufficient
from 2)
 c=15 
 a-b-15>0  ?
b is at least  16, a=17 
 17-16-15 >0  no
a=100, b=20, c=15
 100-20-15>0  yes
2 different questions, insufficient

(A)

jamifahad · Answer

let price of 3 items in increasing order be x1, x2 and x3.

Stmt1: x3=50, x2=20. so x1<20
sum of individual discount= 50*0.2+20*0.1+x1*0.1=12+0.1x1
combined discount = (50+20+x1)*0.15= (70+x1)0.15
Question 12+0.1x1 > (70+x1)0.15
Now we know x1<20. Take x1=10
12+0.1*10>80*0.15
13>12
Take x1=19
12+0.1*19>89*0.15
13.9>13.35
Sufficient.

Stmt2: x1=15
15*0.1+x2*0.1+x3*0.2>(15+x2+x3)*0.15
Take x2=20,x3=30
15*0.1+20*0.1+30*0.2<(15+20+30)*0.15
9.5 < 9.75
Take x2=500, x3=1000
15*0.1+500*0.1+1000*0.2>(15+500+1000)*0.15
251.5>227.25
Yes and no.

OA A.

amit2k9 · Answer

a. price of the third item can be 1 15% of price. sum 2 = 50 + 20 + 1 = 71 15% of 71 = 10.65 discount = 10+2+0.1 = 12.1 discount > 15% of the price. sufficient. b. tells nothing about the prices of the expensive items. Not sufficient. A it is.

sudhir18n · Answer

Hi Ian 
I understand Weighted averages and how it works.. But fail to apply here..
should it be like this ?
(50*1*20%+ 20*1*10%+ 19*10%*1)/3 > ((50+20+19)*15%)/3

is this working correct ?

gmat1220 · Answer

I think the weights here are the prices of the items and not 1, 2 or 3. And the 10% and 20% are the values.

assuming the prices are a, b and c where a > b > c, so the weighted discount will be -

20 * a + 10 * (b + c) / (a + b + c)

a = price of the most expensive item
b and c = price of the less expensive items

If a > b + c then we can be assured that the final discount is moving towards 20%. So in this problem S1 is more than sufficient to answer the question.

jlgdr · Answer

I think that what Ian is trying to explain here is that 15% is halfway between 20 and 10. Therefore if object A is 20% discount then objects B and C are 10% discount each we need to know whether A >= B+C. For instance, let A be 100, therefore discount will be 20. Now let each of B and C be 50. Therefore discount is 5 each total of 10. Total discount will be 30/200 * 100% = 15%. So going back to the question.

Statement 1 tells us that A = 50 and B = 20. Now since C is the least expensive it can't be more than 20. Therefore statement will be sufficient.

Statement 2 is clearly insufficient.

Therefore A is the correct answer

Hope this clarifies
Cheers
J

kashifgolf · Answer

Harry bought 3 items and got 20% discount off regular price of most expensive item and 10% discount off regular price of the other two. Was the total amount of the 3 discounts greater than 15% of the sum of the regular prices of the 3 items?

(1) Regular price of most expensive item was $50 and regular price of next most expensive was $20
(2) Regular price of least expensive item was $15

Solution:-

Lets first look at what the question is asking:- Harry bought 3 items. Harry got a discount of 20% on the most expensive item and 10% discount on the other two.

The question is asking whether the discount Harry received is greater than 15% of the sum of the prices of all 3 items. While we can solve this question algebraically lets try and avoid algebra. We can use the concepts in weighted average to solve this question without making any calculations.

Statement 1 - The most expensive item is 50$ and the second most expensive item is 20$, therefore we can conclude that the value of the third item cannot be greater than 20$. To solve this question using the concepts of weighted average we will club the price of second and third items which cannot be greater than 40$ since the discount on the second and third items are of equal value. We know that the discount on the 50$ item is 20% and 10% on the 40$ items. The total discount on all the three items would have been equal to 15% if the price of the most expensive item and the other two items would have been equal but since the most expensive item is of a greater value (50$) and is discounted by a greater percentage of 20% the weighted average discount would be greater than 15%. Therefore statement 1 is sufficient

Statement 2 - We are only given the price of the least expensive item and therefore there is insufficient information to solve the question. Statement 2 is insufficient.

Answer - A

Tip - One can avoid algebra by simply using LOGIC to solve such questions. Algebraic equations can be time consuming and not the most efficient way to solve such questions.

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Kudos for reply

mcwoodhill · Answer

the simplest is to make a set:

a1, a2, a3, ascending with price

so 10%(a1+a2)+20%a3>15%(a1+a2+a3)?

simplify it you get:

a3>a1+a2?

(1) a3=50, a2=20, so a1+a2<=40

sufficient

(2) a1=15, insufficient, we need to compare a3 with the sum of a1, a2

A

susheelh · Answer

I kind of did this after looking at the question choices.

S1: Most expensive = 50. Next most expensive = 20. Worst case, third item could also cost 20.

So, total discount =  \frac{1}{5}* 50 + \frac{2}{10}*20 = 10 + 4 = 14

Sum of the Max possible regular price  = 50 + 20 + 20 = 90

15% of the max possible sum of regular price =  \frac{3}{20}*90 = 13.5

Clearly  13.5 < 14

So, even with the Max possible price of third item, the total amount of the 3 discounts is  not  greater than 15% of the sum of the regular prices of the 3 items? One answer. A : suff

S2: Has no info on the most expensive item and hence we cannot find the discount. Insuff.

Final answer = A

sarath galavalli · Answer

why did we consider the price of the third item as 1<x<20.
why not 1<=x<=20?

susheelh · Answer

Hello sarath galavalli!

Well, this is based on what the stem has asked us. The question asks us - "Was the total amount of the 3 discounts greater than 15% of the sum of the regular prices of the 3 items?"

This means we need to first find the max discount possible when we take the three items together. To get a highest discount, we need the Maximum possible value of the third most expensive item. Hence, I have taken the extreme value 20 in my solution above. Even if with the "max possible" value of third item the sum of discounts of all three items is less then 15%, then definitely it will be less for any other value.

I am not sure if this was of any help to you. Just tried by best

sarath galavalli

why did we consider the price of the third item as 1<x<20.
why not 1<=x<=20?

amargad0391 · Answer

Hello,

How can we be sure that Item C is less than $20 ? Item C can be between $20 and $50 , isn't it ?
Please explain .

Thank You

susheelh · Answer

Hello  amargad0391 ,

Let me try and attempt to answer.

I think your question is for S1.

The Statement 1 reads as follows

In his explanation jlgdr has denoted the most expensive = A, Second most expensive = B, and Least expensive = C. So, from S1 we get - A = $50, B=$20. If C is the Least expensive among A,B,C then definately C has to be less then $20.

Hope this helps.

minustark · Answer

1) Considering the 3rd item is of $20, total discount = $14, which is greater than 15% of total (90*.15 = 13.5). If the 3rd item is $10, discount = 13, which is also greater than the 15% of total. Using the allegation method (20-15) = (15 - 10) or, 5 = 5, we can infer that whenever the ratio of most expensive to other two will exceed 1: 1, two cases will happen. If the dollar value of 10% discount items is more, the discount will be less than 15%, and when 10% discount items are less, the discount will be more than 15%. Since here the most expensive of the 2 regular items is $20, the discount will always be more than 15%. Sufficient

2) Not sufficinet by itself.

A is the answer.

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Harry bought 3 items and got 20% discount off regular price

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