Five shirts with defects are for sale at Divanly. o

Question

Five shirts with defects are for sale at Divaly. The price of each shirt is a two-digit integer number of euros. If Stefanny buys any two of the shirts, she will pay either €81, €105, €112, or €136. Which of the following could the amount Stefanny will pay if she buys all five shirts?

I €270
II €273
III €285

A. I only
B.II only
C. III only
D. I and III
E. II and III

I €270
II €273
III €285

A. I only 
B.II only 
C. III only 
D. I and III 
E. II and III

paragw · Answer

I could not find any actual set of five two-digit integer shirt prices that satisfies the given condition that every pair sums to only 81, 112, or 136. It seems the total is determinable algebraically, but the individual shirt prices may not exist. Could you please verify the question

kevincan · Answer

It’s surprising that there are only three possible totals

kevincan · Answer

There are actually 4

yukisuki · Answer

Hi, I want to find the fastest way to solve this, many thanks.

Kinshook · Answer

Let the prices of shirts be p1, p2, p3, p4 & p5 in increasing order

p1 + p2 = 81

p4 + p5 = 136

p1 + p2 + p3 + p4 + p5 = 81 + p3 + 136 = 217 + p3

Total number of sums possible = 5C2 = 10

Since there are only 4 sums possible, let us assume that 2 prices are equal

Total number of sums possible = 4C2 + 1 = 7; Number of ways to select 2 numbers out of unique 4 + 1 way to select 2* equal price

Since there are only 4 sums possible, let us assume that 3 prices are equal

Total number of sums possible = 3C2 + 1 = 3 + 1 = 4; Number of ways to select 2 numbers out of unique 3 + 1 way to select 2* equal price

Case 1:
p1 = p2 = p3 = x ; p4, p5
2x = 81; x = 40.5; Not feasible since x is a 2-digit integer

Case 2: 
p1, p2 = p3 = p4 = x, p5
p1 + x = 81; p2 + x = 105
p5 + x = 136; 2x = 112; x = 56; p1 = 25; p5 = 80
p1 + p2 + p3 + p4 + p5 = 25 + 3*56 + 80 = 273

Case 3:
p1; p2; p3 = p4 = p5 = x
2x = 136; x = 68; 
p2 + x = 112; p2 = 44; 
p1 = 81 - 44 = 37; 
44 + 68 = 112; 37 + 68 = 105
p1 + p2 + p3 + p4 + p5 = 37 + 44 + 3*68 = 285

I €270
II €273
III €285

IMO E

egmat · Answer

Hi yukisuki,

The fast route here doesn't need you to find all five prices first. It leans on one clean idea about the  smallest and largest pair sums .

The key shortcut

Sort the five prices as  a ≤ b ≤ c ≤ d ≤ e . Then:

- The  smallest  possible two-shirt sum is  a + b , so a + b =  81 .
- The  largest  possible two-shirt sum is  d + e , so d + e =  136 .

Add those: a + b + d + e =  81  +  136  =  217 .

Since the total of all five is a + b +  c  + d + e, you get:

Total = 217 + c  (c = the middle price)

So each candidate total instantly pins down the middle shirt's price:

- I:  270  - c =  53 
- II:  273  - c =  56 
- III:  285  - c =  68

Then just test feasibility

Now you only need to see whether a valid set exists for each c. Try to build one:

-  III (c = 68):  { 37 ,  44 ,  68 ,  68 ,  68 } - pairs give  81 ,  105 ,  112 ,  136 . ✓
-  II (c = 56):  { 25 ,  56 ,  56 ,  56 ,  80 } - pairs give  81 ,  105 ,  112 ,  136 . ✓
-  I (c = 53):  a + b =  81  with b ≤  53  forces a ≥  28 . Then a + c = a +  53  can only reach  81  (a =  28 , b =  53 ). But that makes b = c =  53 , so b + c =  106  - not one of the allowed sums. No set works. ✗

So only  II and III  are possible -  therefore  the answer is E.

The whole trick is spotting that the min pair and max pair are locked to  81  and  136  - that turns a messy  5 -variable problem into a one-line formula plus a quick build-a-set check.

Answer: E

KarishmaB · Answer

Using logic and reasoning, you can arrive at the answer fairly quickly.

There are 5 shirts but only 4 distinct sums of two prices. Why? If all 5 shirts had distinct pricing, we would have got many more distinct sums. This means some shirts must have identical pricing. 
What if there are only 2 distinct pricing a and b? Then we will get at max 3 distinct sums, a+a, b+b, a+b.
So there must be 3 distinct prices (a, b, c) to get 4 distinct sums. (a+b, b+c, a+c, 2c) So 3 of the shirts have the same price c.

Since one of the sums is 2c, it must be an even number and hence it must be either 112 or 136.

If 2c = 112, then c = 56
Diff of 56 from 81, 105 and 136 is 25, 49 and 80. Out of these which values are a and b? Since 25 + 80 =105 which is one of the sums, it means the prices are 25, 81 and 56.
Total price of all 5 shirts = 25 + 81 + 56*3 = 273.

Hence II is definitely possible. So answer will be (B) or (E). If I am short on time here, I will guess (E) and move on because I know that 2c can be 136 too and it is likely that that sum will also be in there. Let's check though.

If 2c = 136, then c = 68
Diff of 68 from 81, 105 and 136 is 13, 37 and 44. Out of these which values are a and b? Since 37 + 44 = 81 which is one of the sums, it means the prices are 37, 44 and 68.
Total price of all 5 shirts = 37 + 44 + 3*68 = 285

Answer (E)

Five shirts with defects are for sale at Divanly. o

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Five shirts with defects are for sale at Divanly. o

Prep Toolkit

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