The average price of Emily’s 5 meals was $20. Meals with a price of $2

Question

The average price of Emily’s 5 meals was $20. Meals with a price of $25 or more include a free dessert. How many of Emily’s 5 meals included a free dessert?

(1) The most expensive of the 5 meals had a price of $50.
(2) The least expensive of the 5 meals had a price of $10.

nick1816 · Answer

Suppose she spent a,b,c,d and e for her 5 meals.

a+b+c+d+e = 100

where a≤b≤c≤d≤e

Statement 1  e= 50; a+b+c+d=50

Case 1 - d=25; a=b=c= 25/3

2 meals include deserts

Case 2- a=b=c=d= 12.5

1 meal includes desert

Insufficient

Statement 2-  a=10 ; b+c+d+e = 90

Case 1 - b=15; c=d=e= 25

3 meals include deserts

Case 2- b=c=d=e= 22.5

none of the 5 meals includes desert

Insufficient

Combining both statements

a=10; and e=50. Hence, b+c+d=40.

10≤b≤c≤d≤50

max value d can take is (40-10-10=20).

only e is greater than or equal to 25.

Sufficient.

nostrumnihil · Answer

This answer is wrong. We don't have any indication to assume a<b<c<d<e. 
Answer is E

Advait01 · Answer

The question dosen't explicitly mention a<b<c<d<e, but we can assume a  ≤  b  ≤  c  ≤  d  ≤  e, so answer is (C)

vasu1104 · Answer

total 5 meals.
avg price = 20
sum = 100

price for each meal = a,b,c,d,e
a+b+c+d+e =100
assume all num here are in ascending order.

with any meal priced at equal or greater than 25 gets free desserts.
so we need to find how many free desserts?
so reframe it saying how many meal with price >= 25 ?

S-1
e = 50
sum of other 4 = 100.
since there can be different combination of it, we cant get definite ans here.
not sufficient.

S-2
a=10
so sum of other four - 90
here again there can be different combination of value. so not definite ans.
nto sufficient.

S-1&2
a= 10, e=50
sum of other three = 40
the moment any value goes beyond 25, the one value from remaining two have to be in single digit which is not possible since cheapest price is 10.
so with that in mind 10=< b,c,d<25

means only one has price over 25.
so we get definite ans.

choice C

Tinytrotter · Answer

The question does not mention the sequence but combining both statements we know that a = 10 & e = 50.

Which means b+c+d = 40. If one of these 3 were ever to cost 25 or more - let's assume d costs => 25, then b+c<= 15.

Neither of B or C can be below 10 because a is 10 and it is the least expensive dish. So b and c, even minimum cost 20 (where b & c are also assumed 10). That means at max d can cost 20 (40 - 10 - 10).

Which means only e costs more than 25 and that is why she will get only 1 free dessert.

Hence Option (C)

The average price of Emily’s 5 meals was $20. Meals with a price of $2

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The average price of Emily’s 5 meals was $20. Meals with a price of $2

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards