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If Jack buys 5 pieces of three different items priced $20, $80 and $100, does he buy at least one piece of each item?

(1) Three of 5 pieces are priced at $100 each.
(2) Average cost of 5 pieces is greater than $80.


self made

Statement 1: other two items can be priced at $80 each or one can be of $20 & the other of $80. Insufficient

Statement 2: \(Average >80 =>\) total price\(>400\)

Now all the 5 pieces could be of $100 amounting to $500 or 3 can be of $100, one of $80 & the other of $20. Insufficient

Combining 1 & 2 Total price of three items \(=100*3=300\)

so price or remaining two items \(>400-300\)

or price of remaining two items\(>100\).

Now if one item is of 20 and other of 80 then total price of two items will be \(80+20=100\), which is not possible.

Hence we can conclude that the remaining two items are of $80. Sufficient

Option C

Hi niks18,

most of the time you are right there, but here you have missed a point in statement II

(1) Three of 5 pieces are priced at $100 each.
other 2 could be 20 and 80 ..yes
other two can be both same type... No
insuff

(2) Average cost of 5 pieces is greater than $80.
TRICK lies here..
Max average when all three items are there : one each of 20 and 80 and remaining three of 100
, so \(\frac{20+80+3*100}{5}=80\)
BUT the average is >80
so there is no item of $20.. ans is always NO
sufficient
B
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chetan2u
niks18
chetan2u
If Jack buys 5 pieces of three different items priced $20, $80 and $100, does he buy at least one piece of each item?

(1) Three of 5 pieces are priced at $100 each.
(2) Average cost of 5 pieces is greater than $80.


self made

Statement 1: other two items can be priced at $80 each or one can be of $20 & the other of $80. Insufficient

Statement 2: \(Average >80 =>\) total price\(>400\)

Now all the 5 pieces could be of $100 amounting to $500 or 3 can be of $100, one of $80 & the other of $20. Insufficient

Combining 1 & 2 Total price of three items \(=100*3=300\)

so price or remaining two items \(>400-300\)

or price of remaining two items\(>100\).

Now if one item is of 20 and other of 80 then total price of two items will be \(80+20=100\), which is not possible.

Hence we can conclude that the remaining two items are of $80. Sufficient

Option C

Hi niks18,

most of the time you are right there, but here you have missed a point in statement II

(1) Three of 5 pieces are priced at $100 each.
other 2 could be 20 and 80 ..yes
other two can be both same type... No
insuff

(2) Average cost of 5 pieces is greater than $80.
TRICK lies here..
Max average when all three items are there : one each of 20 and 80 and remaining three of 100
, so \(\frac{20+80+3*100}{5}=80\)
BUT the average is >80
so there is no item of $20.. ans is always NO
sufficient
B

Yup agreed. I actually got confused between the earlier average posted 85 & the edited one 80. I changed the value but did not check the approach :-D
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niks18
chetan2u
niks18


Statement 1: other two items can be priced at $80 each or one can be of $20 & the other of $80. Insufficient

Statement 2: \(Average >80 =>\) total price\(>400\)

Now all the 5 pieces could be of $100 amounting to $500 or 3 can be of $100, one of $80 & the other of $20. Insufficient

Combining 1 & 2 Total price of three items \(=100*3=300\)

so price or remaining two items \(>400-300\)

or price of remaining two items\(>100\).

Now if one item is of 20 and other of 80 then total price of two items will be \(80+20=100\), which is not possible.

Hence we can conclude that the remaining two items are of $80. Sufficient

Option C

Hi niks18,

most of the time you are right there, but here you have missed a point in statement II

(1) Three of 5 pieces are priced at $100 each.
other 2 could be 20 and 80 ..yes
other two can be both same type... No
insuff

(2) Average cost of 5 pieces is greater than $80.
TRICK lies here..
Max average when all three items are there : one each of 20 and 80 and remaining three of 100
, so \(\frac{20+80+3*100}{5}=80\)
BUT the average is >80
so there is no item of $20.. ans is always NO
sufficient
B

Yup agreed. I actually got confused between the earlier average posted 85 & the edited one 80. I changed the value but did not check the approach :-D

actually, irrespective of 80 or 85 it is going to be B.
changed it to 80 to make it a bit more tricky for those getting at 80, otherwise 85 is straight B.
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