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If Jack buys 5 pieces of three different items priced $20, $80 and $10

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chetan2u
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If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 05:54
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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.


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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 06:11
chetan2u wrote:
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 $85.


self made


(1) Three of 5 pieces are priced at $100 each.

It can be 3 $100 and 2 items of $20 & $800 Gives Yes

It can be all 5 $100 Gives No

(2) Average cost of 5 pieces is greater than $85.

It says total cost is more than 425

It can be 3 $100 and 2 items of $20 & $800 Gives Yes

It can be all 5 $100 Gives No

Combining both doesn't give any info

Hence E
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If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 06:11
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chetan2u wrote:
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 if all three items are bought then cost of 3 items will be \(=20+80+100=200\)

so the total cost of remaining two items has to be at least 201. But maximum cost of any item is 100, hence we can reach only upto 200, resulting in total cost of 400. But we know total price >400

Hence all three items were not bought. Sufficient

Option B
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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 06:20
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niks18 wrote:
chetan2u wrote:
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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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 06:25
chetan2u wrote:
niks18 wrote:
chetan2u wrote:
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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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   17 Dec 2017, 06:42
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niks18 wrote:
chetan2u wrote:
niks18 wrote:

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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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink] New post   05 Jul 2023, 11:59
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Re: If Jack buys 5 pieces of three different items priced $20, $80 and $10 [#permalink]
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