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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent

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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 05 Nov 2017, 01:04
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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00
[Reveal] Spoiler: OA

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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00

Hello @Bunnel,
I have tried solving this question algebraically, Am getting $16 as the answer. I don't see flaw in my approach. Can you please check the Question. Correct me if I am wong.

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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 05 Nov 2017, 01:17
harikrish wrote:
Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00

Hello @Bunnel,
I have tried solving this question algebraically, Am getting $16 as the answer. I don't see flaw in my approach. Can you please check the Question. Correct me if I am wong.


There was a typo. Edited. Thank you!
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 05 Nov 2017, 01:21
Bunuel wrote:
Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00


Lets assume John's meal costs 120$.
Now Karen's meal will cost 180$
Since Mary's meal costs \(\frac{5}{6}\)th as much, it would be 150$.

The prices of their meals are in the ratio 120:180:150 or 4x:6x:5x(John:Karen:Mary)

Since Mary pays 2$ more than John, 5x - 4x = 2 => x=2
Therefore, the three of them would have paid 15x or 30$(Option B)
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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 05 Nov 2017, 01:23
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Solution:

k=1.5 J----1
m= (5/6J)---2
m-J=2---3

Substituting 1 and 2 in 3, We get k=12, m=10 and J=8.
So the total cost of the lunch is $30. Option B.

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Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 08 Nov 2017, 02:40
Let John's share=J
Karen's share= 50% more of John = (3/2)J
Mary's share = 5/6th of Karen = (5/6)(3/2J) = (5/4)J

Also given, (5/4)J=J+2
or,5J=4J +8
J=8

Therefore,Karen = (3/2)J=12
Mary= (5/4)J= 10

Total expenditure=8+12+10=30

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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 08 Nov 2017, 08:05
let John's meal cost be J.
karen (K) = 0.5J + J= 1.5J
Mary (M) = 5/6*1.5J

As given, 5/6*1.5J = 2+J => J=8

We get, K = 1.5*8 = 12 and M = 5/6*1.5*8 = 10

Hence, total cost = 8+12+10= 30

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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 12 Nov 2017, 07:19
Bunuel wrote:
Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00


We can let M = the cost of Mary’s meal, K = the cost of Karen’s meal, and J = the cost of John’s meal.

Thus:

K = 1.5J

and

M = 5K/6

and

M = J + 2

So, we can first use the first two equations and we have:

M = 5(1.5J)/6

M = 7.5J/6

Finally, we have:

7.5J/6 = J + 2

Multiplying the entire equation by 6, we get:

7.5J = 6J + 12

1.5J = 12

J = 8

So, M = 8 + 2 = 10, and K = 1.5 x 8 = 12.

So, the total for all 3 meals is 8 + 10 + 12 = $30.

Answer: B
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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent [#permalink]

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New post 12 Nov 2017, 08:16
Bunuel wrote:
Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?

(A) $28.33
(B) $30.00
(C) $35.00
(D) $37.50
(E) $40.00

Let cost of John’s meal be 60
So, cost of Karen’s meal will be 90
And,cost of Mary’s meal will be 75

Total cost is 225

Quote:
If Mary paid $2 more than John, how much was the total that the three of them paid

\(\frac{2}{75-60}*225\)

=30 , Answer must be (B) 30
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Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent   [#permalink] 12 Nov 2017, 08:16
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