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Bunuel
A store sells chairs and tables. If the price of 3 chairs and 1 table is 60% of the price of 1 chair and 3 tables, and the price of 1 table and 1 chair is $60, what is the price, in dollars, of 1 table? (Assume that every chair has the same price and every table has the same price.)

(A) 15
(B) 20
(C) 30
(D) 40
(E) 45

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Let, COst of Chair = C, Cost of Table = T

i.e. 3C+T = (60/100)*(C+3T)
i.e. 3C+T = (3/5)*(C+3T)
i.e. 15C + 5T = 3C + 9T
i.e. 12C = 4T
i.e. T = 3C

and T+C = 60
i.e. 3C + C = 60
i.e. C = $15
i.e. T = $45

Answer: Option
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Bunuel
A store sells chairs and tables. If the price of 3 chairs and 1 table is 60% of the price of 1 chair and 3 tables, and the price of 1 table and 1 chair is $60, what is the price, in dollars, of 1 table? (Assume that every chair has the same price and every table has the same price.)

(A) 15
(B) 20
(C) 30
(D) 40
(E) 45

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From the text we know;
3C + 1T = 0.6(1C +3T)
and:
1T + 1C = 60 >>> T = 60 - C

Plug in T in the formula above:
3C + 60 - C = 0.6C + 1.8(60-C)
2C + 60 = 108 - 1.2C
3.2C = 48
C = 15
T = 45

Answer E
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Hi All,

While this question can certainly be solved algebraically, it can also be solved rather easily by TESTing THE ANSWERS and doing a little arithmetic...

We're given a few facts to work with:
1) 3 chairs + 1 table = 60% of the price of (1 chair + 3 tables)
2) 1 chair + 1 table = $60
3) Each chair has the same price and each table has the same price

We're asked for the price of 1 table.

Since 3 chairs and 1 table cost LESS than 1 chair and 3 tables, a chair MUST cost LESS than a table. Since 1 chair + 1 table = $60, that means that 1 chair costs LESS than $30 and 1 table costs MORE than $30. There are only two answers that fit THIS pattern....

Let's TEST Answer D: $40

IF....1 table = $40....
1 chair = $20
3 chairs + 1 table = $100
1 chair + 3 tables = $140
60% of $140 = $84 (not $100....so this CANNOT be the answer).

There's only one answer remaining....

Final Answer:
Here's the proof though....

IF....1 table = $45....
1 chair = $15
3 chairs + 1 table = $90
1 chair + 3 tables = $150
60% of $150 = $90 (this is a MATCH, so this MUST be the answer).

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Rich
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Bunuel
A store sells chairs and tables. If the price of 3 chairs and 1 table is 60% of the price of 1 chair and 3 tables, and the price of 1 table and 1 chair is $60, what is the price, in dollars, of 1 table? (Assume that every chair has the same price and every table has the same price.)

(A) 15
(B) 20
(C) 30
(D) 40
(E) 45

Kudos for a correct solution.

Let T be the price of one table
C be the price of one chair

T + C = 60 --- Equation 1
(3C + T) = (60/100) ( 3T + C)
on solving we get
T - 3C = 0 ---- Equation 2

Solving Equation 1 and 2 simultaneously
we get
T = 45.

Hence Option E is correct.
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3Chair + 1Table = 6/10 (Chair + 3Table)
1Table + 1Chair = 60
3Chair + 1Table = 6/10Chair + 18/10Table
24/10Chair = 8/10Table
Table = 45
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3c + 1t = 60/100 (1c + 3t)
1t + 1c = 60; 1t = ?

2.4c - 0.8t = 0
1c + 1t = 60 (*2.4)
-----------------------
2.4c + 2.4t = 144
2.4c - 0.8t = 0
-----------------------
3.2t = 144
t = 45
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Ans is E
Given,
3C + 1T =60/100(1C+3T)
by solving we get,
3C=T
Also given, C+T=60
substituting value for T , C= 15
T=6-15=45,
hence Ans is E
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