A school supply store sells only one kind of desk and one kind of chai

Question

A school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. If the total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs, then the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?

A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3

PS04502.01

A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3

PS04502.01

ZoltanBP · Accepted Answer

Let d be the price per desk and let c be the price per chair.

3d+c=2(d+3c)
3d+c=2d+6c
d=5c

\frac{4d+c}{d+4c}=\frac{21c}{9c}=\frac{7}{3}

Answer: E

Archit3110 · Answer

GIVEN 
3D+1C=2D+6C
OR SAY 
D=5C
SO
4D+1C= 1D+4C
21=a*9
a= 7/3
IMO E

BrentGMATPrepNow · Answer

Let C = the cost of ONE chair
Let D = the cost of ONE deck

The total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs   
We can write: 3D + C = 2(D + 3C)
Expand to get: 3D + C = 2D + 6C
Subtract 2D from both sides: D + C = 6C
Subtract C from both sides:  D = 5C

The total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?  
This is asking us to determine how many times the price of 1 desk and 4 chairs DIVIDES INTO the price of 4 desks and 1 chair

In other words, we want to find the value of (4D + C)/(D + 4C)

Since we now know that  D = 5C , we can replace  D  with  5C  to get: 
(4D + C)/(D + 4C) =  5C ) + C]/ 5C  + 4C]
=  /9C
= 21C/9C
= 21/9
= 7/3

Answer: E

Cheers,
Brent

EMPOWERgmatRichC · Answer

Hi All,

We're told that a school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. The total cost of 3 desks and 1 chair is TWICE that of 1 desk and 3 chairs. We're asked - the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs. This question is ultimately about ratios - and you have to do a little Algebra to work through it.

The information that we're given can be translated into the following equation: 
3D + C = 2(D + 3C)

We can then do a couple Algebra steps to simplify this equation: 
3D + C = 2D + 6C 
D = 5C

We now know that 1 desk is the same price as 5 chairs. With this information, we can substitute in to the given question: 
Cost of 4 desks and 1 chair = 4D + 1C = 4(5C) + 1C = 21C

Cost of 1 desk and 4 chairs = 1D + 4C = 1(5C) + 4C = 9C

21C/9C = 7/3

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

Let’s let D = the cost of a desk and C = the cost of a chair. We can create the equation:

3D + C = 2(D + 3C)

3D + C = 2D + 6C

D = 5C

Let’s let K = the number of times greater that one desk and 4 chairs costs, compared to the cost of 4 desks and 1 chair. So, we have:

4D + C = K(D + 4C)

Substituting 5C for D, we have:

20C + C = K(5C + 4C)

21C= K(9C)

K = 21C/9C = 7/3

Answer: E

sony1000 · Answer

I used 2 equations:

3D + C = 2 ( D + 3C)....solve for D...D= 5C

4D + C = X (D + 4C)
20C + C = X (5C + 4C)

X= 7/3

hemantbafna · Answer

Desk=d chair=c
3d+1c=2(1d+3c)
3d+1c=2d+6c
d=5c

x=total cost of 4 desks and 1 chair is  how  many times that of 1 desk and 4 chairs
Therefore;
4d+1c=x(1d+4c)
21c=x(9c)
x=7/3
 E

Pritishd · Answer

Lets symbolize desk with  x  and chair with  y  for simplicity.

1.  3x + y = 2 (x + 3y) 
2.  3x + y = 2x + 6y 
3.  x = 5y

1.  4x + y = a (x + 4y) ; let  a  be the unknown factor
2. Lets substitute  x = 5y  in the above equation
3.  20y + y = a (5y + 4y) 
4.  21y = a * 9y 
5.  a = \frac{21y}{9y} = \frac{7}{3}

Ans.  E

CrackverbalGMAT · Answer

Concept: The question tests on the basics of framing and solving equations

Solution:  
Let the price per unit of desks be d and that of chairs be c

Total cost of 3 desks and 1 chair= 3d + c
Total cost of 1 desk and 3 chairs = d + 3c
According to the question,

3d + c = 2(d+3c)

=>d=5c

Total cost of 4 desks and 1 chair =4d + c = X
Total cost of 1 desk and 4 chairs = d +4c =Y
We need to find X is how many times Y??

=>X/Y = 4d + c / d + 4c

=> 4(5c) + c / 5c+ 4c

= 21/9

=7/3

=>X/Y =7/3 = > X is 7/3 times Y  (option e)

For questions on Inequalities, analyze the parts in the question stem and break them down to form an equation.

Devmitra Sen
GMAT SME

ThatDudeKnows · Answer

Plenty of good algebraic solutions above, and that's probably the better way on this question. Just want to show that we can do this one by Plugging In, too.

3D+C = 200
D+3C = 100

Multiply the second equation by 3:
3D+C = 200
3D+9C = 300

Subtract the first from the second:
8C = 100
C = 12.5

Plug C=12.5 into D+3C = 100:
D+37.5 = 100
D = 62.5

4D+C
4(62.5) + 12.5 = 262.5

D+4C
62.5 + 4(12.5) = 112.5

262.5 : 112.5
525 : 225
21 : 9 
7 : 3

Answer choice E.

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A school supply store sells only one kind of desk and one kind of chai

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