Dealership A and Dealership B both sell cars and trucks. If the ratio

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Dealership A and Dealership B both sell cars and trucks. If the ratio of cars to trucks each dealership sold in the last month is the same for both dealerships, did Dealership A sell the same number of cars and trucks last month?

(1) The number of trucks sold by Dealership B was three times the number of cars sold by Dealership A.

(2) The number of cars sold by Dealership B was equal to the number of trucks sold by Dealership A plus twice the number of cars sold by Dealership A.

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CareerGeek · Answer

Let the number of cars & trucks sold by Dealership A be  c_A  &  t_A  respectively and that of Dealership B be  c_B  &  t_B

Given,  \frac{c_A}{t_A} = \frac{c_B}{t_B}

Question: Is  c_A  =  t_A ?

(1) The number of trucks sold by Dealership B was three times the number of cars sold by Dealership A.
-->  t_B  = 3 c_A 
Does not talk anything about  t_A  -->  Insufficient

(2) The number of cars sold by Dealership B was equal to the number of trucks sold by Dealership A plus twice the number of cars sold by Dealership A.
-->  c_B  =  t_A  + 2 c_A 
Does not talk anything about  t_B  -->  Insufficient

Combining (1) & (2),
 t_B  = 3 c_A  &  c_B  =  t_A  + 2 c_A

and we have  \frac{c_A}{t_A} = \frac{c_B}{t_B}

-->  \frac{c_A}{t_A} = \frac{t_A + 2c_A}{3c_A} 
-->  3c_A^2 = t_A^2 + 2c_At_A 
-->  3c_A^2 - 2c_At_A - t_A^2 = 0 
-->  3c_A^2 - 3c_At_A + c_At_A - t_A^2 = 0 
-->  3c_A(c_A - t_A) + t_A(c_A - t_A) = 0 
-->  (3c_A + t_A)(c_A - t_A) = 0 
--> c_A = t_A   -->  Sufficient

IMO Option C

exc4libur · Answer

Dealer A: car/trucks=x/y
Dealer B: car/trucks=z/w
ratio A=B: x/y=z/w
Is x=y?

(1): w=3x; x/y=z/w, x/y=z/3x, 3x^2=yz  insufic

(2): z=y+2x; x/y=z/w, xw=zy, xw=(y+2x)y  insufic

xw=zy, 3x^2=yz, 3x^2=y(y+2x), 3x^2-2xy-y^2=0
(3x+y)(x-y)=0, 3x+y=0, x=-y/3 (invalid), x=y (valid)

Ans (C)

lacktutor · Answer

Dealership A — c_1  and  t_1 
Dealership B— c_2  and  t_2

\frac{c_1}{t_1} = \frac{c_2}{t_2}  
—>  c_1 = t_1  ???

(Statement1):  t_2 = 3c_1 
—>  \frac{c_1}{t_1}= \frac{c_2}{3c_1} 
Clearly insufficient

(Statement2):  c_2 =  t_1 + 2 c_1  
 \frac{c_1}{t _1} = \frac{(t_1+ 2c_1)}{t_2}  
Clearly insufficient

Taken together 1&2, 
 \frac{c_1}{t_1} = \frac{(t_1 + 2c_1)}{3c_1} 
—>  3(c_1)^{2} = (t_1)^{2} + 2c_1*t_1 
 3(c_1)^{2} -2c_1*t_1- (t_1)^{2}=0

(3c_1+ t_1)(c_1 —t_1) = 0 
—>  c_1= t_1  
Sufficient

The answer is C

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dave13 · Answer

VeritasKarishma the highlighted part of solution i dont understand how to derive ...can you pls explain. and what quadratic is used here ? (a-b)^2 if thats the case then why here are two minuses 3c_A^2 - 2c_At_A - t_A^2 = 0 as per (a-b)^2 = a^2-2ab+b^2 thanks!

KarishmaB · Answer

Say A sells c cars and t trucks. Ratio of cars:trucks = c:t B sells cars:trucks in the ratio c:t too. We need to find if c = t or if c/t = 1? (1) The number of trucks sold by Dealership B was three times the number of cars sold by Dealership A. Trucks sold by B = 3c Not sufficient. (2) The number of cars sold by Dealership B was equal to the number of trucks sold by Dealership A plus twice the number of cars sold by Dealership A. Cars sold by B = t + 2c Not sufficient. Using both, (t + 2c)/3c = c/t (Ratio of cars/trucks for B) t^2 + 2ct = 3c^2 Divide by t^2 to get 3\frac{c^2}{t^2} - 2(\frac{c}{t}) - 1 = 0 (assume c/t = x) 3x^2 - 2x - 1 = 0 (3x + 1)(x - 1) = 0 Since x cannot be negative (number of cars and trucks sold both must be positive), x = 1 i.e. c/t = 1 Sufficient. Answer (C) For more on how to solve quadratic equations, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... equations/

jabhatta2 · Answer

Bunuel  - please mark this as manhattan prep

i got this in my manhattan mock

Bunuel · Answer

____________________
Done. Thank you.

bumpbot · Answer

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Dealership A and Dealership B both sell cars and trucks. If the ratio

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GMAT Data Sufficiency (DS) Questions

Dealership A and Dealership B both sell cars and trucks. If the ratio

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