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A certain company that sells only cars and trucks reported
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21 Jul 2010, 15:59
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73% (02:39) correct 27% (02:47) wrong based on 622 sessions
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A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996? (A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3
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Re: Percent and Ratio problem from QR 2nd edition PS 155
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21 Jul 2010, 16:36
jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 Let the revenue from cars in 1996 be \(c\) and revenue from trucks in 1996 be \(t\). Total revenue in 1997 would be \(0.89c+1.07t\). Also as total revenue in 1997 were up 1 percent from 1996, then total revenue in 1997 would be \(1.01(c+t)\). Equate above two: \(0.89c+1.07t=1.01(c+t)\) > \(\frac{c}{t}=\frac{1}{2}\). Answer: A.
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Re: Percent and Ratio problem from QR 2nd edition PS 155
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21 Jul 2010, 23:26
A.. I have probably solved this question 34 times by now.. Remember the answer.. 1:2
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Re: Percent and Ratio problem from QR 2nd edition PS 155
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17 Aug 2010, 07:30
Well, I cannot say its exactly a short method, but looking at the options one can surmise the ans. Take 100 as base figure. Now try the options. You can say that "The drop produced by less revenue by Car sales" MUST BE Surpassed by" The gain from Truck Sales". to achieve the Net Gain Effect. i.e the proportion of Truck Revenue has to be greater than Car Revenue. By this you can directly eliminate C,D and E. Now for remaining options, Say we start from B. Then 4:5 will mean : Rev 400 for cars and Rev 500 for Trucks Calculate 11% @ 400 and 7% @500. We just have to ensure that 7% @ Trucks MUST BE > 11% @ Cars. Right !!!
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A certain company that sells only cars and trucks reported
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30 Nov 2012, 15:49
You could also look at this problem and realize that car sales have a smaller impact on total revenues than truck sales so the ratio has to be smaller than 1. A is the only answer choice that satisfies this relationship.



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Re: A certain company that sells only cars and trucks reported..
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06 Jan 2013, 10:21
Drik wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in1996? [*]1:2 [*]4:5 [*]1:1 [*]3:2 [*]5:3 Its 1/2 Consider the revenue from cars in 1996 x, which reduced to 0.89x in 1997. Consider the revenue from trucks in 1996 y, which rose to 1.07y in 1997. Another piece of information that is given is that the total revenue which was (x+y) in 1996, rose to 1.01(x+y) in 1997. We are required to find x/y. So using these information, we get \(0.89x +1.07y=1.01x +1.01y\). On solving we get x/y as 1/2. +1A
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Re: A certain company that sells only cars and trucks reported
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02 Jan 2014, 22:22
jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 Let us say that the revenues from selling cars was $ 100x and from trucks was $100y in 1996, total revenue: 100x + 100y Revenue in 1997: cars: $ 89x trucks: $ 107y total revenue: 89x + 107y = 101x + 101y Solving the above total revenue equation we get: 12x = 6y i.e. x/y = 1/2 Answer is A.
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Re: A certain company that sells only cars and trucks reported
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25 May 2014, 16:55
Copied pasted this from the closed thread: Quote: Karishma's Post: Cars denoted by 1 and trucks by 2
w1/w2 = (A2  Aavg)/(Aavg  A1) Revenue of cars/Revenue of trucks = (Revenue change of trucks  Avg)/(Avg  Revenue change of cars) Revenue of cars/Revenue of trucks = (7  1)/(1  (11)) Notice the highlighted part above. Revenue of trucks changes by +7% and that of cars by 11%.
Can someone explain this to me please? Why is A2 and A1 switched in this equation? A2A.avg/ A.avg  A1. What am i missing here? Thanks! P.S: I read the "quarter wit" and it makes a lot more sense. Although, how do we tell what are the weights (w1/w2) vs. the actual quantity(avg?)



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Re: A certain company that sells only cars and trucks reported
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08 Aug 2014, 06:50
Straight forward weighted average approach Car Total Truck 11 1 7 126 So Car/ Truck = 6/12 = 1:2 Ans: A
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Re: A certain company that sells only cars and trucks reported
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08 Apr 2017, 01:14
This problem can be solved in less than one minutes by drawing a graph.
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File comment: This problem can be solved in less than one minute by drawing a graph.
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Re: A certain company that sells only cars and trucks reported
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07 Aug 2017, 02:24
Hi Zulu, Took me a while to figure it out the relation ship with the equations 12C=6T, can you please simplyfy this for the readers benefit. I understood the rest of the part.
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Re: A certain company that sells only cars and trucks reported
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10 Aug 2017, 09:37
jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 We can let the revenue from car sales in 1996 = c and the revenue from truck sales in 1996 = t. Since revenue from car sales in 1997 was down 11 percent from 1996, revenue from truck sales in 1997 was up 7 percent from 1996, and total revenue from car and trucks sales in 1997 was up 1 percent from 1996, we can create the following equation: 0.89c + 1.07t = 1.01(c + t) 89c + 107t = 101(c + t) 89c + 107t = 101c + 101t 6t = 12c t = 2c 1/2 = c/t Thus, the ratio of revenue of car sales to truck sales in 1996 is 1 : 2. Answer: A
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Re: A certain company that sells only cars and trucks reported
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18 Nov 2017, 15:57
Is it wrong to approach the question as 11(cars) + 7(trucks) = 1(cars + trucks) > this led me to Trucks = 2 cars... therefore cars/trucks must be 1/2. I am a bit shaky on this type of problem, so I want to confirm if my basis is correct, or if I just got lucky.
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Re: A certain company that sells only cars and trucks reported
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07 Dec 2017, 17:11
Here is an easy solution
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Re: A certain company that sells only cars and trucks reported
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30 Mar 2018, 12:57
Bunuel wrote: jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 Let the revenue from cars in 1996 be \(c\) and revenue from trucks in 1996 be \(t\). Total revenue in 1997 would be \(0.89c+1.07t\). Also as total revenue in 1997 were up 1 percent from 1996, then total revenue in 1997 would be \(1.01(c+t)\). Equate above two: \(0.89c+1.07t=1.01(c+t)\) > \(\frac{c}{t}=\frac{1}{2}\). Answer: A. anybody somedy (c11) + (t+7) = 1 why isnt my equation correct ?



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Re: A certain company that sells only cars and trucks reported
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30 Mar 2018, 13:19
ScottTargetTestPrep wrote: jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 We can let the revenue from car sales in 1996 = c and the revenue from truck sales in 1996 = t. Since revenue from car sales in 1997 was down 11 percent from 1996, revenue from truck sales in 1997 was up 7 percent from 1996, and total revenue from car and trucks sales in 1997 was up 1 percent from 1996, we can create the following equation: 0.89c + 1.07t = 1.01(c + t) 89c + 107t = 101(c + t) 89c + 107t = 101c + 101t 6t = 12c t = 2c 1/2 = c/t Thus, the ratio of revenue of car sales to truck sales in 1996 is 1 : 2. Answer: A hello friends, i dont get if \(\frac{c}{t}\) and \(6t = 12c\) than it should be \(\frac{12}{6}\) and not 6/12 can someone explain please ?



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Re: A certain company that sells only cars and trucks reported
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30 Mar 2018, 13:59
dave13 wrote: hello friends, i dont get if \(\frac{c}{t}\) and \(6t = 12c\) than it should be \(\frac{12}{6}\) and not 6/12 can someone explain please ? Hi Happy to help Let's take it step by step \(6t = 12c\)........divide by t \(6 = 12c/t\).......divide both side by 12 \(6/12 = c/t\) I hope you get now



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Re: A certain company that sells only cars and trucks reported
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30 Mar 2018, 14:49
dave13 wrote: ScottTargetTestPrep wrote: jpr200012 wrote: A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1:2 (B) 4:5 (C) 1:1 (D) 3:2 (E) 5:3 We can let the revenue from car sales in 1996 = c and the revenue from truck sales in 1996 = t. Since revenue from car sales in 1997 was down 11 percent from 1996, revenue from truck sales in 1997 was up 7 percent from 1996, and total revenue from car and trucks sales in 1997 was up 1 percent from 1996, we can create the following equation: 0.89c + 1.07t = 1.01(c + t) 89c + 107t = 101(c + t) 89c + 107t = 101c + 101t 6t = 12c t = 2c 1/2 = c/t Thus, the ratio of revenue of car sales to truck sales in 1996 is 1 : 2. Answer: A hello friends, i dont get if \(\frac{c}{t}\) and \(6t = 12c\) than it should be \(\frac{12}{6}\) and not 6/12 can someone explain please ? Hello, Ratio problems can be solved in a more efficient way that will take you no more than minute to get to the right ans. See youtube video by gmatninja, he explained this method. start from minure 42:00. https://www.youtube.com/watch?v=1R8EQ6IHX9IGood luck with your preparation



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Re: A certain company that sells only cars and trucks reported
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26 Aug 2018, 11:45
Hi guys I have a question. Please kindly help. I use the seesaw method and got ratio of C to T in 1997 as 1/2. However I am under impression that 1/2 is 1997's ratio. So next step, I plug number 300 as 1997 total sales. Hence C 1997 sales is 100; T 1997 sales is 200. Then, using the %change from 1996 to 1997, I deduce 1996 C sales as (100/0.89) and 1996 T sales as (200/1.07). My understanding is that all the sales figure should correlate to one another using the % change of sales. ie. 1996 C sales of (100/0.89) decreases by 11% should get 100 in 1997 sales; 1996 T sales of (200/1.07) increases by 1% should get 200 in 1997. But it doesnt add up because the (100/0.89) + (200/1.07) doesnt get total sales of (300/1.01). Is my understanding not correct? Thank you




Re: A certain company that sells only cars and trucks reported &nbs
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