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Probability, ratio, motion

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Probability, ratio, motion

by Veronica » Tue Sep 28, 2010 4:34 am
6 cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?
a.1/6, b.1/5, c.1/3, d.2/5, e.2/3
I chose B, but answer key is D, please explain!

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7% from 1996. If total revenues from car sales and truck sales in 1997 were up 1% 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
How can I solve this problem 1.5 minutes????

During a trip, F traveled x% of the total distance at an average speed of 40miles/h and the rest of the distance at an average speed of 60miles/h. In terms of x, what was F's average speed for the entire trip?
a.(180-x)/2, b.(x+60)/4, c.(300-x)/5, d.600/(115-x), e.12,000/(x+200)
Please show me the explanation!

Thank you
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by narik11 » Tue Sep 28, 2010 5:21 am
Veronica wrote:6 cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?
a.1/6, b.1/5, c.1/3, d.2/5, e.2/3
I chose B, but answer key is D, please explain!

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7% from 1996. If total revenues from car sales and truck sales in 1997 were up 1% 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
How can I solve this problem 1.5 minutes????

During a trip, F traveled x% of the total distance at an average speed of 40miles/h and the rest of the distance at an average speed of 60miles/h. In terms of x, what was F's average speed for the entire trip?
a.(180-x)/2, b.(x+60)/4, c.(300-x)/5, d.600/(115-x), e.12,000/(x+200)
Please show me the explanation!

Thank you
For 1st one..
Total outcomes for sum=8 are (2,6) (3,5) (4,4) (5,3) (6,2)
Possible outcomes numbered 5 =2

so its 2/5

For 2nd one..

Revenue from Car in 97= Rc97
Revenue from truck in 97=Rt97

so Rc97=0.89Rc96
Rt97=1.07Rt96
and Rc97+Rt97=1.01(Rc96+Rt96) from question

therefore 0.89Rc96+1.07Rt96=1.01Rc96+1.01Rt96

=> .06Rt96=.12Rc96

therefore Rc96/Rt96= 1:2 Option 1
Last edited by narik11 on Tue Sep 28, 2010 5:26 am, edited 1 time in total.
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by GMATGuruNY » Tue Sep 28, 2010 5:25 am
Veronica wrote:
A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7% from 1996. If total revenues from car sales and truck sales in 1997 were up 1% 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
How can I solve this problem 1.5 minutes????


Thank you
This is a weighted average problem.

In a weighted average problem in which we're combining C% with T% in order to get a goal average of G%:

The proportion needed of cars is the positive difference between the truck% and the goal%. (C = T-G)
The proportion needed of trucks is the positive difference between the car% and the goal%. (T = G-C)

In the problem above:
C = -11 (because the car revenue is decreasing by 11%)
T = 7 (because the truck revenue is increasing by 7%)
G = 1 (our goal percentage)

Proportion needed of C = T-G = 7-1 = 6
Proportion needed of T = G-C = 1-(-11) = 12
C:T = 6:12 = 1:2

The correct answer is A.

We also could plug in the answer choices, which represent the ratio of car revenue to truck revenue. We know that we need fewer cars than trucks because the car revenue is decreasing by a greater percentage than the truck revenue is increasing, but the total revenue is increasing. Eliminate C, D and E.

Answer choice A:
Plug in C=100, T=200.
Total = 100+200 = 300.
C down 11% = 100-11= 89
T up 7% = 200+14 = 214
New total = 89+214 = 303
303/300 = 101/100 = 1% increase. Success!
Last edited by GMATGuruNY on Sat Apr 21, 2012 3:24 pm, edited 1 time in total.
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by diebeatsthegmat » Tue Sep 28, 2010 7:55 am
Veronica wrote:6 cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?
a.1/6, b.1/5, c.1/3, d.2/5, e.2/3
I chose B, but answer key is D, please explain!

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7% from 1996. If total revenues from car sales and truck sales in 1997 were up 1% 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
How can I solve this problem 1.5 minutes????

During a trip, F traveled x% of the total distance at an average speed of 40miles/h and the rest of the distance at an average speed of 60miles/h. In terms of x, what was F's average speed for the entire trip?
a.(180-x)/2, b.(x+60)/4, c.(300-x)/5, d.600/(115-x), e.12,000/(x+200)
Please show me the explanation!

Thank you
a quick solustion
question 1:
you select 2 cards whose sum is 8. the cards is 1,2,3,4,5,6
thus for the first 2 card you can select :3,5 or 5,3
the next select is 4,4
last select is 6,2 or 2,6
thus total is 5 time select from which there are 2 times you select the card with number 5
thus 2/5 is correct

question 2

in 1996 C=100% and T=100%
in 1997 C=89% and T=107 %
sum in 1997= 89C+107T=101(C+T)
thus 6T=12C thus C/T=1/2

question 3
distance is D
we know D=T*S
thus for the first S=40 and x% of distance =Dx%=xD/100 we will find t1=xd/4000
the distance we had left is D-Dx/100=(100D-xD)/100 and S=60 so we find t2=(100D-xD)/6000
the total time T=t1+t2=(2Dx+400Dx)/24000=(DX-200DX)/12000

thus the average speech is total D/total T= E the answer
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by GMATGuruNY » Tue Sep 28, 2010 11:09 am
During a trip, F traveled x% of the total distance at an average speed of 40miles/h and the rest of the distance at an average speed of 60miles/h. In terms of x, what was F's average speed for the entire trip?

a.(180-x)/2, b.(x+60)/4, c.(300-x)/5, d.600/(115-x), e.12,000/(x+200)

When there are variables in the answers, plug in.
When the distance in a rate problem is undefined, plug in for the distance.

Plug in x = 50
Plug in distance = 240
50% of 240 = 120
Time = Distance/Rate = 120/40 = 3.
Remaining distance = 240-120 = 120.
Time for remaining distance = Distance/Rate = 120/60 = 2.
Total time = 3+2 = 5.
Rate for whole trip = Total Distance/Total Time = 240/5 = 48.

Now plug x=50 into all the answers to see which gives us our target of 48.

Only answer choice E works:
12000/(50+200) = 12000/250 = 48.

The correct answer is E.
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by pzazz12 » Wed Sep 29, 2010 11:48 pm
diebeatsthegmat wrote:
Veronica wrote:6 cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?
a.1/6, b.1/5, c.1/3, d.2/5, e.2/3
I chose B, but answer key is D, please explain!

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11% from 1996 and revenues from truck sales in 1997 were up 7% from 1996. If total revenues from car sales and truck sales in 1997 were up 1% 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
How can I solve this problem 1.5 minutes????

During a trip, F traveled x% of the total distance at an average speed of 40miles/h and the rest of the distance at an average speed of 60miles/h. In terms of x, what was F's average speed for the entire trip?
a.(180-x)/2, b.(x+60)/4, c.(300-x)/5, d.600/(115-x), e.12,000/(x+200)
Please show me the explanation!

Thank you
a quick solustion
question 1:
you select 2 cards whose sum is 8. the cards is 1,2,3,4,5,6
thus for the first 2 card you can select :3,5 or 5,3
the next select is 4,4
last select is 6,2 or 2,6
thus total is 5 time select from which there are 2 times you select the card with number 5
thus 2/5 is correct

question 2

in 1996 C=100% and T=100%
in 1997 C=89% and T=107 %
sum in 1997= 89C+107T=101(C+T)
thus 6T=12C thus C/T=1/2

question 3
distance is D
we know D=T*S
thus for the first S=40 and x% of distance =Dx%=xD/100 we will find t1=xd/4000
the distance we had left is D-Dx/100=(100D-xD)/100 and S=60 so we find t2=(100D-xD)/6000
the total time T=t1+t2=(2Dx+400Dx)/24000=(DX-200DX)/12000

thus the average speech is total D/total T= E the answer

Thank you very much..........
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by ankur.agrawal » Thu Sep 30, 2010 1:50 am
Plugging in rocks man. Solving algebraically becomes lengthy sumtimes..

Was really helpful for the ration & proportion question.

Thanks Guru.
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