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Questions from Princeton Review Adaptive test 4

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Questions from Princeton Review Adaptive test 4

by pranavc » Mon Aug 04, 2008 9:50 pm
1. Working simultaneously, hay balers 'x' and 'y' can make 74 bales of hay per hour, hay balers 'y' and 'z' can make 62 bales of hay per hour, and hay balers 'x' and 'z' can make 68 bales of hay per hour. How many hours would it take the three balers, working simultaneously, to make 816 bales of hay?

(A) 8
(B) 12
(C) 13 1/2
(D) 18
(E) 22 1/2

2. A certain telephone number has 7 digits. If the telephone number has the digit 0 exactly three times, and the number 1 is not used at all, what is the probability that the phone number contains one or more prime digits?

(A) 1/24
(B) 1/16
(C) 1/2
(D) 15/16
(E) 23/24


3. One-third of a cement mixture that was 20 percent sand by weight was replaced by a second cement mixture of equal weight, resulting in a new mixture that was 25 percent sand by weight. The second cement mixture was what percent sand by weight?

(A) 35%
(B) 25%
(C) 22%
(D) 21%
(E) 17%

4. If 'c' is a non-negative integer, and c(3^c) = c(c^3), then which of the following represents all possible values of 'c'?

(A) 0
(B) 1
(C) 3
(D) 1 and 3
(E) 0 and 3

Crashed and burned on all of the above. Any input would help. Thanks in advance.
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by sudhir3127 » Mon Aug 04, 2008 11:42 pm
Please post the questions individually.. so that everyone can benefit from them...
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Re: Questions from Princeton Review Adaptive test 4

by cornell » Tue Aug 05, 2008 12:20 am
pranavc wrote:1. Working simultaneously, hay balers 'x' and 'y' can make 74 bales of hay per hour, hay balers 'y' and 'z' can make 62 bales of hay per hour, and hay balers 'x' and 'z' can make 68 bales of hay per hour. How many hours would it take the three balers, working simultaneously, to make 816 bales of hay?

(A) 8
(B) 12
(C) 13 1/2
(D) 18
(E) 22 1/2
Let me try to solve no 1.

We have to transpose the information into equations:

Y + X = 74 bales/hr
Y + Z = 62 bales/hr
X + Z = 68 bales/hr

now you can substitute one of the equations

Y + X = 74, it means X = 74 - Y

We go to the third equation

X + Z = 68, so 74 - Y + Z = 67

- Y + Z = 67 - 74

- Y + Z = - 6

Now we can eliminate this equation:

- Y + Z = - 6
Y + Z = 62 +

2Z = 56, then Z = 28 bales/hr

After we found Z, we can calculate X and Y

X + Z = 68
X + 28 = 68, then X = 40 bales/hr

Y + Z = 62
Y + 28 = 62, then Y = 34 bales/hr

The total bales if X Y Z work together = 28 bales/hr + 40 bales/hr + 34 bales/hr = 102 bales/ hr

The question is how many hours would it take for X Y Z to make 816 bales of hay?

Time = 816 bales / (102 bales/hr)
= 8 hours

Hence the answer is A
Life is about choices
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by bluementor » Tue Aug 05, 2008 1:13 am
Similar to cornell's approach, just slightly faster..

Y + X = 74 bales/hr
Y + Z = 62 bales/hr
X + Z = 68 bales/hr

summing up all three equations together you get:

2X + 2Y + 2Z = 204, therefore:

X + Y + Z = 204/2 = 102 bales/hr

If all three worked together, you get 102 bales in an hour.

So to yield 816 bales, they will take 816/102 = 8 hours.

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by bluementor » Tue Aug 05, 2008 2:20 am
My approach to question 2:

In the 7 digit number, three digits are 0. So only the remaining 4 digits are of interest.

Since there are no 1's in the number, so to only possible numbers for the unknown 4 digits are 2, 3, 4, 5, 6, 7, 8 and 9 and by categorizing these:

2, 3, 5 and 7: Prime
4, 6, 8 and 9: Non-prime

Probability of having at least 1 prime digit = 1 - Prob of having no primes at all.

P(no primes in the 4 digits) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16

Therefore:

P(having at least 1 prime digit) = 1 - 1/16 = 15/16

Let me know if this is the right answer. Thanks.

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by bluementor » Tue Aug 05, 2008 2:41 am
Looking at the answer choices, I would choose A as the answer without doing any calculations. The logic to this is along the logic of weighted average. If you have a mixture which was 20% and you want to increase it to 25%, you will have to replace some amount of the original (in this case 1/3 of total) mixture with something higher than 25%, so that it averages down to 25%. The only possible answer choice is 35%.


However, if you are interested in the calculation, here it is:

(1/5)*(2/3) + X*(1/3) = 1/4

here X = %age of second mixture

and by solving for X, you will get 7/20 or 35%.


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by bluementor » Tue Aug 05, 2008 7:20 am
Sorry, for my last post above, its the explanation for question no. 3.

BlueMentor

P.S. Please post each question separately...
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by malolakrupa » Tue Aug 05, 2008 10:29 am
For the last question the answer is E
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by pranavc » Wed Aug 06, 2008 6:18 am
sudhir3127 wrote:Please post the questions individually.. so that everyone can benefit from them...
Thanks for the heads up. I should have posted them individually. I'll definitely do that from now on.
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Re: Questions from Princeton Review Adaptive test 4

by pranavc » Wed Aug 06, 2008 6:22 am
cornell wrote:
pranavc wrote:1. Working simultaneously, hay balers 'x' and 'y' can make 74 bales of hay per hour, hay balers 'y' and 'z' can make 62 bales of hay per hour, and hay balers 'x' and 'z' can make 68 bales of hay per hour. How many hours would it take the three balers, working simultaneously, to make 816 bales of hay?

(A) 8
(B) 12
(C) 13 1/2
(D) 18
(E) 22 1/2
Let me try to solve no 1.

We have to transpose the information into equations:

Y + X = 74 bales/hr
Y + Z = 62 bales/hr
X + Z = 68 bales/hr

now you can substitute one of the equations

Y + X = 74, it means X = 74 - Y

We go to the third equation

X + Z = 68, so 74 - Y + Z = 67

- Y + Z = 67 - 74

- Y + Z = - 6

Now we can eliminate this equation:

- Y + Z = - 6
Y + Z = 62 +

2Z = 56, then Z = 28 bales/hr

After we found Z, we can calculate X and Y

X + Z = 68
X + 28 = 68, then X = 40 bales/hr

Y + Z = 62
Y + 28 = 62, then Y = 34 bales/hr

The total bales if X Y Z work together = 28 bales/hr + 40 bales/hr + 34 bales/hr = 102 bales/ hr

The question is how many hours would it take for X Y Z to make 816 bales of hay?

Time = 816 bales / (102 bales/hr)
= 8 hours

Hence the answer is A
Thanks a ton. That is correct.
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by pranavc » Wed Aug 06, 2008 6:28 am
bluementor wrote:My approach to question 2:

In the 7 digit number, three digits are 0. So only the remaining 4 digits are of interest.

Since there are no 1's in the number, so to only possible numbers for the unknown 4 digits are 2, 3, 4, 5, 6, 7, 8 and 9 and by categorizing these:

2, 3, 5 and 7: Prime
4, 6, 8 and 9: Non-prime

Probability of having at least 1 prime digit = 1 - Prob of having no primes at all.

P(no primes in the 4 digits) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16

Therefore:

P(having at least 1 prime digit) = 1 - 1/16 = 15/16

Let me know if this is the right answer. Thanks.

BlueMentor
Very slick approach. I was with you up until the point where we split the numbers into two categories (i.e. the primes and non-primes) and that is where I went on a whole different tangent. I was trying an approach that involved permutations/combinations. But your method is simple and straightforward. Thanks a ton.
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by pranavc » Wed Aug 06, 2008 6:39 am
bluementor wrote:Looking at the answer choices, I would choose A as the answer without doing any calculations. The logic to this is along the logic of weighted average. If you have a mixture which was 20% and you want to increase it to 25%, you will have to replace some amount of the original (in this case 1/3 of total) mixture with something higher than 25%, so that it averages down to 25%. The only possible answer choice is 35%.


However, if you are interested in the calculation, here it is:

(1/5)*(2/3) + X*(1/3) = 1/4

here X = %age of second mixture

and by solving for X, you will get 7/20 or 35%.


BlueMentor
Thanks. The answer is correct and the explanation makes sense.
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by pranavc » Wed Aug 06, 2008 6:41 am
malolakrupa wrote:For the last question the answer is E
Yes. The OA is E. I made the mistake of canceling out the individual 'c's from the equation and hence picked "C" as the right answer but revisiting it again made me realize I probably shouldn't have done that.
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