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.
Questions from Princeton Review Adaptive test 4
This topic has expert replies
-
- Legendary Member
- Posts: 829
- Joined: Mon Jul 07, 2008 10:09 pm
- Location: INDIA
- Thanked: 84 times
- Followed by:3 members
-
- Junior | Next Rank: 30 Posts
- Posts: 16
- Joined: Thu Jul 03, 2008 5:58 pm
- Location: Indonesia
- Thanked: 2 times
Let me try to solve no 1.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
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
-
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
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.
BlueMentor
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.
BlueMentor
-
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
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
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
-
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
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
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
-
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
Sorry, for my last post above, its the explanation for question no. 3.
BlueMentor
P.S. Please post each question separately...
BlueMentor
P.S. Please post each question separately...
-
- Senior | Next Rank: 100 Posts
- Posts: 58
- Joined: Fri Jul 18, 2008 12:26 pm
- Thanked: 2 times
-
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Feb 26, 2008 4:37 am
- Thanked: 5 times
- Followed by:1 members
Thanks for the heads up. I should have posted them individually. I'll definitely do that from now on.sudhir3127 wrote:Please post the questions individually.. so that everyone can benefit from them...
-
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Feb 26, 2008 4:37 am
- Thanked: 5 times
- Followed by:1 members
Thanks a ton. That is correct.cornell wrote:Let me try to solve no 1.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
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
-
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Feb 26, 2008 4:37 am
- Thanked: 5 times
- Followed by:1 members
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.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
-
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Feb 26, 2008 4:37 am
- Thanked: 5 times
- Followed by:1 members
Thanks. The answer is correct and the explanation makes sense.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
-
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Feb 26, 2008 4:37 am
- Thanked: 5 times
- Followed by:1 members
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.malolakrupa wrote:For the last question the answer is E