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PowerPrep 1 Questions....help

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PowerPrep 1 Questions....help

by medea66 » Mon Feb 25, 2008 8:52 pm
Unfortunately PP doesn't exactly explain the correct answers. I would appreciate it if someone can help with the problems below.

9. "If 75% of the class answered the first question on a certain test correctly, 55% answered the second question correctly, and 20% answered neither one of the questions correctly, then what percent of the class answered both questions correctly?"

a. 10%
b. 20%
c. 30%
d. 50% (correct answer)
e. 65%


13. "A certain car increased its average speed by 5mph in each successive 5-minute interval after the first. If in the first 5-minute interval its average speed was 20mph, then how many miles did the car travel in the third interval?

a. 1.0
b. 1.5
c. 2.0
d. 2.5 (correct answer)
e. 3.0

Thanks.
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Re: PowerPrep 1 Questions....help

by camitava » Tue Feb 26, 2008 12:45 am
medea66 wrote:Unfortunately PP doesn't exactly explain the correct answers. I would appreciate it if someone can help with the problems below.

9. "If 75% of the class answered the first question on a certain test correctly, 55% answered the second question correctly, and 20% answered neither one of the questions correctly, then what percent of the class answered both questions correctly?"

a. 10%
b. 20%
c. 30%
d. 50% (correct answer)
e. 65%
suppose, answer correctly first Qs = a
answer correctly second Qs = b

now total number answer correctly for atleast one of the Qs or both the Qs = 100 - 20 = 80 = a + b - a n b -> This is a formula for set calculation.
and a = 75 and b = 55

so 80 = 75 + 55 - a n b
a n b = 50

This stands for the number who answered correctly for both the Qs.
13. "A certain car increased its average speed by 5mph in each successive 5-minute interval after the first. If in the first 5-minute interval its average speed was 20mph, then how many miles did the car travel in the third interval?

a. 1.0
b. 1.5
c. 2.0
d. 2.5 (correct answer)
e. 3.0

Thanks.
first 5 min interval, the car traveled with avg speed 20mph = 20 * 5 / 60 = 5/3 m
in the second 5 min the car traveled with avg speed 25 mph = 25 * 5 / 60 = 25/12
in the second 5 min the car traveled with avg speed 30 mph = 30 * 5 / 60 = 5/2 = 2.5
so imo d.
Got my response, medea66?
Correct me If I am wrong


Regards,

Amitava
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Re: PowerPrep 1 Questions....help

by Kaunteya » Tue Feb 26, 2008 6:41 am
medea66 wrote:Unfortunately PP doesn't exactly explain the correct answers. I would appreciate it if someone can help with the problems below.

9. "If 75% of the class answered the first question on a certain test correctly, 55% answered the second question correctly, and 20% answered neither one of the questions correctly, then what percent of the class answered both questions correctly?"

a. 10%
b. 20%
c. 30%
d. 50% (correct answer)
e. 65%

An easy way to solve these types of questions, where you are asked about 2 groups, neither and both remember the following equation.

Group1+Group2+Neither-Both=Total

75+55+20-Both=100
150-Both=100
-Both=-50
Both=50%

This equation works for any two groups, where you are given 3 terms above. Another example would be:
Out of a total of 52 students, 32 study French, 25 study Spanish, 10 study neither, how many study both.
32+25+10-Both=52
67-Both=52 Therefore both=15. There are 15 students who study both French and Spanish. If you need to understand this better, and what to do if there are more than two groups I highly suggest studying Kaplan Premier Books, they have a great section covering these types of questions.

Hope that help.
Cheers K

The second question I would have answered as Camitava has showed.
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by medea66 » Tue Feb 26, 2008 6:48 pm
Thank you both so very much, I wish the PP would give explanations to the answers!
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More from PP1

by medea66 » Tue Feb 26, 2008 9:08 pm
Hi, I would greatly appreciate any insight into the following PS problems from PP1:

21. Three machines, individually, can do a certain job in 4, 5, & 6 hours prespectively. What is the greatest part of the job that can be done in 1 hour by two machines working together at their respective rates?

a. 11/30
b. 9/20 (ans)
c. 3/5
d. 11/15
e. 5/6


27. If x does not equal 2, then 3x^2(x-2)-x+2/x-2 =

a. 3x^2 - x+2
b. 3x^2 + 1
c. 3x^2
d. 3x^2-1 (ans)
e. 3x^2-2

37. If 2-digit integers M & N are positive and have the same digits, bt in reverse order, which of the following CANNOT be the sum of M & N?

a. 181 (ans)
b. 165
c. 121
d. 99
e. 44


23. A hiker walked for two days. ON the second day the hiker walked two hours longer and at an average speed on mile faster than he did on the first day. If during the two days he walked for a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?

a. 2mph
b. 3mph (ans.)
c. 4mph
d. 5mph
e. 6mph

Thank you in advance.
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Re: More from PP1

by codesnooker » Wed Feb 27, 2008 5:25 am
medea66 wrote:Hi, I would greatly appreciate any insight into the following PS problems from PP1:

21. Three machines, individually, can do a certain job in 4, 5, & 6 hours prespectively. What is the greatest part of the job that can be done in 1 hour by two machines working together at their respective rates?

a. 11/30
b. 9/20 (ans)
c. 3/5
d. 11/15
e. 5/6
Solution: Lets say A, B and C are three person and they complete a certain job in 4, 5 and 6 hours respectively. As we need to calculate the maximum job can be done any of the two persons in one hour, then pick the persons who can do the job in less number of hours.

Thus for us, A & B are the right candidates.

Job (A) = 1/4
Job (B) = 1/5

Total Job done in one hour = Job (A) + Job (B) = 9/20.
medea66 wrote: 27. If x does not equal 2, then 3x^2(x-2)-x+2/x-2 =

a. 3x^2 - x+2
b. 3x^2 + 1
c. 3x^2
d. 3x^2-1 (ans)
e. 3x^2-2
Sorry, but I can't understand the above equation.
medea66 wrote: 37. If 2-digit integers M & N are positive and have the same digits, bt in reverse order, which of the following CANNOT be the sum of M & N?

a. 181 (ans)
b. 165
c. 121
d. 99
e. 44
Let say M = XY (where X is at tens place and Y is at units place).
Therefore, N = YX

Now when we do simple addition, M + N = AB (where A is tens place and B is ones place).
Therefore, A = X + Y + (1 or 0, depends upon the carry forwards from B).
and B = Y + X - (remove the extra digits greater than once place on addtion of Y + X)

Now starts putting the values from the above results. Lets start from answer 'e'.

44 can be come if M = 22, i.e X = 2 and Y = 2.
99 can be come if M = 45 or 54.
121 can be come if M = 65 or 56.
165 can be come if M = 87 or 78.
181 can't be break down.

Note: in the answer the last digit is always equals to first digits or only less by 1 value. for example for 165, the last number is 15 and first number is 16. However, it is not possible with 181.

medea66 wrote: 23. A hiker walked for two days. ON the second day the hiker walked two hours longer and at an average speed on mile faster than he did on the first day. If during the two days he walked for a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?

a. 2mph
b. 3mph (ans.)
c. 4mph
d. 5mph
e. 6mph

Thank you in advance.
From the question we can derive the following things:-

for 1st day,
Time = T1 = X hours
Av Speed = Y miles/hour
Disatnce = D1 = XY miles

for 2nd day,
Time = T2 = (X + 2) hours
Av Speed = (Y + 1) miles/hour
Disatnce = D2 = (X + 2) (Y + 1) miles

----------------------------
T1 + T2 = 18 hours => X = 8.
D1 + D2 = 64 => Y = 3.

and Y is the average speed of day 1, which is our answer.
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by gabriel » Wed Feb 27, 2008 9:42 am
@medea66 ... Welcome to BTG. Please make sure you follow board rules. You can only post 1 question per thread. Open a new thread for every question.

Regards.
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by medea66 » Wed Feb 27, 2008 7:44 pm
My appologies Gabriel, will keep that in mind.
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wondering if anyone was able to answer #27

by Martian421 » Mon Sep 08, 2008 5:12 am
i think this questions should be labeled as:

if X does not equal 2, then [3x^2(x-2)-(x+2)]/(x-2) =

I'm having some difficulty on this as well... Sorry Mod if this should be listed on a new thread.

Any help is greatly appreciated!
~Martin
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by sumithshah » Mon Sep 08, 2008 9:43 am
Simpler way to do

37. If 2-digit integers M & N are positive and have the same digits, bt in reverse order, which of the following CANNOT be the sum of M & N?

a. 181 (ans)
b. 165
c. 121
d. 99
e. 44
you have AB + BA = X

AB = 10A + B
BA = !0B + A

AB + BA = 11A + 11B = 11(A+B)

Therefore, 11(A+B) = X hence X has to be a multiple of 11.
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by fajoni » Mon Sep 08, 2008 8:21 pm
Hello,

[3X^2(X-2)-(X+2)]/(X-2) =>

[3X^2(X-2)-(X-2)]/(X-2) We factored out the negative 1, then

[(3X^2-1(X-2)]/(X-2) Not quite sure how to put it in there, but it should read three X squared minus one times X-2, all of which is divided by the quantity of X minus 2. We simply used the following rule: 5(4+X) + 4(4+X) is equal to (5+4)(4+X), or 9(4+X).

From here the X-2's cancel out and your left with 3X^2 - 1.

Hope this helps. Good luck with the studying!
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by mcdesty » Wed Jul 09, 2014 11:34 am
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