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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camitava
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suppose, answer correctly first Qs = amedea66 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%
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.
first 5 min interval, the car traveled with avg speed 20mph = 20 * 5 / 60 = 5/3 m13. "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.
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
Regards,
Amitava
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Kaunteya
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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.
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.
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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codesnooker
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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.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
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.
Sorry, but I can't understand the above equation.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
Let say M = XY (where X is at tens place and Y is at units place).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
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.
From the question we can derive the following things:-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.
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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Martian421
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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!
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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sumithshah
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Simpler way to do
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.
you have AB + BA = X37. 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
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.
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!
[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!
