Problem Solving

If k, m, and t are positive integers and + = , do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.



Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.






Six 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

Actually there are 3 ques...need the ans of all 3...

Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.

I think C


training class - T
responding to E-mails - R,
and talking on the phone - P
Total 240 min

(1)
training class - T........................0,9x
responding to E-mails - R............x
and talking on the phone - P........240-1,9x
Total 240 min
P=?
INSUFF

(2)
training class - T........................
responding to E-mails - R............
and talking on the phone - P........
Total 240 min

T=0,6(E+P)
E+P=y
0,6(E+P) + E+P= 0,6y+y=240
1,6y=240
y=150, T=90
P=?
INSUFF

(1)(2)
240-0,9x=150
x=100
1,9x=190
P=50
SUFF

C

OA?

Six 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 am between C and D...
Post OA please

vivek.kapoor83 wrote:If k, m, and t are positive integers and + = , do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

"+ = " what does this mean.

Something is missing for sure.

Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.

total minutes = 240
class, e-mails, phone
phone=?

Statement (1)
C = .90 E
Insufficient.

Statement (2)
C = .60(E+P)
Insufficient.

Combining (1) & (2)
We solve P, Since we know C+E+P = 240
3 equations, 3 variables.

Thus, C.

Six 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

sum of 8

6+2
2+6
4+4
3+5
5+3

total cases = 5
favorable cases = 2

probability = 2/5.

Thus, D.

OAs?

OA's are in order:
1) A
2) C
3) D

Quick question

For problem 3Like 4 meonly i was torn between 1/3 and 2/5

Morgoth,
sum of 8

6+2
2+6
4+4
3+5
5+3


Why dont we consider another 4,4 i.e first draw second draw and second draw first draw?

I think because we cannot distinguish 2 different cases between 4+4 and 4+4. They are the same, thats why there should be 5 different outcomes with 2 desirable outcomes.

vivek.kapoor83 wrote:If k, m, and t are positive integers and + = , do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

---> Didnt get ur question. Would you again post this please?

vivek.kapoor83 wrote:Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.

--> Ans is C(Using both the statements) and the explanation is already given by the friends.

vivek.kapoor83 wrote:Six 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

Ans is 2/5
total no of possible outcomes = (2,6)(6,2)(3,5)(5,3)(4,4) = 5
total no of favorable outcomes = (3,5)(5,3) = 2
So, 2/5

vivek.kapoor83 wrote:
Six 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


Ans is 2/5
total no of possible outcomes = (2,6)(6,2)(3,5)(5,3)(4,4) = 5
total no of favorable outcomes = (3,5)(5,3) = 2
So, 2/5

Why total possible outcomes is 5. Shouldnt it be 6, coz 4,4 is not counted twice. Or should we not count it twice? Since, order doesnt matter, it looks like we need to. Is this reasoning wrong?