Hello...
Could anyone explain this please? (it is ETS question)
If it is 6:27 in the evening on a certain day, what time in the morning was it exactly 2,880,717 minutes earlier? (Assume standard time in one location.)
(A) 6:22
(B) 6:24
(C) 6:27
(D) 6:30
(E) 6:32
Is it option B)6:24
deepakraam SaysIs it option B)6:24
Official Answer is 'D'. Could you explain your solution?
deepakraam SaysHow dd u get 17 for the prime factors?
this was the question:
2. How many positive integers less than 100 have exactly 4 odd factors and no even factors?
a>13
b>14
c>15
d>16
e>17
Explanation :
my answer also works out to 17 ..
For 4 odd factors, N has to be of the form N= a^1 * b^1 or simply a*b
OR a^3 , where a and b are odd prime nos ..
Let us list down prime nos :
3 can pair with 5,7,11,13,17,19,23,29,31 = 9 nos
5 can pair with 7,11,13,17,19 = 5 nos
7 can pair with 11,13 = 2 nos
Only 1 perfect cube of odd no within 100
3^3 = 1 no ...
Hence, total 17 nos within 100 have exactly 4 odd factors ...
Hello...
Could anyone explain this please? (it is ETS question)
If it is 6:27 in the evening on a certain day, what time in the morning was it exactly 2,880,717 minutes earlier? (Assume standard time in one location.)
(A) 6:22
(B) 6:24
(C) 6:27
(D) 6:30
(E) 6:32
Padmalathas SaysOfficial Answer is 'D'. Could you explain your solution?
D is correct.
There are two ways to do it.
long method:
In 24 hours we have 1440 minutes. In 2 days 2880 minutes. In 2000 days 2880000 minutes. So, we have 717 minutes left. If it were to be 720 minutes then it would have been 12 hours behind 6:27 which is 6:27. Since it is 3 minutes short, add back 3 minutes to 6:27.
Hence it is 6:30.
short method:
Look at the question closely. We need to know what 6:27 would be xxxxx717 min ago. Notice that the last digits are 7 in both the numbers. So difference should end in a Zero. And D is the only choice that suffices to it.
Try this simple one Puys...
For each of 6 month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6 month period. If the odds of a light bulb burning out during the first 6 month period following its purchase are 1/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?
A. 5/27
B. 2/9
C. 1/3
D. 4/9
E. 2/3
D. 4/9 ?
Here is my approach. From the info, can draw below:
Odds Of Burning Out Odds of NOT Burning out
0-6 months 1/3 2/3
6-12 months 2/3 1/3 ( half of previous value)
Odds of bulb burning out during 6-12 months= (Odds of NOT Burning out in 0-6 months) * (Odds Of Burning Out in 6-12 months)
= 2/3 * 2/3
=4/9
Try this simple one Puys...
For each of 6 month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6 month period. If the odds of a light bulb burning out during the first 6 month period following its purchase are 1/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?
A. 5/27
B. 2/9
C. 1/3
D. 4/9
E. 2/3
Try this simple one Puys...
For each of 6 month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6 month period. If the odds of a light bulb burning out during the first 6 month period following its purchase are 1/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?
A. 5/27
B. 2/9
C. 1/3
D. 4/9
E. 2/3
Odds of burning bulb from 6 months to 1 year= 2/3 * 2/3=4/9.
Hence option-D.
Someone plz explain.
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6
I think the ans should be below , which is not in given options
(13^4 x 4!) x 48 x 47
Here is my approach:
1st card can be picked in 52 ways
2nd card must be picked from remaining 3 suits. so 39 ways
similarly 3rd and 4th in 26 and 13 ways respectively
5th and 6th card can from any suite and since we are left with 48 cards, we can pick them in 48 and 47 ways respectively
Multiplying all together we get (13^4 x 4!) x 48 x 47
Hello...
Could anyone explain this please? (it is ETS question)
If it is 6:27 in the evening on a certain day, what time in the morning was it exactly 2,880,717 minutes earlier? (Assume standard time in one location.)
(A) 6:22
(B) 6:24
(C) 6:27
(D) 6:30
(E) 6:32
I will go with B 6.24
You must have got confused between B.6.24 and D.6.30.
Approach: If its morning 6.27 , we are 12 hrs earlier. so its 12 * 60 =720 mins earlier.
We need to check if the given minutes are multiple of this. If Yes then it was 6.27 in morning.
But if you try to divide 2,880,717 you will find its not divisible by 12*60.
However 2,880,720 is completely divisible by 12*60.
So its 3 minutes behind 6.27.
So the answer is 6.24.
Someone plz explain.
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6
I think the ans should be below , which is not in given options
(13^4 x 4!) x 48 x 47
Here is my approach:
1st card can be picked in 52 ways
2nd card must be picked from remaining 3 suits. so 39 ways
similarly 3rd and 4th in 26 and 13 ways respectively
5th and 6th card can from any suite and since we are left with 48 cards, we can pick them in 48 and 47 ways respectively
Multiplying all together we get (13^4 x 4!) x 48 x 47
Answer should be A.(13^4) x 48 x 47
I will try to explain:
13 cards represent a suit in a deck of 52 cards.
One thing we should not forget here:Independent choices.
Picking of the 2nd card from a different suit is not dependent on the 1st pick and similarly it has nothing to do with the picking of the 3rd and 4th cards from different suits.
So, you can pick a card from one suit in 13C1 = 13 ways and you can pick 4 cards from 4 different suits in 13C1*13C1*13C1*13C1 = 13^4 ways.
Now you are left with 2 picks and 52-4=48 cards left. You can pick any 2 cards from 48 in 48*47 ways.
[ Here one might try 48C2 =24*47, but we dont care abt any combination. So Dont think abt this ]
So answer is : (13^4) x 48 x 47
I will go with B 6.24
You must have got confused between B.6.24 and D.6.30.
Approach: If its morning 6.27 , we are 12 hrs earlier. so its 12 * 60 =720 mins earlier.
We need to check if the given minutes are multiple of this. If Yes then it was 6.27 in morning.
But if you try to divide 2,880,717 you will find its not divisible by 12*60.
However 2,880,720 is completely divisible by 12*60.
So its 3 minutes behind 6.27.
So the answer is 6.24.
pj02,
Wrong approach.
6:27 7 minutes earlier would be? 6:20 right ?
So, 6:20 after 2880,720 min will be? 6:20 AM
So, 6:20 am will be 6:30 AM after 2880,710 min
pj02,
Wrong approach.
6:27 7 minutes earlier would be? 6:20 right ?
So, 6:20 after 2880,720 min will be? 6:20 AM
So, 6:20 am will be 6:30 AM after 2880,710 min
Thanks for discussing this Vikram.
I have a question :From where you got the number 2880,710?
2880,710 is not divisible by 12*60.
2880,720 is completely divisible by 12*60 which is 3 mins more than 2880,717.
Thanks for discussing this Vikram.
I have a question :From where you got the number 2880,710?
2880,710 is not divisible by 12*60.
2880,720 is completely divisible by 12*60 which is 3 mins more than 2880,717.
Don't get hung up with the divisibility. Think about the ulterior motive.
2880717 is the target. 6:27 would be some thing ending with 0 if it were x7
min before. Period.
Don't get hung up with the divisibility. Think about the ulterior motive.
2880717 is the target. 6:27 would be some thing ending with 0 if it were x7
min before. Period.
Dude..Its all about divisibility. You can not say it was that time around 6.27 am unless the minutes are divisible by 12*60
and if its 3 mins more than 2880717, its 6.27 am, as i have explained 2880720 is nothing but same time in morning.
Let it be any time... if its 4 pm then before 2880720 mins it was 4 am.
if its 9.10 pm then before 2880720 mins it was 9.10 am.
So before 2880720 mins of 6.27 , it was 6.27 am.
Please post the OA.
Answer should be A.(13^4) x 48 x 47
I will try to explain:
13 cards represent a suit in a deck of 52 cards.
One thing we should not forget here:Independent choices.
Picking of the 2nd card from a different suit is not dependent on the 1st pick and similarly it has nothing to do with the picking of the 3rd and 4th cards from different suits.
So, you can pick a card from one suit in 13C1 = 13 ways and you can pick 4 cards from 4 different suits in 13C1*13C1*13C1*13C1 = 13^4 ways.
Now you are left with 2 picks and 52-4=48 cards left. You can pick any 2 cards from 48 in 48*47 ways.
[ Here one might try 48C2 =24*47, but we dont care abt any combination. So Dont think abt this ]
So answer is : (13^4) x 48 x 47
Yes pj.. i also solved this question in the same fashion.
The two cards(after selecting 4 form different suits) are independent of order i.e. we can pick any of the card first.
OA pls.
don't have the options with me as of now for the following question.
Pls try to solve.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random, what is the probability that their product will be of the form x^2-(by)^2, where b is an integer.
don't have the options with me as of now for the following question.
Pls try to solve.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random, what is the probability that their product will be of the form x^2-(by)^2, where b is an integer.
I think there is only 1 way the product will be of that form. x-y and x+y.
Answer should be : 1/6.
don't have the options with me as of now for the following question.
Pls try to solve.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random, what is the probability that their product will be of the form x^2-(by)^2, where b is an integer.
hi,
there is only one combination possible of the given form x^2-(by)^2,which is x^2-y^2,which uses x-y and x+y only...
the total number of ways in which 2 expression can be selected out of these 4 = 4C2 = 6
so the probability is 1/6
hope it is the correct way...
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random, what is the probability that their product will be of the form x^2-(by)^2, where b is an integer.
I think there is only 1 way the product will be of that form. x-y and x+y.
Answer should be : 1/6.
hi,
there is only one combination possible of the given form x^2-(by)^2,which is x^2-y^2,which uses x-y and x+y only...
the total number of ways in which 2 expression can be selected out of these 4 = 4C2 = 6
so the probability is 1/6
hope it is the correct way...
Yes the approach is correct.
Thanks
