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.
Harsh, I agree that the divisibility needs to be considered. I am not denying that. Its the difference of the 3 min that is in question here.
Here is the analogy: If the time is 4:00 AM 20 min ago, then it would be 4:03AM 17 min ago. It will not be 3:57AM. 3:57 will be the time if it was 23 min ago.
Similarly, if it was 6:27 AM 2880720 min ago, then it will be 6:30AM 2880717 min ago. Not 6:24.
Glad we are discussing this.
padmalathas - Can you post the OA and explanation please????
Agreed. They are independent choices and that's where my doubt is. So you choose 1st card from one suit and you have 13 ways. Similarly for 2nd, 3rd and 4th card and hence totally 13^4 ways. But since you have 4 different suits, can not you pick SUITS in 4! ways? That will increase total number ways to 13^4 x 4!
Something like this: S H D C S H C D S C H D ........ ........
S- one of the Spades, H - one of the Hearts, C - one of the clubs, D- one of the diamonds
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
Agreed. They are independent choices and that's where my doubt is. So you choose 1st card from one suit and you have 13 ways. Similarly for 2nd, 3rd and 4th card and hence totally 13^4 ways. But since you have 4 different suits, can not you pick SUITS in 4! ways? That will increase total number ways to 13^4 x 4!
Something like this: S H D C S H C D S C H D ........ ........
S- one of the Spades, H - one of the Hearts, C - one of the clubs, D- one of the diamonds
i told you dude, it has nothing to do with arrangements. Whatever you are doing is arrangements.
Let me make it really simple:
Forget the suits and all... You want to select 4 balls from a bucket. The bucket contains 13 Red,Black,Green and Yellow balls each. Now In how many ways can u select 4 balls of diff colors ?
I bet now you will say 13*13*13*13. There wont be any confusion right :)
Dont get confused by arrangements and combinations.
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list? a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) I just couldnt get 1 confirmed answer.. everytime I solved this to be sure of my prev calculation, I got a new answer .. share your thoughts plz..
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list? a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) I just couldnt get 1 confirmed answer.. everytime I solved this to be sure of my prev calculation, I got a new answer .. share your thoughts plz..
Let Sum of the 20 numbers other than n be: S(20) So Avg of all 21 numbers will be: {S(20) + n } / 21 We are given that 4 times the above avg is equal to n
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) Let the average of 20 numbers is X then sum of all the 20 numbers is 20*X then from the statement in blue we get n = 4*X Kostin asked : n /(20*X +n) ---> 4*X /(20*X+4*X) ----> 4/24 ---> 1/6 :biggrin: P.S : Correct me if i am wrong ..!!!
1/6 should be the ans. Let the avg of 20 numbers be A. Sum of 20 numbers is 20A. Sum of all 21 numbers is 20A+ n= 20A + 4A= 24A. Where n=4A Fraction of n to sum of 21 numbers this is 1/6
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list? a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) I just couldnt get 1 confirmed answer.. everytime I solved this to be sure of my prev calculation, I got a new answer .. share your thoughts plz..
Let Sum of the 20 numbers other than n be: S(20) So Avg of all 21 numbers will be: {S(20) + n } / 21 We are given that 4 times the above avg is equal to n so 4 * {S(20) + n} /21 = n So n / { S(20) + n} = 4/21 Answer is D
Vik :hey buddy howdy doing ..!!!
small correction .... the kostin states that the number n = 4 *AVG(20 numbers) not 4 *AVG(21 numbers)
Let Sum of the 20 numbers other than n be: S(20) So Avg of all 21 numbers will be: {S(20) + n } / 21 We are given that 4 times the above avg is equal to n
so 4 * {S(20) + n} /21 = n
So n / { S(20) + n} = 4/21
Answer is D
Fell FLAT ON MY FACE for this one ! "Missed the avg of O T H E R...." completely. I feel stumped :)
An Empty Metal box weighs ten percent of its total weight when filled with paint. If the weight of a partly filled box is one half that of a completely filled box, what fraction of the box is filled ?
Try this one. I got caught up in the wording and messed it up.
There are approximately 2.2 pounds in a kilogram. To the nearest eighteenth, how many eighteenths of a kilogram are in one pound ?
A) 7 B) 8 C) 9 D) 39 E) 40
hi, one eighteenth of a kilogram means 1000/18 grams... 1 pound means 1000/2.2 grams so number of eighteenths of a kilogram are in one pound = (1000/2.2)/(1000/18 ) = 18 / 2.2 = 8.1818 (8 approx) hope i am correct...
An Empty Metal box weighs ten percent of its total weight when filled with paint. If the weight of a partly filled box is one half that of a completely filled box, what fraction of the box is filled ?
A) 3/5 B) 1/5 C) 1/2 D) 4/9 E) 2/5
let total weight (metal box + paint) is 1000 gms then weight of metal box is 1000/10 = 100 gms so weight of paint = 900 gms
in other case, the weight of the partly filled box = 1/2 * 1000 = 500 gms the paint in this case = 400 gms fraction of the box that is filled is = 400/900 = 4/9 hope it is correct answer...
An Empty Metal box weighs ten percent of its total weight when filled with paint. If the weight of a partly filled box is one half that of a completely filled box, what fraction of the box is filled ?
A) 3/5 B) 1/5 C) 1/2 D) 4/9 E) 2/5
Answer in Bold :biggrin:
let total weight (metal box + paint) is 100% weight of metal box is 1000/10 = 10% so weight of paint = 90%gms
in other case, the weight of the partly filled box = 1/2 * 100% = 50% gms the paint in this case = 40% gms fraction of the box that is filled is = 40/90 = 4/9
thanks for the explanation/solution guys.. the OA is indeed option B ;)
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) I just couldnt get 1 confirmed answer.. everytime I solved this to be sure of my prev calculation, I got a new answer .. share your thoughts plz..
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) Let the average of 20 numbers is X then sum of all the 20 numbers is 20*X then from the statement in blue we get n = 4*X Kostin asked : n /(20*X +n) ---> 4*X /(20*X+4*X) ----> 4/24 ---> 1/6 :biggrin: P.S : Correct me if i am wrong ..!!!
1/6 should be the ans. Let the avg of 20 numbers be A. Sum of 20 numbers is 20A. Sum of all 21 numbers is 20A+ n= 20A + 4A= 24A. Where n=4A Fraction of n to sum of 21 numbers this is 1/6
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
a. (1/20) b. (1/6) c. (1/5) d. (4/21) e. (5/21) I just couldnt get 1 confirmed answer.. everytime I solved this to be sure of my prev calculation, I got a new answer .. share your thoughts plz..
n={4*summation(other 20 nos.)}/20 =>5n=summation(other 20 nos.) summation of 21 nos=summation of 20 nos+n=5n+n=6n =>n=1/6 of summation of 21 nos.
Varun-I believe u might also be getting this as answer.
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?
I know the answer.. and also the means to calculate the same.. this jlt ques ;)
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?
1/4 H H H H H T H T T H T H T T T T H T T H H T T H 2/8
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?
I know the answer.. and also the means to calculate the same.. this jlt ques ;)
P(e)=Probability of coin landing as either Heads H or Tails T. Case 1
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?
I know the answer.. and also the means to calculate the same.. this jlt ques ;)
hi, the probability of landing on heads side is 1/2... the probability of landing on the same side again is 1/2 * 1/2 the probability of landing on the same side third time is 1/2 * 1/2 * 1/2 = 1/8
similarly we get a case for tails too... hence total probability is 1/8 + 1/8 = 1/4 hope it is correct...
.. share your thoughts plz..
