A and B ran a race of 480 m.Inthe first heat,A gives B a headstart of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a headstart of 144 m and is beaten by 1/30th of a minute. What is B's speed in m/s.
1.)12 2.)14 3.)16 4.)18 5.)20
IMO....Answer: A
Solution: Let us say time taken by A is a seconds to finish the first heat of 480 meters. With the first condition, B ran (480 -48 ) meters in (a +60/10) = a+6 seconds So speed of B is: (480-48 )/(a+6) meters/sec
With the Second condition, B ran (480 -144) meters in (a-60/30) = a-2 seconds. So speed of B is: (480-144)/(a-2) meters /sec
Equating both and solving for a, we will get a = 30 sec Substituting and solving in either on of the above will give you B's speed as 12 meters/sec.
Good Problem ! Number crunching took most of the 2 min for me....
I have been following this thread for a lng time ...Thank you guys for your detailed explanations
I have a doubt : A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?
A - 9/10000 B - 81/1000 C - 10/81 D - 1/3 E - 80/81
I have been following this thread for a lng time ...Thank you guys for your detailed explanations
I have a doubt : A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?
A - 9/10000 B - 81/1000 C - 10/81 D - 1/3 E - 80/81
Thanks !!
IMO: A
R: B: Y = 2 : 5 : 3
Probability of getting an R is: 2/10, of getting a B is: 5/10, and of getting a Y is 3/10. Since the number of bulbs lit has to be 6, to suffice this requirement, 2 of each color and 2 only need to be selected.
Total probability therefore is: (2/10 * 5/10 * 3/10 )^2 = 9/10000
Probability of getting an R is: 2/10, of getting a B is: 5/10, and of getting a Y is 3/10. Since the number of bulbs lit has to be 6, to suffice this requirement, 2 of each color and 2 only need to be selected.
Total probability therefore is: (2/10 * 5/10 * 3/10 )^2 = 9/10000
What is the OA? Please advise.
Hey Vikram, I guess it should be 81/1000 ... Yes, we need to select 2 of each kind ...However multiplying individual prob means we have restricted outcome for each color to a specific sequence ...
I mean we could have many sequence like .. RRBBYY OR BBYYRR OR BYYRRB OR so on so forth ...prob for each of these outcomes is 9/10000...
So we need to find no of ways to get 2 of each kind .. No of ways = 6! /(2!*2!*2!) = 90
So total prob = 9/10000 * 90 = 81/1000 ..
However, am still not sure ...any alternate soln Puys ?
1. A certain amount was divided among A, B, C in the ratio of 4:5:6 but by mistake it was divided in such a manner that 4 times of A's share was equal to 5 times B's share and 6 times C's share. As a result A got $154 more than the expected amt. What was the amount divided among them?
2. 45% students in a school are girls. Among the students who have travelled abroad earlier, 25% are boys. If the difference between the number of students who have travelled outside the country, and the number of students who have not is 105. What is the total number of students in that school? a. 852 b. 754 c. 620 d. 525 e. 458
I have been following this thread for a lng time ...Thank you guys for your detailed explanations
I have a doubt : A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?
A - 9/10000 B - 81/1000 C - 10/81 D - 1/3 E - 80/81
Thanks !!
thanks for your explanation guys !! i dont know the OA ..So, is it A or B ?
1. A certain amount was divided among A, B, C in the ratio of 4:5:6 but by mistake it was divided in such a manner that 4 times of A's share was equal to 5 times B's share and 6 times C's share. As a result A got $154 more than the expected amt. What was the amount divided among them?
2. 45% students in a school are girls. Among the students who have travelled abroad earlier, 25% are boys. If the difference between the number of students who have travelled outside the country, and the number of students who have not is 105. What is the total number of students in that school? a. 852 b. 754 c. 620 d. 525 e. 458
Going by options
Travelled outside - Travelled not =105
If above eqn needs to be satisfied the total number of studends should be odd.
Only D satisfies that.
315 and 210 are two numbers but if we take 25% of any of these numbers we get answers in fractions. Am I missing anything????
Originally Posted by friend9921 1. A certain amount was divided among A, B, C in the ratio of 4:5:6 but by mistake it was divided in such a manner that 4 times of A's share was equal to 5 times B's share and 6 times C's share. As a result A got $154 more than the expected amt. What was the amount divided among them?
My approach: A'share as per share division= 4/15x ... x= total amount as per 2nd condition: 4a=5b=6c -----2 again a+b+c=x----2 putting each of b and c in eqn 1. b=4/5a........ c=2/3a.. so a + 4/5a + 2/3a=x... a=15/37x....4 deducting 4/15x -15/37x =154.. solving x= 1850...what's the answer..
2. 45% students in a school are girls. Among the students who have travelled abroad earlier, 25% are boys. If the difference between the number of students who have travelled outside the country, and the number of students who have not is 105. What is the total number of students in that school? a. 852 b. 754 c. 620 d. 525 e. 458
Going by options
Travelled outside - Travelled not =105
If above eqn needs to be satisfied the total number of studends should be odd.
Only D satisfies that.
315 and 210 are two numbers but if we take 25% of any of these numbers we get answers in fractions. Am I missing anything????
I agree with Rockstar ...data is incorrect or question seems to be incomplete ...total students have to be odd such that its 45% is integer ...none of the options work out ..
Finally the day came and I scored 730...looks like a decent one. Thanks to everyone in PG. You really showed enough patience in answering my questions (which were really funny at times). Can someone please redirect me to a PG forum where I can get support for the post GMAT activities?
A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each dat, but that no other piece of clothing is repeated? A) (1/3)^6 (1/2)^3 B) (1/3)^6 (1/2) C) (1/3)^4 D) (1/3)^2 (1/2) E) 5 (1/3)^2
A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each dat, but that no other piece of clothing is repeated? A) (1/3)^6 (1/2)^3 B) (1/3)^6 (1/2) C) (1/3)^4 D) (1/3)^2 (1/2) E) 5 (1/3)^2