another bunch from GRE sept

[sushma123]

1.From the words 'UTAH' and 'MAINE', what is the probability that if we choose 1 letter from each group such that, we wont get the same letter

 

 

2.T IS NO MORE THAN 20%GREATER THAN THE INTEGER X AND T IS NO SMALLER THAN A NUMBER 20% LESS THAN THE INTEGER X. IF T= 60,WHICH OF THE FOLL INDICATES THE SMALLEST POSSIBLE INTERVAL FOR X?

A] 40 LESSTHAN OR EQUAL TO X LESS THAN OR EQUAL TO 80

B] 48LESS THAN OR EQUAL TO X LESS THAN OR EQUAL TO 75

C] 48 LESS THAN OR EEQUAL TO X LESS THAN OR EQUAL TO 72

D] 50 LESS THAN OR EQUAL TO X LESS THAN OR EQUAL TO 78

E] 50 LESS THAN OR EQUAL TO X LESS THAN OR EQUAL TO 75

 

3.for a certain party, the chef prepared a pound of lamb for every 3 guests, a pound of beef for every 4 guests, & a pound of ham for every 6 guests. If the chef prepared a total of 36 pounds of these 3 meats then he prepared meat for how many guests?

a] 24 b] 36 c] 48 d] 52 d] 72

 

 

4.Two vessels contain alcohol and water in the ratio 4:5 and 1:2. Find the ratio in which both the mixtures should be mixed in order to obtain a mixture having alcohol and water in the ratio 2:7

1/2

1/3

2/3

None of the above

[daageep]

#1) UTAH and MAINE.

 

lets say you choose a letter from utah then a letter from maine. its the same if you switched the orders.

 

lets say we randomly chose a letter from UTAH and got U. probability is 1/4. what is the probability that we get a different letter in MAINE? 100%. so the probability of getting U and something not U (from MAINE) is 1/4 * 1 = 1/4.

 

say we chose T. same as above. say we chose H. same thing.

 

say we chose A. what is the probability of not choosing A from MAINE? 4/5. so probability of choosing A from UTAH and choosing M, I, N, or E from MAINE is (1/4)*(4/5)=4/20.

 

summing them up, we have 3*(1/4)+4/20 = 19/20.

 

2) you can interpret what they are saying as T =0.8X. plugging in the given T=60 and solving, we get 50

 

3) ok. he gives a pound of meat X for every 3 guests, a pound of Y meat for every 4 guests, and a pound of Z meat for every 6 guests. lets just say some number (Z) guests showed up. then pounds of X meat would be Z/3. the other cases are similar.

 

the sum is given. So its Z/3+Z/4+Z/6=36. Solving for Z, we get that Z=48.

 

4) this is a simple problem, but i am too stupid to find the right answer. my 'unofficial' way (guessing) is that its none of the above. will someone please chime in with a good explanation? thanks

[Annie9]

good explantaion!:tup:

 

I approached the last problem this way,

4x+5x=9

for1:2;

lets take it as 3:6

3x+6x=9

 

so from both the equations x=1

safter this i got struck......:(

[Hajjat]

4) this is a simple problem, but i am too stupid to find the right answer. my 'unofficial' way (guessing) is that its none of the above. will someone please chime in with a good explanation? thanks

I think this is right, coz there is no total amount of liquid available for both vessels

Suppose

vessel 1 contains 9 litters (4 alco, 5 water)

Vessel 2 contains 3 litters (1 alco, 2 water)

Then

if you add 1:1 from both vessels ==> percentage: 5/7

 

However, if vessel 1 contains 18 litters(8, 10)

if you add 1:1 from both vessels ==> percentage: 9/12 = 3/4

 

The mixtured was formed by mixing 1:1 from both vessels but the alco:water percent was different... So we can not get a 2:7 percent without knowing vissel_1's or vissel_2's total amount of liquid available...

[sarapigasso]

4.Two vessels contain alcohol and water in the ratio 4:5 and 1:2. Find the ratio in which both the mixtures should be mixed in order to obtain a mixture having alcohol and water in the ratio 2:7

1/2

1/3

2/3

None of the above

 

4/5 = 80% of alc

1/2 = 50% of alc

2/7 = 30% of alc

mixing of 80% and 50% of alc will be in the range of 50-80% of alc.

 

answer is D