GMAT Problem Solving Discussions

MBA
3521
If x, y, and z are integers, and x
(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
(2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.


IMO Ans C ..

We need to check if x+z=2y i.e is y avg of x and z ?

St 1 : implies avg of x,y,z is lesser than 4. OR in other words, x+y+zeg If x,y,z is 1,2,3 resp then Avg(1,2,3,4) > Avg(1,2,3) and y is avg of x and z
If x,y,z is 0,2,3 resp then avg(0,2,3,4)>Avg(0,2,3) but y is not avg of x and z ...

Not suff

St 2 : Implies y>4 ...
Again not suff, since y need not necessarily be avg of x and z ...

Not suff ...

Combined : x+y+z4 ....
if y is avg of x and z then x+y+z would be 3y and if y>4 then total would exceed 12 ...So y can never be avg of x and z ...

Suff ...Ans C

OA ?
3522
A certain investment grows at an annual interest rate of 8%, compounded quarterly. Which of the following equations can be solved to find the number of years, x, that it would take for the investment to increase by a factor of 16?

16 = (1.02)x/4
2 = (1.02)x
16 = (1.08 )4x
2 = (1.02)x/4
1/16 = (1.02)4x


Dude, u sure there is no typo error in the options ? have u missed out the raised to sign ?

I solved it as :

16 = ^4x
i.e 16 = 1.02 ^ 4x where x is no of yrs ...dont see this anywhere in the option ....
3523
Hi All,
Need some help with this question...
A container has 4 different varieties of fruits. In how many ways can you choose 6 fruits from the container ? ( Information on number of fruits in the container isn't available)
Thanks!

I saw this question in some material, it says the answer is 33 but doesn't explain how, am clueless about the way to arrive at this number...

Any pointers would help...


Assuming there are are unlimited fruits in each variety (or atleast 6 of each variety) and all fruits of particular variety are identical then the equation reduces to

a+b+c+d = 6 where a,b,c,d >=0 and are integers

1 variety only :

0,0,0,6 ---------4 ways

2 varieties :

0,0,1,5 --------12 ways
0,0,2,4 --------12 ways
0,0,3,3 -------- 6 ways

3 varieties :

0,1,1,4 -------12 ways
0,1,2,3 -------24 ways

4 varieties :

1,1,1,3 -----4 ways
1,1,2,2 ------6 ways

Total = 80 ways ...

Pls correct for errors and silly mistakes ...

First strike, are you sure there is no restriction given i.e we need to select atleast n no of fruits or smtng ? OR atleast 1 of each type or something ?
Could not arrive at the magical number 33 ...are you sure its the correct answer ?
3524

On Target... OA is C..

IMO Ans C ..

We need to check if x+z=2y i.e is y avg of x and z ?

St 1 : implies avg of x,y,z is lesser than 4. OR in other words, x+y+zeg If x,y,z is 1,2,3 resp then Avg(1,2,3,4) > Avg(1,2,3) and y is avg of x and z
If x,y,z is 0,2,3 resp then avg(0,2,3,4)>Avg(0,2,3) but y is not avg of x and z ...

Not suff

St 2 : Implies y>4 ...
Again not suff, since y need not necessarily be avg of x and z ...

Not suff ...

Combined : x+y+z4 ....
if y is avg of x and z then x+y+z would be 3y and if y>4 then total would exceed 12 ...So y can never be avg of x and z ...

Suff ...Ans C

OA ?
3525

OA for this one is B..

You are right.. I for got to put "^" sign.. for all of the options, outside that (1.02) there is "^" sign.. I'll edit that post for all other users.. your answer is right too.. Just need to take 4th root of both sides... thanks


Dude, u sure there is no typo error in the options ? have u missed out the raised to sign ?

I solved it as :

16 = ^4x
i.e 16 = 1.02 ^ 4x where x is no of yrs ...dont see this anywhere in the option ....
3526
2 varieties :

0,0,1,5 --------12 ways
In this line:-
It will effect when that one fruit of third type will appear in arrangement. lets say:-
0,0,1,5 - In 1st pick fruit C appears and in rest 5 picks - fruit D appears. another time fruit C appears in 2nd pick and in rest picks - Fruit D appears. Like this we can get 6 different ways just with 0,0,1,5. then total ways will be 12*6 = 72.

I think when we are picking 1st fruit. there are 4 ways. then, when picking 2nd, again there are 4 ways. 111y for 3rd, 4th, 5th, 6th. so, ans is 4^6.

Answer 33 is definitely wrong..


Assuming there are are unlimited fruits in each variety (or atleast 6 of each variety) and all fruits of particular variety are identical then the equation reduces to

a+b+c+d = 6 where a,b,c,d >=0 and are integers

1 variety only :

0,0,0,6 ---------4 ways

2 varieties :

0,0,1,5 --------12 ways
0,0,2,4 --------12 ways
0,0,3,3 -------- 6 ways

3 varieties :

0,1,1,4 -------12 ways
0,1,2,3 -------24 ways

4 varieties :

1,1,1,3 -----4 ways
1,1,2,2 ------6 ways

Total = 80 ways ...

Pls correct for errors and silly mistakes ...

First strike, are you sure there is no restriction given i.e we need to select atleast n no of fruits or smtng ? OR atleast 1 of each type or something ?
Could not arrive at the magical number 33 ...are you sure its the correct answer ?
3527
2 varieties :
0,0,1,5 --------12 ways
In this line:-
It will effect when that one fruit of third type will appear in arrangement. lets say:-
0,0,1,5 - In 1st pick fruit C appears and in rest 5 picks - fruit D appears. another time fruit C appears in 2nd pick and in rest picks - Fruit D appears. Like this we can get 6 different ways just with 0,0,1,5. then total ways will be 12*6 = 72.

I think when we are picking 1st fruit. there are 4 ways. then, when picking 2nd, again there are 4 ways. 111y for 3rd, 4th, 5th, 6th. so, ans is 4^6.

Answer 33 is definitely wrong..


Good discussion ...Its not a question of arrangement but selection ..

Let us consider one of the ways i.e say 0,0,1,5 ...

Let 4 types of fruits be apple, banana, mango and orange ...
Also let us assume all apples are identical. Likewise for other fruits also ..
And also there are atleast 6 of each kind ...

Now, 0,0,1,5 means u have selected 1 of one kind and 5 of the other kind ...it does not matter in what order u select ...

I mean if the chose 2 kinds are apple and bananas, then

AAAAAB OR AAAABA OR AAABAA or BAAAAA are all the same since it implies 1 banana and 5 apples, which pick gives u banana does not matter...so internal arrangement does not need to be accounted for ...

What we need to check is in how many from 4 varieities can i choose 1 of a particular variety and 5 from other variety ..

Use the rule, if p things out of n are alike then total ways = n! /p! , here 2 varieites have zero selection , hence 4! / 2! = 12 ways ...likewise for each selection ...

Also, 4^6 is not poss ...

Had we to made 6 digit nos using 1,2,3,4 where repetition of digits are allowed then total digits that can be formed is 4^6 ...

Correction for errors welcome !
3528

Hats off buddy!! I got the point.. thanks

Good discussion ...Its not a question of arrangement but selection ..

Let us consider one of the ways i.e say 0,0,1,5 ...

Let 4 types of fruits be apple, banana, mango and orange ...
Also let us assume all apples are identical. Likewise for other fruits also ..
And also there are atleast 6 of each kind ...

Now, 0,0,1,5 means u have selected 1 of one kind and 5 of the other kind ...it does not matter in what order u select ...

I mean if the chose 2 kinds are apple and bananas, then

AAAAAB OR AAAABA OR AAABAA or BAAAAA are all the same since it implies 1 banana and 5 apples, which pick gives u banana does not matter...so internal arrangement does not need to be accounted for ...

What we need to check is in how many from 4 varieities can i choose 1 of a particular variety and 5 from other variety ..

Use the rule, if p things out of n are alike then total ways = n! /p! , here 2 varieites have zero selection , hence 4! / 2! = 12 ways ...likewise for each selection ...

Also, 4^6 is not poss ...

Had we to made 6 digit nos using 1,2,3,4 where repetition of digits are allowed then total digits that can be formed is 4^6 ...

Correction for errors welcome !
3529

HI pUYS. nEED Help with the following questions. Thanks for the trouble.

Take care.




Jean drew a gumball at random from a jar of pink and blue gumballs. Since the
gumball she selected was blue and she wanted a pink one, she replaced it and drew
another. The second gumball also happened to be blue and she replaced it as well. If
the probability of her drawing the two blue gumballs was 9/49
, what is the probability
that the next one she draws will be pink?


A certain cube floating in a bucket of water has between 80 and 85 percent of its
volume below the surface of the water. If between 12 and 16 cubic centimeters of the
cubes volume is above the surface of the water, then the length of a side of the cube
is approximately
options are
4
5
7
8
9



3530

Ques 1)
Jean drew a gumball at random from a jar of pink and blue gumballs. Since the
gumball she selected was blue and she wanted a pink one, she replaced it and drew
another. The second gumball also happened to be blue and she replaced it as well. If
the probability of her drawing the two blue gumballs was 9/49
, what is the probability
that the next one she draws will be pink?


Ques 2)
A certain cube floating in a bucket of water has between 80 and 85 percent of its
volume below the surface of the water. If between 12 and 16 cubic centimeters of the
cube's volume is above the surface of the water, then the length of a side of the cube
is approximately

options are
4
5
7
8
9




Ques 1-

Probability of drawing two blue gumballs= 9/49
since in this case no replacement is done, so the probability of drawing one blue gumball is 3/7. which gives the above result. this implies:
Total number of gumballs = 7
Number of blue gumballs = 3
and number of pink gumballs = 4
Probability of drawing a pink gumball is 4/7.


Ques 2-

is the answer 4?
Not so sure about this one.
If it's right then will post the solution.

Do tell the OA.

Regards,
Neha
3531

@Neha..
I think replacement is there in this case. Lets say total gumballs are 49. then how come blue = 3 and pink = 4.. pink + blue balls -- shd be equal to 49..
According to me, the Q goes like this:-
1st ball i drawn from which turns out to be blue. then it is put back in jar.
then nd ball is drawn and that turns to be blue again. then it is placed back in jar.
prob. of the above case is 9/49. Now, 3rd ball is drawn..
What I can think of is --- answer will be less than 9/49..

I don't know how to solve this one!!!:banghead:

Ques 1-

Probability of drawing two blue gumballs= 9/49
since in this case no replacement is done, so the probability of drawing one blue gumball is 3/7. which gives the above result. this implies:
Total number of gumballs = 49
Number of blue gumballs = 3
and number of pink gumballs = 4
Probability of drawing a pink gumball is 4/7.

Regards,
Neha
3532

hey Neha...
I think you r right on the answer part.
Suppose Blue balls in Jar = x ; Pink balls = y
Probability of 1st ball to be blue = x/(x+y)
Probability of 2nd ball to be blue = x/(x+y)
total prob. of 2 balls blue = {x/(x+y)}^2 = 9/49
Solving, x/(x+y) = 3/7 ==> x=3, y=4
Prob. of next ball to be pink = 4/(3+4) => 4/7
(So, u were right..)


Ques 1-

Probability of drawing two blue gumballs= 9/49
since in this case no replacement is done, so the probability of drawing one blue gumball is 3/7. which gives the above result. this implies:
Total number of gumballs = 49
Number of blue gumballs = 3
and number of pink gumballs = 4
Probability of drawing a pink gumball is 4/7.

Regards,
Neha
3533

0.2*a^3 = 16 => a^3 = 80 ==> a=4.2 ~ 4. So, IMO=A

A certain cube floating in a bucket of water has between 80 and 85 percent of its
volume below the surface of the water. If between 12 and 16 cubic centimeters of the
cubes volume is above the surface of the water, then the length of a side of the cube is approximately.

3534
hey Neha...
I think you r right on the answer part.
Suppose Blue balls in Jar = x ; Pink balls = y
Probability of 1st ball to be blue = x/(x+y)
Probability of 2nd ball to be blue = x/(x+y)
total prob. of 2 balls blue = {x/(x+y)}^2 = 9/49
Solving, x/(x+y) = 3/7 ==> x=3, y=4
Prob. of next ball to be pink = 4/(3+4) => 4/7
(So, u were right..)


Hey Nipun,

I did it this way only but made a typo is writing the solution.
Total number of gum balls= 7
by mistake i wrote: total number of gum balls=49!
Sorry for that.Just saw now. i guess that's why the confusion was there.
Edited it now.


0.2*a^3 = 16 => a^3 = 80 ==> a=4.2 ~ 4. So, IMO=A

A certain cube floating in a bucket of water has between 80 and 85 percent of its
volume below the surface of the water. If between 12 and 16 cubic centimeters of the
cube's volume is above the surface of the water, then the length of a side of the cube is approximately.


I solved it exactly this way. but wasn't sure that we'l take 80% and 16 only or are we supposed also supposed to solve using 85% and 12 and other such cases.
How do we know which value to take since there are ranges involved like 80-85% and 12-16?
How do we know the value we are using is right?


Happy Solving,
Regards,
Neha
3535

Good one.. I found the same issue. then I took both the cases:-
80% and 16 --> 0.2*a^3 = 16 --> a^3 = 80
85% and 12 --> 0.15*a^3 = 12 --> a^3 = 80
It turns out to be same. so, we can pick anyone..
:cheers:

Hey Nipun,

I solved it exactly this way. but wasn't sure that we'l take 80% and 16 only or are we supposed also supposed to solve using 85% and 12 and other such cases.
How do we know which value to take since there are ranges involved like 80-85% and 12-16?
How do we know the value we are using is right?


Happy Solving,
Regards,
Neha
3536

I agree. To cross check,even i solved both ways and zero-downed to answer 4.But what about otherwise? as in other ques?
same way to chose?
if by taking 85% we would have got 5 instead of 4 then which answer is right how do we know that?


Good one.. I found the same issue. then I took both the cases:-
80% and 16 --> 0.2*a^3 = 16 --> a^3 = 80
85% and 12 --> 0.15*a^3 = 12 --> a^3 = 80
It turns out to be same. so, we can pick anyone..
:cheers:
3537
😃 i know by now the OA's must have surfaced. Sorry for the late reply. 😃 Hope I am pardoned. 😃 The OA's to Jean's gumballs and the floating cubes are 4/7 and 4 respectively. 😃 Take care. 😃
3538

a few more questions :)


Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining
jars are brown. Kurt will combine 3 jars of paint into a new container to make a new
color which he will name according to the following conditions:
Brun Y if the paint contains 2 jars of brown paint and no yellow.
Brun X if the paint contains 3 jars of brown paint.
Jaune X if the paint contains at least 2 jars of yellow.
Jaune Y if the paint contains exactly 1 jar of yellow.
What is the probability that the new color will be Jaune?

5/42, 37,42, 1/21, 4/9, 5/9



Within rat colony A, 104 new rats are born every other day and 105 die each day. If rat
colony A has 106 rats (after all deaths and births) at the end of a certain day, then how
long until colony A will have less than 105 members?
(A) exactly 5 days
(B) exactly 9 days
(C) between 8 and 9 days
(D) exactly 10 days
(E) more than 10 days



A die with x sides has consecutive integers on its sides. If the probability of NOT
getting a 4 on either of two tosses is
49
36
, how many sides does the die have?
(A) 4
(B) 5
(C) 7
(D) 8
(E) 13



Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell without
replacing any of them what is the probability that Tarrell will get both a 2 and a 3?

1/5 , 1/45, 1/50, 1/90, 14/45


Take care. :)

3539

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3540

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