GMAT Problem Solving Discussions

MBA
1925
Can sumbdy help me with this...

The ratio of new cars to old cars in a certain city subway system is 5:3.
If 9 tenths of the new cars are considered "Graffiti free", which of the following is the largest possible ratio of "Graffiti free" cars to non-graffiti free cars in the entire system?

Sorry i dont remember the option.

Rahul


To find the greatest possible ratio of a:b, you want to maximize a and minimize b.

So, if we want the largest possible ratio of graffiti free to non-graffiti free cars, we want to maximize the # of cars without graffiti and minimize the number of cars with graffiti.

We know that 9/10 of the new cars are graffiti free, but we're not told anything about the old cars. Since we want to maximize "graffiti free", let's say that ALL of the old cars fall into this category.

Further, let's pick numbers to make this question easier. We want the number of new cars to be divisible by 10, and we have another ratio of 5:3, so let's say we have a total of 80 cars.

Accordingly:

Number of new cars = 50, number of old cars = 30
Of the new cars, .9(50) = 45 are graffiti free, 5 have graffiti
Of the old cars, all 30 are graffiti free.

So, graffiti free:graffiti = 75:5 = 15:1

Regds,
Rishi
1926
Let me add a little detail to the answer I gave for this question.

I start by bracketing John and Susan together, and denote this by (JS).
I treat this bracket as a single entity, and it could occupy one of
seven positions in the line-up. (There are six other people to be
arranged in the line). Now the three remaining boys could be arranged in
3! (=6) ways, and the three girls also in 3! ways. So with the bracket
(JS) plus six others, the total number of acceptable arrangements (i.e.
with boys and girls alternating) is 7 x 3! x 3! = 252. We could of
course have bracketed John and Susan in the order (SJ), and again this
could give a total of 7 x 3! x 3! = 252 ways.

The total of possible arrangements with boys and girls alternating and
with a boy at the start is 4! x 4! = 576. And the total with a girl at
the start is also 4! x 4! = 576 giving a grand total of 1152.

Now to satisfy the condition that John and Susan are NOT together we
subtract the number of ways when they ARE together (=504)

Required number of ways = 1152 - 504 = 648


Hi Preeti,

When u count 7 places available for JS (or SJ), u are essentially doing this:

1. JS B G B G B G

2. B JS G B G B G

3. B G JS B G B G and so on till

7. B G B G B G JS

Out of which only 2, 4 & 6 doesnt satisfy the condition. That leaves us with only 4 ways. And then u can arrange the remaining 6 boys and girls in 3! * 3! ways. And now u can interchange JS into SJ to get the same result.

Hence total no of ways would be 2*4*3!*3!.

In the answer given above I forgot to multiply the second part by 2.

Pls correct me if I am missing sthg.

Thanks.

Now I am again confused !!!

In the first example cant u re-arrange JS into SJ and place it like below:

B SJ G B G B G (3 ways)

So the total no of ways become 4+3 = 7.

Then do the same taking into consideration that a girl is sitting first.
That will make 2*7*3!*3!

Confused !
1927
Inder.14 Says
Is the answer 9/7?

yup...9/7 for me too
1928
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.
(Answer in factorials)



what is the answer for this?
even i am getting 864.

in the solution discussed in previous posts, it should be 4*3!*3! and not 7*3!*3!
1929
yes even i answer the same ..but it is wrong ..n there are no explaination in the test!!
now will sumone tell me y is the answer wrong?


The answer should be A. Not D.

In 2, consider the case, z = 1/3.
The question doesnt say that z is an integer.

Whats the OA ?
1930

I'll try to describe a geometry problem in words. Hope that u guys get a clear picture.

A triangle ABC (A is the top vertex) is inscribed in a circle. The length of the major arc BAC is given to be 24. What is the diameter of the circle.

Choices: 2, 5, 8, 11, 15.

PS: Can someone tell me how to copy screen frm gmatprep.

THanks.

1931
I'll try to describe a geometry problem in words. Hope that u guys get a clear picture.

A triangle ABC (A is the top vertex) is inscribed in a circle. The length of the major arc BAC is given to be 24. What is the diameter of the circle.

Choices: 2, 5, 8, 11, 15.

PS: Can someone tell me how to copy screen frm gmatprep.

THanks.


24/2(pi)r
24/2(pi)r >= 1/2 ( major arc cannot be smaller than half the circle)

solving these two eqns, 14._ _ > r > 7. _ _

so its either 8 or 11
i couldnt narrow it down beyond this , any ideas?
but i have a feeling its 8.
1932
Hi Preeti,

When u count 7 places available for JS (or SJ), u are essentially doing this:

1. JS B G B G B G

2. B JS G B G B G

3. B G JS B G B G and so on till

7. B G B G B G JS

Out of which only 2, 4 & 6 doesnt satisfy the condition. That leaves us with only 4 ways. And then u can arrange the remaining 6 boys and girls in 3! * 3! ways. And now u can interchange JS into SJ to get the same result.

Hence total no of ways would be 2*4*3!*3!.

In the answer given above I forgot to multiply the second part by 2.

Pls correct me if I am missing sthg.

Thanks.

Now I am again confused !!!

In the first example cant u re-arrange JS into SJ and place it like below:

B SJ G B G B G (3 ways)

So the total no of ways become 4+3 = 7.

Then do the same taking into consideration that a girl is sitting first.
That will make 2*7*3!*3!

Confused !


i almost sure the solution will be this


1)no of possible arrangements of 4 boys and 4 girls is:

4!*4!*2

(the 2 in the above case is for two different arrangements,one starting with boy girl..... and the othefr with girl boy.....)


2)the condition rules out the following possibilities'

2*7*3!*3!(as someone rightly pointed out above it'll be 7 and not 4)

3)so the final answer is 4!*4!*2-2*7*3!*3!=648
1933

I got 864 as well.

First I found how many total ways there are to arrange the girls and boys.
GB GB GB GB

4B 3B 2B 1B

G4 G3 G2 G1

Four girls can sit in the first seat. If one girl sits down, 3 girls can sit in the second available seat. so forth . . .
There are 4 x 3 x 2 x 1 ways to arrange the girls
There are 4 x 3 x 2 x 1 ways to arrange the boys
We can swap the orders of each girl-boy pair like this:
BG BG BG BG

So we multiply by 2
So the total is
4!4!2

Now we subtract out the cases where John and Susan sit together.
Think of JS and a single unit.

JS BG BG BG

If JS is in the first slot, we can arrange the rest of the girls and boys in
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
We can not swap them otherwise the alternate order is broken

Now we move JS to the second slot.
BG JS BG BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the third slot
BG BG JS BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the fourth slot
BG BG BG JS
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

So in total we have:
4 x 3!3! = 144 ways

We can also arrange the group with a girl first:
SJ GB GB GB
So in total we have:
4 x 3!3! = 144 ways

So the total ways S and J can sit together is:
3!3!4 + 3!3!4 = 288 ways

My answer would be:
4!4!2 - 3!3!8 = 864

1934

hey, I encountered these problems while giving a test and could not crack them. Can someone help ?

1] 30% of the members in a club use both pool and sauna but 40% of the members who use pool do not use sauna. How many percent members in club use pool ?
2] What is the smallest common multiple of two integers both greater than 250
3] What is the greatest common multiple of two integers less than 144?
4 ] If p is an integer, and m = -p+ (-2)^p. Is m^3 = -1 ?
1.P is even
2.P^3
Urgent help will be appreciated

1935
1] 30% of the members in a club use both pool and sauna but 40% of the members who use pool do not use sauna. How many percent members in club use pool ?

I would make a table such as this:
---------------------sauna-----------no sauna------Totals
pool--------------30=0.6p-------------0.4p--------------P
no pool--------

From the table
We just plug the values into the table 40% who use the pool don't use sauna. Then it follows 60% of the members who use the pool use the sauna. We are given what percentage of the whole this is. It's 30.
So 0.6p = 30
p = 50
My answer would be 50%
1936
hey, I encountered these problems while giving a test and could not crack them. Can someone help ?

1] 30% of the members in a club use both pool and sauna but 40% of the members who use pool do not use sauna. How many percent members in club use pool ?
2] What is the smallest common multiple of two integers both greater than 250
3] What is the greatest common multiple of two integers less than 144?
4 ] If p is an integer, and m = -p+ (-2)^p. Is m^3 = -1 ?
1.P is even
2.P^3
Urgent help will be appreciated



1. 50% use pool.
2. 502. ( 251*2)
3. (143*142*n) while n is any +ve integer.
4. C
for all even p m^3 is not equal to -1 and so it is for all p^3
1937
hey, I encountered these problems while giving a test and could not crack them. Can someone help ?

1] 30% of the members in a club use both pool and sauna but 40% of the members who use pool do not use sauna. How many percent members in club use pool ?
2] What is the smallest common multiple of two integers both greater than 250
3] What is the greatest common multiple of two integers less than 144?
4 ] If p is an integer, and m = -p+ (-2)^p. Is m^3 = -1 ?
1.P is even
2.P^3
Urgent help will be appreciated


Hi Harneet

Answers as follows:-
1] First of all, I am not sure how good a question this is, as it requires an assumption to be made if it needs to be solved. If it can be assumed that every club member uses either pool or sauna, then this question can be solved using the method given by "rmpaes" in the preceding post.

Alternately you can also use the following method:-
P: Number of members who use Pool only
S: Number of members who use Sauna only
X: Number of members who use both Pool and Sauna

Now you can create simple equations using these as follows:-
P+S+X = 100% (quite obvious i hope) --(1)
X=30% (Given) --(2)
40/100(P+X)=P (Number using pool = P+X, number not using Sauna = P ) --(3)

You have 3 equations and 3 variables. Solving should give you the value for P which is required.

2] What is the smallest common multiple of two integers both greater than 250?

For this question, you need to apply common sense instead of mathematics. Again, I do not like the question, as unlike GMAT, it is not definite enough.
The answer to the question given as quoted is 251.
Reason: integer A= 251, integer B = 251.

In the GMAT, this question is more likely to be "What is the smallest common multiple of two integers both greater than 250?" in which case the common sense comes in. Now we know the LCM is at least as big as the least value, which will be 251. Hence logically, the least LCM can be obtained using integer A = 251 and integer B = 502(251*2) giving an LCM of 502.
For questions like these, it is usually much easier to have answer choices present, and evaluating their plausibility.

3]What is the greatest common multiple of two integers less than 144?

Another bad question, as I do not expect to see this in the GMAT either(Can I know the sources for your questions if you don't mind?)

The answer for this question is indeterminable. Reason is that the question says greatest common multiple. Which means greatest number for which an integer is a factor. Hence the answer for this is not 143*142(assuming distinct integers which is again not clarified), as 143*142*n(where n is any positive number) is also a multiple of both 143 and 142.

So I hope you can see why this is an indeterminable solution.

4] This is a DS question, which you may post on the DS board for better referencing by others as well.

Again, this is a dubious question in my opinion, which seems to be testing skills of logarithmic evaluation of an equation, which I have yet to see being tested on the GMAT.

FYI, to answer this qn, you would need to solve:-
p-1=(-2)^p

Unless I have missed a more obvious concept, The answer seems to tend towards C) as 1) and 2) gives you that P is an even negative integer, which more or less ensure LHS is an integer, while RHS is a fraction.
1938

all these are from a kaplan cat.

1939

1. There are two names given JOHNSON and TONY. If one letter is picked from both simultaneously at random, then find the probability that the letter is same?

2. Given the median of two classes are as follows.
Median of a class of 45 students = 53
Median of a class of 34 students = 40
Find the median if both classes are combined?


I dont have OA pls explain.......

1940

Hi, I am new to pagal, and got a question:

There are 40 students in a classroom, 9/20 of them are boys and 4/5 of them are
right-handed. How many right-handed boys are there in the classroom?
(a) Between 10 and 32.
(b) Between 14 and 32.
(c) Between 10 and 18.
(d) Between 14 and 18.
(e) Between 18 and 36


Please answer with exp.
Thanks in advance.
Tony

1941
Hi, I am new to pagal, and got a question:

There are 40 students in a classroom, 9/20 of them are boys and 4/5 of them are
right-handed. How many right-handed boys are there in the classroom?
(a) Between 10 and 32.
(b) Between 14 and 32.
(c) Between 10 and 18.
(d) Between 14 and 18.
(e) Between 18 and 36


Please answer with exp.
Thanks in advance.
Tony

IMO C......pls post OA......
1942
ashishjha100 Says
IMO C......pls post OA......


Hi Ashish,
Here is the OA

The best answer is C.
There are (9/20 x 40 = 1 boys in the class. 80% of them are right-handed, meaning
that (4/5 x 18 = 14.4). Answer C is the best answer

Why cannot that answer be D, what is the meaning ob BETWEEN,
does it also include 14?

Thanks
Tony
1943
Hi, I am new to pagal, and got a question:

There are 40 students in a classroom, 9/20 of them are boys and 4/5 of them are
right-handed. How many right-handed boys are there in the classroom?
(a) Between 10 and 32.
(b) Between 14 and 32.
(c) Between 10 and 18.
(d) Between 14 and 18.
(e) Between 18 and 36


Please answer with exp.
Thanks in advance.
Tony


Hi TOny

The OA is correct, but the explanation should be as follows:-

If you read the question, among 40 students in the class, we are given the following info:-
1) 9/20 are boys, ie, 9/20*40= 18 of the students are boys
2) 4/5 are right handed, ie 4/5 * 40 = 32 students are right handed.

The important difference is in information snippet 2). What seems to be indicated is 4/5 of the students and not 4/5 of the boys are right handed.

So based on 1 and 2, we can find a best case and worst case as follows:-
Best case: Boys are a subset of the right handed students.
In other words, all 18 boys are among the 32 right handed students.
So in this scenario, we can have 18 right handed boys.

Worst case: Assume all the left handed students(40-32=8 ) are boys. So in this case,
the number of right handed boys = (Total boys)- (left handed boys)
= 18-8
=10 right handed boys

Hence we can see that the number of right handed boys can be anywhere between 10 to 18 (Answer choice (c)). The exact number cannot be determined based on the information given in the question, but the answer choices are adequate to enable a unique correct choice.
1944

Can anyone help me solve this problem with the method used?
What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

A. 126

B. 1386

C. 3108

D. 308

E. 13986