GMAT Problem Solving Discussions

MBA
1945
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



solution...
The answer is 13986
The total nos possible are 27. now you have to calculate their sum.
9*100*( 1+5+8 )+ 3*3*10*(1+5+8 ) + (1+5+8 )*3*3 = 13986
Hundreds place Tens place Ones place

Any query.. please reply.....Try to write all the nine possibilities by taking 1 at hundereds place and observe it you will understand how the solution is done...
1946
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



solution...
The answer is 13986
The total nos possible are 27. now you have to calculate their sum.
9*100(1+5+ + 3*3*10*(1+5+ + (1+5+*3*3 = 13986
Hundreds place Tens place Ones place

Any query.. please reply.....Try to write all the nine possibilities by taking 1 at hundereds place and observe it you will understand how the solution is done...
1947

Hi,
Thanks so much for ur reply.But I somehow could not understand it at all:p
SO,that prompted me to find my own solution.
You can let me know if I am correct.Coz this is much simpler and I think I am correct.
OK here goes.We have 27numbers
the smallest if them being 111 and largest being 888
Now we know avg = sum/total
Hence sum = avg * total
avg of a seriesis avg of first and last(approximately)
Hence avg = (111+ 88/2
Givesme 499
Hence I get sum = 499*27 which is 13473....Quite closeto the real value..:smile:

Okay more more toughie .Can someone help me crack with their method?
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?

A. 4 : 1

B. 2 : 1

C. 10 : 1

D. 3 : 1

E. 1 : 2

And one more
In a class of 50 students, 10 did not opt for math. 15 did not opt of science and 2 did not opt for either. How many students of the class opted for both math and science?
I got the ans here as 23but it actually 27.I dont know how?Am I making a sillymistake?
Thanks in advance

1948

Can anyone tell me the approach to solve this:
Thanks in advance.
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3

B. 16

C. 75

D. 24

E. 25

1949

this is wrt the geometric triangle problem...

solving the equation i found to be in between 3 and 7 .. is 5 the answer ?

1950
vinothkumartm Says
there is something wrong with the qn 😃 dont confuse too much on this

Hey.. how did you exactly solve it.. there will always be 1 variable remaining.. am i missing something big here !!
1951
Can anyone tell me the approach to solve this:
Thanks in advance.
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3

B. 16

C. 75

D. 24

E. 25


To solve this question you can use the following approach:-
1) Find all divisible numbers(by 2,3 and 5) between 200 & 300.

This is equivalent to:- ---(a)
Numbers divisible by 2 + Numbers Divisible by 3 + Numbers divisible by 5
-Numbers divisible by 2 and 3(ie 6)
-Numbers divisible by 3 and 5 (ie 15)
-Numbers divisible by 2 and 5(ie 10)
+Numbers divisible by 2,3 and 5(ie 30)

Let me know if you are unclear about how the above formula comes about. It will be helpful if you could draw a venn diagram and take a look.

Now Numbers divisible by 2 = 51(Since it includes both 200 and 300)
Numbers divisible by 3 = 34 (ie,201-300)
Numbers divisible by 5 = 21 (inclusive of both 200 and 300)
Numbers divisble by 6 = 17
Numbers divisible by 10 = 11 (inclusive of both 200 and 300)
Numbers divisible by 15 = 7
Numbers divisble by 30 = 4

Therefore, the value for (a) is:-
51+34+21-(17+11+7)+4
=110-35
=75

2) Subtract this figure from total numbers between 200 and 300 to obtain the answer(numbers not divisible)

Numbers between 200 and 300 = 101

Therefore answer = 101-75 = 26.

This is clearly none of the choices, so am not sure about the validity of this question. Where did you find this question incidentally?
1952

Hi crazycow,
Thanks so much.This is taken from a site called 4gmat.I believe that what you have done is correct and it will help in guessing the right answer which is E(25).
Please explain in detail how you are saying no.s divisible by 2,3 and 5 can be found that way.
Also how do u find nos divisible by 33?Is it by using GP formula?

Thanks.

1953

HI,
Can someone help me solve this problem:
If 5^21 * 4^11=2*10^n,What is the value of n?
A.11
B.21
C.22
D.23
E.32

1954

4^11 = 2^22
so 5^21 * 2^22 = 2 * 10^21

Hope this helps........:smile:

HI,
Can someone help me solve this problem:
If 5^21 * 4^11=2*10^n,What is the value of n?
A.11
B.21
C.22
D.23
E.32
1955

Q3. If set S consists of the numbers 1, 5, -2, 8 & n, is 01. The median of numbers in S is less than 5
2. The median of numbers in S is greater than 1

this question is being dicuseed in the previous threads. Shouldn't the answer be C. This is my understanding:

1. The median of numbers in S is less than 5. So n should be less than 5.
2. The median of numbers in S is greater than 1. so n should be greater than 1.

if we combines these 2 we get 1

1956
Hi crazycow,
Thanks so much.This is taken from a site called 4gmat.I believe that what you have done is correct and it will help in guessing the right answer which is E(25).
Please explain in detail how you are saying no.s divisible by 2,3 and 5 can be found that way.
Also how do u find nos divisible by 33?Is it by using GP formula?

Thanks.



Hi Suruchee

Unfortunately I am unable to post a diagram to help you understand better.

Mods, any help please?
I am unable to upload an image in any form(bmp/jpg/gif). I get a message saying "This is not a valid image file." Am I missing something?

Back to Suruchee
If you are familiar with Set theory, think of this as a 3 intersecting sets problem, and your objective is to figure out the total number of distinct elements in the 3 sets combined.
In other words, for 3 intersecting sets A,B and C, you need to obtain the cardinal value of A U B U C.

If you are not familiar with the set concept, I will need a diagram to explain, which I will do so once I sort out this image upload issue.

Thanks for pointing out the part about numbers divisible by 3. There was an error in my original calculation, which was done on the fly(oops). The correct answer should have been 34. There were similar errors for those divisible by 15 and 30 as well which I have now corrected in my original post.

You do not need to know geometric progression actually. For finding number of 3 multiples, if you can reduce all the numbers as follows:-
201= 67*3
204= 68*3
207= 69*3
...
300=100*3

You need not list down all, just the first and last would do once you have identified the trend. You can clearly see that if u can find numbers between 67 and 100 (both inclusive), you will have the total numbers divisible by 3, which is 34. You can use this approach for other number multiple questions as well, thus eliminating the need to memorise GP formulae.

Why do you say the right answer is (e) 25. Even if your text book or online material indicates so, I disagree and maintain that it is incorrect. The correct answer is 26, irrespective of whether the question read 200 to 300 (both numbers inclusive), or 200 to 300(neither number inclusive) or 200 to 300(only 1 of them inclusive). You can verify this for yourself by writing a simple program, or by using a brute force method to list down all the 101 numbers to see for yourself.

Irrespective of the answer, this is a good question. You should focus on familiarizing yourself with a comfortable approach to solve it(there are bound to be other ways to solve it, mine isn't the only one) so that you can solve it on your own the next time such a problem presents itself. Do not focus too much on the answers unless the source is reliable.
1957

Thanks crazycow,
I got the solution:smile:
Even I am trying to post a question for which I need to upload a doc and it says upload fails.Mods please help(this is a word doc)

1958

Hi,
Can someone help me solve this problem with an explanation?

A list of 100 data items has mean =6 and std deviation = d where d is positive.Which of the foll pairs if added to the list will result in a list of 102 data members,with std dev
1>-6,0
2>0,0
3>0,6
4>0,12
5>6,6

Thanks

1959
Hi,
Can someone help me solve this problem with an explanation?

A list of 100 data items has mean =6 and std deviation = d where d is positive.Which of the foll pairs if added to the list will result in a list of 102 data members,with std dev
1>-6,0
2>0,0
3>0,6
4>0,12
5>6,6

Thanks


ma: 5>6,6: as d=sqr((x-6)^2/102), now for d to decrease we have to not let the numerator increase ie (x-6)^2 has to remain the same as denominator has increased from 100 to 102. Using 6,6 keeps the numerator the same.

: that's me!!! so the above ans was just a conclusion.
"Conclusion" means that i have stoped thinking 😃 Please post the OA.
1960

6,6 is the right answer.

1961

Even I am trying to post a question for which I need to upload a doc and it says upload fails.Mods please help(this is a word doc)

I have sent PM to Moderator , hope he will rectify soon
1962
Q3. If set S consists of the numbers 1, 5, -2, 8 & n, is 01. The median of numbers in S is less than 5
2. The median of numbers in S is greater than 1

this question is being dicuseed in the previous threads. Shouldn't the answer be C. This is my understanding:

1. The median of numbers in S is less than 5. So n should be less than 5.
2. The median of numbers in S is greater than 1. so n should be greater than 1.

if we combines these 2 we get 1


The answer is C. No doubt.
1963
Q3. If set S consists of the numbers 1, 5, -2, 8 & n, is 01. The median of numbers in S is less than 5
2. The median of numbers in S is greater than 1

this question is being dicuseed in the previous threads. Shouldn't the answer be C. This is my understanding:

1. The median of numbers in S is less than 5. So n should be less than 5.
2. The median of numbers in S is greater than 1. so n should be greater than 1.

if we combines these 2 we get 1

OA stands for Original Answer

IMO stands for In My Opinion

IMHO stands for In My Honest Opinion...

LOL stands for Laughs Out Loud !!!

any more acronyms u wish to know !!
1964
Q3. If set S consists of the numbers 1, 5, -2, 8 & n, is 01. The median of numbers in S is less than 5
2. The median of numbers in S is greater than 1

this question is being dicuseed in the previous threads. Shouldn't the answer be C. This is my understanding:

1. The median of numbers in S is less than 5. So n should be less than 5.
2. The median of numbers in S is greater than 1. so n should be greater than 1.

if we combines these 2 we get 1


I think it should be C!