GMAT Problem Solving Discussions

MBA
2901
My take is B

here for max possible range the numbers should be.
x,x+1, 55,56,3x+20

as the mean is 55
5x+132/5 =55
=> x=28.6

so rage is x---------3x+20
=> 2x+20 =77.2= 77 1/5



huw dd u come to this conclusion

here for max possible range the numbers should be.
x,x+1, 55,56,3x+20
2902

can you pls provide am explanation for the same???

thanks

Ans D
Let the population of the least populated district be x. And that of remaining be 1.1x. Forming tha equation we get 12x = total population. Hence x is 11000
2903
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

a. 78
b. 77 1/5
c. 66 1/7
d. 55 1/7
e. 52

IMO E


Lemme know OA.. if it's right, i'll explain!
2904

@ndhadha & @rohitforever:
wrong answer.. u wanna try again or want me to post the OA..??..;)

My take is B

here for max possible range the numbers should be.
x,x+1, 55,56,3x+20

as the mean is 55
5x+132/5 =55
=> x=28.6

so rage is x---------3x+20
=> 2x+20 =77.2= 77 1/5

IMO E


Lemme know OA.. if it's right, i'll explain!
2905

I got a) 78

2906

I'm also getting the ans as 78.Here is huv I worked it out.

Lets take the 5 nos as x,x,55,55,y where x = smallest no and y = largest no. Avg of all 5 nos is 55.So sum of all 5 nos is 55*5

We get 2x+y = 165

Also, y = 3x+20.

solving above eqn we get x = 29 and y = 107.So range is 107-29 = 78.

The nos are 29,29,55,55,107.

OA please

2907

that's correct.. the OA is option A indeed

Riva_M Says
I got a) 78

I'm also getting the ans as 78.Here is huv I worked it out.

Lets take the 5 nos as x,x,55,55,y where x = smallest no and y = largest no. Avg of all 5 nos is 55.So sum of all 5 nos is 55*5

We get 2x+y = 165

Also, y = 3x+20.

solving above eqn we get x = 29 and y = 107.So range is 107-29 = 78.

The nos are 29,29,55,55,107.

OA please
2908
I'm also getting the ans as 78.Here is huv I worked it out.

Lets take the 5 nos as x,x,55,55,y where x = smallest no and y = largest no. Avg of all 5 nos is 55.So sum of all 5 nos is 55*5

We get 2x+y = 165

Also, y = 3x+20.

solving above eqn we get x = 29 and y = 107.So range is 107-29 = 78.

The nos are 29,29,55,55,107.

OA please


Question that most of the PUy's will ask you is:

What made you take first two numbers as x,x and 4th number as 55?
2909
Deepak you are a math guru... He solves the problems with shortcuts.
Some Puy's wont understand why he took first two numbers as x,x and 4th number as 55? Its not mentioned in the problem.. right guys?

I will try to explain why Deepak took the numbers like that:

In short: A more theoretical solution ( but half.. Complete it with Deepak's explanation )


Let's choose the numbers as x, y, 55, z, 3x+20 ... :)

So range = largest number - smallest number
=3x + 20 - x
= 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now for x be maximum, only when y and z are minimum.

So
1. Lets find the the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Obviously it has to be equal to x. It can't be less than x..

So the five numbers are x,x,55,55,3x+20.

I'm also getting the ans as 78.Here is huv I worked it out.

Lets take the 5 nos as x,x,55,55,y where x = smallest no and y = largest no. Avg of all 5 nos is 55.So sum of all 5 nos is 55*5

We get 2x+y = 165

Also, y = 3x+20.

solving above eqn we get x = 29 and y = 107.So range is 107-29 = 78.

The nos are 29,29,55,55,107.

OA please
2910

yes the formula is correct. Unfortunately, this formula can best explained with the help of Mr Venn's diagram but i dont know how to draw one and paste the image in the post :p

I would suggest you try drawing the diagram in such a way that each region contains its own amount. For example of the sets A, B and C, the region of A can divided into 4 sections:
a: amount ONLY in A
b: amount in BOTH A and B but NOT in C
c: amount in BOTH A and C and NOT in B
d: amount COMMON to A, b and C

Similarly if you divide the Sets B and C with their values as above , you can form the formula in question.

Hope this helps.

Need Some help -

I get a lil confused with 3 sets.
Is the below formula correct
n(AUBUC)=n(A)+n(B)+n(C)-n(AnB)-n(AnC)-n(BnC)+n(AnBnC)???

It'd be nice if someone can explain!!

Thanks in advance!
2911

ok. we have a condition which says no two city can have population greater than 10% of the population of any city. Keeping that in mind we have to minimize the population of one of the cities.

Lets say the population of the city with min population is x.
x will be least when rest of the cities will have max allowable population which is 1.1x( not more than 10% greater than x)

Forming the equation we get:
x + 10(1.1 x) = total population
x + 11 x = total population
12 x = 132000
x= 11000

HTH.


can you pls provide am explanation for the same???

thanks
2912
Deepak you are a math guru... He solves the problems with shortcuts.
Some Puy's wont understand why he took first two numbers as x,x and 4th number as 55? Its not mentioned in the problem.. right guys?

I will try to explain why Deepak took the numbers like that:

In short: A more theoretical solution ( but half.. Complete it with Deepak's explanation )


Let's choose the numbers as x, y, 55, z, 3x+20 ... :)

So range = largest number - smallest number
=3x + 20 - x
= 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now for x be maximum, only when y and z are minimum.

So
1. Lets find the the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Obviously it has to be equal to x. It can't be less than x..

So the five numbers are x,x,55,55,3x+20.



PJ,
You read my mind rite..Thx a ton for the crystal clear explanation
2913

Try this

Which of the following numbers divides 23^4 + 18^2 + 72^2 + 2116^2?


1)2
2)3
3)5
4)7
5)None of these


^ = power

2914
Try this

Which of the following numbers divides 23^4 + 18^2 + 72^2 + 2116^2?


1) 2
2) 3
3) 5
4) 7
5) None of these


^ = power

Find the unit digits of all numbers-
1+4+4+6=15
so answer is 3.
2915
Find the unit digits of all numbers-
1+4+4+6=15
so answer is 3.


Even I got the same answer, using the same method. Is it correct?
2916
Find the unit digits of all numbers-
1+4+4+6=15
so answer is 3.


A number ending with 3 will not necessarily be divisible by 3.
E.g : 13,23
2917

The last digit is 3 so divisibility by 2 and 5 eliminated. On expanding the expression and checking for 7 and 3 you get not divisible. Hence E.

2918
Riva_M Says
The last digit is 3 so divisibility by 2 and 5 eliminated. On expanding the expression and checking for 7 and 3 you get not divisible. Hence E.

shinjini Says
Even I got the same answer, using the same method. Is it correct?

Find the unit digits of all numbers-
1+4+4+6=15
so answer is 3.

Try this

Which of the following numbers divides 23^4 + 18^2 + 72^2 + 2116^2?


1)2
2)3
3)5
4)7
5)None of these


^ = power


Let's do it together, one term at a time,
23^4 + 18^2 + 72^2 + 2116^2

23^4 =23*23*23*23
When we divide each 23 by 3, we get remainder 2.
R(23^4) = 2*2*2*2 = 16.
Total remainder 16 when divided by 3 gives final remainder = 1.

Similarly, 18^2 when divided by 3 will give remainder = 0; 72^2 hen divided by 3 will give remainder = 0; 2116^2 when divided by 3 will give remainder = 1.

Hence, 23^4 + 18^2 + 72^2 + 2116^2 when divided by 3 will give the remainder,
1+0+0+1 = 2. Therefore 3 does not divide the expression given.

Work out for 2, 5 and 7 and you will see that remainder with 5 = 1+4+4+1 = 10 ... again 10 divided by 5 leaves remainder = 0! That's your answer!
2919
D is correct.
Re: GMAT Problem Solving Discussions - 04-08-2009, 11:32 AM - Add Post To Favorites



Quote:
Originally Posted by Padmalathas View Post
Hello...

Could anyone explain this please? (it is ETS question)

If it is 6:27 in the evening on a certain day, what time in the morning was it exactly 2,880,717 minutes earlier? (Assume standard time in one location.)
(A) 6:22
(B) 6:24
(C) 6:27
(D) 6:30
(E) 6:32

Quote:
Originally Posted by Padmalathas View Post
Official Answer is 'D'. Could you explain your solution?
D is correct.

There are two ways to do it.

long method:

In 24 hours we have 1440 minutes. In 2 days 2880 minutes. In 2000 days 2880000 minutes. So, we have 717 minutes left. If it were to be 720 minutes then it would have been 12 hours behind 6:27 which is 6:27. Since it is 3 minutes short, add back 3 minutes to 6:27.
Hence it is 6:30.

short method:
Look at the question closely. We need to know what 6:27 would be xxxxx717 min ago. Notice that the last digits are 7 in both the numbers. So difference should end in a Zero. And D is the only choice that suffices to it.
QUOTE]

A simple method puys:

Just divide 2,880,717 by 60...the remainder is 3....add this to 6:27 and u get the answer....
2920

got to be (3) 5.
unit digits sum up to 5 at the end. Any number ending with 5 or 0 has to be divisible by 5.

Try this

Which of the following numbers divides 23^4 + 18^2 + 72^2 + 2116^2?


1) 2
2) 3
3) 5
4) 7
5) None of these


^ = power