GMAT Problem Solving Discussions

MBA
2621
This one is straight from OG 11, and I was not convinced with their explanation. Please HELP !
________________________________________

The inside dimensions of a wooden box are 6 in x 8 in x 10 in. A Cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has maximum volume?

A) 3
B) 4
C) 5
D) 6
E) 8

Please explain.


buddy this is indeed a very good question, almost every one will go wrong in first attempt, unless the person came across such tricky ones.

here is the exp.

take a box with 6* 8* 10

base dimensions can be
6*8 , 6*10, 8*10

if i chose 10 , tin will come out of the box,
so tin can have radius so is the max radius

rgds-
sri
2622

Hi,
Can anyone help me in solving this problem.

Q :In how many ways can a mixed doubles tennis game be arranged from eight married couples, If no husband and wife play in the same game.
A. 840
B.1680
C.420
D.None of these.

2623
Hi,
Can anyone help me in solving this problem.

Q :In how many ways can a mixed doubles tennis game be arranged from eight married couples, If no husband and wife play in the same game.
A. 840
B.1680
C.420
D.None of these.


same game or same team..??..
2624

same game !!!

2625
Hi,
Can anyone help me in solving this problem.

Q :In how many ways can a mixed doubles tennis game be arranged from eight married couples, If no husband and wife play in the same game.
A. 840
B.1680
C.420
D.None of these.


Is the answer A?
2626

I second you..
IMO A is the answer..

x 2 = 840..


Is the answer A?

Hi,
Can anyone help me in solving this problem.

Q :In how many ways can a mixed doubles tennis game be arranged from eight married couples, If no husband and wife play in the same game.
A. 840
B.1680
C.420
D.None of these.

2627

Hi Swap25,

Thanks for OA.
Can you please elaborate it further.
It will be of gr8 help !

Regards.



I second you..
IMO A is the answer..

x 2 = 840..
2628
Riva_M Says
Is the answer A?


yes.. The OA is A ...

And whatever SWap explained... its correct ...

Good Job !!!
2629

here comes one more on Probability:

If the integer m and n are chosen at random from 1 to 100 , then the probability that a number of the form 7^m + 7^n divisible by 5 is
A.1/4
B.1/2
C.1/16
D.1/8

2630
here comes one more on Probability:

If the integer m and n are chosen at random from 1 to 100 , then the probability that a number of the form 7^m + 7^n divisible by 5 is
A.1/4
B.1/2
C.1/16
D.1/8


Is B the answer? Probability of having both m and n odd or even.
2631
Riva_M Says
Is B the answer? Probability of having both m and n odd or even.


Thats wrong... this is not that simple Riva !!!
2632
Riva_M Says
Is B the answer? Probability of having both m and n odd or even.


here is one hint for you...

7^m + 7 ^n

it should be divisible by 5.

Now :
7^1=7 Last digit =7
7^2=49 Last digit =9
7^1= Last digit =3
7^1= Last digit =1

As the number should be divisible by 5.

Either it can be 7 + 3 = __0
OR 9 + 1 = __ 0

We need to take these conditions.
Now try to find the answer...
2633
here comes one more on Probability:

If the integer m and n are chosen at random from 1 to 100 , then the probability that a number of the form 7^m + 7^n divisible by 5 is
A.1/4
B.1/2
C.1/16
D.1/8


Example wud be something as below;
x = (7^m + 7^n) is divisible by 5

1=7
2=49
3=343
4=2401
5=16807
6=117649
7=823453
8=.........1
9=.........7
10=........9

for m=1, and n =3; x = 343+7 = divisible by 5

for m=1, and n =7; x = 343+16807 = divisible by 5

for m=1, and n =11; = divisible by 5

for m=3, and n =5; x = 343+16807 = divisible by 5

for m=3, and n =9; = divisible by 5

then d a vice versa.. I mean interchange the values for m and n above.. u should be able to find out all the occurences.. and then the answer ;)

my gut feeling tells me its 1/2.. lemme know the OA plz.. 😉
2634
Example wud be something as below;
x = (7^m + 7^n) is divisible by 5

1=7
2=49
3=343
4=2401
5=16807
6=117649
7=823453
8=.........1
9=.........7
10=........9

for m=1, and n =3; x = 343+7 = divisible by 5

for m=1, and n =7; x = 343+16807 = divisible by 5

for m=1, and n =11; = divisible by 5

for m=3, and n =5; x = 343+16807 = divisible by 5

for m=3, and n =9; = divisible by 5

then d a vice versa.. I mean interchange the values for m and n above.. u should be able to find out all the occurences.. and then the answer ;)

my gut feeling tells me its 1/2.. lemme know the OA plz.. ;)



nice attempt... but OA is 1/4
2635
pj02 Says
nice attempt... but OA is 1/4


probability of m being even - 1/2
probability of n being even - 1/2
now both m and n even - 1/2*1/2 = 1/4

for m and n odd again we get 1/4
The answer is for having both m and n even and odd hence 1/2

I dont have the pleasure of understanding the mistake in my approach .. can anyone throw some light?!

Harsh if you have the OE can u post it plz
2636

Hi,

mixed doubles = (1M+1F) vs (1M+1F)

so for each game we need 2M and 2F..
2M can be selcted in 8C2 ways..and den 2F, as per d cond, in 6C2 ways..
so 8C2x6C2..
moreover, each 2M and 2F selected combn can b arrainged in 2 difft ways..
so 8C2x6C2x2

HTH..

Hi Swap25,

Thanks for OA.
Can you please elaborate it further.
It will be of gr8 help !

Regards.

2637
probability of m being even - 1/2
probability of n being even - 1/2
now both m and n even - 1/2*1/2 = 1/4

for m and n odd again we get 1/4
The answer is for having both m and n even and odd hence 1/2

I dont have the pleasure of understanding the mistake in my approach .. can anyone throw some light?!

Harsh if you have the OE can u post it plz


i dont have OE , but i have solved this problem.
I will post it after some time.
Riva... Why are u thinking about even and odd? can you tell me what made you calculate probability of m even and odd ?
We should not forget number should be divisible by 5.
How can you be sure that for all even/odd m,n the sum will be divisible by 5?
2638
here comes one more on Probability:

If the integer m and n are chosen at random from 1 to 100 , then the probability that a number of the form 7^m + 7^n divisible by 5 is
A.1/4
B.1/2
C.1/16
D.1/8

probability of m being even - 1/2
probability of n being even - 1/2
now both m and n even - 1/2*1/2 = 1/4

for m and n odd again we get 1/4
The answer is for having both m and n even and odd hence 1/2

I dont have the pleasure of understanding the mistake in my approach .. can anyone throw some light?!

Harsh if you have the OE can u post it plz


Yes..ans should be 1/4 ..

Riva..one small error in your approach ...my attempt to correct it ..
Cyclicity of 7 is 4 ...hence not every odd power of 7 has the same units digit nor every even power of 7 has the same units digit ...

to elaborate ...we are only interested in units digit . If the sum of units digit ends in 0 or 5 ...our job done ...

Pattern : (only the units digit )
7^1 =7
7^2 =9
7^3 =3
7^4 =1

And it repeats ....Now, only 2 poss :
7+3 = 0 OR 9+1 = 0

So, 7^(4n+1) + 7^(4n+3) = mult of 5
And, 7^(4n) + 7^(4n+2) = mult of 5 ..

Though, ur spotting of even or odd is partially correct, sum of any 2 even powers or sum of any 2 odd powers wont be mult of 5

Hence we cannot simply say that when both m and n are either odd or even , no is mult of 5 (7^1 +7^1 is nt mult of 5 , even 7^2 + 7^2 is not mult of 5)

The way i looked at it :
For any chosen value of m, 1 out 4 values of n would work ..
Hence, prob is 1/4 ...

OR
Your way ( using odd, even )

total poss = 100 * 100

When m is even , half of the even values of n would work i.e 50*25
when m is odd, half of the odd values of n would work i.e 50*25

Hence, prob = (50*25) + ( 50* 25) / (100*100) = 1/4 ..
Hope that helps !!

If required, can elaborate a lil more ..
2639
Yes..ans should be 1/4 ..

Riva..one small error in your approach ...my attempt to correct it ..
Cyclicity of 7 is 4 ...hence not every odd power of 7 has the same units digit nor every even power of 7 has the same units digit ...

to elaborate ...we are only interested in units digit . If the sum of units digit ends in 0 or 5 ...our job done ...

Pattern : (only the units digit )
7^1 =7
7^2 =9
7^3 =3
7^4 =1

And it repeats ....Now, only 2 poss :
7+3 = 0 OR 9+1 = 0

So, 7^(4n+1) + 7^(4n+3) = mult of 5
And, 7^(4n) + 7^(4n+2) = mult of 5 ..

Though, ur spotting of even or odd is partially correct, sum of any 2 even powers or sum of any 2 odd powers wont be mult of 5

Hence we cannot simply say that when both m and n are either odd or even , no is mult of 5 (7^1 +7^1 is nt mult of 5 , even 7^2 + 7^2 is not mult of 5)

The way i looked at it :
For any chosen value of m, 1 out 4 values of n would work ..
Hence, prob is 1/4 ...

OR
Your way ( using odd, even )

total poss = 100 * 100

When m is even , half of the even values of n would work i.e 50*25
when m is odd, half of the odd values of n would work i.e 50*25

Hence, prob = (50*25) + ( 50* 25) / (100*100) = 1/4 ..
Hope that helps !!

If required, can elaborate a lil more ..


here is my sol

we know cyclicity of 7 is 4
and
m & n can be of from 4k,4k+2 of 4k+1,4k+3 for 7^m+7^n to be div by 5.
values of 4k (from k=1 to k=25) =25
values of 4k+2 ( from k=0 to k=24) = 25

similarly there are 25 values of 4k+1 and 4k+3

so

(25C1*25C1) *(25C1*25C1) / 100 * 100 = 1/8
2640
here is my sol

we know cyclicity of 7 is 4
and
m & n can be of from 4k,4k+2 of 4k+1,4k+3 for 7^m+7^n to be div by 5.
values of 4k (from k=1 to k=25) =25
values of 4k+2 ( from k=0 to k=24) = 25

similarly there are 25 values of 4k+1 and 4k+3

so

(25C1*25C1) *(25C1*25C1) / 100 * 100 = 1/8


Hey Sri,
yes, when m = 4k then n = 4k+2 but reverse is true as well
i.e when m = 4k + 2 , then n = 4k

Likewise, when m= 4k+1 then n= 4k+3 and reverse is true as well
i.e when m=4k+3 then n=4k+1

Hence,
(25C1*25C1) + (25C1*25C1) + (25C1*25C1) + (25C1*25C1) /100*100 = 1/4