GMAT Problem Solving Discussions

MBA
2601
A room has the following dimensionz..

Length : 6 metres
Breadth : 6 metres
Height : 5 metres

an Ant which is @ one end of the longest diagonal needs to reach the other end of the diagonal in the shortest possible path ..!!

whats the length of the shortest path ?


the answer is not this --> square root(36+36+25)

reason : Itz an Ant n it cant fly ..keep trying will post the asnwer soon ..:biggrin:


Riva_M --> even thats not the answer :biggrin:
2602

-----------------------------
There were 36,000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441,000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45,000
(B) 360,000
(C) 364,500
(D) 392,000
(E) 396,900
-----------------------------


let the no of HArdbacks be H, then the number of paper backs = 9H

so total = H + 9 H +36000 = 441000

10 H = 405000
9 H = P = 9* 405000/10 = 364500
Ans --> 364500

2603
Hi Bhavin,
We must care abt the launch time.

Ur calculations can also yield another result when it is done this way : 9/10 *(441000-36000) = 364500

The ratio of 9:1 holds only for certain window and not for total sales period.So they have clearly specified the timeframe they are starting to measure.Else they cud have st mentioned the total sales and ratio.


When i meant launch time does not matter, it was coz initial sales of 36000 is already included in total sales of 441000 ...
So we, shd not subtract that in any case ...
2604
A room has the following dimensionz..

Length : 6 metres
Breadth : 6 metres
Height : 5 metres

an Ant which is @ one end of the longest diagonal needs to reach the other end of the diagonal in the shortest possible path ..!!

whats the length of the shortest path ?


the length of the shortest path will be (6^2 + (5+6)^2)^1/2
= sqrt(36+121)
= sqrt (157)
2605
the length of the shortest path will be (6^2 + (5+6)^2)^1/2
= sqrt(36+121)
= sqrt (157)


Could u just explain ..!!!

U were very near but here it is ...!!

Ans = Square Root((6+6)^2 + 5 ^2)

how : Imagine the room to be a carton (cubiod) , Split the box open at one edge,
it will be in the form of a rectangle of lenth (6+6+6+6) * 5
-----------------------------------B
....______________________________________________________
....__________________________ |____________ |____________
...._____________|_____________ |____________ |____________
5..._____________|_____________ |____________ |____________
...._____________|_____________ |____________ |____________
...._____________|_____________ |____________ |____________
A..______ 6 ____ |______ 6 ______|______6 ____ ______ 6 ____

it look like this .....!!

the ant needs to reach from A to B which is in the form of a right angled triangle with perpendicular sides as 12 (6+6) and 5 ..!!

So the answer is 13 ( Sqaure Root( 12^2 + 5 ^2))
P.S : 6 , 5 in the diagram are the measurements of length breadth n height :D
2606
We are not asked the longest diagonal, but the longest possible distance that can be traversed between any 2 points.

Why is that not the case?

I am not sure of ur approach.But longest distance as per yours calculations can be any combinations of length.10,10,5,5,10 etc.may be you can also repeat the path to travel the longest distance.Though it sounds foolish but possible as per your approach.
2607
I again thought of this in different light.
The answer will be little less then the main diagonal.I calculate by using maxima and minima funda and landed with 9.4 as answer.
here its assumed that ant will travel only through end to end diagonal but one possibility exists whr its travelling through a different path digonally.


Hi nee can u tell whats wrong with my approach.I thnk answer will be less then 13.
Hi neel thx man i understand the concept.I was doing exactly the same thng but in a more complicated way.
2608
guy with guts Says
Hi nee can u tell whats wrong with my approach.I thnk answer will be less then 13.


guy with guts :hello :biggrin:

I am not sure of the maxima n minima approach:embarssed::splat::oops::banghead: so cant help u in explaining why your solution is wrong .. but l;et me know if my solution is wrong :biggrin:
2609
TOday Question on Equations :

Two Non-Zero numbers a and b satisfy the following equations:

4*x + 6*y + x*y + 37*y*y = 0
(y/x) + (x/y) = 24

if 2*x + 3*y not equal to 0 find value of y :biggrin:
2610
Could u just explain ..!!!

U were very near but here it is ...!!

Ans = Square Root((6+6)^2 + 5 ^2)

how : Imagine the room to be a carton (cubiod) , Split the box open at one edge,
it will be in the form of a rectangle of lenth (6+6+6+6) * 5
-----------------------------------B
....______________________________________________________
....__________________________ |____________ |____________
...._____________|_____________ |____________ |____________
5..._____________|_____________ |____________ |____________
...._____________|_____________ |____________ |____________
...._____________|_____________ |____________ |____________
A..______ 6 ____ |______ 6 ______|______6 ____ ______ 6 ____

it look like this .....!!

the ant needs to reach from A to B which is in the form of a right angled triangle with perpendicular sides as 12 (6+6) and 5 ..!!

So the answer is 13 ( Sqaure Root( 12^2 + 5 ^2))
P.S : 6 , 5 in the diagram are the measurements of length breadth n height :D



I think i have the shortest! (6+5)^2+5^2 = a lil over 12 😁

This was a good one !
2611
I think i have the shortest! (6+5)^2+5^2 = a lil over 12 😁

This was a good one !



Request you to provide detailed explanation :biggrin:
2612
@nee! Says
Request you to provide detailed explanation :biggrin:


Im sorry its not (6+5)^2 + 5^2, its (6+5) + 6^2 = (157)^0.5

Imagine a cube with A and B as the two edges of the main diagonal with B on one corner of the floor of the cube and A on the roof at the diagonally opposite corner. Now if you fall down that face which has an edge as A and is perpendicular to B, youll get a 11x6 rectangle with the diagonal as AB, whose length is the answer.

Hope i didnt go wrong anywhere!
2613
Im sorry its not (6+5)^2 + 5^2, its (6+5) + 6^2 = (157)^0.5

Imagine a cube with A and B as the two edges of the main diagonal with B on one corner of the floor of the cube and A on the roof at the diagonally opposite corner. Now if you fall down that face which has an edge as A and is perpendicular to B, youll get a 11x6 rectangle with the diagonal as AB, whose length is the answer.

Hope i didnt go wrong anywhere!



Awesum , u r right ..sorry fr my ignorance ...!!! but explanation seems apt :biggrin: PG raaks
2614
guy with guts Says
I am not sure of ur approach.But longest distance as per yours calculations can be any combinations of length.10,10,5,5,10 etc.may be you can also repeat the path to travel the longest distance.Though it sounds foolish but possible as per your approach.


Yeah, may be you are right.
Ans should be 15 then!
2615

In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

I am lost.. Got several methods and several different answers..
bhavin I think u r d one to rescue.. :

Possible answers:
1)13C1 x 13C1 x 13C1 x 13C1 x 48C2 (not in option)
2)13C1 x 13C1 x 13C1 x 13C1 x 48 x 47
3) 1 x 1 x 1 x 1 x 48C2

2616

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.

2617
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.


Three cases : The closed box rests on area ..

6 * 8 , 8 * 10, 10 * 6..

Case 2 ..if itz 6*8 then height of the cylinder would be 10 and radius 3 .... Volume = pi * 3 * 3 * 10 = 90*pi

Case 2 ..if itz 10*8 then height of the cylinder would be 6 and radius 4 .... Volume = pi * 4* 4 * 6 = 96*pi

Case 3 ..if itz 6*10 then height of the cylinder would be 8 and radius 3 .... Volume = pi * 3 * 3 * 8 = 72* pi


Maximum volume --> CAse 2 -> 96 * pi --> radius is 4
Option 2

:banghead:
2618
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

I am lost.. Got several methods and several different answers..
bhavin I think u r d one to rescue.. :

Possible answers:
1)13C1 x 13C1 x 13C1 x 13C1 x 48C2 (not in option)
2)13C1 x 13C1 x 13C1 x 13C1 x 48 x 47
3) 1 x 1 x 1 x 1 x 48C2


Hey ..nice to know u trust my abilities , coz i dont, am just average at maths ..

well, amazing question in the first place ...however, even d way i solved it, none of the options feature in it ...i am an engg..more often than not i go backwards frm options to questions ...n thats how i eliminated all ..(I may be wrong ..)

So i begin fault finding in ans choices ..
a. (13^4) x 48 x 47 ...it means order is imp for last 2 selections after selecting 1 frm each suit , we cannot apply 48P2 for last 2 selections ..incorrect

b. (13^4) x 27 x 47 ...(27 does not make sense frm anywer ...for last 2 selections it is not same as 48C2 ..incorrect)

c. 48C6 ...it does not satisfy criteria of atleast 1 from each suit ..incorrect

d. 13^4 ...only 4 cards have been selected ..no mention about the next 2 ..incorrect

e. (13^4) x 48C6 ....does not make much sense ..so i eliminated it as well .

Your options :

13C1 x 13C1 x 13C1 x 13C1 x 48C2 ....multiple counting ..same scenario may repeat with this counting

Here ,we are first selecting 1 from each suit and then remaining any 2 from left out 48 cards

eg of multiple counting ..
case 1 : ace of every suit as 1 from each suit and say spade and club 2 as next 2 cards

case 2 : club 2 and ace of remaining suit as 1 card from every suit and ace club and spade 2 as next 2 cards ...

Same cards, counted twice ...hence not correct

The way i thought of the problem :
Let us list the no of ways restriction is met :
1) 3 cards of 1 suit and 1 card from remaining 3 suites ...

OR

2) 2 cards from 2 suites and 1 card from remaining 2 suites ..

These are the only 2 possibilities so that we have atleast 1 card from each suite..Let us count the no of ways each of the 2 conditions can be met...

1) we can select 1 out of 4 suites in 4C1 ways and 3 cards from that suite in 13C3 ways and then 1 card from remaining 3 suites in 13C1 ways each..

Hence, total ways = 4C1 * 13C3 *13C1 * 13C1 *13C1 =( 4*13^4 *12*11) / ( 3*2*1)

2) we can select 2 out of 4 suites in 4C2 ways and 2 cards from each of these suites in 13C2 ways each and 13C1 ways for remaining 2 suites..

Total ways = 4C2 * 13C2 *13C2 *13C1 * 13C1 = ( 6*13^4 * 12^2 ) / 4

Hence, grand total = 1 + 2
= ( 4*13^4 *12*11) / ( 3*2*1) + ( 6*13^4 * 12^2 ) / 4
= 13^ 4 ( 88 + 216 )
= 13^4 * 304 ...ANS

and we can observe it is lesser than 13^4 * 48C2 ...

But, this seems unlikely answer ...very laborious as well (5 mins to calculate and 20 mins to type)

Pls correct fr error , and if somebody has a better, shorter and correct approach, please share

P.S : sorry for such a long post ...i liked this sum, so cud not resisit posting a long one !!
2619

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.

Hi Vikram,

Let me try to contribute.

In order to maximize the volume of the canister, you'll have to maximize the radius (going by the volume of the cylinder formula). One may be persuaded into thinking that 10 in. is the longest inside dimension of the cylinder, hence 5 (option C) is the answer.

However, the catch here is that you will not be able to fit a cylindrical canister of radius 5 in. because you'll need a box of 10 in. x 10 in. in base area in order to satisfy that criteria. Hence, option C is ruled out.

Option E is a teaser. You could possibly misread the 8 in. in the answer choices to be the 8 in. in the question.

The remaining dimensions are either 3 in. (A) or 4 in. (B)

Hence, Option B ( 4 in. ) is the correct answer.

HTH
2620

Hey Bhavin,

its 2 out of 2 for you and dat too good ones..
so hav no other way but to trust ur abilities..
kidding dude.. nd to clear ur misconcpetion I dont find u avg at maths..
may b more @ it on PM..

coming back to question..
yeah dats d correct way u solved it.. yest I spent lots of time to solve nd finally ended with dat soln.. (though it took not 5 bt 50 mins for me)

thnks again buddy !! keep it up

Hey ..nice to know u trust my abilities , coz i dont, am just average at maths ..
well, amazing question in the first place ...however, even d way i solved it, none of the options feature in it ...
The way i thought of the problem :
Let us list the no of ways restriction is met :
1) 3 cards of 1 suit and 1 card from remaining 3 suites ...
OR
2) 2 cards from 2 suites and 1 card from remaining 2 suites ..

In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6
I am lost.. Got several methods and several different answers..
bhavin I think u r d one to rescue.. :

Possible answers:
1)13C1 x 13C1 x 13C1 x 13C1 x 48C2 (not in option)
2)13C1 x 13C1 x 13C1 x 13C1 x 48 x 47
3) 1 x 1 x 1 x 1 x 48C2