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Tanya prepared 4 different letters to 4 different addresses.

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Joined: 24 Sep 2008
Posts: 189
Schools: MIT / INSEAD / IIM - ABC

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08 Sep 2009, 09:28
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65% (hard)

Question Stats:

44% (02:30) correct 56% (01:16) wrong based on 55 sessions

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Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

OPEN DISCUSSION OF THIS QUESTION IS HERE: tanya-prepared-4-different-letters-to-4-different-addresses-88626.html
[Reveal] Spoiler: OA
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Re: GMAT Prep CAT 2 [#permalink]

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08 Sep 2009, 10:09
Is the reply for the second question 500/x% ...............??
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Joined: 24 Sep 2008
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Re: GMAT Prep CAT 2 [#permalink]

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08 Sep 2009, 10:23
ank wrote:
Is the reply for the second question 500/x% ...............??

Yes tht's OA...plz explain
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Re: GMAT Prep CAT 2 [#permalink]

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08 Sep 2009, 10:59
For the 2nd question-

Total dist=x miles

First 5 miles in 30miles/hr. so say T1=5/30 which is 1/6 ..(speed=distance/time)

Remaining distance=x-5 miles in 60 miles/hr so say T2= x-5/60

Total time taken T3= T1 + T2 = 1/6 + x-5/60 , on solving we get it to be T3= 5+x/60

Now, if Don traveled total distance of x miles in 60 miles/hr then the total time would be =T= x/60

So the question asks us for the % difference so, T3-T/T * 100

(5+x/60 - x/60) / (x/60) * 100 = 500/x%

Last edited by ank on 12 Sep 2009, 11:30, edited 3 times in total.
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Re: GMAT Prep CAT 2 [#permalink]

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08 Sep 2009, 11:00
For the Tanya question... Ans is 1/3

Lets take 4 letters L1 L2 L3 L4 and 4 envelopes E1 E2 E3 E4 ... L1 should go into E1 and so on...

The Question asks us to find the prob of only one letter going into the correct envelope, which means the other 3 go into wrong envelopes.

Initially, Lets find the Total no of ways of arranging 4 letters in 4 diff envelopes which is 4! = 24

so, L1 to go into E1(correct envelope)...Thus..... 1 choice
L2 can go into E3 or E4(wrong envelopes)........ 2 choices
L3 can go into only E2 or E4(wrong envelopes)...2 choices
L4 can go only to E2 or E3(wrong envelopes).....2 choices

Thus probability that only 1 letter will be put into the envelope with its correct address= 1*2*2*2/4!=8/24 => 1/3
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Re: GMAT Prep CAT 2 [#permalink]

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08 Sep 2009, 11:17
ank wrote:
For the Tanya question... Ans is 1/3

Lets take 4 letters L1 L2 L3 L4 and 4 envelopes E1 E2 E3 E4 ... L1 should go into E1 and so on...

The Question asks us to find the prob of only one letter going into the correct envelope, which means the other 3 go into wrong envelopes.

Initially, Lets find the Total no of ways of arranging 4 letters in 4 diff envelopes which is 4! = 24

so, L1 to go into E1(correct envelope)...Thus..... 1 choice
L2 can go into E3 or E4(wrong envelopes)........ 2 choices
L3 can go into only E2 or E4(wrong envelopes)...2 choices
L4 can go only to E2 or E3(wrong envelopes).....2 choices

Thus probability that only 1 letter will be put into the envelope with its correct address= 1*2*2*2/4!=8/24 => 1/3

OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???
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Re: GMAT Prep CAT 2 [#permalink]

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12 Sep 2009, 11:21
ank wrote:
For the 2nd question-

Total dist=x miles

First 5 miles in 30miles/hr. so say T1=5/30 which is 1/6 ..(speed=distance/time)

Remaining distance=x-5 miles in 60 miles/hr so say T2= x-5/60

Total time taken T3= T1 + T2 = 1/6 + x-5/60 , on solving we get it to be T3= 5+x/60

Now, if Don traveled total distance of x miles in 60 miles/hr then the total time would be =T= x/60

So the question asks us for the % difference so, T3-T/T * 100

(5+x/50 - x/60) / (x/60) * 100 = 500/x%

in the last step it should be 5+x/60...
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Re: GMAT Prep CAT 2 [#permalink]

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12 Sep 2009, 13:24
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GODSPEED wrote:
OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???

The number of ways to put exactly one letter in the right envelope is not equal to 4C1. There are 4C1 ways to choose which letter goes in the right envelope, but you then need to work out how many ways the remaining letters can be placed in the wrong envelope. For the first of these three letters, there are 2 wrong envelopes you could choose. Now, you still have one letter left which has its matching envelope unused; you must put this letter in the wrong envelope, so you have only 1 choice for this letter, and finally for the last letter you have only 1 choice for where to put it. So you have 2*1*1 = 2 ways to assign the remaining letters incorrectly, which is why you need to multiply your answer by 2.

I posted a slightly different solution to BTG a while ago, which I'll paste here:

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.
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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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09 Jul 2012, 08:00
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Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

1) $$\frac{1}{24}$$
2) $$\frac{1}{8}$$
3)$$\frac{1}{4}$$
4) $$\frac{1}{3}$$
5)$$\frac{3}{8}$$
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Re: Tanyz prepared 4 different letters to be sent to 4 different [#permalink]

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09 Jul 2012, 08:02
Stiv wrote:
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
1) $$\frac{1}{24}$$
2) $$\frac{1}{8}$$
3)$$\frac{1}{4}$$
4) $$\frac{1}{3}$$
5)$$\frac{3}{8}$$

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Re: GMAT Prep CAT 2 [#permalink]

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31 Jul 2012, 02:39
IanStewart wrote:
GODSPEED wrote:
OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???

The number of ways to put exactly one letter in the right envelope is not equal to 4C1. There are 4C1 ways to choose which letter goes in the right envelope, but you then need to work out how many ways the remaining letters can be placed in the wrong envelope. For the first of these three letters, there are 2 wrong envelopes you could choose. Now, you still have one letter left which has its matching envelope unused; you must put this letter in the wrong envelope, so you have only 1 choice for this letter, and finally for the last letter you have only 1 choice for where to put it. So you have 2*1*1 = 2 ways to assign the remaining letters incorrectly, which is why you need to multiply your answer by 2.

I posted a slightly different solution to BTG a while ago, which I'll paste here:

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.

Slightly old post, but very interesting explanation...

Just would like to add my 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
No#2. W - C - W - W
No#3. W - W - C - W
No#4. W - W - W - C

There are 4 ways, it can be done

So, probability of 1 letter going in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3
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Posts: 39728

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31 Jul 2012, 02:50
GODSPEED wrote:
Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

OPEN DISCUSSION OF THIS QUESTION IS HERE: tanya-prepared-4-different-letters-to-4-different-addresses-88626.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: tanya-prepared-4-different-letters-to-4-different-addresses-88626.html

TOPIC LOCKED.
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Re: Tanya prepared 4 different letters to 4 different addresses.   [#permalink] 31 Jul 2012, 02:50
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