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# Three friends, A, B and C have houses along a straight road, in that

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Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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18 Jul 2018, 23:29
00:00

Difficulty:

55% (hard)

Question Stats:

44% (01:01) correct 56% (01:10) wrong based on 38 sessions

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Three friends, A, B and C have houses along a straight road, in that order. If all three want to meet up at one of their houses, what is the minimum combined distance they need to travel?

(1) The distance between A’s house and B’s house is 2 miles.
(2) The distance between A’s house and C’s house is 8 miles

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Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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Updated on: 19 Jul 2018, 03:33
Lets move set by set
statement first , I only knew distance b/w a & b (not sufficient)
Statement 2 , I ony knew distance b/w a & c (still not sufficient)

now lets consider both statements :

a-->b = 2
a-->c = 8

Now if we know positions also we can analyse this
case 1 :
a-----b--------c
|--2--||---6---|

|------8--------|

now we can find minimum combined distance if they meet at b (that is logically be the place where everybody has to travel min distance): that is 8

Case 2:
b----- a---------c

|--2--| |----8---|

|------10--------|

now we can find minimum combined distance if they meet at a : that is 10

NOTE : answer to this should be C , my bad i didn't read the "in order phrase".

Originally posted by GMATbuster92 on 18 Jul 2018, 23:49.
Last edited by GMATbuster92 on 19 Jul 2018, 03:33, edited 3 times in total.
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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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18 Jul 2018, 23:55
kshitizbansal92 wrote:
Lets move set by set
statement first , I only knew distance b/w a & b (not sufficient)
Statement 2 , I ony knew distance b/w a & c (still not sufficient)

now lets consider both statements :

a-->b = 2
a-->c = 8

Now if we know positions also we can analyse this
case 1 : a b c
|-----||-----|
2 6
|-------------|
8
now we can find minimum combined distance: that is 8

Case 2: b a c
|----| |-------- |
2 8
|----------------|
10
now we can find minimum combined distance: that is 10

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Hi kshitizbansal92

The question is about minimum distance so one ans 8 is possible..which you have calculated earlier
10 is not minimum so ans will be c.
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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 00:00
in question you are not given the position of the houses , how can you decide what distance b has to travel to reach c, that's why 2 cases to show the difference
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Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 00:14
kshitizbansal92 wrote:
in question you are not given the position of the houses , how can you decide what distance b has to travel to reach c, that's why 2 cases to show the difference

Bunuel please correct me if i am wrong
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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 01:25
Bunuel wrote:
Three friends, A, B and C have houses along a straight road, in that order. If all three want to meet up at one of their houses, what is the minimum combined distance they need to travel?

+1 for B.

Given,
Three friends, A, B and C have houses along a straight road, in that order.
Minimum combined distance they need to travel = ?

(1) The distance between A’s house and B’s house is 2 miles.

(2) The distance between A’s house and C’s house is 8 miles
Shortest way for three of them to meet would be at B's house
Now,
Total distance = AB+BC = AC i.e, 8 miles
Sufficient

Hence, B.
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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 01:40

Solution

Given:

• A, B and C have houses along a straight road, in that order.
• All three decided to meet up at one of their houses.

To find:

• Minimum combined distance all friends need to travel.

Let us assume the:
• Distance between the house of A and B is x
• And, distance between the house of B and C is y.

Then, there can be three cases:

1. All friends decide meet at the house of A.
a. Then, distance travelled by B= x
b. And, distance travelled by C= Distance from C to B + Distance from B to A= y+x
i. Hence, total distance covered= 2x+y

2. All friends decide meet at the house of B.
a. Then, distance travelled by A= x
b. And, distance travelled by C= Distance from C to B = y
i. Hence, total distance covered= x+y

3. All friends decide meet at the house of C.
a. Then, distance travelled by A= x+y
b. And, distance travelled by B= y
i. Hence, total distance covered= x+2y

From all the three cases, it is very apparent that minimum distance is in case 2.
Hence, we only need distance from house of A to C to find minimum distance covered by all the friends.

Statement-1: “The distance between A’s house and B’s house is 2 miles.“

x= 2, but we do not have the value of y.
Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2: “The distance between A’s house and C’s house is 8 miles “

Distance from A’s house to C’s house= x+ y= 8
Hence, minimum distance covered is 8 miles.
Statement 2 alone is sufficient to answer the question.

Hence, the correct answer is option B.

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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 03:27
Bunuel wrote:
Three friends, A, B and C have houses along a straight road, in that order. If all three want to meet up at one of their houses, what is the minimum combined distance they need to travel?

(1) The distance between A’s house and B’s house is 2 miles.
(2) The distance between A’s house and C’s house is 8 miles

kshitizbansal92 wrote:
in question you are not given the position of the houses , how can you decide what distance b has to travel to reach c, that's why 2 cases to show the difference

kshitizbansal92 you have been given the order of houses. Please check again.
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Re: Three friends, A, B and C have houses along a straight road, in that  [#permalink]

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19 Jul 2018, 03:36
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Re: Three friends, A, B and C have houses along a straight road, in that &nbs [#permalink] 19 Jul 2018, 03:36
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