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If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 [#permalink ]
03 Apr 2011, 07:04

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Please post a simple solution for this question. I could do it by number picking strategy but then it takes a lot of time.

Hello,

I recently gave a CAT and there was a question for which the answer explanation was not very clear.

If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 ,What is the smallest possible value of 2a+b ? OPEN DISCUSSION OF THIS QUESTION IS HERE: the-integers-a-b-and-c-are-positive-a-b-5-2-and-a-c-128150.html _________________

---Jimmy Life`s battles dont always go, To the stronger or faster man; But sooner or later the man who wins, Is the man who THINKS HE CAN . KUDOS me if you feel my contribution has helped you.

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
03 Apr 2011, 07:12

jimmy86 wrote:

Please post a simple solution for this question. I could do it by number picking strategy but then it takes a lot of time.

Hello,

I recently gave a CAT and there was a question for which the answer explanation was not very clear.

If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 ,What is the smallest possible value of 2a+b ? a/b= 5/2

2a=5b ---1

a=(5/2)b

a/c=7/5

5a=7c

5(5/2*b)=7c

25b=14c

b is a multiple of 14

c is a multiple of 25.

LCM of b and c = 350

25b=350

b=14

14c=350

c=25

5a=7c

5*a=7*25

a=35

2a+b = 2*35+14=84.

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
03 Apr 2011, 09:07
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jimmy86 wrote:

Please post a simple solution for this question. I could do it by number picking strategy but then it takes a lot of time.

Hello,

I recently gave a CAT and there was a question for which the answer explanation was not very clear.

If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 ,What is the smallest possible value of 2a+b ? When you have two positive integers a and b, and the ratio of a to b is 5 to 2, then a must always be a multiple of 5, and b must always be a multiple of 2. This will always be true when you have a ratio of two integers, provided the ratio is completely reduced (so if you knew, say, the ratio of a to b was 10 to 4, you'd need to reduce that ratio to 5 to 2 first before drawing any conclusions about multiples).

So here, we know that the ratio of a to b is 5 to 2, so a is a multiple of 5. We also know the ratio of a to c is 7 to 5, so a is a multiple of 7. Thus a is a multiple of both 5 and 7, and the smallest possible value of a is 35. If a is 35, then since a/b = 5/2, b would be 14, and 2a + b = 84.

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
03 Apr 2011, 18:02
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We can see that :

a:b:c = 35:14:10

because a has 7 and 5 as factors, so least value of a = 35, and hence b = 14 ( and c = 25)

So 2a + b = 70 + 14 = 84

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
03 Apr 2011, 21:09

a:b = 5:2 a:c = 7:5 So to form a:b:c we a needs to be the LCM of 7 and 5 i.e. 35 So Multiply first ratio by 7 and second by 5 a:b:c = 35:14:25 So 2*35+14 = 84

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
04 Apr 2011, 00:26

thanks everyone.... Specially liked Ian`s explanation....the answer is a lot simpler than i thought....

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
02 Feb 2013, 16:19

fluke wrote:

jimmy86 wrote:

Please post a simple solution for this question. I could do it by number picking strategy but then it takes a lot of time.

Hello,

I recently gave a CAT and there was a question for which the answer explanation was not very clear.

If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 ,What is the smallest possible value of 2a+b ? a/b= 5/2

2a=5b ---1

a=(5/2)b

a/c=7/5

5a=7c

5(5/2*b)=7c

25b=14c

b is a multiple of 14

c is a multiple of 25.

LCM of b and c = 350

25b=350

b=14

14c=350

c=25

5a=7c

5*a=7*25

a=35

2a+b = 2*35+14=84.

--------------------------------------------------------------------------------------------------------------------------------------------

Hey Fluke,

Can you please explain me where my method went wrong.

a/b=5/2---> a=2.5*b

a/c=7/5--->a=1.4*C

2.5*b = 1.4*c

c=1.8*b

As B and C have to be integers, the least integer value for B that would make C an integer is '5'.

From the stem, we know that 2a=5b

2a+b=5b+b=6b

As we know that b=5. 6b=30.

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Re: PS : Algebra/Ratio - High Difficulty [#permalink ]
04 Feb 2013, 03:42
ulabrevolution wrote:

fluke wrote:

jimmy86 wrote:

Please post a simple solution for this question. I could do it by number picking strategy but then it takes a lot of time.

Hello,

I recently gave a CAT and there was a question for which the answer explanation was not very clear.

If integers a,b,c are positive and a/b= 5/2 and a/c=7/5 ,What is the smallest possible value of 2a+b ? a/b= 5/2

2a=5b ---1

a=(5/2)b

a/c=7/5

5a=7c

5(5/2*b)=7c

25b=14c

b is a multiple of 14

c is a multiple of 25.

LCM of b and c = 350

25b=350

b=14

14c=350

c=25

5a=7c

5*a=7*25

a=35

2a+b = 2*35+14=84.

--------------------------------------------------------------------------------------------------------------------------------------------

Hey Fluke,

Can you please explain me where my method went wrong.

a/b=5/2---> a=2.5*b

a/c=7/5--->a=1.4*C

2.5*b = 1.4*c

c=1.8*b As B and C have to be integers, the least integer value for B that would make C an integer is '5'.

From the stem, we know that 2a=5b

2a+b=5b+b=6b

As we know that b=5. 6b=30.

From 2.5b=1.4c it follows that c=25/14*b. Now, 25/14 does not equal to 1.8.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-integers-a-b-and-c-are-positive-a-b-5-2-and-a-c-128150.html _________________

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Re: PS : Algebra/Ratio - High Difficulty
[#permalink ]
04 Feb 2013, 03:42

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