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For all x, the expression x* is defined to be ax + a, where a is......

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For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 22 May 2016, 23:00
1
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A
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D
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For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)
Spoiler: OA
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Re: For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 23 May 2016, 00:02
1
nalinnair wrote:
For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)


x* = ax + a = a (x + 1)
2* = ?

Statement 1: 3* = 2
or 3a + a = 2
a = 1/2

Hence 2* = (1/2)*(2 + 1) = 3/2
SUFFICIENT

Statement 2: 5* = 3
Or 5a + a = 3
a = 1/2

Hence 2* = (1/2)*(2 + 1) = 3/2
SUFFICIENT

Correct Option: D
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Re: For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 24 May 2016, 04:54
2
Trying to solve these equations is unnecessary and in 90% of DS questions a waste of time

A) 3a + a = 2

Linear eqn with 1 variable, 1 solution. sufficient

2) 5 a + a = 6

Same reasoning as above

Answer : D
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For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post Updated on: 19 Dec 2017, 11:31
nalinnair wrote:
For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)


Given
=>x*=ax+a
=>x*=a(x+1)
Therefore
=>2*=a(2+1)
=>2*=3a ----(1)
so if we can find the value of 'a' we can hav the value of 2*

Now
Statement 1 says 3*=2
3*=4a from (1)
=> 4a=2
=> a=2/4=1/2
sufficient

Statement 2 says 5*=3
5*=6a from (1)
=> 6a=3
=> a=3/6=1/2
sufficient

Therefore option 'D'

Thanks

Originally posted by dineshril on 19 Dec 2017, 07:47.
Last edited by dineshril on 19 Dec 2017, 11:31, edited 1 time in total.
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Re: For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 19 Dec 2017, 08:50
1
dineshril wrote:
nalinnair wrote:
For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)


Given
=>x*=ax+a
=>x*=a(x+1)
Therefore
=>2*=a(2+1)
=>2*=3a ----(1)
so if we can find the value of 'a' we can hav the value of 2*

Now
Statement 1 says 3*=2
3*=4a from (1)
=> 4a=2
=> a=2/4=1/2
sufficient

Statement 2 says 5*=3
5*=6a from (1)
=> 6a=3
=> a=3/6=1/2
sufficient

Therefore option 'E'

Thanks


Hi

Correct solution. But when each statement alone is sufficient to answer a question, the answer is D

I guess that was a typo :)
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Re: For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 19 Dec 2017, 11:30
amanvermagmat wrote:
dineshril wrote:
nalinnair wrote:
For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)


Given
=>x*=ax+a
=>x*=a(x+1)
Therefore
=>2*=a(2+1)
=>2*=3a ----(1)
so if we can find the value of 'a' we can hav the value of 2*

Now
Statement 1 says 3*=2
3*=4a from (1)
=> 4a=2
=> a=2/4=1/2
sufficient

Statement 2 says 5*=3
5*=6a from (1)
=> 6a=3
=> a=3/6=1/2
sufficient

Therefore option 'E'

Thanks


Hi

Correct solution. But when each statement alone is sufficient to answer a question, the answer is D

I guess that was a typo :)


Hi Aman

Thanks for the correction. Its 'D'
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Re: For all x, the expression x* is defined to be ax + a, where a is......  [#permalink]

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New post 29 Jan 2019, 23:17
nalinnair wrote:
For all \(x\), the expression \(x^*\) is defined to be \(ax + a\), where \(a\) is a constant. What is the value of \(2^*\)?

(1) \(3^* = 2\)
(2) \(5^* = 3\)


So basically

2^* will be 2a + a => 3a

Now if i can get value of a, i will be good

(1) \(3^* = 2\)
3a + a = 2
a = 1/2

We are good

(2) \(5^* = 3\)

6a = 3
a = 1/2

We are good

Both statements are Sufficient

D

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Re: For all x, the expression x* is defined to be ax + a, where a is......   [#permalink] 29 Jan 2019, 23:17
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