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Two natural numbers a and b are such that a = 3b + b^2

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Two natural numbers a and b are such that a = 3b + b^2  [#permalink]

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New post 06 Jan 2019, 09:21
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Question Stats:

54% (01:39) correct 46% (01:29) wrong based on 71 sessions

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Two natural numbers a and b are such that a = 3b + b^2. Is 'a' a multiple of 5?

(1) b is a multiple of 2

(2) b is a factor of 10
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Re: Two natural numbers a and b are such that a = 3b + b^2  [#permalink]

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New post 07 Jan 2019, 22:38
From statement 1:

b is a multiple of 2.
b can be 0, 2, 4, 6, etc.
If b is 0 and 2, then a will be a multiple of 5.
If b is 4, then a will not be a multiple of 5.
Hence statement 1 is insufficient.

From statement 2:

b is a factor of 10.
b is 1, 2, 5, 10.
If b is 2, 5 or 10. Then a will be a multiple of 5.
If b is 1, then a will not be a multiple of 5.
Insufficient.

Combining both gives the value of b as 2 or 10.
For 2 or 10, a will be a multiple of 5.
Hence sufficient.

C is the answer.
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Re: Two natural numbers a and b are such that a = 3b + b^2  [#permalink]

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New post 08 Jan 2019, 00:46
1
Hi,


Given, “a” and “b” are natural numbers.

That is, “a” and “b” are positive integers.

Such that, a = 3b + b^2.

Question:

Is “a” multiple of 5 ?

i.e., whether a = 5, 10, 15…..?

Statement I is insufficient:

b is a multiple of 2.

“b” is a positive integer and multiple of 2..

So, “b” can be 2,4,6,8,10…

If b = 2, then

a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES.

But if b = 4, then

a = 3b + b^2 = 12 + 16 = 28 and it is not a multiple of 5. Answer to the question is NO.

Statement II is insufficient:

b is a factor of 10

So, “b” values are 1, 2, 5 or 10.

If b = 2, then

a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES.

But if b = 1, then

a = 3b + b^2 = 3 + 1 = 4 and it is not a multiple of 5. Answer to the question is NO.

Together it is sufficient.

Values of “b” which satisfy both statement I and II is 2 and 10.

So,

If b = 2, then

a = 3b + b^2 = 6 +4 = 10 and is a multiple of 5. Answer to the question is YES.

And if b = 10, then

a = 3b + b^2 = 30 + 100 = 130 and it is also a multiple of 5. Answer to the question is again YES.

So together it is sufficient.

Answer is C.

Hope this helps.
Regards,
Junaid.
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Re: Two natural numbers a and b are such that a = 3b + b^2  [#permalink]

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New post 02 Feb 2019, 09:38
akurathi12 wrote:
Two natural numbers a and b are such that a = 3b + b^2. Is 'a' a multiple of 5?

(1) b is a multiple of 2

(2) b is a factor of 10


Natural numbers are 1,2,3,4,5,6,7,8

from 1, b will be 2,4,6,8,10

a = 3b + b^2, just substitute the values to get
Yes when b = 2
No when b = 4

from 2, b= 1,2,5,10
Yes when b=2,5,10
No when b=1

Combine both to get b= 2,10, both will give Yes

C
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Re: Two natural numbers a and b are such that a = 3b + b^2  [#permalink]

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New post 27 Mar 2019, 22:42
Given , a=3b+b^2 or a=b(3+b)
To find a/5 = Integer or b(3+b)/5 = Integer ?

Statement 1, b/2 = Integer
Insufficient, since b can be 2 (sufficient) , but if b is 4 (insufficient)

Statement 2: B is a factor of 10 i.e 1 ,2, 5, 10
Alone insufficient, since b can be 1 also.

After combining : Sufficient to say Yes, since possible values of b can be 2 and 10.

Ans : C
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Re: Two natural numbers a and b are such that a = 3b + b^2   [#permalink] 27 Mar 2019, 22:42
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