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If the sum of squares of two numbers is 97, then which one of the foll

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If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 23 Feb 2019, 05:11
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If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
1. 64
2. 16
3. −32
4. 48
5. 24
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 24 Feb 2019, 01:38
Can someone please explain me the answer?
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 24 Feb 2019, 11:09
Is the question like which one could be their product?

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If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 24 Feb 2019, 18:53
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OhsostudiousMJ wrote:
Can someone please explain me the answer?


This question is out of scope. To solve it, we have to know Inequality of arithmetic and geometric means.

a + b ≥ 2√ab

So, let's assume these two numbers are a and b. Then we have:

a² + b² = 97 ≥ 2√a² b² = 2ab
=> 2ab ≤ 97, ab ≤ 97/2 = 48.5

A.
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 25 Feb 2019, 01:17
I tried to solve it by even-odd concept however i failed. is it possible to solve these question by that concept?
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 27 Feb 2019, 21:26
OhsostudiousMJ wrote:
Can someone please explain me the answer?

Let the two nos. be a and b
Suppose we don't know the relation between square of two nos. and its product.

Standard mathematics says that sum of a number and its reciprocal is always >=2

Thus
a/b + b/a > =2

Let us multiply the two sides by ab
The
a^2 + b^2 >= 2ab

Thus sum of squares of two nos. is always >= Twice its product.
Thus
97 >= 2*ab
In option A , Product = 64
Thus 2* Product = 128 > 97
This cant be true
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If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post Updated on: 28 Feb 2019, 15:12
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I see this as a special quadratics problem.

\(A^2+B^2=97\), and we're asked about \(AB\). This should ring your special quadratics alarm bells for sure.

Note that \((A \pm B)^2 = A^2 \pm 2AB+B^2 \geq 0\). (Squares are never negative.)
This means that \(A^2+B^2 \geq \mp 2AB\), so \(\frac{97}{2} \geq |AB|\).
And since this means that \(AB\) definitely cannot be \(64\), we're done.

(For those of you citing the AM-GM inequality, yes, AM-GM is out of GMAT scope, and yes, I basically just re-derived the two-unknown case of it. However, this doesn't mean that one needs to know AM-GM, per se, to solve this problem, so I'm not entirely convinced that this problem is out of GMAT scope.)

Originally posted by AnthonyRitz on 28 Feb 2019, 01:55.
Last edited by AnthonyRitz on 28 Feb 2019, 15:12, edited 3 times in total.
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 28 Feb 2019, 02:47
AnthonyRitz wrote:
I see this as a special quadratics problem.

\(A^2+B^2=97\), and we're asked about \(AB\). This should ring your special quadratics alarm bells for sure.

Note that \((A \pm B)^2 = A^2 \pm 2AB+B^2 \geq 0\). (Squares are never negative.)
This means that \(A^2+B^2 \geq \mp 2AB\), so \(\frac{97}{2} \geq |AB|\).
And since \(AB\) definitely cannot be \(64\), we're done.

(Yes, for those of you citing the AM-GM inequality, it is out of GMAT scope, and I basically just re-derived the two-unknown case of it. However, this doesn't mean that one needs to know AM-GM, per se, to solve this problem, so I'm not entirely convinced that this problem is out of GMAT scope.)


The best explanation, thanks. :thumbup:
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Re: If the sum of squares of two numbers is 97, then which one of the foll  [#permalink]

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New post 28 Feb 2019, 04:01
raghavrf wrote:
If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
1. 64
2. 16
3. −32
4. 48
5. 24


Given a constant sum of two numbers, the product takes the maximum value when the numbers are equal.
Since a^2 + b^2 is constant,
a^2*b^2 takes the maximum value when a^2 = b^2 = 97/2 = 48.5

Maximum value of a^2b^2 = (48.5 * 48.5) = 48.5^2
So maximum value of ab = 48.5

Hence the product cannot be 64.
Answer (A)

For more on this, check: https://www.veritasprep.com/blog/2015/0 ... at-part-v/
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Re: If the sum of squares of two numbers is 97, then which one of the foll   [#permalink] 28 Feb 2019, 04:01
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