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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:

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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post Updated on: 18 Jul 2013, 13:52
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A
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C
D
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\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.

Originally posted by mneeti on 18 Jul 2013, 13:25.
Last edited by Bunuel on 18 Jul 2013, 13:52, edited 1 time in total.
Renamed the topic and edited the question.
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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 18 Jul 2013, 13:57
1
4
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.



Square the given expression applying \((a+b)^2=a^2+2ab+b^2\):

\((\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2=\)
\(=24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}=48+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}\).

Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(48+2\sqrt{24^2-25*23}=48+2\sqrt{1}=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)

Answer: D.

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Hope it helps.
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 18 Jul 2013, 14:19
1
Bunuel wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.



Square the given expression applying \((a+b)^2=a^2+2ab+b^2\):

\((\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2=\)
\(=24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}\).

Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)

Answer: D.

Similar questions to practice:
square-root-126567.html
root-126569.html
sqrt-15-4-sqrt-14-sqrt-15-4-sqrt-146872.html
topic-101735.html
root-of-138296.html
root-9-root-80-root-9-root-95570.html
if-2-sqrt-5-x-1-then-x-105374.html
if-2-root-5-x-1-then-x-99663.html
which-of-the-following-is-equal-to-3-squar-98531.html#p759352

Hope it helps.

Yup, thanks! It helps.
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 18 Jul 2013, 20:35
2
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9


I took a different approach. The \(\sqrt{23}\) is between 4 and 5. As 23 is closer to 25 than 16, the square root will be just under 5. So \(5\sqrt{23}\) will be just under 25. That makes the first part of the equation approximately \(\sqrt{49}\) and the second part of the equation \(\sqrt{1}\).
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 19 Jul 2013, 00:00
MzJavert wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9


I took a different approach. The \(\sqrt{23}\) is between 4 and 5. As 23 is closer to 25 than 16, the square root will be just under 5. So \(5\sqrt{23}\) will be just under 25. That makes the first part of the equation approximately \(\sqrt{49}\) and the second part of the equation \(\sqrt{1}\).


Actually this part is not correct. You can see that the question itself has this term : \(\sqrt{24-5\sqrt{23}}\) and the term under the root has to be positive. Thus, \(24\geq5\sqrt{23}\). So, it is just less than 24. With this you can approximate the first part as\(\approx \sqrt{48}\) and not as \(\sqrt{49}\).

IMO, now you wouldn't know whether the second part is approximately 0 or a little more than 0. So I guess, with this method, one can get stuck between C and D.

I would follow what Bunuel has suggested above.
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 04 Dec 2013, 17:26
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.


First expression is almost 7 since square root of 23 is pretty close to 5 and square root of 49 = 7

By the same token expression two should be very close to 1 since it can't be negative

So we have 'Almost 7' + 1 = 'Almost 8'

D works just fine

Cheers
J :)
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 17 Feb 2014, 15:19
I am not able to understand how (24^2 - 25*23)^1/2 = 2.

As How I am doing it:

(24^2 - (24+1)(24-1))^1/2
(24^2 - (24^2 - 1^2))^1/2
(24^2 - 24^2)^1/2 by approximation as 24^2-1 would be almost equal to 24^2
=0

Can anyone please help me where I am going wrong?

Many Thanks!!!
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 17 Feb 2014, 21:09
jlgdr wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.


First expression is almost 7 since square root of 23 is pretty close to 5 and square root of 49 = 7

By the same token expression two should be very close to 1 since it can't be negative

So we have 'Almost 7' + 1 = 'Almost 8'

D works just fine

Cheers
J :)


How do you get that expression two is very close to 1?

\(\sqrt{24-5\sqrt{23}}\)
If we assume that 23 is close to 25 so its square root is 5, (24 - 25) gives you -1. But that is not possible since the whole thing is under a root. It must be slightly positive. Since \(\sqrt{23}\) is less than 5, \(5*\sqrt{23}\) must be less than 25, in fact it must be slightly less than 24 since the root has to be 0 or +ve.

So \(\sqrt{24-5\sqrt{23}}\) will become slightly more than 0.

This means you have 'Almost 7' to which you are adding 'Almost 0'. Can you say whether the sum will be less than or more than 7? No. Hence you get stuck between (C) and (D).

Now to get to the answer you will have to use Bunuel's method.
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 17 Feb 2014, 21:15
vivekisthere wrote:
I am not able to understand how (24^2 - 25*23)^1/2 = 2.

As How I am doing it:

(24^2 - (24+1)(24-1))^1/2
(24^2 - (24^2 - 1^2))^1/2
(24^2 - 24^2)^1/2 by approximation as 24^2-1 would be almost equal to 24^2
=0

Can anyone please help me where I am going wrong?

Many Thanks!!!


You are misreading something.

\(\sqrt{24^2 - 25*23} = \sqrt{576 - 575} = 1\)

The 2 with it is the 2 from the 2ab term of \((a+b)^2 = a^2 + b^2 + 2ab\)

Put a = \(24 + 5\sqrt{23}\)
Put b = \(24 - 5\sqrt{23}\)

Now find \((a + b)^2\). You get this as 50.
Then what is (a+b)? It is \(\sqrt{50}\)
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 21 Feb 2014, 12:51
For me the best thing to do is note that 5\sqrt{23} is equal to 25*23 doing this you note that the answer to your question will be ˜\sqrt{50} and is between 7 and 8
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 04 Mar 2014, 00:10
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.


\(\sqrt{23}\) >>>> considered it as \(\sqrt{25}\)
First expression would come up as 7
So 7 + something would be the Answer = D
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 24 Dec 2014, 04:41
If we USE approximation then first value we get near 7 and the second value just near 1 so the final value between 7 and 8.
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 26 Dec 2016, 08:24
VeritasPrepKarishma wrote:
jlgdr wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9

On the face the problem looks very easy but I can't get the answer. Please help! Somewhere OA is correct, I think if OA is correct there is certainly some defect in question itself. Still, I want others' opinion. Thanks.


First expression is almost 7 since square root of 23 is pretty close to 5 and square root of 49 = 7

By the same token expression two should be very close to 1 since it can't be negative

So we have 'Almost 7' + 1 = 'Almost 8'

D works just fine

Cheers
J :)


How do you get that expression two is very close to 1?

\(\sqrt{24-5\sqrt{23}}\)
If we assume that 23 is close to 25 so its square root is 5, (24 - 25) gives you -1. But that is not possible since the whole thing is under a root. It must be slightly positive. Since \(\sqrt{23}\) is less than 5, \(5*\sqrt{23}\) must be less than 25, in fact it must be slightly less than 24 since the root has to be 0 or +ve.

So \(\sqrt{24-5\sqrt{23}}\) will become slightly more than 0.

This means you have 'Almost 7' to which you are adding 'Almost 0'. Can you say whether the sum will be less than or more than 7? No. Hence you get stuck between (C) and (D).

Now to get to the answer you will have to use Bunuel's method.



Hi,

I solved the question using same method, however it was difficult to judge and so went with C. It might have taken extra time in exam. What to do in similar situations? Any other way around?
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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 17 Apr 2018, 02:37
Bunuel Hi. I've read your explanations and seems like I understood it but can you please tell me what was wrong with my approach?

x=\((\sqrt{24+5([square_root]23}[/square_root]))^(\sqrt{1/2)+(24-5([square_root]23}))[/square_root])^(1/2)\)
1: square both sides --> \(X^2\)=\(24+5(\sqrt{23})+24-5(\sqrt{23})\)
2: positive and negative \(5(\sqrt{23})\)=0 so \(x^2\)=24+24=48
3: make a square root from both sides --> x=\((\sqrt{48})\) and it is less than 7
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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 10 Jul 2018, 10:02
Bunuel wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9




Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)

Answer: D.

Hope it helps.


Bunuel, I understood your explanation until this point:

Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

Is like you decided not to show some of your solving, but I got lost. I don't know how you arrived at the answer.
Please help, thanks!
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Re: [24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 10 Jul 2018, 14:00
MrJglass wrote:
Bunuel wrote:
mneeti wrote:
\(\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}}\) lies between:

(A) 4&5
(B) 5&6
(C) 6&7
(D) 7&8
(E) 8&9




Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)

Answer: D.

Hope it helps.


Bunuel, I understood your explanation until this point:

Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

Is like you decided not to show some of your solving, but I got lost. I don't know how you arrived at the answer.
Please help, thanks!


Added extra steps:

\((\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2=\)
\(=24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}=48+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}\).

Apply \((a+b)(a-b)=a^2-b^2\) to the second term (\(2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}\)):

\(48+2\sqrt{24^2-25*23}=48+2\sqrt{1}=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)
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[24+ 5(23)^1/2)]^1/2 + [24- 5(23)^1/2)]^1/2 lies between:  [#permalink]

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New post 10 Jul 2018, 17:52
Bunuel wrote:
MrJglass wrote:
Bunuel wrote:


Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)

Answer: D.

Hope it helps.


Bunuel, I understood your explanation until this point:

Apply \((a+b)(a-b)=a^2-b^2\) to the second term:

\(24+5\sqrt{23}+2\sqrt{24^2-25*23}+24-5\sqrt{23}=48+2=50\).

Is like you decided not to show some of your solving, but I got lost. I don't know how you arrived at the answer.
Please help, thanks!


Added extra steps:

\((\sqrt{24+5\sqrt{23}} +\sqrt{24-5\sqrt{23}})^2=(\sqrt{24+5\sqrt{23}})^2+2\sqrt{24+5\sqrt{23}}*\sqrt{24-5\sqrt{23}}+(\sqrt{24-5\sqrt{23}})^2=\)
\(=24+5\sqrt{23}+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}+24-5\sqrt{23}=48+2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}\).

Apply \((a+b)(a-b)=a^2-b^2\) to the second term (\(2\sqrt{(24+5\sqrt{23})(24-5\sqrt{23})}\)):

\(48+2\sqrt{24^2-25*23}=48+2\sqrt{1}=50\).

So, our expression equals to: \(\sqrt{50}\approx{7.something}\)


Thanks a lot!
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