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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.

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 \((a+b)^2=a^2+2ab+b^2\):

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 \((a+b)^2=a^2+2ab+b^2\):

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 \(\sqrt{1}\). _________________

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. _________________

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

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. _________________

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

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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