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(\sqrt25/\sqrt10)
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13 Jul 2019, 01:59
13 Jul 2019, 01:59
1
Expert Reply
3
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Question Stats:
68% (01:23) correct
31% (01:57) wrong based on 143 sessions
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Quantity A |
Quantity B |
\(\frac{\sqrt{25}}{\sqrt{10}}\frac{\sqrt{8}}{\sqrt{15}}\) |
\(\frac{\sqrt{51}}{\sqrt{46}}\frac{\sqrt{23}}{\sqrt{34}}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
(\sqrt25/\sqrt10)
[#permalink]
08 Mar 2024, 08:56
08 Mar 2024, 08:56
1
Quantity A
This can be easily dealt with because the numbers are simple
\(\frac{\sqrt{25}}{\sqrt{10}} \frac{\sqrt{8}}{\sqrt{15}}\)
\(\frac{5}{\sqrt{5 \times 2}} \frac{2\sqrt{2}}{\sqrt{5 \times 3}}\)
\(\frac{5}{\sqrt{5} \sqrt{2}} \frac{2\sqrt{2}}{\sqrt{5} \sqrt{3}}\)
\(\frac{2}{\sqrt{3}}\)
\(\frac{2\sqrt{3}}{3}\)
This gives us approximately \(\frac{3.4}{3}\) which is greater than \(1\)
Quantity B
\(\frac{\sqrt{51}}{\sqrt{46}}\frac{\sqrt{23}}{\sqrt{34}}\)
This may seem intimidating until you realize that it can be rewritten as
\(\sqrt{\frac{51 \times 23}{ 46 \times 34}}\)
which yields
\(\sqrt{\frac{51}{68}}\)
We stop right here as \(\frac{51}{68}\) is less than \(1\) and taking its square root is not going to change this fact, in fact it will only going to make it smaller
Quantity A: greater than 1
Quantity B: less than 1
Quantity A is greater.
This can be easily dealt with because the numbers are simple
\(\frac{\sqrt{25}}{\sqrt{10}} \frac{\sqrt{8}}{\sqrt{15}}\)
\(\frac{5}{\sqrt{5 \times 2}} \frac{2\sqrt{2}}{\sqrt{5 \times 3}}\)
\(\frac{5}{\sqrt{5} \sqrt{2}} \frac{2\sqrt{2}}{\sqrt{5} \sqrt{3}}\)
\(\frac{2}{\sqrt{3}}\)
\(\frac{2\sqrt{3}}{3}\)
This gives us approximately \(\frac{3.4}{3}\) which is greater than \(1\)
Quantity B
\(\frac{\sqrt{51}}{\sqrt{46}}\frac{\sqrt{23}}{\sqrt{34}}\)
This may seem intimidating until you realize that it can be rewritten as
\(\sqrt{\frac{51 \times 23}{ 46 \times 34}}\)
which yields
\(\sqrt{\frac{51}{68}}\)
We stop right here as \(\frac{51}{68}\) is less than \(1\) and taking its square root is not going to change this fact, in fact it will only going to make it smaller
Quantity A: greater than 1
Quantity B: less than 1
Quantity A is greater.
General Discussion
GRE Prep Club Tests Editor
Joined: 13 May 2019
Affiliations: Partner at MyGuru LLC.
Posts: 186
Given Kudos: 5
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE:Education (Education)
Re: (\sqrt25/\sqrt10)
[#permalink]
13 Jul 2019, 08:12
13 Jul 2019, 08:12
2
Expert Reply
Carcass wrote:
Quantity A |
Quantity B |
\(\frac{\sqrt{25}}{\sqrt{10}}\frac{\sqrt{8}}{\sqrt{15}}\) |
\(\frac{\sqrt{51}}{\sqrt{46}}\frac{\sqrt{23}}{\sqrt{34}}\) |
There are so many ways to do this problem!
The first is the most certain : Use the calculator. It has a √ button, so you can actually use it to compare these values if you carefully work through the proper order of operations. If you have difficulty with exponents and radicals spend the 1-2 minutes on the certain, methodical approach.
However, there is also a relatively simple method for logically estimating the relationship based on the idea that if all values in a system of multiplication or division have the same radical or exponent you can combine the terms through multiplication or division while keeping the exponent or radical.
So, let's look at that logical estimation method. First, eliminate Choice D, because there are no variables so the relationship must be definitive.
Now, let's look at Quantity A and imagine that the radicals are gone, so we'd simply be looking at (25/10) x (8/15).
Estimate roughly to convert that into Quantity A is 2.5 x .5 ≈ 1.25.
Sure, there is a √ but we can basically ignore it because it's in Quantity B as well.
Next, let's look at Quantity B and imagine that the radicals are gone, so we'd simply be looking at (51/46) x (23/34).
Estimate roughly to convert that into Quantity A is 1.1 x .667 ≈ .75.
Once again ignoring the √ we can easily see that roughly .75 < roughly 1.25 so Quantity A is greater and we can select Choice A.
