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Which of the following best approximates the value of q if \(5^{28}+3^{11}=5^q\)?

A. 39 B. 30 C. 28 D. 27 E. 17

Note that we need approximate value of \(q\). Now, \(5^{28}\) is much, much, much bigger than \(3^{11}\). So adding \(3^{11}\) to \(5^{28}\) will be closer to \(5^{28}\) then to \(5^{29}\), basically \(3^{11}\) is negligible in this case. So \(q\approx{28}\).

thx i guess i need to go review my exponents haha. i was trying to actually do the problem until it started to get out of hand....sigh...
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Which of the following best approximates the value of q if 5^28 + 3^11 = 5^q ? (A) 39 (B) 30 (C) 28 (D) 27 (E) 17

The answer has to be at least 28 since you are adding something to 5^28.

Note that 3^11 is much much smaller than 5^28 so adding it will not change the value of 5^28 much. If it is hard to visualize, think of it this way:

The next possible answer is 5^30 which is greater than 5^28 5^30 = 25*5^28

But 5^28 + 5^28 = 2*5^28 Even if we add 5^28 to 5^28, all we get is 2*5^28 which is much much smaller than 25*5^28 In fact, what we are actually adding is only 3^11 so 5^28 + 3^11 must be much closer to 5^28. Hence its value must be approximately 5^28.
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Which of the following best approximates the value of q if \(5^{28}+3^{11}=5^q\)?

A. 39 B. 30 C. 28 D. 27 E. 17

Note that we need approximate value of \(q\). Now, \(5^{28}\) is much, much, much bigger than \(3^{11}\). So adding \(3^{11}\) to \(5^{28}\) will be closer to \(5^{28}\) then to \(5^{29}\), basically \(3^{11}\) is negligible in this case. So \(q\approx{28}\).

Answer: C.

Bunuel, my approach for this question was:

I approximated \(3^{11}\) to \(30^{10}\) and thus \(3^{11}\) = \(5*6^{10}\) ----> factor ---> \(5^{10}\) (\(5^{18}\) + 6) .. Since the addition in the parenthesis is minescule, we can ignore it. And thus we have \(5^{10}\) * \(5^{18}\) = \(5^{28}\) .... so q = 28

But obviously 30^10 is about a third of the size of 3^11 so I do not know if these operations work in general. Do they?

Which of the following best approximates the value of q if \(5^{28}+3^{11}=5^q\)?

A. 39 B. 30 C. 28 D. 27 E. 17

Note that we need approximate value of \(q\). Now, \(5^{28}\) is much, much, much bigger than \(3^{11}\). So adding \(3^{11}\) to \(5^{28}\) will be closer to \(5^{28}\) then to \(5^{29}\), basically \(3^{11}\) is negligible in this case. So \(q\approx{28}\).

Answer: C.

Bunuel, my approach for this question was:

I approximated \(3^{11}\) to \(30^{10}\) and thus \(3^{11}\) = \(5*6^{10}\) ----> factor ---> \(5^{10}\) (\(5^{18}\) + 6) .. Since the addition in the parenthesis is minescule, we can ignore it. And thus we have \(5^{10}\) * \(5^{18}\) = \(5^{28}\) .... so q = 28

But obviously 30^10 is about a third of the size of 3^11 so I do not know if these operations work in general. Do they?

It seems that you need to brush up fundamentals on this topic, because unfortunately very little above is correct.

\(3^3=27\) which is pretty close to \(5^2\) \(3^1^1\) is 3 times \(3^3\) and one time \(3^2\) thus we can approximate to 3 times \(5^2\) and one 5

\(5^2^8+5^7=5^7(5^2^1+1)\).

we can conclude that the best approximation is \(5^2^8\) since \(5^2^1+1\) is a little bit more than \(5^2^1\). and \(5^7\) times (a little bit more than \(5^2^1\)) = around \(5^2^8\)

hope it helps.
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Last edited by gmat6nplus1 on 29 Dec 2013, 03:18, edited 1 time in total.

Re: Which of the following best approximates the value of q if [#permalink]

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29 Dec 2013, 02:35

vwjetty wrote:

Which of the following best approximates the value of q if 5^28+3^11=5^q ?

A. 39 B. 30 C. 28 D. 27 E. 17

We have: 5^28+3^11=5^q ==> because 3^11 > 0 --> 5^q MUST be equal or greater than 5^28 ==> q MUST be equal or greater than 28 ==> Option D and E are out immediately.

Divide both sides by 5^q and q >= 28 We have: 5^(28-q) + 3^11/5^q = 1

Because q >= 28 ==> 3^11/5^q = 0.0000xyz that is very small, we can ignore it. Thus, 5^(28-q) must be approximate 1 ==> 28-q = 0 ==> q is approximate 28 C is the answer.

Hope it helps.
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Which of the following best approximates the value of q if \(5^{28}+3^{11}=5^q\)?

A. 39 B. 30 C. 28 D. 27 E. 17

Note that we need approximate value of \(q\). Now, \(5^{28}\) is much, much, much bigger than \(3^{11}\). So adding \(3^{11}\) to \(5^{28}\) will be closer to \(5^{28}\) then to \(5^{29}\), basically \(3^{11}\) is negligible in this case. So \(q\approx{28}\).

Answer: C.

I thought it this way:\(3^11=5^q-5^28\).Now if we take\(q>28\),we'll be able to take out \(5^28\) common from RHS;that cannot equal \(3^11\).And if we take \(q<28\),We'll get -ve RHS;that again cannot equal \(3^11\). So the closest value can be \(q=28\) only.But can this question be there in the GMAT?Even the closest/aprox. value in this case is quite far away!

Re: Which of the following best approximates the value of q if [#permalink]

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20 Feb 2016, 10:13

VeritasPrepKarishma wrote:

Smita04 wrote:

Which of the following best approximates the value of q if 5^28 + 3^11 = 5^q ? (A) 39 (B) 30 (C) 28 (D) 27 (E) 17

The answer has to be at least 28 since you are adding something to 5^28.

Note that 3^11 is much much smaller than 5^28 so adding it will not change the value of 5^28 much. If it is hard to visualize, think of it this way:

The next possible answer is 5^30 which is greater than 5^28 5^30 = 25*5^28

But 5^28 + 5^28 = 2*5^28 Even if we add 5^28 to 5^28, all we get is 2*5^28 which is much much smaller than 25*5^28 In fact, what we are actually adding is only 3^11 so 5^28 + 3^11 must be much closer to 5^28. Hence its value must be approximately 5^28.

great explanation...I tried to solve it by logic..and eliminated right away D and E. A is way too much. Between B and C, went with B..now I see why I did the mistake..

Which of the following best approximates the value of q if 5^28 + 3^11 = 5^q ? (A) 39 (B) 30 (C) 28 (D) 27 (E) 17

The answer has to be at least 28 since you are adding something to 5^28.

Note that 3^11 is much much smaller than 5^28 so adding it will not change the value of 5^28 much. If it is hard to visualize, think of it this way:

The next possible answer is 5^30 which is greater than 5^28 5^30 = 25*5^28

But 5^28 + 5^28 = 2*5^28 Even if we add 5^28 to 5^28, all we get is 2*5^28 which is much much smaller than 25*5^28 In fact, what we are actually adding is only 3^11 so 5^28 + 3^11 must be much closer to 5^28. Hence its value must be approximately 5^28.

great explanation...I tried to solve it by logic..and eliminated right away D and E. A is way too much. Between B and C, went with B..now I see why I did the mistake..

B could've been easily eliminated as for increasing another power of 5, you will need to atleast multiply that number by 5 atleast. 3^11 can be approximated to (3^6)^2 and 3^6 will give you 5^4 so you will end up getting almost 5^8 and this when added to 5^28 will still not do anything.
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