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Since you are asked to find the power of 5 i.e. m, you should be looking for 5 on the right hand side of the equation raised to some power. There is 10 on the denominator which is 2 * 5 raised to the power 35. The rest of the solution is rearranging to compare the powers of 5 on both sides.
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Originally posted by hb on 10 Aug 2013, 18:53.
Last edited by hb on 10 Aug 2013, 19:07, edited 1 time in total.
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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23 Aug 2014, 22:26
Just need to break the equations a bit: 1. (1/5)^m (1/4)^18 = (1/5)^m (1/2)^36 = so now we have 36 powers of 1/2 and need to find for 5 , what we need is the relation between this equation and other so we will try to sync them up. 2. 1/(2(10)^35) = 1/(2(2*5)^35 = so now we have 36 powers of 2 and 35 powers of 5 Finally what we need is how many powers of 5?? its 35 , OA:D. Hope its clear _________________
Comparing left and right side (1/2)^36 is same on both sides . For equation to be true (1/5)^35 should equal to (1/5)^m which is possible when m=35
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Mar 2015, 05:17
devilbart wrote:
annaleroy wrote:
Hey!
I had this question on the GMATPrep 1 and I can't figure it out, could someone please help
(1/5)^m (1/4)^18 = 1/(2(10)^35)
m= 35
This was a problem solving question but i do not understand how they got m=35!
Thanks for your help!!
Pretty Straight forward. forget about the LHS and re arrange the R.H.S to match the LHS fractions
RHS:
1/(2(2*5)^35)
1/(2^36*5^35)
1/(4^18 * 5^35) ...... [ 2^36 is same as saying (2^2)^18]
break the denominator as per the LHS
(1/4)^18 * (1/5)^35
and Whalaa you have Mr. m
I got a few stupid questions:
1) What is RHS / LHS? 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"?
3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m \(1^m/5^m?\)
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Mar 2015, 05:27
erikvm wrote:
devilbart wrote:
annaleroy wrote:
Hey!
I had this question on the GMATPrep 1 and I can't figure it out, could someone please help
(1/5)^m (1/4)^18 = 1/(2(10)^35)
m= 35
This was a problem solving question but i do not understand how they got m=35!
Thanks for your help!!
Pretty Straight forward. forget about the LHS and re arrange the R.H.S to match the LHS fractions
RHS:
1/(2(2*5)^35)
1/(2^36*5^35)
1/(4^18 * 5^35) ...... [ 2^36 is same as saying (2^2)^18]
break the denominator as per the LHS
(1/4)^18 * (1/5)^35
and Whalaa you have Mr. m
I got a few stupid questions: no doubts are stupid and its always better to clear your basic doubt then to remain with them
hi erik,
1) What is RHS / LHS? RHS is right hand side of equality sign and LHS is left hand side ... 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"? 4^18=(2*2)^18=\((2^2)\)^18=2^(2*18)=2^36..... 5^4=(5^2)^2=25^2
3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m \(1^m/5^m?\).. 1^m=1 for all values of 1.... if we divide or multiply 1 any number of times, the result will be 1 so m can be removed from power..
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Mar 2015, 05:36
erikvm wrote:
Thanks for the quick response. I actually know these rules but I forget to apply them, its really frustrating
doesnt matter erik... it happens with everyone if we don't refresh basics.... i am sure you will overcome all this by putting in some hardwork.. all the best..
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Oct 2015, 04:13
1
jst6059 wrote:
Question from the practice test that has me dumbfounded:
If \((\frac{1}{5})^m * (\frac{1}{4})^18 = (\frac{1}{2(10)^35})\), what does m equal?
A) 17 B) 18 C) 34 D) 35 E) 36
Format your question properly and search for a question before you post. The question has already been discussed. Look above for the solution.
Topics merged.
As for your question, for all such questions, you need to break the bigger numbers down to their respective prime factors by noting that \(4=2^2\) and \(10=2*5\), After you are doing breaking both sides of the equation down to the respective prime factors, equate the powers of similar 'bases' to get to your final answer.
Also \(1/x = x^{-1}\) and \(x^a*x^b = x^{(a+b)}\)
Applying these relations to the question at hand you get:
---> \(5^{-m}*2^{-36} = 2^{-1}*(2^{-35}*5^{-35})\) ---> \(5^{-m}*2^{-36} = 2^{-36}*5^{-35}\), equate both sides now to see that m=35. D is the correct answer.
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Oct 2015, 05:52
erikvm wrote:
I got a few stupid questions:
1) What is RHS / LHS? 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"?
3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m \(1^m/5^m?\)
1) What is RHS / LHS? LHS and RHS are the acronyms of Left hand side and RIgh hand side respectively. 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"? We can go from 4^18 --> 2^36, because 4 is a power of two, (2^2 = 4) Whereas 10 is not a power of 5. You cannot express 10 just by using 5. Divide and multiply by 2 rules means to divide and multiply an expression by 2 to make it simpler.
For example: 18/5 = 18*2/5*2 = 36/10 = 3.6 By doing this, we can solve easily
3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m 1m/5m?.. Yes (1/5)^m = 1^m/5^m, but however many times you multiply 1, the result will always be 1. Hence we can write 1^m simply as 1.
gmatclubot
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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04 Oct 2015, 05:52
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80% (01:24) correct 
