Math Question

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kashewman: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Mon Aug 14, 2006 10:52 am; Location: Washington D.C.

by kashewman » Sun Sep 10, 2006 10:58 am

I had one question that I don't seem to understand.

Algebra: [(1/5)^m] * [(1/4)^18] = 1/[2(10)^35]
what is m?

Any insight would be appreciated, thanks.

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Source: Beat The GMAT — Problem Solving | Sun Sep 10, 2006 10:58 am

kashewman: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Mon Aug 14, 2006 10:52 am; Location: Washington D.C.

One more question..

by kashewman » Sun Sep 10, 2006 11:00 am

Does anyone have a good formula for mixture problems? Thanks.

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aim-wsc: Legendary Member; Posts: 2469; Joined: Thu Apr 20, 2006 12:09 pmFollowed by: 14 members; Location: BtG Underground; Thanked: 85 times

...err can you make some things clear?

like: 2(10)^35 did you mean [(2(10))^35]=20^35
or 2 X (10)^35

OK i presume its latter one.

Its a tricky one & i must confess i had to used my instincts

before i explain the answer, you must understand the following rule

given n>m
then
(a^m) X (b^n) = ((aXb)^m) X (b^(n-m))

eg. (5^4) X (3^6) = ((5X3)^4) X (3^2)
= (15 ^4) X 9

OK back to problem.
LHS:
(1/5)^m * (1/4)^18
this can be also wirtten as
:arrow: (1/5)^m * (1/2)^36

RHS:
1/[2 X 10^35]= 1/[2 X (2X5)^35)]
:arrow: =1/[(2^36) X (5^35)]

ANSWER:
therefore
m=35

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beatthegmat: Site Admin; Posts: 6778; Joined: Mon Feb 13, 2006 8:30 amFollowed by: 994 members; Location: Los Angeles, CA; Thanked: 1249 times

Re: One more question..

by beatthegmat » Sun Sep 10, 2006 7:34 pm

kashewman wrote:Does anyone have a good formula for mixture problems? Thanks.

Check out page 48 of my GMAT Flashcards: https://www.beatthegmat.com/viewtopic.php?t=32

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abby_g: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Sat Sep 30, 2006 1:59 am; Location: Mumbai; Thanked: 1 times

MAth

by abby_g » Sat Sep 30, 2006 2:06 am

The answer is : m = 35

RHS: 1/2(10)^35 = (1/5^35 )* (1/2^36)
= (1/5^35) * (1/4^18)

equate on both sides and woyla you get m = 35. Correct?

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