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Exponents ; operations with rational #s

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by bhumika.k.shah » Sat Feb 06, 2010 1:36 am
If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
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by harsh.champ » Sat Feb 06, 2010 1:59 am
bhumika.k.shah wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
This question seems to be too tough to appear on GMAT.
Asking the INSTRUCTORS and EXPERTS,can such type of questions come in exam which involves concepts of recurring and terminating decimals??


Bhumika,can you post the answer under the spoiler mode??
Also,do you have the problem approach??
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by bhumika.k.shah » Sat Feb 06, 2010 2:06 am
Hi harsh,

This question already appears in OG 12th edition PS section...So i think yes these kinda questions do appear on GMAT

i think the problem looks difficult. But shouldnt be that difficult.

Answer will be provided after few comments.

I dont know how to approach these kinda sums . Thats d problem .

Regards,
Bhumika
harsh.champ wrote:
bhumika.k.shah wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
This question seems to be too tough to appear on GMAT.
Asking the INSTRUCTORS and EXPERTS,can such type of questions come in exam which involves concepts of recurring and terminating decimals??


Bhumika,can you post the answer under the spoiler mode??
Also,do you have the problem approach??
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by shashank.ism » Sat Feb 06, 2010 2:20 am
This question seems to be too tough to appear on GMAT.
Asking the INSTRUCTORS and EXPERTS,can such type of questions come in exam which involves concepts of recurring and terminating decimals??


Bhumika,can you post the answer under the spoiler mode??
Also,do you have the problem approach??
Harsh.champ I think the problem of recurring and terminating decimal can come in exams if the problem is not lengthy but tricky and involves a little calculation..I have seen a few problems of similar types in GMAT reference books.
I don't know how much tough is this problem.. but after checking out that this problem is lengthy or tricky we can come up to the conclusion that this type of problem can be asked in exam or not...

I think INSTRUCTORS and EXPERTS would surely throw some light in this regard..
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by shashank.ism » Sat Feb 06, 2010 2:29 am
bhumika.k.shah wrote:
If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
ok coming up with solution of the problem, It was much easier than I expected when I started solving
t = (1) / (2^9 * 5^3)
Now what u have to do is to make the terms of denominator in terms of 10's, 100's,1000's ....if possible..

now here 2^9 * 5^3 = 2^6 * (5*2)^3=2^6 * (10)^3 = 64* 10^3
now 1/2^6 = 0.01.... (divide upto n decimals where u get some value apart from 0)
also 1/10^3=0.001


so 1/(2^9*5^3)= 1/2^6 * 1/10^3 = (0.01..)*0.001=0.00001

so you are going to get 4 zeros between the decimal point and the first nonzero digit to the right of the decimal point
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by ajith » Sat Feb 06, 2010 3:15 am
bhumika.k.shah wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
The fraction is (1/(10^3*2^6)

= 1/64000

10^-5 >f>10^-4

Any fraction between 10^-5 and 10^-4 will have 4 zeros preceding a digit , after decimal point
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by bhumika.k.shah » Sat Feb 06, 2010 4:53 am
ajith how did u get the highlighted part ??

ajith wrote:
bhumika.k.shah wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
The fraction is (1/(10^3*2^6)

= 1/64000

10^-5 >f>10^-4

Any fraction between 10^-5 and 10^-4 will have 4 zeros preceding a digit , after decimal point
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by ajith » Sat Feb 06, 2010 5:05 am
bhumika.k.shah wrote:ajith how did u get the highlighted part ??

1/(2^9 * 5^3 )

=1/(2^6*2^3*5^3) [a^x = a^y*a^(x-y)]

=1/(2^6*10^3) [a^x*b^x = (ab)^x]


The fraction is (1/(10^3*2^6)
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by harshavardhanc » Sat Feb 06, 2010 5:39 am
bhumika.k.shah wrote:
I dont know how to approach these kinda sums . Thats d problem .

Regards,
Bhumika

Here are my two cents ( a different harsh though ;) ) :

whenever you see 2 & 5 in a GMAT PS, in most of the cases, remember that you will have to do something with 10 or powers of 10.

here we have been given 2^9 * 5^3 . Ask yourself, what is the max. power of 10 you can get ? it will be = (2*5)^3 = 1000.

now after segregating this part, you are left with 2^6 in the denominator, i.e 64.

Ask yourself again, when I divide 1 by a two-digit number > 10 (long division method), how many zeros do I put before I get any non-zero digit in the quotient ( isn't it 1?). for e.g if you divide 1 by 11 your quotient will be .099.... OR when you divide 1 by 99, your quotient will be .01 .

wait a min! remember that you have 1000 already present in the denominator (our first step). This thousand will add another 3 zeros before the decimal and the non-zero digit.

In our question we have 1/64. don't do the complete division. you just have to know that there is one zero between the decimal and the first non-zero digit and then, three more are coming from division by 1000. Hence, a total of 4 zeros.

Hope this helps!
Regards,
Harsha
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by komal » Tue Feb 16, 2010 10:22 am
bhumika.k.shah wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

different approaches?
1/10^3 * 1/2^6 = 0.001 * 0.01 = 0.00001 = 4 zeroes

(B) is correct
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