If p=125×243×16/405, how many digits are in p?

Question

A. 1
B. 2
C. 3
D. 4
E. 5

Balajikarthick1990 · Answer

Its D...4 digits. Value will be in 1xxx.

Abhishek009 · Answer

p = 125×243×\frac{16}{405} p = 25×243×\frac{16}{81} p = 25×3×16 p = 25×48 p = 1200 { Shortcut calculation is 4800/4 , since 25 = 100/4; add 2 zeroes to 48 then divide by 4} Thus answer must be D. 4 :lol:

chetan2u · Answer

We can or rather we SHOULD do some approximation..
the fraction 16/405 in a way is given to approximate, otherwise it could be 19/405 or so..
so lets take 405 as 400..

p=125×243×\frac{16}{400} = \frac{25 * 5 *243 *4*4}{4*4*25} ..
=> 5*243 = slightly >1000 so 4 digits
D

KarishmaB · Answer

Approximation is your best bet here. Note that 405 is a bit less than twice of 243. It's between 1.5 and 2 times of 243.

p = 125 * 16/2 = 125 * 8 = 1000

So p is at least 1000. It is a bit more than 1000. So it must have 4 digits.

Answer (D)

Abkhazian · Answer

p = 125 * 243 * 16 / 405
p= 5^3 * 3 * 9^2 * 4^2 / (5 * 9^2)
p= 5^2 * 3 * 4^2
p = 20^2 * 3 = 1200

answer D

I've been wrong few times while working with powers of 10 or with approximations... in this example it's quite easy but sometimes the numbers are much complexe, so I don't feel comfortable if I d'ont go thru the numbers.. And considering that these type of questions are supposed to be the "easy time saving question" the test day, the risk of inattention errors increased...

merajul · Answer

P = 125*243*(16/405) = (5^3 *3^5*2^4)/(5*3^4) = 5^2*3*2^4=10^2*12

For 10, Digits 2+1 =3
For 12, Digits,1+0 =1
Thus total digits would be (3+1) = 4

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If p=125×243×16/405, how many digits are in p?

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If p=125×243×16/405, how many digits are in p?

Prep Toolkit

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