PS OG 12 #32

[sudi760mba]

sqrt[(16)(20) + (8)(32)]?

Other ways of solving this other than what's in the book?

Thanks!

[LFalken]

Sqrt((16*20) + (8*32))

= sqrt (( 320 + 256 ))

= sqrt (576)

= 24


hope this helps.

[rish]

IMO..this might be quicker

sqrt[(16*20)+(8*32)]
sqrt[16(20+16)]
sqrt[16*36]
4*6
24

[bharathh]

Use factors of 2

sqrt(2^4*2^2*5 + 2^3*2^5)
= sqrt(2^6(5+2^2))
=sqrt(2^6*3^2)
= 2^3*3 = 24

[Kevdog2834]

IMO..this might be quicker

sqrt[(16*20)+(8*32)]
sqrt[16(20+16)]

rish how do you know to take 16 out of both sides?

In the OG it says 8 * 32 is same as 16 * 16. This step confuses me on how they got there and it looks like you did the same thing. Hopefully this is something easy if explained for us. Just want to make sure if we see another problem with different numbers we will know how to quickly solve it correctly.

[indiantiger]

sqrt[(16)(20) + (8)(32)]
= sqrt[(16)*(20) + (8*2)*(16)]
= sqrt[16*(20+16)]
= 4 sqrt(36)
= 4*6 = 24

[krazy800]

IMO..this might be quicker

sqrt[(16*20)+(8*32)] - (1)
sqrt[16(20+16)]

rish how do you know to take 16 out of both sides?

In the OG it says 8 * 32 is same as 16 * 16. This step confuses me on how they got there and it looks like you did the same thing. Hopefully this is something easy if explained for us. Just want to make sure if we see another problem with different numbers we will know how to quickly solve it correctly.

32 = 2* 16

therefore 8*32 = 8*2*16 = 16*16


HTH!!!

[Patrick_GMATFix]

Don't waste your time multiplying and adding everything under the root. Instead it makes sense to keep things as products of factors in order to extract them from the square root more easily. The answer is B. Solution & Take-Away are attached.

[Kevdog2834]

32 = 2* 16

therefore 8*32 = 8*2*16 = 16*16


HTH!!!

Thank you for showing the simplicity of the problem. It was very easy but my mind was not catching it until now.