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Some difficult ones .

Expert replies
by rockeyb » Mon Nov 30, 2009 9:55 am
1.The value of (10^8 - 10^2)/(10^7 - 10^3) is closest to which of the following ?

a. 1
b. 10
c. 10^2
d. 10^3
e. 10^4


2.For which of the following functions f is f(x) = f(1-x) for all x ?

a.f(x) = 1-x
b.f(x) = 1-x^2
c.f(x) = x^2 -(1-x)^2
d.f(x) = x^2(1-x)^2
e.f(x) = x/(1-x)

Please provide explanation for your answers .



[spoiler]1. B[spoiler]
[spoiler] 2. D [spoiler][/spoiler]
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Source: — Problem Solving |

by jewbagel » Mon Nov 30, 2009 10:15 am
ok heres for number 1

(10^8-10^2)=10^2*(10^6-1)
(10^7-10^3)=10^3*(10^4-1)

if you approx 10^6-1 as about 10^6 and 10^4-1 as about 10^4

the equation becomes (10^6)/(10^5)

6-5=1 so its 10^1 or 10
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by jewbagel » Mon Nov 30, 2009 10:37 am
if you plug in (1-x) into x^2 you get (1-x)^2 and if you plug it into (1-x)^2 you get (-x)^2 which is the same as (x)^2

if you plug in (x) into x^2 and x into (1-x)^2 you get x^2*(1-x)^2

so with d , you get the same thing if you plug x or 1-x in
answer is d
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by rockeyb » Mon Nov 30, 2009 10:38 pm
I got the first explanation and makes sense , but I am not clear with the second one can you please explain it a bit more .
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by AlexBee » Tue Dec 01, 2009 7:14 am
jewbagel wrote:ok heres for number 1

(10^8-10^2)=10^2*(10^6-1)
(10^7-10^3)=10^3*(10^4-1)

if you approx 10^6-1 as about 10^6 and 10^4-1 as about 10^4

the equation becomes (10^6)/(10^5)

6-5=1 so its 10^1 or 10


Could you please explain this:
the equation becomes (10^6)/(10^5)
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by jewbagel » Tue Dec 01, 2009 7:18 am
ok

10^2/10^3 = (1/10^1)

10^6-1 we approx as 10^6
10^4-1 we approx as 10^4

(1/10)*(10^6/10^4) = (10^5/10^4)=10
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by rockeyb » Tue Dec 01, 2009 10:08 pm
jewbagel wrote:if you plug in (1-x) into x^2 you get (1-x)^2 and if you plug it into (1-x)^2 you get (-x)^2 which is the same as (x)^2

if you plug in (x) into x^2 and x into (1-x)^2 you get x^2*(1-x)^2

so with d , you get the same thing if you plug x or 1-x in
answer is d
Can you explain this part please .
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by JasLamba » Wed Dec 02, 2009 4:38 am
1.The value of (10^8 - 10^2)/(10^7 - 10^3) is closest to which of the following ?

a. 1
b. 10
c. 10^2
d. 10^3
e. 10^4

For this one we can factor 10^2. So,
Step 1) 10^2(10^6-1)/10^2(10^5 - 10^1)
Step 2) Cancel out the 10^2. This results in: (10^6)-1/(10^5)-10
Step 3) Since this question uses the key word closest, we can deduce that some type of approximation is involved. So basically, it boils down to this: 10^6/10^5 = 10^1
IMO - B
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by JasLamba » Wed Dec 02, 2009 4:44 am
1.The value of (10^8 - 10^2)/(10^7 - 10^3) is closest to which of the following ?

a. 1
b. 10
c. 10^2
d. 10^3
e. 10^4

For this one we can factor 10^2. So,
Step 1) 10^2(10^6-1)/10^2(10^5 - 10^1)
Step 2) Cancel out the 10^2. This results in: (10^6)-1/(10^5)-10
Step 3) Since this question uses the key word closest, we can deduce that some type of approximation is involved. So basically, it boils down to this: 10^6/10^5 = 10^1
IMO - B
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by Gmater730 » Wed Dec 02, 2009 10:23 am
rockeyb wrote:
jewbagel wrote:if you plug in (1-x) into x^2 you get (1-x)^2 and if you plug it into (1-x)^2 you get (-x)^2 which is the same as (x)^2

if you plug in (x) into x^2 and x into (1-x)^2 you get x^2*(1-x)^2

so with d , you get the same thing if you plug x or 1-x in
answer is d
Can you explain this part please .
when you plug x , or 1-x in for the variable in the original equation, in this case (x), the answer is the same. Hence f(x)=f(1-x)
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by grockit_jake » Mon Dec 14, 2009 10:00 am
For the first question, you dont really need to calculate anything:

(10^8 - 10^2)/(10^7 - 10^3)

Relative to 10^8, 10^2 is very tiny. Think 100,000,000 - 100 = 99,999,900. Not that different from 10^8.

Relative to 10^7, 10^3 is very tiny.

So, in essence, you may think of (10^8 - 10^2)/(10^7 - 10^3) as 10^8/10^7, which equals 10^1 = 10.
Jake Becker
Academic Director
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