1. ## Easy Stuff?

I hope this is easy for you guys, because it is most certainly not for me...

1) Brandon Filop had seven times as many pennis as Billy Wang. When Brandon gave Billy 9 pennies, he still had 36 more pennies than Billy. How many pennies did Brandon have originally?

2)The number of digits in the number $\displaystyle (2^{1995})(5^{1992})$ is:
a)3885
b)1992
c)1993
d)1994
e)1995

3)The express $\displaystyle 6\sqrt{2}+\sqrt{32}$ is equivalent to
a) $\displaystyle 7\sqrt{34}$
b) $\displaystyle 20$
c) $\displaystyle 10\sqrt{2}$
d) $\displaystyle 6\sqrt{34}$
e) $\displaystyle 6\sqrt{64}$

4) If $\displaystyle 7^{d-4}=\frac{1}{49}$, then d=
a) 6
b) 4
c) 2
d) -2
e) -3

5) A rectangular tank contains water to a depth of 11 cm. The base of the tank measures 5 cm by 7 cm. If all the water in the bank is poured into an empty cylindrical tank of base radius 5 cm and there is no water overflow, find the depth of water in the cylindrical tank.

Thanks for helping me guys!

2. 1) Let F denote Brandon Filop's pennies and W denote Billy Wang's pennies.
$\displaystyle F=7W$ ...[1]
$\displaystyle F-9=(W+9)+36$ ...[2]
By subtraction we have
$\displaystyle 9=7W-(W+9)-36$
$\displaystyle 9=6W-45$
$\displaystyle 6W=54$
$\displaystyle W=9$
So $\displaystyle F=63$

2) The number of digits of a number N is $\displaystyle \lfloor \log N \rfloor +1$

So the number of digits in $\displaystyle (2^{1995})(5^{1992})$ is $\displaystyle \lfloor \log{(2^{1995})5^{1992})}\rfloor+1=\lfloor \log{2^{1995}}+\log{5^{1992}}\rfloor+1=\lfloor 1995\log{2}+1992\log{5} \rfloor +1$

Then get out a calculator and that's 1993 digits.

3) $\displaystyle 6\sqrt{2}+\sqrt{32}=6\sqrt{2}+\sqrt{16 \cdot 2}=6\sqrt{2}+4\sqrt{2}=10\sqrt{2}$

4) $\displaystyle 7^{d-4}=\frac{1}{49}$

$\displaystyle 7^{d-4}=\frac{1}{7^2}$

$\displaystyle 7^{d-4}=7^{-2}$

$\displaystyle d-4=-2$

$\displaystyle d=2$

5) The rectangular tank contains $\displaystyle 11 \cdot 5 \cdot 7=385cm^3$ of water.

The volume of the cylindrical tank is $\displaystyle V=\pi 5^2 \times h=25\pi h$ where h is the height.

Then we substitute in 385: $\displaystyle 385=25 \pi h$

$\displaystyle h=\frac{385}{25 \pi}$

$\displaystyle h=\frac{77}{5 \pi}cm$

3. Hehe thanks!! But I just have a few more questions

1) $\displaystyle \frac{88}{(77)(66)}$ has the same value as
a) $\displaystyle \frac{0.88}{(7.7)(6.6)}$
b) $\displaystyle \frac{0.88}{(0.77)(0.66)}$
c) $\displaystyle \frac{8.8}{(7.7)(6.6)}$
d) $\displaystyle \frac{8.8}{(0.77)(0.66)}$

2) If $\displaystyle \frac{2}{5}x=16-\frac{2}{3}x$, then $\displaystyle \frac{5x}{2x+10}$ equals

Just two more Hope you could help me man.

4. Originally Posted by Rocher
Hehe thanks!! But I just have a few more questions

1) $\displaystyle \frac{88}{(77)(66)}$ has the same value as
a) $\displaystyle \frac{0.88}{(7.7)(6.6)}$
b) $\displaystyle \frac{0.88}{(0.77)(0.66)}$
c) $\displaystyle \frac{8.8}{(7.7)(6.6)}$
d) $\displaystyle \frac{8.8}{(0.77)(0.66)}$
You can do this by hand with a little inspiration, but you can do this very easily with a calculator.

$\displaystyle \frac{88}{77 \cdot 66} = \frac{88}{77 \cdot 66} \cdot \frac{\frac{1}{100}}{\frac{1}{100}}$

$\displaystyle = \frac{\frac{88}{100}}{\left ( 77 \cdot \frac{1}{10} \right ) \left ( 66 \cdot \frac{1}{10} \right )}$

$\displaystyle = \frac{0.88}{7.7 \cdot 6.6}$

-Dan

5. Originally Posted by Rocher
2) If $\displaystyle \frac{2}{5}x=16-\frac{2}{3}x$, then $\displaystyle \frac{5x}{2x+10}$ equals
The simplest idea here is to solve for x and plug x into your expression:

$\displaystyle \frac{2}{5}x=16-\frac{2}{3}x$

$\displaystyle \left ( \frac{2}{5}x \right ) \cdot 15 = \left ( 16-\frac{2}{3}x \right ) \cdot 15$

$\displaystyle 6x = 240 - 10x$

$\displaystyle 16x = 240$

$\displaystyle x = \frac{240}{16} = 15$

So
$\displaystyle \frac{5x}{2x+10} = \frac{5(15)}{2 \cdot 15 + 10} = \frac{75}{40} = \frac{15}{8}$

-Dan