GRE Quantitative Reasoning - How to find a solution to a compound fraction

can be rewritten as times what?

$10^{14}$

$10^{11}$

$10^{12}$

$10^{13}$

$10^{15}$

Explanation

To solve this problem, realize that a decimal may be placed at the very end of this integer:

Now, count how many spaces the decimal will need to move to the left to reach:

It must move a total of spaces, so:

$3.83490102847190(10^{14})=383490102847190$

Simplify: $\frac{\frac{x}{2}+\frac{x}{3}+\frac{x}{4}-\frac{x}{5}}{6}$

$\frac{53}{360}x$

$\frac{77}{360}x$

$\frac{53}{60}x$

$\frac{77}{60}x$

$\frac{77}{10}x$

Explanation

To solve this problem, begin with simplifying the numerator. This can be done by first finding a common denominator. For

$\frac{x}{2}+\frac{x}{3}+\frac{x}{4}-\frac{x}{5}}$

a common denominator would be :

$\frac{30}{30}(\frac{x}{2})+\frac{20}{20}(\frac{x}{3})+\frac{15}{15}(\frac{x}{4})-\frac{12}{12}(\frac{x}{5}})$

$\frac{30x}{60}+\frac{20x}{60}+\frac{15x}{60}-\frac{12x}{60}}$

Which combines into:

$\frac{53}{60}x$

But recall that this is just the numerator, and there is still a in the denominator:

$\frac{\frac{53}{60}x}{6}$

So, the final answer is:

$\frac{53}{360}x$

is a repeating decimal. What digit is in the $57^{th}$ place?

Explanation

Examination of the value reveals that after the sequence , the decimal repeats, and that the sequence has a length of eight values.

A longer answer would be to write the sequence out and count down the digits until the value was found. However, this is a time-consuming process and one that is prone to error.

Rather, notice how since the numbers repeat, it's possible to skip most of the counting:

$1^{st}$ digit:

$9^{th}$ digit:

And so on. Since is the closest multiple to , we can subtract the two to find the digit that matches the $57^{th}$ digit.

$57^{th}-56^{th}=1^{st}$

So the $57^{th}$ digit is .

Simplify the following:

$\frac{z^5-9z^4+27z^3-27z^2}{z^3-6z^2+9z}$

Explanation

Looking at this equation, note that since all terms in the numerator and denominator contain a , it is possible to rewrite it as follows:

$\frac{z^5-9z^4+27z^3-27z^2}{z^3-6z^2+9z}=\frac{z^2(z^3-9z^2+27z-27z)}{z(z^2-6z+9)}$

$\frac{z(z^3-9z^2+27z-27z)}{(z^2-6z+9)}$

Now the parenthetical terms must be addressed. The problem statement and answer choices give a clue that they are some sort of multiples of .

In fact, Pascal's triangle reveals that the top and bottom are the cube and square of this term respectively:

$\frac{z(z-3)^3}{(z-3)^2}$

Cancelling terms, we are left with:

Which of the following is equal to $\frac{1}{4}$ of the reciprocal of percent?

Explanation

The first step will be to find the reciprocal of percent. Note that as a percent, this should be converted to a decimal form: .

The reciprocal of a number is given by divided by that number, so the reciprocal of is given as:

$\frac{1}{.00025}= 4000$

Therefore:

$(\frac{1}{4})4000=1000$

Edward rolls three dice; two are six-sided, and one is twenty-sided, with 1 through 6 represented on the six-sided dice, and 1 through 20 represented on the twenty-sided die.

What is the probability that the sum of his roll will equal 5?

$\frac{1}{120}$

$\frac{1}{144}$

$\frac{1}{36}$

$\frac{1}{240}$

$\frac{1}{720}$

Explanation

The first step will be to calculate how many total different potential rolls are possible. This is given by the product of the number of possible rolls for each of the dice: 6, 6, 20:

$Number of Possible Rolls = 6 \cdot 6\cdot 20=720$

Next it is necessary to account for the total number of rolls that will sum up to 5. Writing them out can help, such as in the format of:

$(1^{st} die, 2^{nd} die,3^{rd} die)$

Knowing that there are 6 different rolls that sum up to 5, the probability of rolling a 5 can be found by taking the number of events that satisfy this sum and dividing it by the number of total possible events:

$Probability_{Rolling 5} =\frac{6}{720} = \frac{1}{120}$