How to find decimal fractions - GRE Quantitative Reasoning

With the numerator having more decimal spots than the denominator, we need to move the decimal point in the numerator two places to the right.

Then in the denominator, we move the decimal point also two to the right. Since there's only one decimal place we just add one more zero.

Then we can reduce by dividing top and bottom by .

Since there are four decimal places, we shift the decimal point in the numerator four places to the right.

For the denominator, since there is no decimal point, we just add four more zeroes.

Then reduce by dividing top and bottom by .

We need to convert the sentence into a math expression. Anytime there is "of" means we need to multiply. Let's first convert the decimal to a fraction. We need to move the decimal point two places to the right.

Since is the same as we can add two more zeroes to the denominator.

We can reduce the to a and the to a .

Then reduce the to and the to .

.

Then dividing into and we get .

We need to convert this sentence into a math expression. Anytime there is "of" in a sentence it means we need to multiply. Let's convert into a decimal which is .

Thus our mathematical expression becomes:

.

Divide both sides by .

Move decimal point two places to the right. The numerator will become . Then simplify by dividing top and bottom by .

Let's convert the decimal into a fraction.

If we multiply everything by , we should have an easier quadratic.

Remember, we need to find two terms that are factors of the c term that add up to the b term.

This is the only value.

Let's actually simplify the top of the fraction. divides into .

We should have:

.

Then move the decimal two spots to the right and add two zeroes to the denominator.

Let's actually multiply top and bottom by to get: .

Now we want to eliminate those zeroes. By dividing, the decimal point in the numerator moves to the left three places to get an answer of or .

Since each decimal has two digits, we can convert easily to integers.

Then multiply top and bottom by to get:

is reduced to and is reduced to

Then and can be divided by to get and respectively.

Let be . Let's multiply that value by . The reason is when we subtract it, we will get us an integer instead and the repeating decimals will disappear.

If we subtract, we get .

Divide both sides by and we get .

If you divide by on top and bottom, you should get the answer. Otherwise, just divide top and bottom by three times based on the divisibility rules for . If the sum is divisible by , then the number is divisible by .

To begin, it can be useful to convert the values in the fraction

into a modified scientific notationnotation:

Now multiply the ten terms (adding exponents together) and the non-ten terms:

From here, reduce the terms, subtracting the bottom tens exponent from the top tens exponent:

To compare these two quantities, we'll want to simplify Quantity A.

The fraction

may be a bit daunting; let's convert it to scientific notation:

Now multiply the non-ten terms, and the ten terms (add the exponents together):

Now cancel like factors in the numerator and denominator:

The two quantities are equal.

For this problem, realize that the store cannot buy part of a crate of socks. If they only have enough to pay for part of a crate, they might as well not have any money at all.

For the amount of money listed, figure out how many crates can be purchased:

Quantity A

So six crates can be purchased.

Quantity B:

Not quite enough for seven; only six crates can be purchased.

The two quantities are equal.

With the numerator having more decimal spots than the denominator, we need to move the decimal point in the numerator two places to the right.

Then in the denominator, we move the decimal point also two to the right. Since there's only one decimal place we just add one more zero.

Then we can reduce by dividing top and bottom by .

Whenever there are decimals in fractions, we remove them by shifting the decimal place over however many it takes to make number an integer.

In this case we have to move the decimal in the numerator to the right one place.

Then, we add just one zero to the denominator.

Final answer becomes:

.

Since there are four decimal places, we shift the decimal point in the numerator four places to the right.

For the denominator, since there is no decimal point, we just add four more zeroes.

Then reduce by dividing top and bottom by .

We need to convert the sentence into a math expression. Anytime there is "of" means we need to multiply. Let's first convert the decimal to a fraction. We need to move the decimal point two places to the right.

Since is the same as we can add two more zeroes to the denominator.

We can reduce the to a and the to a .

Then reduce the to and the to .

.

Then dividing into and we get .

We need to convert this sentence into a math expression. Anytime there is "of" in a sentence it means we need to multiply. Let's convert into a decimal which is .

Thus our mathematical expression becomes:

.

Divide both sides by .

Move decimal point two places to the right. The numerator will become . Then simplify by dividing top and bottom by .

Let's convert the decimal into a fraction.

If we multiply everything by , we should have an easier quadratic.

Remember, we need to find two terms that are factors of the c term that add up to the b term.

This is the only value.

Let's actually simplify the top of the fraction. divides into .

We should have:

.

Then move the decimal two spots to the right and add two zeroes to the denominator.

Let's actually multiply top and bottom by to get: .

Now we want to eliminate those zeroes. By dividing, the decimal point in the numerator moves to the left three places to get an answer of or .

Since each decimal has two digits, we can convert easily to integers.

Then multiply top and bottom by to get:

is reduced to and is reduced to

Then and can be divided by to get and respectively.