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Non-integer values come in two primary formats: decimals and fractions. That said, the Executive Assessment of Fractions and Decimals section is all about learning to solve non-integer problems. Let’s break it down below.

A fraction is a numerator divided by a denominator that doesn’t cancel out into an integer value. When evaluating fractions, these are rational values where the denominator divides the numerator. For instance, ¾ of x divides x by four and counts 3 of those 4 parts.

Alternatively, you can use the Greatest Common Factor to reduce a fraction to its Least Common Denominator.

For instance: 4/8 = ½ because 4/8 divided by 4/4= ½.

The above example shows what happens when you do what is commonly known as canceling out. You’re in essence dividing 1 as the numerator divided by the denominator as the same values.

Taking Fractions of Wholes

For example ¾ of 120.

120 ÷ 4 = 30

Step 2: Multiply the quotient by the numerator

30 x 3 = 90

Hence, you have an easy way of getting ¾ of 120 rather than multiplying 120 by 3 and then dividing the product by 4. This method is efficient because you’re going to work with smaller values. Since multiplication and division have pretty much the same order of operation, you can start by dividing first and then multiplying instead of processing out the numerator then the denominator with larger values.

This is one of the most effective methods used when adding, subtracting, or comparing fractions.

For example \(⅖ + 3/7 =?\)

Step 1. Multiply the denominators to get the common denominator. \(5 \times 7 = 35\) (Common denominators remain constant throughout the addition operation)

Step 2. Cross-multiply the numerators to get equivalent numerators for the two fractions.

\((2 \times 7)\) and \((3 \times 5)\). 14 and 15 are hence the two numerators.

\(14/35 + 15/35\)

Step 3. Sum the numerators only

\(14/35 + 15/35 = 29/35\)

The denominator remains constant as you add the numerator.

Hence \(⅖ + 3/7 = 29/35\)

This method is applicable for addition, subtraction, and comparison because you’ll need to find the common denominator to solve the given problems.

Multiplying and dividing fractions is easier than adding or subtracting because they don’t need a common denominator.

Multiplying: \(⅖ \times 3/7\)

This will require you to multiply straight across \(\frac{(2 \times 3)}{(5 \times 7)}\) to get \(6/35\). This is a way easier operation to go by because you’ll only need to multiply the numerators and the denominators

Division: \(⅖ ÷ 3/7\)

Multiply the first fraction with the reciprocal of the second fraction.

(Reciprocal is the inverse of the numerator and the denominator).

Rewrite the fraction as \(⅖ \times 7/3\)

Multiply the fractions straight across \((2 \times 7) ÷ (5 \times 3)\)

\(⅖ ÷ 3/7 = 14/15\)

In some instances, you will be required to convert common fractions to decimals or vice versa. This is important when you’re required to give the answer in a form that’s different from the form of the question. For instance, you might get a question in decimals that requires you to give the final answer as a fraction.

You want to be good at these conversions especially when time is limited like during an exam. Hence, you will need to convert the non-integers into a like form so that working out is faster.

Examples of common fractions and their decimal forms:

\(½ = 0.5\)

\(⅓ = 0.333, ⅔ = 0.667\)

\(¼ = 0.25, ¾ = 0.75\)

\(⅕ = 0.2\)

\(⅙ = 0.167, ⅚ = 0.833\)

\(⅛ = 0.125, ⅜ = 0.375, ⅝ = 0.625, ⅞ = 0.875\)

\(1/9 = 0.111\)

\(1/10 = 0.1\)

The first 4 fractions can be further extrapolated to solve bigger problems. For example, \(⅙\) is half of \(⅓\). Hence, \(⅙\) into decimals will be half of 0.333 which is 0.167. Likewise, \(⅗\) would be \(⅕ \times 3\). The same applies to its decimal form, \(0.2 \times 3 = 0.6\)

When required to convert such a fraction to a decimal: 25/100, you would need to consider the number of zeros in the denominator and that would give you the number of decimal points. This is because every decimal point is based off by ^10.

This would normally take longer.

For instance: \(3/11\)

You’d have to use the long-division method

0.272

11)3.0

2.2

.80

.77

.30

.22

Since 11 is bigger than 3, add one decimal point to get 3.0.

\(1.1 \times 2= 2.2\) which when subtracted from 3.0 gives \(0.8. 11 \times 0.07 = 0.77. 0.8 – 0.77 = 0.003\). To this point, the pattern repeats forever, hence it is not a terminating decimal.

Hence, \(0.272\) is the decimal form of \(3/11\).

Only divisions that involve factors of 5 and 2 will terminate. If you have any other prime number besides 5 and 2 as your denominator, you’re going to have a repetitive pattern.

Take the following number as an example:

1405.249

1 has a place value of thousands

4 – hundreds

0 – tens

5 – units of ones

2 – tenths, hence you have 2 tenths

4 – hundredths, 4 hundredths

9 – thousandths

To handle any decimal question, you will need to first off eliminate the decimals by multiplying the entire equation by a power of 10. This makes it easy to operate the number when it is an integer rather than when it’s not.

\(0.2y = 4x\)

Multiply the whole equation by 10

\(2y = 40x\)

Hence \(y = 20x\)

Use scientific notation to manipulate decimal expressions (multiplying the decimal number by a negative power)

\(0.0009 = 0.0001(9)\)

\(0.0001 = 1/1000 = 10^{-4}\)

The number hence becomes \(10^{-4} \times 9\) because every negative power shifts the decimal to the left. This becomes a simpler way of manipulating a complex non-integer.

Fractions and decimals can be further exploited as concepts in the executive assessment.

For instance, if given the following question:

Which of the following values is greatest?

A). \(⅔\)

B). \(⅗\)

C). \(5/7\)

D). \(9/13\)

E). \(11/17\)

In such a case, you’d need to use elimination or estimation to eliminate some of the fractions. You have common fractions such as ⅔ and ⅗.

⅔ = 0.667 while ⅗ is 0.6

Choice B is eliminated because it is smaller than choice A.

For the remaining fractions, converting them into decimals would take longer than the Bowtie method.

Comparing choices C and D:

\(5/7 \times 9/13\)

The common denominator becomes \(7 \times13 = 91\)

\(5/7 \times 91\) = 65, \(9/13 \times 91= 63\)

Since choice D is smaller than C, you’re now left with choices C and E

Using the Bowtie method again:

\(5/7\) and \(11/17\)

The common denominator is \(119 (7 \times 17)\)

\(5/7 \times \(117 = 85\) while \(11/17 \times 119 = 77\)

Hence, \(11/17\) is ruled out.

The greatest value in this question hence becomes 5/7 (choice C).

Therefore, consider the properties of fractions and decimals to get a seamless approach to handling non-integer values on the executive assessment by practicing some problems on your own.

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