Loading...
Which of the following fractions has a decimal equivalent that is a terminating decimal?
Let's start with understanding what we're looking for. A terminating decimal is simply a decimal that ends - like \(0.5\) or \(0.25\) or \(0.125\). This is different from a repeating decimal like \(0.333...\) or \(0.166...\) that goes on forever.
Here's the key insight: A fraction will give us a terminating decimal only when its denominator (after we've simplified the fraction completely) can be built using only the numbers 2 and 5 multiplied together. Why? Because our decimal system is based on 10, and \(10 = 2 \times 5\). So denominators that are made from powers of 2 and 5 work nicely with our decimal system.
For example:
Process Skill: TRANSLATE - Converting the concept of "terminating decimal" into a concrete rule about denominators
Before we can check the denominators, we need to make sure each fraction is in its simplest form. This means finding any common factors between the top and bottom numbers and canceling them out.
Let's work through each option:
Option A: \(\frac{10}{189}\)
We need to find the greatest common divisor of 10 and 189.
\(10 = 2 \times 5\)
\(189 = 3^3 \times 7 = 27 \times 7\)
Since 10 and 189 share no common factors, \(\frac{10}{189}\) is already in lowest terms.
Option B: \(\frac{15}{196}\)
\(15 = 3 \times 5\)
\(196 = 14^2 = (2 \times 7)^2 = 4 \times 49 = 2^2 \times 7^2\)
Since 15 and 196 share no common factors, \(\frac{15}{196}\) is already in lowest terms.
Option C: \(\frac{16}{225}\)
\(16 = 2^4\)
\(225 = 15^2 = (3 \times 5)^2 = 3^2 \times 5^2\)
Since 16 and 225 share no common factors, \(\frac{16}{225}\) is already in lowest terms.
Option D: \(\frac{25}{144}\)
\(25 = 5^2\)
\(144 = 12^2 = (4 \times 3)^2 = (2^2 \times 3)^2 = 2^4 \times 3^2\)
Since 25 and 144 share no common factors, \(\frac{25}{144}\) is already in lowest terms.
Option E: \(\frac{39}{128}\)
\(39 = 3 \times 13\)
\(128 = 2^7\)
Since 39 and 128 share no common factors, \(\frac{39}{128}\) is already in lowest terms.
Now we need to look at each denominator and see what prime numbers it's made of. Remember, for a terminating decimal, the denominator can ONLY contain factors of 2 and 5.
Option A: Denominator = \(189 = 3^3 \times 7\)
This contains 3 and 7, so it will NOT give a terminating decimal.
Option B: Denominator = \(196 = 2^2 \times 7^2\)
This contains 7, so it will NOT give a terminating decimal.
Option C: Denominator = \(225 = 3^2 \times 5^2\)
This contains 3, so it will NOT give a terminating decimal.
Option D: Denominator = \(144 = 2^4 \times 3^2\)
This contains 3, so it will NOT give a terminating decimal.
Option E: Denominator = \(128 = 2^7\)
This contains ONLY powers of 2, so it WILL give a terminating decimal!
Process Skill: INFER - Recognizing that only denominators with factors of 2 and 5 exclusively will work
Looking at our analysis, only Option E has a denominator that contains exclusively factors of 2 (and no factors of 5, which is also allowed).
Let's verify: \(\frac{39}{128} = 39 \div 128\)
Since \(128 = 2^7\), when we perform this division, we'll get a terminating decimal.
\(\frac{39}{128} = 0.3046875\) (this decimal terminates!)
All other options contain prime factors other than 2 and 5 in their denominators, so they will produce repeating decimals.
The answer is E. \(\frac{39}{128}\)
This fraction has a denominator of \(128 = 2^7\), which contains only powers of 2. Therefore, it will produce a terminating decimal when converted to decimal form.
Faltering Point 1: Students may confuse terminating and repeating decimals, thinking that any fraction with a "nice looking" numerator or denominator will automatically terminate. For example, they might assume \(\frac{25}{144}\) terminates because both 25 and 144 are perfect squares, without understanding the core rule about prime factorization.
Faltering Point 2: Students often misremember or misapply the terminating decimal rule. They might think the rule is about the numerator and denominator both being even, or that any denominator that's a power of 10 works, rather than understanding that the denominator must contain ONLY factors of 2 and 5.
Faltering Point 3: Students may forget that fractions need to be reduced to lowest terms before applying the terminating decimal test. They might analyze the original fraction without simplifying first, which could lead to incorrect conclusions about the prime factorization of the effective denominator.
Faltering Point 1: Students frequently make errors in prime factorization, especially with larger numbers. For example, they might incorrectly factor 189 as \(9 \times 21\) instead of recognizing it as \(3^3 \times 7\), or miscalculate \(196 = 14^2 = 2^2 \times 7^2\) by dropping exponents or missing prime factors.
Faltering Point 2: When checking if numbers share common factors for reducing fractions, students may miss common divisors or incorrectly identify common factors that don't exist. This leads to working with non-simplified fractions and analyzing the wrong denominator for the terminating decimal test.
Faltering Point 3: Students may correctly identify prime factors but incorrectly conclude whether the fraction terminates. For instance, seeing that \(225 = 3^2 \times 5^2\), they might focus only on the \(5^2\) and conclude it terminates, forgetting that the presence of ANY prime factor other than 2 or 5 makes the decimal non-terminating.
No likely faltering points - once students have correctly identified which denominator contains only factors of 2 and/or 5, the answer selection is straightforward since only one option will satisfy this condition.