How to divide integers

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Algebra › How to divide integers

Questions 1 - 10

1

Divide the numbers: $98\div 32$

$\frac{49}{16}$

$\frac{11}{4}$

$\frac{32}{7}$

$\frac{12}{5}$

$\frac{27}{8}$

Explanation

Rewrite the expression into a fraction.

$98\div 32 = \frac{98}{32}$

Write the numbers in terms of their common factors.

$\frac{98}{32} =\frac{2\times 49}{2 \times 16}$

Cancel the common terms. There are no common factors with the numbers 49 and 16.

The answer is: $\frac{49}{16}$

2

Divide sixty with twenty five.

$\frac{12}{5}$

$\frac{5}{2}$

$\frac{24}{11}$

$\frac{15}{4}$

$\frac{25}{3}$

Explanation

Write the expression as a fraction.

$\frac{60}{25}$

Rewrite the numerator and denominator with common factors.

$\frac{60}{25} = \frac{12\times 5}{5 \times 5}$

Cancel the common terms on the top and bottom.

The answer is: $\frac{12}{5}$

3

Divide: $48 \div 14$

$\frac{24}{7}$

$\frac{17}{4}$

$\frac{14}{3}$

$\frac{13}{4}$

Explanation

Rewrite the expression as a fraction.

$\frac{48}{14}$

Write out the multiples of the numerator and denominator.

$\frac{48}{14} =\frac{2 \times 3\times 8}{2 \times 7}$

The twos can be cancelled. Simplify what remains of the fraction.

$\frac{3\times 8}{7}= \frac{24}{7}$

The answer is: $\frac{24}{7}$

4

Divide: $\frac{81}{18}$

$\frac{9}{2}$

$\frac{3}{2}$

$\frac{12}{7}$

$\frac{8}{3}$

$\frac{9}{4}$

Explanation

Instead of using long division to solve, we can rewrite the numerator and denominator by their common factors.

$\frac{81}{18} = \frac{3\times 3 \times 3 \times 3}{3\times 3 \times 2}$

Cancel the common terms in the numerator and denominator.

Write the remaining terms.

$\frac{3 \times 3}{2} = \frac{9}{2}$

The answer is: $\frac{9}{2}$

5

Divide eighty four with sixteen.

$\frac{21}{4}$

$\frac{26}{3}$

$\frac{7}{2}$

$\frac{16}{3}$

$\frac{23}{4}$

Explanation

Write the problem statement with an expression as a fraction.

$\frac{84}{16}$

Rewrite the numerator and denominator by common factors. Both numbers are divisible by four.

$\frac{84}{16} = \frac{4\times 21}{4\times 4}$

Notice that we can cancel the common fours on the top and bottom.

The answer is: $\frac{21}{4}$

6

Divide the integers: $38 \div 14$

$\frac{19}{7}$

$\frac{9}{7}$

$\frac{11}{7}$

$\frac{7}{2}$

Explanation

Rewrite the expression as a fraction.

$38 \div 14 = \frac{38}{14}$

Rewrite the fraction by common factors of two.

$\frac{38}{14} = \frac{2 \times 19}{2\times 7}$

Divide and cancel the twos, which will become one.

The answer is: $\frac{19}{7}$

7

Divide the following integers: $1080\div 320$

$\frac{27}{8}$

$\frac{3}{2}$

$\frac{9}{2}$

Explanation

Rewrite the expression as a fraction.

$\frac{1080}{320}$

Cancel out the zeros on the ones place. This is the same as dividing both numbers by ten.

The result is: $\frac{108}{32}$

Both numbers are divisible by two. Write the fractions in terms of their common factors.

$\frac{108}{32} = \frac{2\times 54}{2\times 16} = \frac{2\times 2\times 27}{2\times 2\times 8}$

The common twos can be cancelled on the numerator and denominator.

The answer is: $\frac{27}{8}$

8

Divide: $\frac{32}{84}$

$\frac{8}{21}$

$\frac{4}{11}$

$\frac{4}{13}$

$\frac{8}{23}$

$\frac{8}{19}$

Explanation

Write the factors of the two numbers.

$\frac{32}{84}= \frac{4\times 8 }{4\times 21}= \frac{8}{21}$

9

Divide $126 \div 36$ .

$\frac{7}{2}$

$\frac{21}{8}$

$\frac{14}{3}$

$\frac{23}{8}$

$\frac{13}{3}$

Explanation

Rewrite the expression in fraction form.

$126 \div 36= \frac{126}{36}$

Write out the common factors of both the numerator and denominator.

$\frac{126}{36} = \frac{6\times 21}{6\times 6} =\frac{6\times 3\times 7}{6\times 3 \times 2}$

We can then see that the fraction can be reduced to $\frac{7}{2}$ without having to conduct long division.

The answer is $\frac{7}{2}$ .

10

If a number is divisible by , which of the following is always true?

I. Divisible by

II. Divisible by

III. Divisible by

I and II

I only

II only

All true

None true

Explanation

If a number is divisble by some number , then it is also divisible by all of its factors. Hence, if a number is divisible by , it is also divisible by , and . Thus, I and II are both true.

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