Question

4 Solve the given division problem using repeated subtraction.
a. 33÷11
b. 70÷14
c.56÷8
d. 48÷12
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Answer 1

Answer:

a. 3

b. 5

c. 7

d. 4

Step-by-step explanation:

Repeated Subtraction is a method that subtracts the equal number of items from a group, also known as division. Using this method, the same number is subtracted repeatedly from another larger number until the remainder is zero, or smaller than the number being subtracted.

a. $\frac{33}{11}$

Subtract the numerator from the denominator repeatedly,

$33-11=22$

$22-11=11$

$11-11=0$

To get the remainder zero, we subtracted the number three times.

So the value of the division will be $3$.

⇒$\frac{33}{11}=3$

b. $\frac{70}{14}$

Subtract the numerator from the denominator repeatedly,

$70-14=56\\56-14=42\\42-14=28\\28-14=14\\14-14=0$

To get the remainder zero, we subtracted the number five times.

So, the value of the division will be $5$.

⇒$\frac{70}{14}=5$

c. $\frac{56}{8}$

Subtract the numerator from the denominator repeatedly,

$56-8=48\\48-8=40\\40-8=32\\32-8=24\\24-8=16\\16-8=8\\8-8=0$

To get the remainder zero, we subtracted the number seven times.

So, the value of the division will be $7$.

⇒$\frac{56}{8}=7$

d. $\frac{48}{12}$

Subtract the numerator from the denominator repeatedly,

$48-12=36\\36-12=24\\24-12=12\\12-12=0$

To get the remainder zero, we subtracted the number four times.

So, the value of the division will be $4$.

⇒$\frac{48}{12}=4$

Answer 2

Given,

We have to solve the given division problems by using repeated subtraction method.

• Repeated Subtraction is a method that subtracts the equal number of items from a group, also known as division.
• Using this method, the same number is subtracted repeatedly from another larger number until the remainder is zero, or smaller than the number being subtracted.

a) $\frac{33}{11}$

Subtract $11$ from $33$ repeatedly until we get zero

$33-11=22$

$22-11=11$

$11-11=0$

Therefore, to get the remainder zero, we have subtract the given number three times

b) $\frac{70}{14}$

Subtract $14$ from $70$ repeatedly until we get zero

$70-14=56$

$56-14=42$

$42-14=28$

$28-14=14$

$14-14=0$

Therefore, to get the remainder zero, we have subtract the given number five times

c) $\frac{56}{8}$

Subtract $8$ from $56$ repeatedly until we get zero

$56-8=48$

$48-8=40$

$40-8=32$

$32-8=24$

$24-8=16$

$16-8=8$

$8-8=0$

Therefore, to get the remainder zero, we have subtract the given number seven times

d) $\frac{48}{12}$

Subtract $12$ from $48$ repeatedly until we get zero

$48-12=36$

$36-12=24$

$24-12=12$

$12-12=0$

Therefore, to get the remainder zero, we have subtract the given number four times.