Guessing how many times a smaller number goes into bigger number

Question

For example when diving 105 / 148. After you add a number 0 to the numerator, the division becomes 1050 / 148.

The answer becomes a decimal with 1050 / 148. The two numbers are not divisible by a common number so the first step i have to do is guess how many times 148 goes into 1050.

My approach is to round 148 to 200 and since 200 * 5 = 1000 the first number in the quotient must be above 5. I then had to guess 148 * 5, then 148 * 6, then 148 * 7. I'm studying for a standardized test that does not allow calculators or else i would just use a calculator.

Is there a better or smarter strategy to guessing how many times 148 goes into 1050 other than guessing like i did?

Answer 1

The strategy depends on the experience someone has. I would round up 148 to 150. And then divide 1050 by 150. The result is 7. Therefore $\frac{105}{148}\approx 0.7$. To round up 148 to 200 is too imprecise, but calculating with 1050 instead of 105 was good idea.

Another possibility is to divide 1000 (rounded down) by 150 (rounded up). The fraction becomes $\frac{1000}{150}=\frac{100}{15}=\frac{20}{3}\approx 7$, because $21/3=7$. Dividing the result by 10 is 0.7.

Answer 2

For this particular problem:

I recognize $105$ as $3 \times 5 \times 7$.

Then I wonder: Are any of these factors shared by $148$?

No, unfortunately not: We can quickly see the latter is not divisible by $3$ or $5$.

However, its predecessor $147 = 7 \times 21 = 3 \times 7 \times 7$.

So: I might just estimate by replacing the denominator:

$$\frac{105}{148} \approx \frac{105}{147} = \frac{3 \times 5 \times 7}{3 \times 7 \times 7} = \frac{5}{7}$$

If you happen to know that $\frac{1}{7} = 0.\overline{142857}$, then you might recognize $\frac{5}{7}$ as just over $0.71$.

Finally: Since we decreased the denominator, the original ratio is a bit less than our adjusted one. Since our adjusted ratio is just over $0.71$, this seems like a pretty good guess.

(Indeed: $\frac{105}{148} = 0.709459\ldots$)