Chapter 12 — Another Peek Beyond the Point

The chapter is titled 'Another Peek Beyond the Point' and covers multiplication and division of decimals, extending the place-value procedures students already know for counting numbers to work with decimal numbers. It also applies decimal arithmetic to understand the design of leap years in the Gregorian calendar.

Remove the decimal points and multiply the resulting whole numbers. Then count the total number of digits after the decimal point in both the multiplier and multiplicand — place the decimal point in the product so that it has exactly that many decimal digits. For example, 5.96 × 24.8: since 596 × 248 = 147808 and there are 2 + 1 = 3 decimal places total, the answer is 147.808.

When dividing a decimal by 10, 100, 1000, etc., simply move the decimal point to the left by as many places as there are zeroes in the divisor. For example, 18.7 ÷ 10 = 1.87, and 18.7 ÷ 100 = 0.187.

Convert the decimal divisor into a whole number by multiplying both the dividend and the divisor by the same power of 10. For example, 4.68 ÷ 0.13 becomes 468 ÷ 13 (multiplying both by 100), and then you apply the standard long-division procedure.

It is the standard long-division procedure extended past the ones place: after dividing the ones, regroup any remainder ones into tenths, then regroup remainder tenths into hundredths, and so on. Each time you move from ones to tenths in the quotient, place a decimal point there. For instance, 237 ÷ 8 = 29.625.

No. If both decimals are greater than 1, the product is greater than both. If both are between 0 and 1, the product is less than both. If one is between 0 and 1 and the other is greater than 1, the product is less than the larger number but greater than the smaller one — for example, 0.25 × 8 = 2, which is greater than 0.25 but less than 8.

A non-terminating decimal is a quotient that never ends because each step of long division leaves a non-zero remainder. The chapter shows that 10 ÷ 3 = 3.333… because every step leaves a remainder of 1, so the process never stops.

142857 is the repeating block of digits that appears when you divide 1 by 7 (1 ÷ 7 = 0.142857142857…). It is called a cyclic number because multiplying it by any number from 1 to 6 gives the same digits cycled around in a different order. The chapter notes that dividing 1 by 17 gives another such cyclic number, and the Austrian mathematician Emil Artin conjectured in 1927 that there are infinitely many such numbers — a question that remains unsolved.

Earth takes 365.2422 days to orbit the Sun. Over 100 calendar years of 365 days each, Earth falls 24.22 days behind. To compensate, one extra day is added every fourth year (a leap year), giving 365 calendar days × 100 years + 25 extra days = 36,525 calendar days, while Earth actually needs 36,524.22 days. This slight over-compensation led to further rules: skip the leap day in century years (divisible by 100) but restore it for years divisible by 400.

Using the full Gregorian leap-year rules, there are (750 × 365) + (240 × 366) + (8 × 365) + (2 × 366) = 365,242 calendar days in 1000 years, compared to 365,242.2 days that Earth actually needs — a difference of just 0.2 days.

Find an equivalent fraction whose denominator is of the form 1, 10, 100, or 1000. For example, 1/2 = 5/10 = 0.5, and 29/4 = 725/100 = 7.25 (multiply numerator and denominator by 25, since 4 × 25 = 100). If no such conversion is possible (e.g., denominator 3), use the long-division procedure instead.

Yes, the PDF is completely free to view and download on cbseprepmaster.com. No sign-up or account is required.