Mathematics · Professor Pi

Every card in Number, Year 10: standard form, recurring decimals and error intervals

The whole deck, in order — so you can read it through before your child ever sees it.

1 GCSE-MATH-NUM-0016

Work out (3 × 10⁴) × (2 × 10⁵). Give your answer in standard form.

6 × 10⁹

HintDeal with the front numbers and the powers of ten separately.

WhyMultiply the front numbers: 3 × 2 = 6. Multiply the powers of ten by adding the indices: 10⁴ × 10⁵ = 10⁹. The order of a multiplication can be rearranged, which is why the two parts can be handled apart.

2 GCSE-MATH-NUM-0017

48 × 10⁵ is not in standard form, because 48 is not between 1 and 10. Written correctly, it is 4.8 × 10 to the power ____.

6

HintMaking the front number ten times smaller has to be balanced in the power of ten.

Why48 = 4.8 × 10, so 48 × 10⁵ = 4.8 × 10 × 10⁵ = 4.8 × 10⁶. The value has not changed (4 800 000 both ways); only how it is shared between the front number and the power.

3 GCSE-MATH-NUM-0018

Work out (2 × 10⁸) ÷ (8 × 10³). Give your answer in standard form.

2.5 × 10⁴

HintDivide the front numbers, divide the powers of ten, then check the front number is at least 1.

Why2 ÷ 8 = 0.25 and 10⁸ ÷ 10³ = 10⁵, giving 0.25 × 10⁵. That is not standard form, because 0.25 is less than 1. Make the front number ten times bigger and the power one smaller: 2.5 × 10⁴ (which is 25 000).

4 GCSE-MATH-NUM-0019

Work out (3 × 10⁵) + (4 × 10⁴). Give your answer in standard form.

3.4 × 10⁵

HintThe two numbers have different powers of ten, so the front numbers cannot simply be combined.

WhyWrite both with the same power of ten, or as ordinary numbers: 300 000 + 40 000 = 340 000 = 3.4 × 10⁵. Adding the front numbers only works when the powers of ten already match.

5 GCSE-MATH-NUM-0020

After a calculation, a calculator or spreadsheet shows 4.2E7. This is its way of showing 4.2 × 10⁷. Write the value as an ordinary number. (digits only, no spaces or commas)

42000000

HintThe power says how many places the digits move, not how many zeros to write.

Why4.2 × 10⁷ means 4.2 multiplied by 10 seven times. The digits move seven places, and one of those places is filled by the 2, so six zeros follow: 42 000 000. Other calculators show the same thing as 4.2 ×10 with a small raised 7.

6 GCSE-MATH-NUM-0021

A decimal in which a digit, or a block of digits, repeats for ever without ending

A recurring decimal

HintThe word describes something that keeps coming back, like a dream you have again and again.

Why1/3 = 0.3333… and 3/11 = 0.272727… are examples. A decimal that stops, such as 0.375, is called terminating. Every fraction gives one kind or the other.

7 GCSE-MATH-NUM-0022

A decimal is written as 0.16 with a dot above the 6 only. Write the number out to six decimal places, without rounding. (type 0. followed by six digits)

0.166666

HintOnly the digit under the dot repeats.

WhyA dot above a single digit means that digit repeats for ever, and digits with no dot appear once. So the 1 appears once and the 6 goes on: 0.16666… This is the decimal for 1/6.

8 GCSE-MATH-NUM-0023

When a block of several digits repeats, such as 0.142857142857…, dots are written above only the first and the ____ digit of the repeating block.

last

HintThe two dots work like a pair of bookends.

WhyFor 0.142857142857… the dots go above the 1 and the 7, marking where the repeating block starts and stops. Dots above the 3 and the 2 of 0.312 would mean 0.312312312…

9 GCSE-MATH-NUM-0024

Use division to write 2/11 as a decimal. Which digits recur?

0.181818…, with the digits 1 and 8 recurring

HintCarry out 2 ÷ 11 as a short division and watch for a remainder you have already had.

Why20 ÷ 11 = 1 remainder 9, then 90 ÷ 11 = 8 remainder 2, and the remainder 2 is where the division began, so the digits 1, 8 repeat for ever. In dot notation there is a dot above the 1 and a dot above the 8.

10 GCSE-MATH-NUM-0025

Which one of these fractions gives a recurring decimal: 3/8, 5/12 or 7/20?

5/12

HintLook at the prime factors of each denominator.

WhyA fraction in its simplest form terminates only if its denominator has no prime factors other than 2 and 5, the factors of 10. 8 = 2 × 2 × 2 and 20 = 2 × 2 × 5 pass, but 12 = 2 × 2 × 3 contains a 3. So 5/12 = 0.41666…, while 3/8 = 0.375 and 7/20 = 0.35.

11 GCSE-MATH-NUM-0026

A length, L cm, is 7.4 cm when rounded to 1 decimal place. Write the error interval for L.

7.35 ≤ L < 7.45

HintGo half of one tenth below the stated value and half of one tenth above it.

WhyRounding to 1 decimal place means to the nearest 0.1, so the true value can be up to 0.05 away on either side. Anything from 7.35 up to, but not including, 7.45 rounds to 7.4.

12 GCSE-MATH-NUM-0027

A number, n, is truncated to 1 decimal place. The result is 3.6. Write the error interval for n.

3.6 ≤ n < 3.7

HintChopping digits off can never make a positive number bigger.

WhyTruncating simply removes the digits after the first decimal place, so 3.6, 3.65 and 3.699 all become 3.6. The smallest possible value is 3.6 itself and the number must stay below 3.7.

13 GCSE-MATH-NUM-0028

Cutting a number off after a chosen digit, simply dropping the digits that follow, with no rounding up

Truncating the number

HintA related word describes a cone with its top sliced off.

WhyTruncated to 1 decimal place, 4.78 becomes 4.7, whereas rounding would give 4.8. Truncating always gives a value that is equal to or below the original positive number.

14 GCSE-MATH-NUM-0029

The error interval for a length rounded to 7.4 cm is 7.35 ≤ L < 7.45. Why is the sign at the upper end "<" and not "≤"?

Because 7.45 itself would round up to 7.5

HintTry rounding the top value of the interval to 1 decimal place.

WhyA 5 in the next place rounds up, so 7.45 belongs with 7.5. The length can be 7.449 or 7.4499, as close to 7.45 as you like, but never equal to it. The lower end, 7.35, does round to 7.4, so it is included.

15 GCSE-MATH-NUM-0030

The mass of a parcel is 250 g, measured to the nearest 10 g. What is the smallest mass, in grams, that the parcel could have? (number only)

245

HintHalve the size of the measuring step, then take it away.

WhyTo the nearest 10 g, the true mass can be up to 5 g either side of 250 g. The least it can be is 250 − 5 = 245 g, and it must be less than 255 g. These limits of accuracy are the two ends of the error interval 245 ≤ m < 255.

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