-
Non-calculator 4 marks(a)
Find the value of \(83 \times 100\)
(b)Find the value of \(92000 \div 1000\)
(c)Find the value of \(7.87 \times 10\)
(d)Find the number that goes in the box.
\(\square \times 100 = 5900\)
Show answer
(a) \(8300\)
(b) \(92\)
(c) \(78.7\)
(d) \(59\)Multiplying by 100 moves every digit two columns to the left, and two zeros fill the empty columns.
\(83 \times 100 = 8300\)
Dividing by 1000 moves every digit three columns to the right, so the three zeros come off the end.
\(92000 \div 1000 = 92\)
Multiplying by 10 moves every digit one column to the left, so the decimal point ends up one place further along.
\(7.87 \times 10 = 78.7\)
Multiplying and dividing undo each other, so divide by 100 instead.
\(5900 \div 100 = 59\)
-
Non-calculator 3 marks
One pack holds 516 tiles.
A builder buys 29 packs.
Find the total number of tiles.
Show answer
\(14964\)Set the calculation out as a long multiplication.
Multiply by the units digit: \(516 \times 9 = 4644\)
Multiply by the tens digit: \(516 \times 20 = 10320\)
Add the two rows together:
\(4644 + 10320 = 14964\)
- M1\(516 \times 9 = 4644\)
- M1\(516 \times 20 = 10320\)
- A1\(4644 + 10320 = 14964\)
-
Non-calculator 4 marks(a)
Work out \(1236 \div 6\)
You must use a written method.
(b)Work out \(1224 \div 24\)
You must use a written method.
Show answer
(a) \(206\)
(b) \(51\)Set the division out using the bus stop method and work from the left.
6 does not go into 1, so carry it across: 6 into 12 goes 2, remainder 0.
6 into 3 goes 0, remainder 3.
6 into 36 goes 6, remainder 0.
\(1236 \div 6 = 206\)
Check by multiplying back: \(206 \times 6 = 1236\)
Take away whole lots of 24 until nothing is left.
\(24 \times 50 = 1200\), and \(1224 - 1200 = 24\)
\(24 \times 1 = 24\), and \(24 - 24 = 0\)
Altogether that is \(50 + 1 = 51\) lots of 24.
\(1224 \div 24 = 51\)
- (a) M1correct short division method, first quotient digit \(2\) with remainder \(0\) carried on
- (a) A1answer \(206\)
- (b) M1subtracting multiples of \(24\), such as \(1200\) from \(1224\), leaving \(24\)
- (b) A1answer \(51\)
-
Non-calculator 3 marks
A factory is packing pencils.
There are 87 pencils altogether. Each packet holds 6 pencils.
(a)Calculate how many packets can be filled completely.
(b)Calculate how many packets are needed to hold all 87 pencils.
Show answer
(a) \(14\ \text{packets}\)
(b) \(15\ \text{packets}\)Divide the number of pencils by the number that fits in one packet:
\(87 \div 6 = 14\) remainder 3
So 14 packets are filled completely, and 3 pencils are left over.
The 3 pencils left over still have to be packed, so one more packet is needed.
\(14 + 1 = 15\)
This is the same division as part (a), but the remainder is dealt with the other way round: 14 packets are full, and 15 packets are needed.
- (a) M1\(87 \div 6\) attempted
- (a) A114 packets filled completely
-
Non-calculator 3 marks
You are given that
\(24 \times 22 = 528\)
(a)State the value of \(2.4 \times 22\)
(b)State the value of \(24 \times 0.22\)
(c)State the value of \(528 \div 22\)
Show answer
(a) \(52.8\)
(b) \(5.28\)
(c) \(24\)\(2.4\) is \(24\) divided by \(10\), so the answer is \(528\) divided by \(10\).
\(528 \div 10 = 52.8\)
\(0.22\) is \(22\) divided by \(100\), so the answer is \(528\) divided by \(100\).
\(528 \div 100 = 5.28\)
Multiplying and dividing undo each other. The given fact says \(24 \times 22 = 528\), so dividing \(528\) by \(22\) takes you back to \(24\).
-
Non-calculator 4 marks(a)
A reel of wire holds 3.2 metres.
Calculate how much 6 reels hold altogether.
(b)A different job needs 30.8 metres, shared equally between 7 reels.
Calculate how much each reel accounts for.
Show answer
(a) \(19.2\ \text{metres}\)
(b) \(4.4\ \text{metres}\)Ignore the decimal point and multiply: \(32 \times 6 = 192\).
There was one decimal place in \(3.2\), so there is one in the answer:
\(3.2 \times 6 = 19.2\)
Divide as usual, keeping the decimal point in line:
\(308 \div 7 = 44\) tenths, so
\(30.8 \div 7 = 4.4\)
- (a) M1multiplying the digits, \(32 \times 6 = 192\)
- (a) A1decimal point placed correctly, \(19.2\)
- (b) M1dividing the digits, \(308 \div 7 = 44\)
- (b) A1decimal point placed correctly, \(4.4\)
-
Non-calculator 3 marks
A warehouse packs 714 bolts into crates of 18.
Joel says that makes 39 full crates with 12 bolts left over.
Is Joel right?
Check by multiplying, and show your working.
Show answer
Yes — \(39 \times 18 + 12 = 714\), and the number started with was 714.Multiply the number of full crates by how many go in each crate, then add the bolts left over.
\(39 \times 18 = 702\)
\(702 + 12 = 714\)
The number started with was 714, so Joel is right.
- M1multiply, \(39 \times 18 = 702\)
- M1add the remainder, \(702 + 12 = 714\)
- A1Yes, comparing 714 with 714
-
Non-calculator 3 marks
A sack holds 672 seeds.
They are split equally into 6 bags, and each bag is then split equally into 7 packets.
(a)Calculate how many seeds are in each packet.
(b)Write down the single division that would give the same answer in one step.
Show answer
(a) \(16\)
(b) \(672 \div 42\)\(672 \div 6 = 112\) seeds in each bag.
\(112 \div 7 = 16\) seeds in each packet.
Dividing by 6 and then by 7 is the same as dividing by \(6 \times 7 = 42\).
\(672 \div 42 = 16\)
- (a) M1\(672 \div 6 = 112\) seeds in each bag
- (a) A1\(112 \div 7 = 16\)
-
Non-calculator 3 marks(a)
Work out \(88 \times 200\).
(b)Use your answer to part (a) to work out \(88 \times 199\).
Show answer
(a) \(17600\)
(b) \(17512\)\(88 \times 200 = 17600\)
199 is one away from 200, so \(88 \times 199\) is \(88 \times 200\) with one lot of 88 take away.
\(17600\) take away \(88\) gives \(17512\).
- (b) M1\(17600\) with one lot of 88 take away
- (b) A1\(17512\)
-
Non-calculator 3 marks
Work out \(0.35 \times 0.33\)
Give your answer as a decimal.
Show answer
\(0.1155\)Ignore the decimal points and multiply \(35 \times 33\) as whole numbers.
Multiply by the units digit: \(35 \times 3 = 105\)
Multiply by the tens digit: \(35 \times 30 = 1050\)
Add the rows: \(105 + 1050 = 1155\)
Count the decimal places in the two factors: \(2 + 2 = 4\), so the answer has 4 decimal places, counted in from the right of \(1155\) (with zeros put in where needed).
\(0.35 \times 0.33 = 0.1155\)
- M1both partial products of \(35 \times 33\) found: \(105\) and \(1050\)
- M1whole-number product \(1155\)
- A1point placed to give \(0.1155\)