Full Grade 2 test

One ten equals 10 ones. When one ten is exchanged for 10 ones, the tens digit decreases by 1 and the number of ones increases by 10. The value of the number does not change.

Multiplication is a quick way to add equal groups. The first number can show how many groups there are, and the second how many objects are in each group.

For example, 3 × 4 = 4 + 4 + 4 = 12. Use the times table or add the equal amounts.

A digit’s value depends on its place in the number. In a three-digit number, the places from left to right are hundreds, tens and ones.

For example, in 426 the digit 4 means 400, the digit 2 means 20, and the digit 6 means 6. A highlighted 0 has a value of 0.

One metre equals 100 centimetres: 1 m = 100 cm.

From 0 to 1 metre on a ruler, there are one hundred steps of one centimetre.

Changing the order of the factors does not change the product. Find the other factor.

Compare the hundreds digits first. If they are the same, compare the tens; if those are also the same, compare the ones.

Choose > if the left number is greater, < if it is smaller, and = if they are equal. The open side of the sign faces the greater number.

Add the ones digits. Subtract ten from their sum: these are the ones left after exchanging ten ones for a ten. The question asks for the ones digit, not the full sum.

Combine the hundreds, tens and ones given. The hundreds give the first digit, the tens the second, and the ones the third.

For example, 300 + 40 + 7 = 347. If there are no tens or ones, put 0 in that place: 300 + 0 + 7 = 307.

Count the counters and divide by 2. Each child receives half of the whole set.

A cube has six identical flat faces. Each has four equal sides and four right angles.

Split the second addend. Add enough to reach the next multiple of ten, then add the rest. Check that the two parts add up to the original addend.

Add the ones. If there are at least ten, exchange ten ones for one ten. Include this extra ten when adding the tens.

A jump's length is its end value minus its start value. Here the second jump starts at a multiple of ten. Count intervals, not tick marks.

If one addend changes and the other stays the same, the sum changes by the same amount in the same direction.

Frequency means the number of occurrences or choices, not price, duration or rank.

Move the stated number of cells right, then down. Do not count the starting cell as a step.

A full day includes daytime and night-time. It lasts 24 hours.

There are 12 hours from midnight to midday and another 12 to the next midnight: 12 + 12 = 24.

Look at the ones digit.

• If it is 0, 1, 2, 3 or 4, round to the lower ten.
• If it is 5, 6, 7, 8 or 9, round to the higher ten.

For example, 243 rounds to 240, while 247 rounds to 250. The rounded number ends in 0.

Add the known frequencies and subtract their sum from the total votes.

Count the rows and the counters in one row. Add one to the number of rows, then multiply by the counters per row.

Add tens to tens and ones to ones. Exchange 10 ones for one ten; 10 tens make one hundred.

For example, 47 + 68 = 40 + 60 + 7 + 8 = 100 + 15 = 115.

Moving left decreases the number. The first jump reaches a multiple of ten; the second moves further left. Subtract the size of the second jump from the intermediate number.

One ten equals 10 ones. To exchange ones for whole tens, group them in tens: divide the number of ones by 10. To exchange tens for ones, multiply the number of tens by 10. The representation changes, but the total amount stays the same.

Compare the old and new prices. Subtract the smaller price from the larger one to find the change in cents.

The perimeter is the sum of all the side lengths. Opposite sides of a rectangle have equal lengths.

Add the length and width, then multiply the sum by 2. For example, sides of 5 and 3 give a perimeter of 5 + 3 + 5 + 3 = 16.

Use the side lengths stated in the question or labelled in the drawing.

One minute lasts 60 seconds: 1 min = 60 s.

For these time units, use 60 rather than 10 or 100.

Find the missing subtrahend by subtracting the remainder from the starting number. Check by adding it to the remainder: you should get the starting number.

The divisor is missing, not the result of the division. Divide the starting number by the result shown.

For example, for 12 ÷ ? = 3, the missing number is 12 ÷ 3 = 4. Check: 12 ÷ 4 = 3.

The fence goes around the whole rectangular garden, so find its perimeter. There are two equal lengths and two equal widths.

Add the length and width, then multiply by 2. The answer is in metres. Include all four sides, rather than calculating the area.

Check each number separately. A card equal to the boundary does not qualify.

All four sides of a square are equal. The perimeter is their total length.

To find one side length, divide the perimeter by 4. For example, a perimeter of 20 gives a side of 20 ÷ 4 = 5. Check by multiplying by 4.

Look at the operation sign. For +, add the numbers; for −, subtract the second number from the first.

It can help to split the second number into tens and ones. For example, 27 − 13 = 27 − 10 − 3 = 14. When adding, you can also add the tens first and then the ones.

Find the missing factor by dividing the product by the known factor.

For example, for 3 × ? = 12, calculate 12 ÷ 3 = 4. Check that multiplying by your answer gives the product shown.

Work out the subtraction inside the brackets first.

Then multiply the difference by the number outside the brackets. For example, (9 − 6) × 4 = 3 × 4 = 12.

A rectangle’s perimeter includes two lengths and two widths.

1. Divide the perimeter by 2 to get the sum of one length and one width.
2. Subtract the known side length from that sum.

For example, a perimeter of 20 cm and one side of 4 cm give the other side as 20 ÷ 2 − 4 = 6 cm. Check by adding all four sides.

Count the vertices of each polygon separately, then add the results.

A vertex is where two sides meet. Count the vertices in each figure.

Test each proposed input using both the condition and its corresponding action.

Follow the outline and count each straight side once.

A pair has two passengers. If the number of passengers is even, everyone can be put into pairs with no remainder.

Look at the last digit: 0, 2, 4, 6 or 8 means even. In that case, the answer to making pairs with no remainder is “Yes”.

Count the stickers and divide them into four equal groups. One group is a quarter of the set.

Every bag must contain the same number of sweets. Divide the total number of sweets by the number of bags.

For example, sharing 18 sweets into 3 bags gives 18 ÷ 3 = 6 in each. Check by multiplying the amount in one bag by the number of bags.

Think about whether the gifted money stayed with the same two children. Did their total amount change, or only how much each child has?