This is the marking copy and it carries every answer. Her sheet is the other file — same numbering, same margin codes, so you can mark straight down the page.
Find the sum of all the factors of 81.
answer 121
method List the factors in pairs: 1 × 81, 3 × 27, 9 × 9. So the factors are 1, 3, 9, 27 and 81 — the 9 appears only once because it pairs with itself. Sum = 1 + 3 + 9 + 27 + 81 = 121.
watch Writing 9 twice because 9 × 9 = 81. A factor is listed once however many times it appears in a pair.
I am a 3-digit odd number. All my three digits are different. All my three digits are multiples of 3. The digit in the ‘hundreds’ place is twice the digit in the ‘ones’ place. What number am I?
answer 693
method The digits that are multiples of 3 are 3, 6 and 9. The number is odd, so the ones digit is 3 or 9. If the ones digit were 9, the hundreds digit would be 18 — not a digit. So the ones digit is 3 and the hundreds digit is 2 × 3 = 6. The tens digit must be a different multiple of 3, which leaves 9. The number is 693.
watch Counting 0 as a multiple of 3 and offering 603 as well. In primary work the multiples of 3 are 3, 6, 9, … — they start at 3, not 0.
Alice’s present age is a 2-digit number which is also a multiple of 4. 5 years ago, her age was a factor of 54. What is Alice’s age 5 years later?
answer 37
method The factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Adding 5 to each gives the possible present ages: 6, 7, 8, 11, 14, 23, 32, 59. Of those, only 32 is both a 2-digit number and a multiple of 4. So Alice is 32 now, and 5 years later she is 32 + 5 = 37.
watch Answering 32 — that is her age NOW. The question asks for five years later.
Mike had 72 oranges and 45 pears. He packed the fruits such that each bag contained an equal number of oranges and an equal number of pears. He packed the fruits into as many bags as possible. How many fruits were there in each bag?
answer 13
method The number of bags must divide both 72 and 45 exactly, and we want as many bags as possible — so it is the largest common factor. 72 = 8 × 9 and 45 = 5 × 9, so the highest common factor is 9. With 9 bags: 72 ÷ 9 = 8 oranges and 45 ÷ 9 = 5 pears in each, giving 8 + 5 = 13 fruits per bag.
watch Answering 9 — that is the number of BAGS. Or using a smaller common factor like 3, which works but does not give as many bags as possible.
Kenny has 3 pieces of rope of length 24 cm, 60 cm and 96 cm. He wants to cut them into smaller pieces of equal length without any leftovers. Each smaller piece must be as long as possible. How many such pieces of equal length can he get?
24 cm |------------| 60 cm |------------------------------| 96 cm |------------------------------------------------|
answer 15
method The piece length must divide 24, 60 and 96 exactly, and be as long as possible — so it is the highest common factor. 24 = 12 × 2, 60 = 12 × 5, 96 = 12 × 8, and no bigger number divides all three, so each piece is 12 cm. Counting the pieces: 2 + 5 + 8 = 15 pieces.
watch Answering 12 — that is the LENGTH of each piece, and the paper's answer line unhelpfully prints "cm" even though the question asks how many. Read the question, not the blank.
Using all the digits 8, 0, 9, 2, form the smallest multiple of 5.
(1) 2890 (2) 2980 (3) 8920 (4) 9820
answer (1) — 2890
method A multiple of 5 ends in 0 or 5. There is no 5 among the digits, so the number must end in 0. That leaves 8, 9 and 2 for the first three places, and the smallest arrangement of those is 2, 8, 9. So the number is 2890.
watch Forgetting that ALL four digits must be used, or trying to put the 0 first to make the number small.
After a rectangular piece of paper was cut into a maximum of 18 squares of sides 4 cm each, an L-shaped strip of paper was left over as shown. Given that the length of the rectangular piece of paper is 26 cm, what is the perimeter of the rectangular piece of paper before it was cut?
+--+------------------------------+ | | | | | | | | the 18 squares of 4 cm | | | | | | | +--+------------------------------+ | strip | -+- 1 cm +---------------------------------+ -+- |<------------ 26 cm ------------>| the leftover L-strip is the narrow band down the left edge plus the 1 cm band along the bottom
answer 78
method Fit the 4 cm squares along the 26 cm length: 26 ÷ 4 = 6 remainder 2, so 6 squares fit across and a 2 cm strip is wasted down one side. With 18 squares in rows of 6, there must be 18 ÷ 6 = 3 rows, using 3 × 4 = 12 cm of the breadth. The leftover strip along the bottom is 1 cm, so the breadth is 12 + 1 = 13 cm. Perimeter = 2 × (26 + 13) = 2 × 39 = 78 cm.
watch Trying to get the breadth by dividing the squares' area (18 × 16 = 288 cm²) by 26. The L-strip is wasted paper, so the squares' area is NOT the paper's area.
What is the smallest number greater than 1 that leaves a remainder of 1 when it is divided by 2, by 3 and by 4?
(A) 7 (B) 9 (C) 13 (D) 25 (E) 49
answer (C) — 13
method Take away the remainder: the number minus 1 must divide by 2, 3 and 4, so it is a multiple of 12. The smallest is 12, giving 13. (25 and 49 also work, but they are not the smallest.)