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1. In the number 4073, what is the value of the digit 7?[3]
2. What is 1 + 2 + 3 + ... + 20?[3]
3. Which of these fractions lies between one third and one half?[3]
4. What is the remainder when 7 to the power 100 is divided by 5?[3]
5. The four-digit number 27_4 is divisible by 3. Which digit could go in the gap?[4]
6. The mean of five numbers is 12. One of the numbers is removed and the mean of the remaining four is 11. What number was removed?[4]
7. The two digits of a number are swapped and the number gets bigger by 36. Which of these could be the number?[4]
8. What is 2 + 4 + 6 + ... + 100?[4]
A 1275B 2450C 2500D 2550E 51009. What is the units digit of 3 multiplied by itself 100 times, that is 3 to the power 100?[4]
A 1B 3C 7D 9E it cannot be found without a calculator10. A and B are different digits. The two-digit number AB added to the two-digit number BA gives 132. What is A + B?[5]
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1. C — 70 KA-0055 Place value and digit positions
A I gave the last digit of the number instead of looking at the 7 at all.B I gave the digit itself rather than what it is worth in the position it sits in.D I counted the positions from the left, so I read the 7 as being in the hundreds place.E I gave the value of the largest digit instead of the value of the 7.2. C — 210 KA-0057 Clever regrouping (pair to round numbers)
A I paired the numbers from the outside in but left the 20 out of every pair, so I added only up to 19.B I estimated the total as a round number instead of working it out.D I used eleven pairs of 20 instead of ten pairs of 21.E I multiplied 20 by 21 and forgot that pairing counts every number twice.3. C — 2/5 KA-0141 Number lines and betweenness
A I compared the bottom numbers and assumed a bigger denominator means a bigger fraction.B I saw that 4 lies between 3 and 2 and assumed one quarter therefore lies between the two fractions.D I checked that it is bigger than one third but never checked it against one half.E I compared the top numbers only and ignored what the bottoms do to the size.4. A — 1 KA-0169 Remainders and modular cycles
B I reduced 7 to 2 correctly but then used a remainder of 1 when dividing 100 by 4.C I read the cycle backwards and landed on the entry before the right one.D I used 100 divided by 4 giving 25 and read off the 25th entry as if the cycle had length 25.E I assumed a power that large must be divisible by 5.5. B — 2 KA-0011 Divisibility rules: 2, 3, 4, 5, 9, 10
A I added the known digits to get 13 and then checked whether the missing digit itself was a multiple of 3.C I chose 3 because the question mentioned dividing by 3, without testing the digit sum at all.D I tested the last two digits, which is the rule for 4, not the rule for 3.E I added the digits but left out the 4 at the end, so my sum was four too small.6. D — 16 KA-0033 Means, and working backwards from a mean
A I subtracted the two means from each other and gave the difference as the removed number.B I assumed the removed number must be the new mean, since the mean dropped to 11.C I assumed removing a number equal to the old mean is what changes the mean.E I added the two totals instead of subtracting one from the other.7. B — 15 KA-0056 Place value and digit positions
A I checked that the digits differ and stopped, without working out how much the swap actually changes the number.C I saw a difference of 2 between the digits and expected that to give a change of 36.D I ignored the direction: swapping this number makes it smaller, not bigger.E I picked a number whose digits differ by 4 without noticing that the larger digit is already in front.8. D — 2550 KA-0058 Clever regrouping (pair to round numbers)
A I found the sum of 1 to 50 and forgot that every term here is twice as big.B I used 49 pairs instead of 25, losing one pair from the count.C I estimated the answer as a round number rather than pairing properly.E I paired the terms but forgot that pairing counts every number twice, so I never halved.9. A — 1 KA-0149 Last-digit behavior of products and powers
B I assumed every power of 3 ends in 3.C I found the cycle 3, 9, 7, 1 but counted its positions from zero, landing one step early.D I used a remainder of 2 instead of 0 when dividing 100 by the cycle length of 4.E I assumed a number this large has no reachable last digit, rather than looking for a repeat.10. D — 12 KA-0060 Operation puzzles and cryptarithms
A I added the digits of 132 instead of working out what A and B must be.B I found A + B correctly and then halved it, as if the question wanted one digit.C I spotted the 11 in the working and gave that instead of the sum it multiplies.E I found a pair of digits by trial and misadded them by one.