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1. Red and blue counters are in the ratio 3 to 5. There are 40 counters altogether. How many are red?[4]
2. 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]
3. What is the remainder when 2 to the power 30 is divided by 7?[4]
4. What is 2 + 4 + 6 + ... + 100?[4]
A 1275B 2450C 2500D 2550E 51005. 24 pencils and 36 erasers are shared into identical gift bags, using every item. What is the largest number of bags possible?[4]
6. A price goes up by 20%, then down by 20%. Compared with the original price, the final price is:[4]
A 20% lowerB 4% lowerC the sameD 4% higherE 40% higher7. The mean of four numbers is 9. Three of them are 5, 8 and 12. What is the fourth?[4]
8. 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 calculatorKangaroo Atlas · https://kangaroo-atlas.vercel.app · seed 1 · CC BY-NC-SA 4.0 · Independent project, not affiliated with Math Kangaroo in USA, NFP.
1. D — 15 KA-0024 Ratio as parts of a whole
A I found what one part is worth and gave that as my answer instead of multiplying it back up.B I divided 40 by the five blue parts instead of by the eight parts that make the whole.C I read the ratio 3 to 5 as meaning 3 out of every 5 and worked from that.E I found the value of one part correctly but then multiplied by 5 instead of 3, giving the blue counters.2. 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.3. A — 1 KA-0053 Remainders and modular cycles
B I found the cycle 2, 4, 1 but read the remainder of 30 divided by 3 as one instead of zero.C I counted the cycle positions from zero, which shifted my answer one step along.D I divided 30 by 7 and used that remainder rather than the remainder of the powers.E I assumed the remainder would be one less than the divisor because the power is large.4. 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.5. C — 12 KA-0062 Factors, multiples, LCM and GCD by listing
A I found a number that divides both but not the largest one, stopping at the first that worked.B I used the difference between 36 and 24 divided by 2 rather than a common factor.D I used a factor of 36 without checking that it also divides 24.E I found the least common multiple instead of the greatest common divisor.6. B — 4% lower KA-0065 Percentages and simple discounts
A I applied only the decrease and forgot that the price had already risen.C I assumed a rise and a fall of the same percentage cancel each other out exactly.D I got the size of the change right but the direction wrong.E I added the two percentages together instead of applying them one after the other.7. C — 11 KA-0066 Means, and working backwards from a mean
A I assumed the missing number must be the mean itself.B I averaged the three known numbers instead of working from the total.D I repeated one of the numbers already given rather than computing the missing one.E I used a total of 37 instead of 36, adding the mean in once too often.8. 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.