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Worksheet generator
Pick what you want, print the page. The answer key comes with it, on its own sheet, and it tells you the mistake behind every wrong choice — not just which letter is right.
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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]
A5
B8
C13
D15
E24
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]
A1
B11
C12
D16
E23
3.What is the remainder when 2 to the power 30 is divided by 7?[4]
A1
B2
C4
D5
E6
4.What is 2 + 4 + 6 + ... + 100?[4]
A1275
B2450
C2500
D2550
E5100
5.24 pencils and 36 erasers are shared into identical gift bags, using every item. What is the largest number of bags possible?[4]
A4
B6
C12
D18
E72
6.A price goes up by 20%, then down by 20%. Compared with the original price, the final price is:[4]
A20% lower
B4% lower
Cthe same
D4% higher
E40% higher
7.The mean of four numbers is 9. Three of them are 5, 8 and 12. What is the fourth?[4]
A9
B10
C11
D12
E13
8.What is the units digit of 3 multiplied by itself 100 times, that is 3 to the power 100?[4]
A1
B3
C7
D9
Eit cannot be found without a calculator
Math Kangaroo Star · https://kangaroo-atlas.vercel.app · seed 1 · CC BY-NC-SA 4.0 · Independent project, not affiliated with Math Kangaroo in USA, NFP.
Answer key — Math Kangaroo practice
Each wrong choice carries the thinking that produces it. When a student picks C, this is what they were doing.
1.D — 15KA-0024Ratio as parts of a whole
AI found what one part is worth and gave that as my answer instead of multiplying it back up.
BI divided 40 by the five blue parts instead of by the eight parts that make the whole.
CI read the ratio 3 to 5 as meaning 3 out of every 5 and worked from that.
EI found the value of one part correctly but then multiplied by 5 instead of 3, giving the blue counters.
2.D — 16KA-0033Means, and working backwards from a mean
AI subtracted the two means from each other and gave the difference as the removed number.
BI assumed the removed number must be the new mean, since the mean dropped to 11.
CI assumed removing a number equal to the old mean is what changes the mean.
EI added the two totals instead of subtracting one from the other.
3.A — 1KA-0053Remainders and modular cycles
BI found the cycle 2, 4, 1 but read the remainder of 30 divided by 3 as one instead of zero.
CI counted the cycle positions from zero, which shifted my answer one step along.
DI divided 30 by 7 and used that remainder rather than the remainder of the powers.
EI assumed the remainder would be one less than the divisor because the power is large.
4.D — 2550KA-0058Clever regrouping (pair to round numbers)
AI found the sum of 1 to 50 and forgot that every term here is twice as big.
BI used 49 pairs instead of 25, losing one pair from the count.
CI estimated the answer as a round number rather than pairing properly.
EI paired the terms but forgot that pairing counts every number twice, so I never halved.
5.C — 12KA-0062Factors, multiples, LCM and GCD by listing
AI found a number that divides both but not the largest one, stopping at the first that worked.
BI used the difference between 36 and 24 divided by 2 rather than a common factor.
DI used a factor of 36 without checking that it also divides 24.
EI found the least common multiple instead of the greatest common divisor.
6.B — 4% lowerKA-0065Percentages and simple discounts
AI applied only the decrease and forgot that the price had already risen.
CI assumed a rise and a fall of the same percentage cancel each other out exactly.
DI got the size of the change right but the direction wrong.
EI added the two percentages together instead of applying them one after the other.
7.C — 11KA-0066Means, and working backwards from a mean
AI assumed the missing number must be the mean itself.
BI averaged the three known numbers instead of working from the total.
DI repeated one of the numbers already given rather than computing the missing one.
EI used a total of 37 instead of 36, adding the mean in once too often.
8.A — 1KA-0149Last-digit behavior of products and powers
BI assumed every power of 3 ends in 3.
CI found the cycle 3, 9, 7, 1 but counted its positions from zero, landing one step early.
DI used a remainder of 2 instead of 0 when dividing 100 by the cycle length of 4.
EI assumed a number this large has no reachable last digit, rather than looking for a repeat.