Class 4 Maths Olympiad Questions with Answers – 22 Starter Problems with Worked Solutions
Class 4 Maths Olympiad Questions with Answers — 22 Starter Problems with Worked Solutions
Math Olympiad › Class 4 › Problems
Next set: 28 harder problems (levels 4 to 5)
Original problems for Class 4, ordered from warm-up to genuinely hard, and every one of them comes
with a full worked solution. This is also called Grade 4 in many schools, and these questions suit
anyone looking for Olympiad practice, a maths challenge, enrichment or harder problem solving at this level.
What makes this different from a free worksheet: every wrong option has a written reason explaining
why a student would be tempted to pick it. Reading those is where the real learning happens.
This is the first half of the Class 4 bank. When the Think and Challenge levels stop feeling hard, move to the 28 harder problems, or sit the full paper against a clock.
real school Olympiad paper. Attempt every question — there is no penalty for guessing.
Think · level 2 of 5 · 6 problems
A staircase is built with blocks. The top row has 1 block, the next row has 3, then 5, then 7. Each new row has 2 more blocks than the row above. How many blocks are in the sixth row?
Correct answer: (b) 11
Row 1 has 1 block, row 2 has 3, row 3 has 5, row 4 has 7. The step is always 2.
Row 5 has 9 blocks and row 6 has 11 blocks.
So the sixth row holds 11 blocks.
Usual mistake: Starting an arithmetic pattern from zero instead of from the first given term.
Why the other options look tempting
- (a) Starts the pattern at 0 blocks and then adds 2 each time, as if the top row were empty.
- (c) Keeps adding 2 but counts one row too many, reaching 12 instead of 11.
- (d) Carries on adding 2 and then adds a 3 as well, double-counting the last step.
A ribbon 1 metre long is cut into 4 equal pieces. One of those pieces is cut in half again. What fraction of the original ribbon is the smallest piece now?
Correct answer: (d) 1/8
Cutting into 4 equal pieces makes each piece 1/4 of the whole ribbon.
Cutting one of those pieces in half makes the new piece 1/2 of 1/4.
Half of a quarter is 1/8 of the original ribbon.
Usual mistake: Stopping after the first cut, or halving 1/4 by 4 instead of by 2.
Why the other options look tempting
- (a) Answers after the first cut only, and forgets that one piece was halved again.
- (b) Cuts the piece into 4 rather than 2, which would give 1/16 of the whole.
- (c) Reports the size of one of the pieces that were not cut again, which is still 1/2.
Two different odd numbers are added together and the sum is written on the board. Then one of the two original numbers is added to that sum. What must be true about the number now on the board?
Correct answer: (d) odd
Adding two odd numbers always gives an even number.
The number added next is one of the original odd numbers.
An even number plus an odd number is always odd.
Usual mistake: Stopping after the first step, or inventing a divisibility rule from the number of terms.
Why the other options look tempting
- (a) Works out the first step correctly, then forgets to do the second addition.
- (b) Thinks that adding three numbers always gives a multiple of 3, which depends on the digits and not on the count.
- (c) Assumes an even number plus an odd number must be a multiple of 6, which is not a rule of parity.
A shopkeeper buys 12 identical notebooks for Rs.180. He then sells 8 of those notebooks for Rs.160. What is his profit on the 8 notebooks he sold?
Correct answer: (b) ₹40
Cost of 1 notebook is 180 divided by 12, which is Rs.15.
So the cost of the 8 notebooks sold is 8 times 15, which is Rs.120.
Profit is what he received minus what those 8 notebooks cost: 160 − 120 = Rs.40.
Usual mistake: Using the cost of the whole lot against the sale of only part of it.
Why the other options look tempting
- (a) Subtracts the Rs.160 earned on 8 notebooks from the Rs.180 spent on all 12, mixing the two lots.
- (c) Adds the Rs.120 cost to the Rs.160 received instead of taking one from the other.
- (d) Works out the cost of one notebook as Rs.15 and the money received per notebook as Rs.20, then subtracts in the wrong order.
A ribbon is 2 metres 40 centimetres long. A piece 75 centimetres long is cut off. How long is the ribbon that is left?
Correct answer: (a) 165 cm
1 metre is 100 centimetres, so 2 m 40 cm is 240 cm.
Cutting 75 cm off leaves 240 − 75 centimetres.
What remains is 165 cm long.
Usual mistake: Converting units once and then borrowing, which changes the arithmetic without being noticed.
Why the other options look tempting
- (b) Borrows 1 metre without adjusting, leaving 30 cm − 75 cm and then adding 125 to make it look sensible.
- (c) Adds the length that was cut off to the total instead of removing it.
- (d) Subtracts 105 cm, which is the 75 cm cut plus the extra 30 cm that should have been borrowed from the metres.
Meena buys 4 notebooks at Rs.15 each and 2 pens at Rs.8 each. She pays with a Rs.100 note. How much change does she get?
Correct answer: (b) ₹24
The 4 notebooks cost 4 times 15, which is Rs.60.
The 2 pens cost 2 times 8, which is Rs.16.
Total cost is 60 + 16 = Rs.76, so the change is 100 − 76 = Rs.24.
Usual mistake: Working out the running total in one line without separating the two purchases.
Why the other options look tempting
- (a) Adds up only the pens and forgets the notebooks entirely.
- (c) Prices every notebook at Rs.5, so the four notebooks cost only Rs.20.
- (d) Adds the pen cost to the notebook cost a second time, double counting Rs.16.
Challenge · level 3 of 5 · 16 problems
Triangles are laid out in rows so that the first row has 1 triangle, the second has 3, the third has 6 and the fourth has 10. If the same pattern continues, how many triangles are there in the eighth row altogether?
Correct answer: (c) 36
Row n holds 1 + 2 + … + n triangles, which is n multiplied by n + 1, halved.
For the eighth row that is 8 times 9, halved.
8 times 9 is 72, and 72 halved is 36 triangles.
Usual mistake: Carrying the same step forward instead of letting it increase by 1 each time.
Why the other options look tempting
- (a) Keeps adding 5, which was the last step used, instead of letting the step grow.
- (b) Adds 5, 6, 7, 8 and 9 to the fourth row, which overshoots the eighth row.
- (d) Uses n times n instead of n times n + 1, halved, so the answer is 8 points too small.
Ravi had Rs.60. He spent 1/3 of it on a book, and then spent 1/4 of what was left on a notebook. How much money does he have now?
Correct answer: (c) ₹30
1/3 of Rs.60 is Rs.20, so Rs.40 is left after the book.
1/4 of the Rs.40 that is left is Rs.10, spent on the notebook.
So Ravi has 40 − 10 = Rs.30 left.
Usual mistake: Applying the second fraction to the starting amount instead of the amount remaining.
Why the other options look tempting
- (a) Takes 1/4 of the original Rs.60, which is Rs.15, instead of 1/4 of the Rs.40 that remained.
- (b) Reports the amount spent on the book rather than the amount left at the end.
- (d) Stops after the first purchase and reports Rs.40, ignoring the notebook.
A number leaves no remainder when it is divided by 6, by 9, or by 15. What is the smallest such number that is greater than 100?
Correct answer: (c) 180
6, 9 and 15 are made from the prime factors 2, 3 and 5, and 9 needs 3 twice.
Their least common multiple is 2 times 9 times 5, which is 90.
90 is not greater than 100, so the next multiple of 90 is 180.
Usual mistake: Giving the least common multiple when a larger one is asked for, or checking only some of the divisors.
Why the other options look tempting
- (a) Gives the least common multiple itself, but the question asks for a number greater than 100.
- (b) Skips a multiple: 90 times 3 is the second one above 100, not the first.
- (d) Picks 6 times 20, which is divisible by 6 and by 15 but leaves remainder 3 when divided by 9.
A square of side 2 cm is cut into 4 equal smaller squares, and then one diagonal is drawn in each of the 4 small squares. How many triangles are there in total?
Correct answer: (a) 8
One diagonal cuts a square into exactly 2 triangles.
There are 4 small squares, and each gives 2 triangles.
So there are 4 times 2 = 8 triangles altogether.
Usual mistake: Forgetting that a single diagonal produces two triangles, not one.
Why the other options look tempting
- (b) Counts one triangle per small square instead of the two that a diagonal makes.
- (c) Counts the 8 small triangles and then adds 4 more that are not actually formed.
- (d) Treats a square as having four halves along diagonals, when it has only two.
A train leaves at 9:15 in the morning and arrives at 11:40 the same morning. It stands still at one station for 20 minutes on the way. For how many minutes was the train actually moving?
Correct answer: (c) 125 minutes
From 9:15 to 11:40 is 2 hours 25 minutes, which is 145 minutes.
The train was not moving for 20 of those minutes.
So the moving time is 145 − 20 = 125 minutes.
Usual mistake: Reading a journey time as moving time when a stoppage is mentioned.
Why the other options look tempting
- (a) Gives the total time from departure to arrival and forgets the stop.
- (b) Takes the stop off twice, or removes 40 minutes instead of 20.
- (d) Adds the stop to the travelling time instead of removing it.
A number leaves remainder 3 when it is divided by 7, and it also leaves remainder 3 when it is divided by 5. What is the smallest such number that is greater than 40?
Correct answer: (c) 73
Subtracting 3 from the number leaves something divisible by both 7 and 5.
A number divisible by both 7 and 5 is a multiple of 35.
So the number is 3 more than a multiple of 35. The values are 38 and 73, and 38 is too small.
So the smallest such number above 40 is 73.
Usual mistake: Finding the smallest match and stopping, without checking that it is above the stated limit.
Why the other options look tempting
- (a) This is the smallest such number, but it is not greater than 40.
- (b) 63 is a multiple of 7, so dividing it by 7 leaves remainder 0 and not 3.
- (d) 75 is a multiple of 5, so dividing it by 5 leaves remainder 0 and not 3.
A 3 by 3 grid is drawn on a sheet of paper, making 9 small squares. How many squares of any size are there altogether, counting the bigger ones made from several small squares?
Correct answer: (a) 14
Squares of size 1 by 1: there are 9 of them.
Squares of size 2 by 2: there are 4 of them.
Square of size 3 by 3: there is 1. So the total is 9 + 4 + 1 = 14.
Usual mistake: Counting only unit squares when the question says squares of any size.
Why the other options look tempting
- (b) Counts only the smallest squares and ignores the larger ones built from them.
- (c) Misses the single largest square, giving 13 instead of 14.
- (d) Also counts the grid lines themselves as though they formed extra squares.
Four bags hold 12, 15, 9 and 14 marbles. How many marbles must be added to the bag with the fewest marbles so that all four bags hold the same number?
Correct answer: (c) 6
To make all four equal, every bag has to reach the largest amount, which is 15.
The bag with the fewest marbles has 9.
So 15 − 9 = 6 marbles must be added.
Usual mistake: Raising the smallest to the second smallest rather than to the largest.
Why the other options look tempting
- (a) Raises the smallest bag to 12, which is only the second smallest and would not make all four equal.
- (b) Reports the number of marbles in the smallest bag instead of the number added.
- (d) Reports the number in the largest bag instead of the difference between largest and smallest.
A number is multiplied by 4, then 7 is subtracted, and the result is 29. What was the number to start with?
Correct answer: (d) 9
Work backwards from 29. The last thing done was subtracting 7, so add 7: 29 + 7 = 36.
Before that the number was multiplied by 4, so divide by 4: 36 divided by 4 is 9.
So the number was 9.
Usual mistake: Reversing a subtraction as a subtraction instead of an addition.
Why the other options look tempting
- (a) Reverses one step too many and divides by 4 a second time.
- (b) Stops after multiplying back and forgets to undo the subtraction of 7.
- (c) Subtracts 7 from the result while working backwards, when the step was to add 7.
Using the digits 4, 7 and 9 exactly once each, how many different three-digit numbers can be written?
Correct answer: (b) 6
The first place can take any of the 3 digits.
The second place then has 2 choices and the last place has 1 choice left.
So there are 3 times 2 times 1 = 6 different numbers.
Usual mistake: Choosing digits for one place and forgetting that the choices shrink as you go.
Why the other options look tempting
- (a) Only changes the first digit and treats the remaining two as fixed.
- (c) Gives 3 choices in each of the three places, forgetting that a digit cannot be used twice.
- (d) Counts arrangements in which a digit may be repeated, which is a different question.
What is the smallest number which, when 1 is added to it, is divisible by 2, 3, 4 and 6 all at the same time?
Correct answer: (d) 11
2, 3, 4 and 6 are made from the prime factors 2 and 3, and 4 needs 2 twice while 6 needs one each.
Their least common multiple is 12.
We want the number that is 1 less than a common multiple, so it is 12 − 1 = 11.
Usual mistake: Answering with the common multiple rather than the number that is one away from it.
Why the other options look tempting
- (a) Gives the least common multiple itself, when the number needed is 1 less than it.
- (b) Gives 1 more than the least common multiple instead of 1 less.
- (c) 8 is divisible by 2 and by 4 but not by 3 and not by 6.
How many different factors does the number 24 have? Remember that a factor is a number that divides 24 exactly, and that 1 and 24 are both counted.
Correct answer: (d) 8
24 can be written as 2 multiplied by 2 multiplied by 2 multiplied by 3, that is 2 cubed times 3.
For a prime raised to power 3, the factors can take 1, 2, 4 or 8, which is 4 choices.
For the prime 3 the choices are 1 or 3, which is 2 choices. The factors are 1, 2, 3, 4, 6, 8, 12 and 24.
So 4 times 2 gives 8 factors.
Usual mistake: Counting factor pairs instead of single factors, or stopping at the square root.
Why the other options look tempting
- (a) Lists 1, 2, 3, 4, 6 and 8, then forgets that 12 and 24 also divide 24 exactly.
- (b) Counts only the factor pairs 1 times 24, 2 times 12, 3 times 8 and 4 times 6, and so leaves out 24 itself.
- (c) Counts each factor twice, once counting up and once counting down.
How many three-digit numbers can be written using only the digits 5, 0 and 9, if no digit may be used more than once?
Correct answer: (a) 4
The first place cannot be 0, because then the number would only have two digits. So the first place has 2 choices: 5 or 9.
For the second place there are 2 digits left, and then 1 for the last place.
So there are 2 times 2 times 1 = 4 such numbers.
Usual mistake: Forgetting that 0 cannot start a multi-digit number.
Why the other options look tempting
- (b) Allows 0 in the first place, which gives numbers that are not three-digit at all.
- (c) Gives only one arrangement for each possible first digit and ignores the choices left over.
- (d) Allows digits to be repeated, giving 2 choices in each of the three places.
A number is written as 4 tens, 3 hundreds and 6 ones. What is the number?
Correct answer: (a) 346
3 hundreds is 300 and 4 tens is 40, and 6 ones is 6.
Adding them in order of size gives 300 + 40, which is 340, and 340 + 6.
The number is 346.
Usual mistake: Writing the place values in the order they are spoken instead of in order of size.
Why the other options look tempting
- (b) Swaps the 3 and the 4, so 4 sits in the hundreds place and 3 in the tens place.
- (c) Puts the 6 in the tens place, but 6 is the ones digit and 4 belongs there in the tens.
- (d) Puts 4 in the hundreds place and 6 in the tens place, changing both of the stated digits.
All roses are flowers. All flowers need water. Which of these statements must be true?
Correct answer: (a) All roses need water.
Every rose is a flower.
Every flower needs water.
Chaining the two, a rose is a flower and so needs water. Therefore all roses need water.
Usual mistake: Reversing a statement of the form all A are B into all B are A.
Why the other options look tempting
- (b) Reverses the first statement. We know every rose is a flower, not that every flower is a rose.
- (c) Contradicts the chain, because a rose is a flower and every flower needs water.
- (d) May well be true or false. Nothing in the two statements says whether every flower is a rose.
The average of four numbers is 15. Three of the numbers are 12, 18 and 20. What is the fourth number?
Correct answer: (d) 10
An average of 15 across 4 numbers means the four numbers add up to 4 times 15, which is 60.
The three known numbers add up to 12 + 18 + 20, which is 50.
The fourth number is 60 − 50 = 10.
Usual mistake: Thinking the average is another value in the list rather than a balance of the total.
Why the other options look tempting
- (a) Subtracts only 12 and 18 from the total of 60.
- (b) Assumes the missing number is also the average, which is not what an average means.
- (c) Reports the largest of the three given numbers.
Skills covered
- Arithmetic operations — 2 problems
- Factors and multiples — 2 problems
- Fractions and decimals — 2 problems
- Money — 2 problems
- Number patterns — 2 problems
- Number puzzles — 2 problems
- Counting — 1 problem
- Data handling — 1 problem
- Geometry and symmetry — 1 problem
- Logical reasoning — 1 problem
- Measurement — 1 problem
- Numbers and place value — 1 problem
- Odd and even — 1 problem
- Remainders — 1 problem
- Time and speed — 1 problem
- Working backwards — 1 problem
Common mistakes in these topics
- Starting an arithmetic pattern from zero instead of from the first given term.
- Stopping after the first cut, or halving 1/4 by 4 instead of by 2.
- Stopping after the first step, or inventing a divisibility rule from the number of terms.
- Using the cost of the whole lot against the sale of only part of it.
- Converting units once and then borrowing, which changes the arithmetic without being noticed.
- Working out the running total in one line without separating the two purchases.
- Carrying the same step forward instead of letting it increase by 1 each time.
- Applying the second fraction to the starting amount instead of the amount remaining.
- Giving the least common multiple when a larger one is asked for, or checking only some of the divisors.
- Forgetting that a single diagonal produces two triangles, not one.
- And 12 more, each shown under the problem it belongs to.
Related practice
Ready to be timed? Sit the full Class 4 practice test — 35 questions, 60 minutes, no negative marking, which uses the same
three-section format as the real paper. You can also go back to the
Class 4 Maths Olympiad hub.
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