Grade 6 math puzzles — free practice & printable teacher quizzes

• A percent reduction tells you how much is taken off the original price, not the new price. Find the discount in dollars, then subtract it from the original price.
• Use 25% = 1/4 to find the discount on the $80 jacket. Then subtract that discount from $80.

Answer: D — $60

Since 25% = 1/4, the discount is $80 ÷ 4 = $20. The new price is $80 − $20 = $60. The choice $20 is the amount saved, not the new price.

• Use 0 °C as a checkpoint. Work out whether the rise reaches it, and then consider what happens to any part of the rise left over.
• How many degrees does it take to rise from −7 °C to 0 °C? Compare that with the total rise of 12 degrees.

Answer: A — 5 °C

It takes a rise of 7 degrees to go from −7 °C to 0 °C. That leaves 12 − 7 = 5 degrees of the rise, so the temperature at noon was 5 °C. The trap in choosing −5 °C is keeping the negative sign even though the rise carried the temperature past zero.

• Think of halfway as a distance: find the gap between the two fractions, then move half that gap from the smaller fraction.
• The gap is 1/2 − 1/3 = 1/6. Split that gap in half, then add it to 1/3.

Answer: C — 5/12

Half of the gap 1/6 is 1/12, so the halfway point is 1/3 + 1/12 = 4/12 + 1/12 = 5/12. The fraction 2/5 is between 1/3 and 1/2, but it is not halfway: it is 1/15 above 1/3 and 1/10 below 1/2.

• The gift changes each person's share, but not their combined total. Compare Lina's share before and after.
• Lina starts with 3/8 of the total and ends with 1/3 of it. What fraction of the total did she give away?

Answer: A — 27

Lina gives away 3/8 − 1/3 = 1/24 of the unchanged total. Since that is 3 stickers, the total is 72. Lina started with 3/8 of 72, or 27 stickers. The tempting 24 is how many she has after giving away 3.

• In a run of consecutive numbers, the middle one is the average of them all.

Answer: C — 15

Five consecutive numbers are symmetric around the middle one, so the middle is 65 ÷ 5 = 13, and the numbers are 11, 12, 13, 14, 15 — largest 15. Exploiting symmetry (instead of setting up x + (x+1) + …) is faster and shows WHY it works.

• Start with a rectangle of unknown length and width. Describe how each percentage change affects a side, then consider how the changes combine when you multiply to find area.
• Let the original sides be L and W. Write the changed sides as 1.1L and 0.9W, then compare their product with the original area, LW.

Answer: A — it shrinks by 1%

Let the original length be L and the width be W, so the original area is LW. The new area is (1.1L)(0.9W) = 0.99LW, or 99% of the original area. Therefore, it shrinks by 1%. The tempting answer “it stays the same” is wrong: a 10% increase in one side and a 10% decrease in the other do not cancel when the side lengths are multiplied.

• Start with the square roots. What can you learn about their final digits from the squares’ final digits?
• After checking division by 3, use (root − 1) × (root + 1) to check division by 7.

Answer: C — 2

A four-digit square has a root from 32 to 99. The square must be odd and leave 1 when divided by 5, so it ends in 1; its root must therefore end in 1 or 9. The possible roots are 39, 41, 49, 51, 59, 61, 69, 71, 79, 81, 89, 91, and 99. Removing roots divisible by 3 leaves 41, 49, 59, 61, 71, 79, 89, and 91. Because root² − 1 = (root − 1)(root + 1), the test for 7 leaves only 41 and 71. Their squares are 1681 and 5041; subtracting 1 gives 1680 and 5040, both divisible by every number from 2 through 7. The answer is 2. The tempting answer 8 forgets the test for 7.

• Before counting knights, track who would have to be the same kind or different kinds.
• Suppose Ada tells the truth. What would Drew be, and could Ben and Cora then be different kinds?

Answer: D — 2 knights

If Ada were a knight, Drew would be a liar. Cora’s claim would then be true exactly when Ben is a knight, so Ben and Cora would be the same kind. But Ada’s claim requires exactly one of them to be a knight—a contradiction. Thus Ada is a liar and Drew is a knight. Ben’s claim now means Ben is a knight exactly when Cora is a liar, so one of them is a knight: 2 knights in all. “It cannot be determined” is tempting because either Ben or Cora could be the knight, but the total is fixed.

• Think about what each digit is worth before and after a swap. The other two digits do not change.
• Swapping the thousands and tens places changes the number by 990 times the difference between those digits. For the hundreds and ones places, the multiplier is 99.

Answer: E — 2341

The first swap shows that the tens digit is 1,980 ÷ 990 = 2 greater than the thousands digit. The second shows that the hundreds digit is 198 ÷ 99 = 2 greater than the ones digit. Using 1, 2, 3, and 4 once each gives two possibilities: 1432 and 2341. Although 1432 satisfies both swaps, it is even. The odd number is 2341.

• Pair the tallest bar with the shortest, then pair the two leftover bars. What useful comparison do those pairs give you?
• The four-week average is 42.5, so a two-week pair needs a total greater than 85.

Answer: E — 2

The four weeks total 170 tickets, giving an average of 42.5. Only weeks 2 and 4 (45 + 55 = 100) and weeks 3 and 4 (40 + 55 = 95) total more than 85, so the answer is 2. Weeks 1 and 4 and weeks 2 and 3 each total exactly 85; counting them would mistake an equal average for a greater one.

• Draw a 3-by-3 grid and trace where each position moves in a half-turn. Use the unchanged color pattern to work out which positions must match.
• Tiles that trade places must have the same color. Think about how many green tiles those matching groups can contribute to the total of five.

Answer: E — Center tile

A half-turn exchanges the eight outer tiles in four pairs and leaves the center tile in place. Each pair contains either zero or two green tiles, so the outer tiles contribute an even number of green tiles. Since the coaster has five green tiles, the center tile must be green. The top-left tile is not required to be green: the center, top-middle, bottom-middle, middle-left, and middle-right tiles could be the five green tiles, leaving the top-left tile white.

• Count cubes rather than pieces.
• How many cubes could 8 pieces hold at most?

Answer: A — 8

The number of cubes stays at 5 × 2 + 7 = 17. Each piece contains at most 2 cubes, so 8 pieces could hold at most 16 cubes: 8 is impossible. Joining zero, one, two, or three pairs of singles gives 12, 11, 10, or 9 pieces. The trap is 9: although 17 cubes cannot all become rods, they can form 8 rods and 1 single.