Maths Puzzles Test
A marble statue is being carved by a sculptor. In the first week 35% is cut away, in the second week 20% of the remainder is cut away and in the third week 25% of the remainder is cut away and polished to produce the final statue. The weight of the final
  • 125
  • 135
  • 115
  • 105
Explanation: Reversing the calculations works perfectly: • Start with the final weight: 48.75 • Divide by 0.75 (undo the 25% cut): 48.75÷0.75=65 • Divide by 0.8 (undo the 20% cut): 65÷0.8=81.25 • Divide by 0.650.65 (undo the 35% cut): 81.25÷0.65=125 The original marble weighed 125 lbs.
If a sheet of cardboard has an area of 186 square inches, and two pieces each measuring 6 inches x 3 are cut out, what is the area of the remaining cardboard?
  • 110
  • 150
  • 160
  • 200
Explanation: • Area of the original cardboard = 186 square inches186 • Area of one piece cut out = 6×3=18 square inch • Area of two pieces cut out = 2×18=36 square inches • Remaining area = 186−36=150 square inch Final Answer: 150 square inches
A test was taken by 60 students and was scored on a scale from 0 to 100. Only 21 students scored higher than or equal to 80. What is the smallest possible average score of all 60 students?
  • 26
  • 28
  • 18
  • 30
Explanation: • 21 students scored at least 80: To minimize the average, assume all 21 scored exactly 80. 21×80=1680 • The other 39 students scored 0: 39×0=039 • Total score: 1680+0=1680 • Average score: 1680/60=28 Smallest possible average = 28.
Siva is 10 years old, and his mother is four times as old as he is. When John is 15 years old, how old will his mother be? (Assuming she is not dead.)
  • 40
  • 45
  • 50
  • 55
Explanation: • Siva's current age: 10 years old. • His mother's current age: 4 times his age: 4×10=40 years old. • When Siva is 15 years old, 5 years will have passed (since 15−10=5) • Add 5 years to his mother's age: 40+5=45 years old. Answer: When Siva is 15 years old, his mother will be 45 years old.
An aircraft flew from A to B at an average speed of 230 m.p.h. It returned from B to A at an average speed of 300 m.p.h. what was its average speed for the two journeys?
  • 260
  • 270
  • 265
  • 250
Explanation: The formula for the average speed when different speeds are involved over the same distance: Average speed = 2 × Speed1 × Speed2 / Speed1 + Speed2 Here, Speed1=230 m.p.h. (from A to B), and Speed2 = 300m.p.h. (from B to A). Substituting these values into the formula: Calculate average speed Average_speed = (2 * speed_1 * speed_2) / (speed_1 + speed_2) Average speed = 2 × 230 × 300 / 230 + 300 The average speed for the entire journey is approximately 260.38 m.p.h.
I am four times as old as my son. In 20 year's time, I shall be twice as old as him. How old are we today?
  • 30
  • 40
  • 45
  • 50
Explanation: • Let your son's age be x. • Your age is 4x because you are four times as old as him. In 20 years: • Your son's age will be x+20. • Your age will be 4x+20. The problem says that in 20 years, you'll be twice as old as your son: 4x+20 = 2(x+20) Simplifying the equation: 4x+20 = 2x+40 2x=20 x=10 So, your son is 10 years old and you are 4x10 = 40 years old today.
Midway through his round a golfer hits a 203 yard drive which brings his average length per drive for the round up to now from 176 to 179 yards. How far would he have had to hit the drive to bring his average length of drive up from 176 to 181 yards?
  • 221
    • Explanation: • The golfer's average drive before hitting the 203-yard drive was 176 yards. • After hitting the 203-yard drive, his average increased to 179 yards. We need to find the length of the drive needed to increase his average from 176 to 181 yards. Determine the number of previous drives Let the number of previous drives be nnn. The total distance of the previous drives would be 176n yards. After hitting the 203-yard drive, the total number of drives becomes n+1, and the new total distance becomes 176n+203. Now, the new average is 179 yards, so: 176n+203/n+1 = 179 Let's solve for n. Step 2: Solve for n 176n + 203 = 179(n+1) 176n + 203 = 179n + 179 203 − 179 = 179n − 176n 24 = 3n n = 8 So, the golfer had hit 8 drives before the 203-yard drive. Find the required drive length for the average of 181 yards Now, we need to find the length of the drive required to raise the average from 176 to 181 yards after 9 drives. Let the new drive length be x. After this drive, the total number of drives is 9, and the total distance becomes 176 × 8 + x The new average will be 181 yards: 176 × 8 + x / 9 = 181 1408 + x / 9 = 181 Multiply both sides by 9: 1408+x=1629 x=1629−1408 x=221 Answer: The golfer would need to hit a 221-yard drive to bring his average length up to 181 yards. 2nd Type The simplified summary: • 8 holes, average 176 yards = 1,408 yards. • 9 holes, average 179 yards = 1,611 yards. • 9 holes, average 181 yards = 1,629 yards. To increase the average to 181 yards, the golfer would need a drive of: 1,629−1,408=221 yards. So, a 221-yard drive is needed to bring the average to 181 yards.