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Q11. What is the volume of soil in an open pit of size 2m x 2m x 10cm?
This is a logical trick question. Because it is an 'open pit', the soil has already been excavated and removed. Therefore, the volume of soil currently inside the pit is zero.
Q12. Three boxes are coloured red, blue and green and so are three balls. In how many ways can one put the balls one in each box such that no ball goes into the box of its own colour?
This is a derangement problem for 3 items (d3). The only possible configurations are (R-ball in B-box, B-ball in G-box, G-ball in R-box) and (R-ball in G-box, B-ball in R-box, G-ball in B-box). There are exactly 2 ways.
Q13. Four small squares of side x are cut out of a square of side 12 cm to make a tray by folding the edges. What is the value of x so that the tray has the maximum volume?
The volume V of the tray is Length * Width * Height = (12 - 2x)(12 - 2x)x. To maximize volume, we find the derivative dV/dx and set it to zero, which yields x = 2 and x = 6. Since x = 6 results in a volume of zero, the maximum volume occurs when x = 2.
Q14. A rectangular flask of length 11 cm, width 8 cm and height 20 cm has water filled up to height 5 cm. If 21 spherical marbles of radius 1 cm each are dropped in the flask, what would be the rise in water level?
The volume of a single spherical marble is (4/3)*pi*r^3 = (4/3)*(22/7)*(1)^3 = 88/21 cm^3. The volume of 21 marbles is 21 * (88/21) = 88 cm^3. This displaced volume causes the water to rise by a height 'h'. The volume of the displaced water column is Base Area * h = 11 * 8 * h = 88h. Equating the two: 88h = 88, so h = 1 cm.
Q15. Moriarty stole the treasure and hid it in one of ten pillars. The note handed to Sherlock read: 'The clue is hidden in this statement'. Which pillar is it?
The statement 'The clue is hidden in this statement' consists of exactly 9 words. In logic puzzles of this type, the count of the key indicator (words) often points to the solution. Therefore, it is the 9th pillar (IX).
Q16. The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips, the probability that he/she will be caught during at least one of the trips is:
The probability of not being caught in a single trip is 1 - 0.1 = 0.9. The probability of not being caught in all 4 trips is (0.9)^4. The probability of being caught at least once is the complement: 1 - (0.9)^4.
Q17. A polynomial f(x) divided by x-5, x-3, or x-2 leaves a remainder of 1. Which of the following could be the polynomial?
The polynomial is of the form k(x-5)(x-3)(x-2) + 1. Expanding (x-5)(x-3)(x-2) gives x^3 - 10x^2 + 31x - 30. Adding the remainder 1 results in x^3 - 10x^2 + 31x - 29.
Q18. A 2 m long ladder is to reach a wall of height 1.75 m. The largest possible horizontal distance of the ladder from the wall could be:
The ladder forms the hypotenuse of a right-angled triangle. Using Pythagoras theorem: Horizontal Distance = sqrt(2^2 - 1.75^2) = sqrt(4 - 3.0625) = sqrt(0.9375). Since sqrt(1) is 1, sqrt(0.9375) must be slightly less than 1 meter.
Q19. A plate of 5m x 2m size with uniform thickness, weighing 20 kg, is perforated with 1000 holes of 5cm x 2cm size. What is the weight of the plate (in kg) after perforation?
The initial area of the plate is 500 cm x 200 cm = 100,000 cm^2. The area of one hole is 5 cm x 2 cm = 10 cm^2. 1000 holes remove an area of 10,000 cm^2. This is exactly 10% of the total area. Removing 10% of the area removes 10% of the weight (2 kg). The remaining weight is 20 - 2 = 18 kg.
Q20. A frog hops exactly 1 meter at a time. What is the least number of hops required to reach a point 10 cm away?
The frog can hop 1m forward and then 1m back at an angle such that the final displacement is only 10cm. This is possible using the triangle inequality where two sides are 1m and the third is 0.1m.