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Interesting Riddles with Their Solutions – You Must be Curious

If you are sharp enough, full with presence of mind, interesting thinker, then try to solve the following riddles with your own sweet logic. Unfurling the riddles will definitely make you funnier and help you build your logic stronger.

Riddles and Puzzles

Below you are given 10 interesting riddles. Try to solve them yourself:

Q1. He who makes it, doesn’t need it actually.
He who buys it, doesn’t use it anyway.
Who finally use it, does neither see or feel it.
What is it?

Answer

Answer

A Coffin

Source:

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Q2. If you want to make a building made of bricks, then how many bricks you required to complete it?

Answer

Answer

Only “One” brick – the last “one”

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Q3. CONCERNING A CHECK

A man went into a bank to cash a check. In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He pocketed the money without examining it, and spent a nickel on his way home. He then found that he possessed exactly twice the amount of the check. He had no money in his pocket before going to the bank. What was the exact amount of that check?

Answer

Answer

The amount must have been $31.63. He received $63.31. After he had spent a nickel there would remain the sum of $63.26, which is twice the amount of the check

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Q4. DOLLARS AND CENTS

A man entered a store and spent one-half of the money that was in his pocket. When he came out he found that he had just as many cents as he had dollars when he went in and half as many dollars as he had cents when he went in. How much money did he have on him when he entered?

Answer

Answer

The man must have entered the store with $99.98 in his pocket

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Q5. LOOSE CASH

What is the largest sum of money-all in current coins and no silver dollars-that I could have in my pocket without being able to give change for a dollar, half dollar, quarter, dime, or nickel?

Answer

Answer

The largest sum is $1.19, composed ofa half dollar, quarter, four dimes, and four pennies.

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Q6. GENEROUS GIFTS

A generous man set aside a certain sum of money for equal distribution weekly to the needy of his acquaintance. One day he remarked, “If there are five fewer applicants next week, you will each receive two dollars more.” Unfortunately, instead of there being fewer there were actually four more persons applying for the gift. “This means,” he pointed out, “that you will each receive one dollar less.” How much did each person receive at that last distribution?

Answer

Answer

At first there were twenty persons, and each received $6.00. Then fifteen persons (five fewer) would have received $8.00 each. But twenty-four (four more) appeared and only received $5.00 each. The amount distributed weekly was thus $120.00.

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Q7. BUYING BUNS

Buns were being sold at three prices: one for a penny, two for a penny, and three for a penny. Some children (there were as many boys as girls) were given seven pennies to spend on these buns, each child to receive exactly the same value in buns. Assuming that all buns remained whole, how many buns, and of what types, did each child receive?

Answer

Answer

There must have been three boys and three girls, each of whom received two buns at three for a penny and one bun at two for a penny, the cost of which would be exactly 7¢.

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Q8. THE FALSE SCALES

Picture Riddle

A pudding, when put into one of the pans of these scales, appeared to weigh four ounces more than nine-elevenths of its true weight, but when put into the other pan it appeared to weigh three pounds more than in the first pan. What was its true weight?

Answer

Answer

If the scales had been false on account of the pans being unequally weighted, then the true weight of the pudding would be 154 oz., and it would have weighed 130 oz. in one pan and 178 oz. the other. Half the sum of the apparent weights (the arithmetic mean) equals 154. But the illustration showed that the pans weighed evenly and that the error was in the unequal lengths of the arms of the balance. Therefore, the apparent weights were 121 oz. and 169 oz., and the real weight 143 oz. Multiply the apparent weights together and we get the square of 143-the geometric mean. The lengths of the arms were in the ratio 11 to 13.

If we call the true weight x in each case, then we get the equations:

answer

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Q9. THE FIVE CARDS

Picture Riddle 1

I have five cards bearing the figures 1, 3, 5, 7, and 9. How can I arrange them in a row so that the number formed by the first pair multiplied by the number formed by the last pair, with the central number subtracted, will produce a number composed of repetitions of one figure? Thus, in theexample I have shown, 31 multiplied by 79 and 5 subtracted will produce 2444, which would have been all right if that 2 had happened to be another 4. Of course, there must be two solutions, for the pairs are clearly interchangeable.

Answer

Answer

The number is either 3 9 1 5 7 or 5 7 1 3 9. In either case the product of the two pairs 39 and 57, minus 1, results in 2222.

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Q10. SQUARING THE DIGITS

Picture Riddle 2

Take nine counters numbered I to 9, and place them in a row as shown. It is required in as few exchanges of pairs as possible to convert this into a square number. As an example in six pairs we give the following: 78 (exchanging 7 and 8), 8 4, 4 6, 6 9, 9 3, 3 2, which gives us the number 139,854,276, which is the square of 11,826. But it can be done in much fewer moves.

Answer

Answer

In four moves 73,34,48,25, we can get 157,326,849, which is the square of 12,543. But the correct solution is I 5, 84, 46, which gives us the number 523,814,769, the square of 22,887, which is in three moves only.

Source: Q3 – Q10

 

About Puzzleduniya

I am a quiz, puzzle lover, who always intended to post different type of study materials, which will help the all the competitive job seekers all around the world.

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