Memory Challenge - Working Memory Test

Memory Challenge - Working Memory Test

Study the following picture which includes 20 items for 2 minutes; you should remember what are included in it!

Memory Challenge - Working Memory Test

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Now, take a pen and paper, try to remember what are included in the above illustration, and write them (names) on the paper. You have 2 minutes to remember!

Memory Challenge - Visual Test

Memory Challenge - Visual Test

For one minute (60 seconds), study the following picture where 7 letters are connected by 10 lines:

Figure 1 - Memory Challenge - Visual Test

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The following is a copy of the above picture but one line is missing. You should remember what two letters were connected by that line.

Figure 2 - Memory Challenge - Visual Test

Memory Challenge - Visual String Study

Memory Challenge - Visual String Study

Study the following figure for 2 minutes and then answer the questions that follow:

Memory Challenge - Visual String Study

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Questions:

1. Which of the following letters is not depicted in the figure?

a) H
b) V
c) Y
d) Q

2. Which string of letters is located inside the central circle?

a) BGHF
b) CARW
c) FGHB
d) CRAW

3. Which of the following letters is not depicted in the circle located at the left of the central circle?

a) U
b) V
c) O
d) Q

Equivalent to the original rotated in three dimensions!

Which of A or B is equivalent to the original block rotated in three dimensions?

source: Brain Flexing IQ Tests by Fraser Simpson.

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Answer:

A

Piles with Pebbles

You are given three piles with 5, 49 and 51 pebbles respectively. Two operations are allowed: (a) merge two piles together or (b) divide a pile with an even number of pebbles into two equal piles. Is there a sequence of operations that would result in 105 piles with one pebble each?

source: From a compendium of Math competition problems.

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Answer:


Since all three piles are odd, the first operation must be merger of two existing piles, resulting in an even-sized pile and an odd-sized pile. Let g denote the greatest common divisor of these two pile sizes. Since one pile is odd sized and the other pile is even sized, g must be odd. For odd g, no matter what sequence of operations is carried out (merger of two existing piles or division of an even pile into two equal piles), every pile size will continue to be a multiple of g. If g > 1, then we can never reach a configuration where all piles are size 1. Now, starting with {5, 49, 51}, three states are possible: {54, 51}, {5, 100} and {49, 56}. The GCD is 3, 5 and 7, respectively, in these three states. So we can never get 105 piles with 1 pebble each.

Forgeries

You have 10 bags full of coins, in each bag are 1,000 coins. But one bag is full of forgeries, and you can't remember which one. However, you know that a genuine coin weigh 1 gram, but forgeries weigh 1.1 grams. To hide the fact that you can't remember which bag contains forgeries, you plan to go just once to the central weighing machine to get one accurate weight. How can you identify the bag with the forgeries with only one weighing?

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Answer:

Take 1 coin from the first bag, 2 coins from the second bag, . . ., and 10 coins from the tenth bag and weigh the picked coins together! If there were no forgeries, you know that the total weight should be (1+2+3+ . . . +10) = 55 grams. However, as an example, if the weight is 55.3 grams, then you know that 3 coins are forgeries, so that must be the third bag.

Hard Brain Teaser: The Counterfeit Coin

Hard Brain Teaser: The Counterfeit Coin

One of twelve coins is counterfeit: it is either heavier or lighter than the rest. Is it possible to identify the counterfeit coin in three weighings on a beam balance?


Solution:

This solution is first presented by "BRIAN D. BUNDY":

We may label the coins 1, 2, ..., 12 so that we can distinguish between and identify them using these labels.

1.            1. Weigh 1, 2, 3, 4 against 5, 6, 7, 8.

2. They balance, so 9, 10, 11, 12 contain the odd coin. Weigh 6, 7, 8 against 9, 10, 11.
1. They balance, therefore 12 is the odd coin and so weigh 12 against any other to discover whether it is heavy or light.
2. 9, 10, 11 are heavy and so they contain an odd heavy coin. Weigh 9 against 10. If they balance, 11 is the odd heavy coin, otherwise the heavier of 9 and 10 is the odd coin.
3. If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light coin.
3. 5, 6, 7, 8 are heavy and so either they contain an odd heavy coin or 1, 2, 3, 4 contain an odd light coin. Weigh 1, 2, 5 against 3, 6, 10.
1. They balance, so the odd coin is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy coin.
2. 3, 6, 10 are heavy, so the odd coin can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light coin.
3. 3, 6, 10 are light, so the odd coin is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.
4. If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

Which figure should replace the question mark?

Which figure should replace the question mark?

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C: The dot moves round inside the pentagon 3, 4, 5, … places clockwise at each stage and alternates black/white.

Words in the Direction of Compass

Place all the words listed below in the grid. Each word goes in the direction of a compass point, is in a straight line and starts and finishes in one of the shaded squares.

source: The Great IQ Challenge

Answer:

Number and Kind of Bills

John buys a used car for $2,000 (no tax). He pays in cash as a certain number of $1 bills, 15 times as many $5 bills, and a certain number of $10 bills and 3 times as many $100 bills. How many bills of each kind does John pay for the car?

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Answer:


10 $1 bills, 150 $5 bills, 4 $10 bills, and 12 $100 bills.

(10 * 1) + (150 * 5) + (4 * 10) + (12 * 100) = 10 + 750 + 40 + 1200 = 2000 

Rotation of Pinion P4

The pinions P1, P2, P3, and P4 have 18, 22, 10, and 30 teeth respectively. If pinion P1 rotates 5 times how many times will pinion P4 rotate?

Answer: 3

(5 * 18) / 30 = 3

Odd Year!

Which year does not confirm with the others?

1958

1955

1865

1856

Answer: 1958

1958 >>> 1 + 9 + 5 + 8 = 23

where:

1955 >>> 1 + 9 + 5 + 5 = 20

1865 >>> 1 + 8 + 6 + 5 = 20

1856 >>> 1 + 8 + 5 + 6 = 20

Last Two Terms

What are the last two terms in these series?

Answer: H and 40

Note:

Z is the 26th letter >>> 26 * 5 = 130

26 - 3 = 23; W is the 23rd letter >>> 23 * 5 = 115

23 - 4 = 19; S is the 19th letter >>> 19 * 5 = 95

19 - 5 = 14; N is the 14th letter >>> 14 * 5 = 70

14 - 6 = 8; H is the 8th letter >>> 8 * 5 = 40

Missing a Letter

Which letter will complete this word?

Answer: U. The word is TUTELARY (tutelary: having the guardianship or charge of protecting a person or a thing; guardian; protecting; as, "tutelary goddesses").

Identify the Next Domino

Identify the next domino in the following sequence:

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Answer:

m (+5)            >>> r
68 - (5) * 4      >>> 48
r - 48 == 18 - 48 >>> -30

Five Card Trick

A mathematician asks a volunteer to give him five cards drawn from a pack of fifty-two. He hands one card back to the volunteer and arranges the remaining four in some sequence he chooses. He then hands the sequence to a second volunteer and leaves the room. His assistant enters. The assistant asks the second volunteer to read out aloud the sequence handed to him. The assistant ponders a little and correctly announces the identity of the card held by the first volunteer. How could this be done? In general, how large a deck of cards can be handled if n cards are drawn initially?

Solution: Click here for the solution!

X and Y

Two numbers are such that if the first gets 24 from the second they will be in the ratio 2 to 1, but if the second receives 33 from the first, their ratio will be 1 to 5. What are these two numbers?

Answer: 52 and 62

What are the four numbers?

The sums of four numbers, omitting each of the numbers in turn, are 27, 23, 21 and 19. What are the four numbers?

Answer: 3, 7, 9, and 11
7 + 9 + 11 = 27
3 + 9 + 11 = 23
3 + 7 + 11 = 21
3 + 7 + 9 = 19

Replace the question mark!

Which of the boxes should be used to replace the question mark?

Answer: A