The terms pole and throw are used to describe switch contact variations. A pole is a set of contacts or terminals that are connected to a single circuit. A throw is one of two or more positions that the switch can adopt. A single-throw switch has one position that closes contacts, a double-throw switch has two positions.

“Single-pole, double-throw” switches, such as the ones illustrated in this puzzle, are called “two-way” switches in British English, whereas they are generally called “three-way” switches in American terminology. A “two-way” switch in American English refers to a “single pole, single throw” switch.

Terminal D must connect to terminal I, and terminal E must connect to terminal J.   D to J and E to I will also work. The light bulb and the power source must be connected in series, A to G, for example. Terminal F then connects to C and terminal B connects to H.   F to H and B to C will also work. Either switch can then turn the bulb on or off.

The area of a right triangle is computed by multiplying the base times the height and dividing by two. The pieces of the puzzle have an area of 32 square units. Although they can be assembled to form what appears to be a 13×5 right triangle, they actually form quadrilaterals that are slightly smaller or slightly bigger than a 13×5 right triangle. The long side of the these “triangles”, i.e, what seems to be the hypotenuse, is not a straight line.

The top figure has an area of 32 square units. The bottom figure, including the empty square, has an area of 33 square units. A real 13×5 right triangle would have an area of 32.5 square units. The distortion is difficult to see because one square of the picture is approximately 3% of the area.

The distortion can be seen more clearly when the empty square constitutes a larger percentage of the area, as in the figure below where 1 square represents 13% of the area.

To calculate the proportion of the constituents of the blend, we need to find out the fatty acid profiles of canola oil and olive oil, and then determine what combination would yield the fatty acid profile of the mixture. We start by looking at the fatty acid composition table:

We calculate the percentage of saturated (%sat), monounsaturated (%mono), and polyunsaturated (%poly) fat for the mixture, and tabulate the saturation profiles for the three oils.

Calculation of the fatty acid percentages for the oil mixture:

%sat = 100×1/14 = 7%
%mono = 100×9/14 = 64%
%poly = 100×4/14 = 29%

We notice that the percentages do not add to 100% for olive oil due to rounding and other constituents not listed. This just means that our answer will not be as accurate as it could be with better analytical data.

If X represents the fraction of olive oil in the mixture, (1-X) is the fraction of canola oil. The proportion of the components can be calculated for the saturated, monounsaturated, and polyunsaturated components separately to arrive at three different answers.

This is a classic problem that many people get wrong because they reason that half of a hen cannot lay an egg, and a hen cannot lay half an egg. However, we can get a satisfactory solution by treating this as a purely mathematical problem where the numbers represent averages.

To solve the problem, we first need to find the rate at which the hens lay eggs. The problem can be represented by the following equation, where RATE is the number of eggs produced per hen·day:

1½ hens × 1½ days × RATE = 1½ eggs

We convert this to fractions thus:
3/2 hens × 3/2 days × RATE = 3/2 eggs

Multiplying both sides of the equation by 2/3, we get:
1 hen × 3/2 days × RATE = 1 egg

Multiplying both sides of the equation again by 2/3 and solving for RATE, we get:
RATE = 2/3 eggs per hen·day

Now that we know the rate at which hens lay eggs, we can calculate how many hens (H) can produce 12 eggs in six days using the following equation:

H hens × 6 days × 2/3 eggs per hen·day = 12 eggs

Solving for H, we get:
H = 12 eggs /(6 days × 2/3 eggs per hen·day) = 12/4 = 3 hens

Therefore, the farmer needs 3 hens to produce 12 eggs in 6 days.

Fiji and American Samoa are in different time zones and are separated by the International Date Line. When it is 11:30 PM on Tuesday in American Samoa, it is 10:30 PM on Wednesday in Fiji. Since the flight takes two hours, you will arrive in Fiji at 12:30 AM on Thursday.

If you went from Fiji to American Samoa at 11:30 AM on Tuesday, you would arrive at 2:30 PM on Monday (the day before you left). The International Date Line was adjusted in 1995 so that all the islands of Kiribati, which span three time zones, would have the same date.

Since every statement is false, let us convert them into true statements, and number each statement:

• Ann: 1) One of us took the painting. 2) The painting was gone when I left.
• Bob: 3) I arrived first, third, or fourth. 4) The painting was still here.
• Chuck: 5) I arrived first, second, or fourth. 6) The painting was gone when I arrived.
• Tom: 7) Whoever stole the painting arrived after me. 8) The painting was still here.

According to statement #7, Tom is not the thief. #8: Since the painting was there when Tom arrived, he could not have been the last to arrive. Tom must have gone there first, second, or third. #6: The painting was gone when Chuck was there, so he didn’t arrive first. #5: So Chuck got there second or fourth. #4 and #8: As two other members (Bob and Tom) arrived to see the painting, Chuck didn’t get there second, either. So Chuck arrived fourth. #3: This means Bob arrived first or third. #2: Since the painting was gone when Ann left, she didn’t arrive first. Otherwise, no member after her would have seen the painting. So Ann went there second or third and Chuck arrived fourth. But since two other members (Bob and Tom) saw the painting when they arrived, Ann didn’t go there second, either. So Ann arrived third. #3: Therefore, Bob arrived first, and Tom arrived second.

In summary, Bob arrived first. Tom got there next and the painting was still there, so Bob was not the thief, and neither was Tom. When Ann arrived, the painting was still there, but it was gone when she left. So Ann was the one who stole the painting. Chuck arrived last and discovered that the painting was gone.

The cost of 200 shares at $50.00 Dollars each is $10,000.00 Dollars.
The 50% gain on $10,000 is $5,000, making the new total $15,000.00 Dollars.
The 40% loss on $15,000 is $6,000, reducing the total amount to $9,000.00 Dollars.
This is a $1000.00 Dollar loss, plus $10 Dollars for the Buy transaction, and $10 Dollars for the Sell transaction.

The investor lost a total of $1020.00 Dollars.

It seems counterintuitive that the investor should lose money when the percentage of the loss is less than the percentage of the gain. However, the percentage of the loss is calculated on a bigger amount than the percentage of the gain.

She has a minimum of 15 stamps.
Mexico: 1 blue, 2 red, 4 unknown color
France: 2 blue, 3 red, 1 unknown color
Unknown country: 1 purple, 1 red

Handshakes are used traditionally as greetings, but they are also used to seal an agreement when a business transaction has been mutually accepted. A handshake is sometimes used to characterize the personality of an individual. A firm handshake is interpreted to indicate an assertive person or an extroverted personality, whereas a less firm or limp handshake is viewed as a sign of weakness and lack of confidence. Some diseases, such as influenza, can be spread by shaking hands with an infected individual and then touching one’s face. (See Hygiene)

With two people (A and B), there is one handshake
(A with B).

With three people (A, B, and C), there are three handshakes
(A with B and C;   B with C).

With four people (A, B, C, and D), there are six handshakes
(A with B, C, and D;   B with C and D;   C with D).

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ … + n.

Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.

This is the quadratic equation n²+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +… etc. until the total is 66. The last number that you entered (11) is n.

There is a saying that you cannot add apples and oranges. If you have 3 apples and 2 oranges do you have 5 apples? No. Do you have 5 oranges? No. You have five fruits, but the number of apples and oranges has not changed. Similarly, you cannot add real money and “what they think they paid”.

When we count only real money, the students have $3, the clerk has $2, and the manager has $25. That is $30 total.