Since 4 quarters make a dollar, $13.35 requires 53 quarters and 1 dime.
It is not possible to replace a quarter with dimes; however, it is possible to replace 2 quarters with 5 dimes.
The different combinations can be represented by ordered pairs where Q denotes quarters and D denotes dimes as follows:
(53Q, 1D), (51Q, 6D), (49Q, 11D), ..., ..., (1Q, 131D).
So, there are 27 different combinations.

S(12) = 1 + 2 + 3 + ... + 10 + 11
E(25)/2 = (2 + 4 + 6 + ... + 22 + 24)/2 = 1 + 2 + 3 + ... + 11 + 12

Adding the two equations gives 6z = 54 or z = 9.

If y = 1, then Quantity A equals Quantity B.
If y > 1, then Quantity A is greater, e.g., if y = 4, then Quantity A = 4 and Quantity B = 2.
If y < 1, then Quantity B is greater, e.g., if y = 1/4, then Quantity A = 1/4 and Quantity B = 1/2.

Let the radius of the large circle be R and the radius of each of the four small circles be r.
Total area of four identical small circles = 4 (π r²).
Area of colored region = π R² − 4 (π r²).
Since these areas are equal, π R² − 4 π r² = 4 π r² or π R² = 8 π r².
Thus, R = √8 r = √2 (2r) = 1.414 (diameter of the small circle).

Let a be the length of the side of the square.
Area of colored square = a².
The diagonal of the square is the radius of the circle.
By Pythagoras' Theorem, a² + a² = r² or 2 a² = r².
Area of colored square = a² = 0.5r².

One could use the calculator. Then, 5/6 = 0.833
(5/6)⁴ = (5 x 5 x 5 x 5)/(6 x 6 x 6 x 6) = 625/1296 = 0.482
It is worth noting that if 0 < y < 1, then yⁿ will always be less than y for n > 1.

Note that p, q, r and s are negative.
Let p = −8, q = −5, r = −3 and s = −1.
Then, pq = 40 and rs = 3.
In fact, pq is always greater than rs in this case.

Let l be the length of the rectangle and b its breadth.
Then, Perimeter of rectangle = 2 (l + b) and Area of rectangle = l b.
If perimeter (of rectangle A) = 24, then l + b = 12. Let l = 11 and b = 1, then area = 11. On the other hand, if l = 7 and b = 5, then area = 35.
If perimeter (of rectangle B) = 36, then l + b = 18. Let l = 10 and b = 8, then area = 80. On the other hand, if l = 17 and b = 1, then area = 17.
So, Quantity B is greater if area of rectangle A is 11 and area of rectangle B is 80. On the other hand, Quantity A is greater if area of rectangle A is 35 and area of rectangle B is 17.

Let C be the total number of crayons.
Number of blue crayons = C/8.
Number of red crayons = C/24.
Number of blue and red crayons = (C/8) + (C/24) = (3C +C)/24 = 4C/24 = C/6.
Number of green crayons = C − (C/6) = 5C/6.

(1/3) + (1/6) = (2/6) + (1/6) = 3/6 = 1/2
Also, (0.2)² = 0.2 x 0.2 = 0.04

The number of male employees in the New York office is 13 or more (i.e., more than half of 25), which is greater than the total number of employees in the Los Angeles office (which is 12).

It is important to note that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
So, 5 + 12 > z or z < 17.
Also, 5 + z > 12 or z > 7.
So, 7 < z < 17.

If David had earned a 70 on each his first three tests, his average would have obviously been 70. But, he earned 10 marks more in his third test, so his average after three tests will be 73.33 (i.e., 70 + 10/3).
Since David gets a 75 on each of the fourth and fifth tests (which is higher than 73.33), he will end up with a higher average after five tests.
A quick way to compute the average after five tests (which is not necessary) is to use the method followed earlier. Taking 70 as the reference, David earned 10 marks more in his third test, 5 marks more in his fourth test, and 5 marks more in his fifth test. So, his average after 5 tests would be 74 (i.e., 70 + 20/5).
Note that the average (arithmetic mean) of n numbers is their sum divided by n.

This problem can be solved without actually calculating the probabilities. The probability of getting three heads in five coin tosses is the same as the probability of getting two tails. Further, by a symmetry argument, the probability of getting two heads is identical to the probability of getting two tails.

58⁵⁹ + 59⁵⁹ < 59⁵⁹ + 59⁵⁹ = 2 (59⁵⁹) < 59 (59⁵⁹) = 59⁶⁰ < 60⁶⁰.
Note your calculator will give an error message because the numbers are too big.
The only other alternative is to use the method of induction (or reason by analogy) as follows.
1² + 2² = 4 which is less than 3³ = 27.
2³ + 3³ = 35 which is less than 4⁴ = 256.

Because lines A and B are perpendicular,
(slope of line A) x (slope of line B) = −1.
Slope of line A = (2 − 4)/(3 − 1) = −1.
So, slope of line B = 1.
Alternatively, a quick sketch will show that line A is sloping downwards and has a negative slope. A line perpendicular to it will slope upwards and have a positive slope.

Let P be the regular price of the computer system at XYZ Store.
Then, the regular price of the computer system at ABC Store is 0.8 P.
Quantity A = 0.95 (0.8 P) = 0.76 P and Quantity B = 0.75 P.

At 65 miles per hour, it will take less than two hours to drive 110 miles.
At 55 miles per hour, it will take more than two hours to drive 130 miles.
It is not necessary to actually compute the time using the formula:
Time = Distance/Speed.
Quantity A = 110/65 < 2 hours and Quantity B = 130/55 > 2 hours.

Taking the cube of Quantity A gives 54/(3³) = 54/27 = 2.
Taking the cube of Quantity B gives (2³)/4 = 8/4 = 2.

A circle whose radius is 19 has a diameter equal to 38. A circle with a larger diameter will have a larger circumference, because circumference C is proportional to diameter D (note C = π D).

When y is negative, −3y is necessarily positive and y³ is necessarily negative, e.g., when y = −2, −3y = 6 and y³ = −8.

Let the area of the square be a². Then, the side of the square equals a, and the perimeter of the square equals 4a.
Now, area of circle = a² = π r² where r is the circle's radius.
So, Circumference of circle = 2 π r = 2  π (a/√π) = 2 √π a = 3.545 a.

If a is the length of an edge of the cube, then
volume = a³ and surface area = 6a².
Now, a³ = 2 (6a²) or a = 12.

A little visualization with a sketch may help solve this problem.
There are 8 small cubes with no blue faces (the 2" x 2" x 2" core of the 4" cube).
There are 24 small cubes with one blue face (4 each at the center of each of the 6 faces of the 4" cube).
There are 24 small cubes with two blue faces (2 each at the center of each of the 12 edges of the 4" cube).
There are 8 small cubes with three blue faces (at the corners of the 4" cube).