Speed = Distance / Time.
So, Time = Distance / Speed = (28/70) hr = 0.4 hr = 0.4 x 60 minutes = 24 minutes.

Cross-multiplication gives a = 42 and b = 6/5 = 1.2
Thus, (a + 6) / (b + 6) = (42 + 6) / (1.2 + 6) = 48/7.2 = 2/0.3 = 20/3.

Effectively, the ship sails 12 nautical miles due west and 9 nautical miles due north. These form the sides of a right-angled triangle, whose
hypotenuse is to be determined. By the Pythagorean Theorem,
Hypotenuse = (12² + 9²)^½= (144 + 81)^½ = √225 = 15.

The prime factors of 2310 are 2, 3, 5, 7 and 11. Their product is 2310.
Since r = 5, s = 7 and t = 11, the value of
rs − 3t = (5 x 7) − (3 x 11) = 35 − 33 = 2.

2 + 3 + 5 + 7 + 31 = 48.
So, √48 = √(16 x 3) = √16 x √3 = 4 √3.

In one hour, the four carpenters can individually complete one-third, one-fourth, one-fifth and one-seventh of the task. For completion of the maximum fraction of the task, the rates of the three quickest carpenters must be added. Thus,
(1/3) + (1/4) + (1/5) = (20 + 15 + 12)/60 = 47/60.
In 1 hour, three carpenters together can complete (47/60)^th of the task.
In ¾ hr (i.e., 45 minutes), they can complete (47/80)^th of the task.
Note that (47/60) x ¾ = 47/80.

Diane needs to drive the minimum distance to reach home if the store, college, and her husband's office are on exactly the same route as her drive from her office to home. In that case, the minimum value of s is 3 (i.e., 13 − 2 − 3 − 5).
On the other hand, Diane needs to drive the maximum distance to reach home if the store, college and her husband's office are on a route essentially away from her drive from office to her home. In that case, she retraces her way to her office and then to home. Thus, the maximum value of s is 23 (i.e., 13 + 2 + 3 + 5).

(25^1/3 x 25^1/3) / 25^1/2 = 25^2/3/25^1/2 = 25^{2/3 − 1/2} = 25^1/6 = (5²)^1/6 = 5^1/3

15¹⁸ = (5 x 3)¹⁸ = 5¹⁸ x 3¹⁸
If 3ⁿ is a factor of 15¹⁸, then the maximum possible value of n is 18.

The equations provided are:
p + q − r = s ...(1)
p − q + r = t ...(2)
Adding equations (1) and (2), one obtains p = (s + t) / 2
Subtracting equation (2) from equation (1), one obtains q − r = (s − t) / 2
∴ (q − r) / p = (s − t) / (s + t ).

Let the $8000 amount be divided into two parts x and y such that x + y = 8000.
The annual return on x is 5% and that on y is 3%; so, 5x / 100 = 3y / 100 or y = 5x / 3.
On eliminating y, one obtains x + 5x / 3 = 8000 or 8x / 3 = 8000.
Thus, x = 3000 and y = 5000.
So, 5x / 100 = 150 and 3y / 100 = 150.
∴ Total income after one year is $150 + $150 = $300.

Total price of the mixture is (80 x 10) + (120 x 5) = $800 + $600 = $1400.
Total weight of the mixture is 200 pounds.
∴ The price of the resulting mixture is 1400 / 200 = $7 per pound.

Let y be the original number. After a reduction of 20%, the result is 0.8y.
Now, this number is increased by 25%. The result is 1.25 x 0.8y.
Note that 1.25 = 5/4 and 0.8 = 4/5; so, 1.25 x 0.8y = y.
Since a reduction by 20% and a subsequent increment by 25% gives the original number, the correct answer is 1.643 x 10¹².

Let xy be the two-digit number,
with x being an integer from 1, 2, ..., 9 and y being an integer from 0, 1, ..., 9.
Thus, 10x + y = q(x + y)
So, (q − 1) y = (10 − q) x
The largest possible value that q can take is 10 (since x and y cannot be negative).
If q = 10, then the above equation gives (9) y = (0) x or y = 0 for any x.
Thus, q = 10 = 90/9 = 80/8 = 70/7 = ... = 10/1.

a − p = 6b − 6q
Substituting p = 6q and simplifying gives a = 6b or a / b = 6.
Alternatively, given that (a − p) / (b − q) = 6 = p / q
By the property of equal ratios, we have ((a − p) + p) / ((b − q) + q) = 6 or a / b = 6.
Note that the property of equal ratios states that
given a set of equal ratios, a / b = c / d = e / f = ... = k (a constant),
the ratio (a + c + e + ...) / (b + c + d + ...) = k = a / b = c / d = e / f = ...

2xy is necessarily divisible by 10 because its factors are 2 and 5.
The other quantities are not necessarily divisible by 10 as shown by the examples below.
If x = 15 and y = 10, then x + y = 25.
If x = 15 and y = 10, then x − y = 5.
If x = 15 and y = 5, then xy = 75.
If x = 15 and y = 10, then x² + y² = 225 + 100 = 325.

The general equation of a line is y = mx + c (where m is the slope of the line and c is its y-intercept).
The slope of the line x + 2y − 15 = 0 is −½ (obtained on rearranging the equation as y = −½ x + 15/2).
The slope of a line perpendicular to this line is 2 (since the product of the slopes of the two lines must be −1).
For the line perpendicular to the given line, m = 2 and c = 0 (because it passes through the origin).
So, the required equation of the perpendicular is y = 2x.
The intersection point of the lines (y = 2x & x + 2y = 15) is found by solving the equations simultaneously.
Substitute y = 2x in x + 2y = 15 to get x + 4x = 15 or 5x = 15. So, x = 3 and y = 6.
∴ The required co-ordinates are (3,6).
Note: The correct point must lie on the given line. This fact can be used to eliminate a few of the choices.

The values of x that satisfy |x| < 9 are −9 < x < 9.
There are 17 integers (−8, −7, ..., −1, 0, 1, ..., 7, 8) in this interval.
If both sides of (4 / x) > x are multiplied by a positive value of x, then x² < 4.
The only (positive) integer that satisfies −2 < x < 2 is 1.
If both sides of (4 / x) > x are multiplied by a negative value of x, then x² > 4.
The six integers that satisfy x < −2 are −8, −7, −6, −5, −4 and −3.
Since only 7 out of 17 integers satisfy (4 / x) > x, the probability is 7 / 17.

There are totally 7 balls in the bag.
The probability of drawing a black ball (first from the bag) is 3/7.
Now, 6 balls remain in the bag, and the probability of drawing a white ball is 4/6 or 2/3.
Since the two events are independent, the probability of drawing a black ball and then a white ball is
(3/7) x (2/3) = 2/7.
Note that the order in which the balls are drawn makes no difference to the final answer in this case.

Haan, Jango and Lando receive the money in the ratio 3 : 2 : 1.
Thus, Jango receives (2/6) x $600 = $200.
Alternatively, if the shares of Lando, Jango and Haan are x, y and z dollars respectively, then
x + y + z = $600.
Since Jango receives $2 for every $3 that Haan receives, y = 2z/3
Since Lando receives half a dollar for every dollar Jango receives, x = y/2
Eliminating x and z, one obtains y/2 + y + 3y/2 = $600
∴ 3y = $600 or y = $200.

14 = 2 x 7; 18 = 2 x 3 x 3; 30 = 2 x 3 x 5.
So, Least Common Multiple (LCM) of 14, 18 and 30 = 2 x 3 x 3 x 5 x 7 = 630.
Required number = 630 + 3 = 633.

When the integer n is divided by 3, the quotient is p and the remainder is 1. So, n = 3p + 1
When the integer n is divided by 5, the quotient is q and the remainder is 7. So, n = 5q + 7
Eliminating n, one obtains 3p + 1 = 5q + 7.
∴ 5q − 3p = 1 − 7 = −6.