Faucet A fills 1/3 of the tank in 1 hour.
Faucet B drains 1/9 of the tank in 1 hour.
Fraction of tank filled in 1 hour = 1/3 − 1/9 = 3/9 − 1/9 = 2/9
∴ Fraction of tank filled in 1½ hours = 2/9 x 3/2 = 1/3

Substitute n = 0 and n = 2 in the given equation (rule). This gives the following two equations:
a(2) = 4 − 3 a(0)
a(0) = 4 − 3 a(2)
Eliminating a(2), one obtains a(0) = 4 − 3 [4 − 3 a(0)] = −8 + 9 a(0).
Thus, the solution is: a(0) = 1; a(2) = 1

From the definition of ~X,
~ 4.3 = 4 ; ~ 6.7 = 7 ; ~ (−0.9) = −1 ;
The value of the given expression is: 4 + 7 − 1 = 10.

x² < 9 implies −3 < x < 3. Also −2 < y < 1.
Clearly xy < 6 for all possible values of x and y. This product xy approaches 6 as x approaches −3 and y approaches −2.
All other options are false as can be seen by substituting x = y = 0.

Multiply the given expression by −2/3.
Hence, −(2/3)(4x − 3y) = 2y − 8(x / 3) = −(2/3)24 = −16.

The speeds at which the train travels in successive hours are the terms of the arithmetic progression 10, 20, 30,... To find the time (say, t hours) taken by the train to complete the 450-mile journey, the sum of this sequence to the appropriate term is considered as follows.
Time x Speed = Distance = 450 miles.
1 hour x (10 + 20 + 30 + ... to t terms) = 450
Dividing by 10, one obtains 1 + 2 + 3 + ... + t = 45
Now, 1 + 2 + 3 + ... + t = t (t + 1)/2 (a well-known result worth remembering).
So, t (t + 1) = 90 giving t = 9.

Given that [n^1/2] = 17, it is necessary that 17 < n^1/2 < 18.
Squaring throughout, 289 < n < 324.
Clearly there are 35 values of n (natural numbers from 289 to 323) that satisfy [n^1/2] = 17.

Since Anakin bought the watch at a discount of x%, he paid $100(1 − (x/100)) for the watch.
Luke bought the watch from Anakin at discount of y%.
So, Luke paid Anakin 100(1 − (x/100))(1 − (y/100)) = 100 − x − y + (xy/100) = 100 − (x + y − (xy/100)).

0.6313131... = (6/10) + (31/1000) + (31/10⁵) + (31/10⁷) + ...
= (6/10) + (31/10³) x (1 + (1/10²) + (1/10⁴) + ...)
= (6/10) + (31/10³) x (1/(1 − (1/100))) on summing the infinite geometric series
= (6/10) + (31/990) = (594 + 31)/990 = 625/990 = 125/198.

The given numbers are 11/2, 21/4, 33/2, and 42/5.
The required number that would be exactly divisible by all the above fractions would be the least common multiple (LCM) of 11, 21, 33 and 42.
The LCM may be obtained by prime factorization as follows:
21 = 3 x 7; 33 = 3 x 11; and 42 = 2 x 3 x 7 giving LCM = 2 x 3 x 7 x 11 = 462.

Let N be the natural number and L be the larger of the two individual parts.
Then, the given relation is N / L = L / (N − L).
∴ N ² − LN = L².
Let the required ratio be R = N / L. Dividing the above equation by L² gives
R² − R − 1 = 0.
The solution to the above quadratic equation isR = (1 + 5^1/2) / 2.
Since N is a natural number, only the positive root is considered for R.

Since a and b are the roots, x = a or x = b.
So, (x − a) (x − b) = 0.
∴ x² − (a + b) x + ab = 0.
On comparing with given quadratic equation, − (a + b) = a and ab = b.
Since a and b are nonzero, the solution is a = 1 and b = −2.

Rearranging the given expression as follows:
(a² + 9b² − 6ab + (b² + c² − 2bc) = 0
Thus, (a − 3b)² + (b − c)² = 0
Since the square of a quantity is always non-negative, the only possibility is that a = 3b and b = c.

On substituting b = a and c = 3a in abc = 384, one obtains (a) (a) (3a) = 384 or 3a³ = 384.
Thus, a³ = 128 or a = 128^1/3= (64^1/3) (2^1/3) = 4 (2^1/3).

Clearly, 172, 174, 176 and 178 are divisible by 2 and hence not prime.
175 is divisible by 5.
171 and 177 are divisible by 3 (because the sum of their digits is divisible by 3).
The numbers that remain are 173 and 179, which are prime.

Multiply both sides of the inequality by xy; so, left-hand side = y and right-hand side = x.
If xy > 0, then the inequality is unchanged; so, y < x.
But, if xy < 0, then the inequality is reversed; so, y > x.

Let the original radius be R. Then, Original perimeter = 2πR; Original area = πR²; Original ratio = 2/R.
Now, new radius = 1.5R; New perimeter = 3πR; New area = 2.25πR²; New ratio = 2/(1.5R) = 4/(3R).
Fractional change = (4/(3R) − 2/R) / (2/R) = −1/3.
Percentage decrease = 1/3 x 100% = 33.33%.

a, b and m are integers.
Now, 2m − 5 is odd, which implies that a − b is odd.
So, if a is odd, then bis even or vice versa. This eliminates options "a" and "b".
The option "a + 2b" is eliminated since it may be odd or even depending on whether a is odd or even.
ab must be even because either a or b has to be even.
Hence, (ab − 1) must be odd.

The problem statement can be mathematically written as 21 < 7N < 64.
∴ 3 < N < 9.143
The integers that satisfy this inequality are 4, 5, 6, 7, 8 and 9. Their sum is 39.

When two lines are perpendicular to each other, the product of their slopes is −1.
Since the slope of line p is 1/8, the slope of line q must be −8.
Hence, the sum of the slopes of the two lines is (−8) + (1/8) = −63/8.

This problem requires calculation of the number of ways in which 8 different objects can be combined into groups of 3.
The number of ways in which r objects can be chosen from n objects is given by the expression:
ⁿC_r = n! / (r!(n − r)!), where ! stands for the factorial notation.
Since n = 8 and r = 3, the required number of ways is 8! / (3! 5!) = 8 x 7 x 6 / (3 x 2 x 1) = 56.
Note that 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1; 3! = 3 x 2 x 1; and 5! = 5 x 4 x 3 x 2 x 1.

Adding the given equations, 5x + 5y = 5
Dividing by 5, x + y = 1.