Quantitative Aptitude and Reasoning

When all three taps are on, the total draining rate is R_A + R_B + R_C = k + 2k + 3k = 6k.
It takes 1 hour (60 minutes) to drain the full tank.
So, V = (Total Rate) * Time = (6k) * 60 = 360k.

When taps A and C are on, and B is off, the total draining rate is R_A + R_C = k + 3k = 4k.
Let T be the time in minutes it takes to empty the full tank with A and C on.
The volume drained is V.
V = (Rate A+C) * T = (4k) * T.

We know V = 360k.
So, 360k = 4k * T.
Divide both sides by 4k (assuming k > 0):
T = 360k / 4k = 360 / 4 = 90 minutes.

If car A doubles its speed (2*S_A), the new time taken is T_A’ = D/(2*S_A).
According to the second condition: T_A’ = T_B – 1 => D/(2*S_A) = D/60 – 1 (Equation 2)

Multiply Equation 2 by 2:
2 * [D/(2*S_A)] = 2 * [D/60 – 1]
D/S_A = D/30 – 2

Now we have two expressions for D/S_A:
From Eq 1: D/S_A = D/60 + 1
From the modified Eq 2: D/S_A = D/30 – 2

Equating the two expressions:
D/60 + 1 = D/30 – 2
Rearrange the terms to solve for D:
1 + 2 = D/30 – D/60
3 = (2D – D) / 60
3 = D/60
D = 180 km.

Now substitute the value of D back into Equation 1 to find S_A:
180/S_A = 180/60 + 1
180/S_A = 3 + 1
180/S_A = 4
S_A = 180 / 4
S_A = 45 km/hr.

The average of the first four numbers is 25.
$(n_1 + n_2 + n_3 + n_4) / 4 = 25$
The sum of the first four numbers is $n_1 + n_2 + n_3 + n_4 = 25 \times 4 = 100$. (Equation 1)

The average of the last four numbers is 20.
$(n_2 + n_3 + n_4 + n_5) / 4 = 20$
The sum of the last four numbers is $n_2 + n_3 + n_4 + n_5 = 20 \times 4 = 80$. (Equation 2)

We need to find the difference between the first and the last number, which is $n_1 – n_5$.

Subtract Equation 2 from Equation 1:
$(n_1 + n_2 + n_3 + n_4) – (n_2 + n_3 + n_4 + n_5) = 100 – 80$

Expanding the left side:
$n_1 + n_2 + n_3 + n_4 – n_2 – n_3 – n_4 – n_5 = 100 – 80$

The terms $n_2, n_3, n_4$ cancel out:
$n_1 – n_5 = 20$

The difference between the first and the last number is 20.

Work done = Rate × Time.
Rate = Work / Time.

A and B together complete the work in 10 days.
Their combined rate is $R_A + R_B$.
$(R_A + R_B) \times 10 = W$
$R_A + R_B = \frac{W}{10}$

B alone completes the work in 15 days.
B’s rate is $R_B$.
$R_B \times 15 = W$
$R_B = \frac{W}{15}$

We want to find the time it takes for A alone to finish the work. Let this time be $T_A$.
$R_A \times T_A = W$
$T_A = \frac{W}{R_A}$

Substitute the value of $R_B$ into the combined rate equation:
$R_A + \frac{W}{15} = \frac{W}{10}$

Solve for $R_A$:
$R_A = \frac{W}{10} – \frac{W}{15}$

Find a common denominator for the fractions (LCM of 10 and 15 is 30):
$R_A = \frac{3W}{30} – \frac{2W}{30}$
$R_A = \frac{3W – 2W}{30} = \frac{W}{30}$

Now, calculate the time taken for A alone:
$T_A = \frac{W}{R_A} = \frac{W}{\frac{W}{30}}$
$T_A = W \times \frac{30}{W} = 30$

A alone can finish the work in 30 days.

Part 2: Distance $D_2 = D – D_1 = D – \frac{3}{4}D = \frac{1}{4}D$. Speed $S_2 = v \text{ km/hr}$.
Time taken for Part 2: $T_2 = \frac{D_2}{S_2} = \frac{\frac{1}{4}D}{v} = \frac{D}{4v}$ hours.

Total distance for the journey is $D$.
Total time for the journey is $T_{total} = T_1 + T_2 = \frac{D}{80} + \frac{D}{4v}$.

The average speed for the full journey is given as $50 \text{ km/hr}$.
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$
$50 = \frac{D}{\frac{D}{80} + \frac{D}{4v}}$

We can factor out $D$ from the denominator:
$50 = \frac{D}{D\left(\frac{1}{80} + \frac{1}{4v}\right)}$
$50 = \frac{1}{\frac{1}{80} + \frac{1}{4v}}$

Taking the reciprocal of both sides:
$\frac{1}{50} = \frac{1}{80} + \frac{1}{4v}$

Now, solve for $v$:
$\frac{1}{4v} = \frac{1}{50} – \frac{1}{80}$

Find a common denominator for the right side, which is 400:
$\frac{1}{4v} = \frac{8}{400} – \frac{5}{400}$
$\frac{1}{4v} = \frac{8 – 5}{400}$
$\frac{1}{4v} = \frac{3}{400}$

Cross-multiply:
$4v \times 3 = 1 \times 400$
$12v = 400$

Divide by 12:
$v = \frac{400}{12}$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 4:
$v = \frac{400 \div 4}{12 \div 4} = \frac{100}{3}$

The value of $v$ is $\frac{100}{3} \text{ km/hr}$.