Mathematics

Let tap A be turned off after $x$ hours. Tap B runs for the whole 12 hours, while tap A runs only for $x$ hours.

Rates: tap A fills $\frac{1}{30}$ and tap B fills $\frac{1}{15}$ of the tank per hour.

$\frac{x}{30}+\frac{12}{15}=1$ $\frac{x}{30}=1-\frac{4}{5}=\frac{1}{5}$ $x=6$

Tap A was turned off after 6 hours.

Check: A fills $\frac{6}{30}=\frac15$, B fills $\frac{12}{15}=\frac45$, total $=1$ tank.

Let the cost price of the first bicycle be Rs $x$ and the second bicycle be Rs $y$.

First bicycle: sold at 40% profit
$SP = 1.40x = 7140$ $x = \frac{7140}{1.40} = 5100$

Profit on first bicycle: $7140 - 5100 = 2040$

Second bicycle: sold at 15% loss
$SP = 0.85y = 7140$ $y = \frac{7140}{0.85} = 8400$

Loss on second bicycle: $8400 - 7140 = 1260$

Total cost price: $CP = 5100 + 8400 = 13500$

Total selling price: $SP = 7140 + 7140 = 14280$

Overall profit: $14280 - 13500 = 780$

Overall profit percentage: $\frac{780}{13500} \times 100 = 5.78\% \text{ (approx.)}$

Answer: Arjun made an overall profit of 5.78% (approximately).

Given: Total amount = Rs. 840, ratio = 2 : 3 : 7

Sum of ratio parts: $2 + 3 + 7 = 12$

Value of one part: $\text{One part} = \frac{840}{12} = \text{Rs. } 70$

Shares:

• First share $= 2 \times 70 =$ Rs. 140
• Second share $= 3 \times 70 =$ Rs. 210
• Third share $= 7 \times 70 =$ Rs. 490

Check: $140 + 210 + 490 = 840$ ✓

Answer: The amount is divided as Rs. 140, Rs. 210 and Rs. 490.

Total balls $=2+3+2=7$

Non-blue balls (red + green) $=2+3=5$

Two balls are drawn at random.

Favourable cases: both balls are non-blue.

$P(\text{none is blue})=\frac{{}^5C_2}{{}^7C_2}$

$=\frac{\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}$

$=\frac{10}{21}$

Answer: $\boxed{\frac{10}{21}}$