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We know that, any factor of the (2⁵ × 3³ × 5²) can be expressed as 2^a × 3^b × 5^c, where a ranges from 0 to 5, b ranges from 0 to 3 and c ranges from 0 to 2.

Here, we need to find odd factors of 2⁵ × 3³ × 5², so there will be no factor of 2 in the given number. So, a will be 0.

As a result, a can take only 1 value, b can take 4 values and c can take 3 values.

∴ total number of odd factors = 1 × 4 × 3 = 12

We know that, Average = Sum of all quantities/number of quantities

Average of 7 numbers = 29

∴ sum of all numbers = 29 × 7 = 203

Average of first 3 numbers is 23.

Sum of first 3 numbers = 3 × 23 = 69

Average of last 3 numbers is 42.

Sum of last 3 numbers = 3 × 42 = 126

4^th number = 203 – 69 – 126 = 8

Let’s assume he invested Rs. x each in both firms.

We know that Simple interest = (P × R × T)/100

Where, P = principal, R = % rate of interest, T = time in years

For both investments, P = x, R = 15

T = 3.5 for first investment and T = 5 for second investment.

As per given information:

\(\begin{array}{l} \frac{{x \times 15 \times 5}}{{100}} - \frac{{x \times 15 \times 3.5}}{{100}} = 270\ \Rightarrow \frac{{x \times 15 \times 1.5}}{{100}} = 270 \end{array}\)

⇒ x = 270 × 100/22.5 = 1200

∴ He invested Rs. 1200 in each firm.

Therefore, he invested Rs. 1200 + Rs. 1200 = Rs. 2400 in total.

Let’s assume that every person consumes x units of food every day.

Initially, army of 1800 men has provision for 50 days.

∴ total supplies = 1800 × 50 × x

Let’s assume n men left after 10 days. So for first 10 days, there were 1800 men, but for next 50 days, there were (1800 – n) men.

∴ 1800 × 50 × x = 1800 × 10 × x + (1800 – n) × 50 × x

⇒ 1800 × 40 = (1800 – n) × 50

⇒ 1800 – n = 1440

⇒ n = 360

A finishes the work in 25 days, so A does 1/25^th part of work in one day.

B finishes the work in 25 days, so B does 1/25^th part of work in one day.

A and B together work for 5 days.

∴ Part of work finished by both in 5 days \(= 5 \times \left( {\frac{1}{{25}} + \frac{1}{{25}}} \right) = \frac{{10}}{{25}} = \frac{2}{5}\)

Part of work remaining \(= 1 - \frac{2}{5} = \frac{3}{5}\)

Number of days taken by A to finish remaining work \(= \frac{3}{5} \div \frac{1}{{25}} = 15\)

\(\begin{array}{l} 5{\left( {\sqrt 5 } \right)^{x + 5}} = {\left( {\sqrt 5 } \right)^{2x + 7}}\ {\left( {\sqrt 5 } \right)^2}{\left( {\sqrt 5 } \right)^{x + 5}} = {\left( {\sqrt 5 } \right)^{2x + 7}}\ {\left( {\sqrt 5 } \right)^{x + 7}} = {\left( {\sqrt 5 } \right)^{2x + 7}} \end{array}\) 

From laws of indices: x + 7 = 2x + 7

⇒ x = 0

Speed of train = 36 km/hr = 36000/3600 m/s = 10 m/s

When a train crosses a bridge, it travels a distance equal to its own length and the length of the bridge.

Since the train crosses a bridge in 30 sec,

Distance traveled by train = 30 × 10 = 300 m

∴ Length of train + Length of bridge = 300

⇒ Length of bridge = 300 – 130 = 170 m

Let the three shares be a, b and c.

As per given information, 4a = 3b = 2c = k

⇒ a = k/4, b = k/3 and c = k/2

Now, a + b + c = 1833

\(\Rightarrow \frac{k}{4} + \frac{k}{3} + \frac{k}{2} = 1833\) 

\(\Rightarrow \frac{{\left( {3 + 4 + 6} \right)k}}{{12}} = 1833\) 

⇒ k = 1833 × 12/13 = 1692

∴ 2^nd share = b = k/3 = 1692/3 = 564

Let’s assume the speed in still water is x km/hr and speed of the stream is y km/hr.

Now, speed with the stream = x + y km/hr

And speed against the stream = x – y km/hr

As per given information, x + y = 8 and x – y = 4

Solving the two equations simultaneously, we get: x = 6 and y = 2

Hence, speed of boat in still water = 6 km/hr

y exceeds x by 10%.

∴ y = 1.1x

x = y/1.1

Required percentage \(= \frac{{y - \frac{y}{{1.1}}}}{y} \times 100 = \frac{{1.1 - 1}}{{1.1}} \times 100 = 9\frac{1}{{11}}\)