Step 1: Work Done by A.
A completes \( \frac{7}{10} \) of the work in 15 days. The work done by A in 1 day is:
\[
\text{Work rate of A} = \frac{7}{10} \div 15 = \frac{7}{150}
\]
Step 2: Work Done by A and B Together.
The remaining work is \( 1 - \frac{7}{10} = \frac{3}{10} \). A and B together complete this work in 5 days, so their combined work rate is:
\[
\text{Work rate of A and B} = \frac{3}{10} \div 5 = \frac{3}{50}
\]
Since A's work rate is \( \frac{7}{150} \), the work rate of B is:
\[
\text{Work rate of B} = \frac{3}{50} - \frac{7}{150} = \frac{9}{150} - \frac{7}{150} = \frac{2}{150} = \frac{1}{75}
\]
Step 3: Time for A and B Together to Complete the Work.
The combined work rate of A and B is:
\[
\text{Work rate of A and B together} = \frac{7}{150} + \frac{1}{75} = \frac{7}{150} + \frac{2}{150} = \frac{9}{150} = \frac{3}{50}
\]
The total time to complete the entire work is:
\[
\text{Time} = \frac{1}{\text{Work rate of A and B together}} = \frac{50}{3} = 13 \frac{1}{3} \, \text{days}
\]
Final Answer: \[ \boxed{13 \frac{1}{3} \, \text{days}} \]
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: