Solution:
Number of questions answered correctly = 40% of 125 = 50
For 60% grade, number of questions to be answered correctly = 60% of 250 = 150
Remaining number of questions to be answered correctly
= 150 - 50
= 100
∴ Required percentage
$$\eqalign{ = \left( {\frac{{100}}{{125}} \times 100} \right)\% \cr = 80\% \cr} $$

Solution:
60% of 264
$$\eqalign{ = \left( {\frac{{60}}{{100}} \times 264} \right) \cr = 158.40 \cr} $$
10% of 44
$$\eqalign{ = \left( {\frac{{10}}{{100}} \times 44} \right) \cr = 4.40 \cr} $$
15% of 1056
$$\eqalign{ = \left( {\frac{{15}}{{100}} \times 1056} \right) \cr = 158.40 \cr} $$
30% of 132
$$\eqalign{ = \left( {\frac{{30}}{{100}} \times 132} \right) \cr = 39.60 \cr} $$
∴ 60% of 264 = 15% of 1056

Solution:
Let the amount taxable purchases be Rs. x
$$\eqalign{ {\text{Then}},6\% \,\,{\text{of}}\,x = \frac{{30}}{{100}} \cr \Rightarrow x = {\frac{{30}}{{100}} \times \frac{{100}}{6}} = 5 \cr \therefore {\text{Cost of tax free items}} \cr = {\text{Rs}}{\text{.}}\left[ {25 - \left( {5 + 0.30} \right)} \right] \cr = {\text{Rs}}{\text{.}}\,19.70 \cr} $$

Solution:
$$\eqalign{ {\text{Let the positive number be x}}. \cr {\text{According}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr x \times x = x + \frac{{\left( {x \times 2000} \right)}}{{100}} \cr {x^2} = x + 20x \cr {x^2} - 21x = 0 \cr {\text{Either}}, \cr x = 0,\,\,or \cr x = 21 \cr {\text{21}}\,{\text{is}}\,{\text{the}}\,{\text{possible}}\,{\text{value}}{\text{.}} \cr {\text{Then}}\,{\text{square}}\,{\text{of}}\,21\,{\text{is}}\,441 \cr} $$

Solution:
Let 23% of 8040 + 42% of 545 = x% of 3000
Then,
$$\left( {\frac{{23}}{{100}} \times 8040} \right) + \left( {\frac{{42}}{{100}} \times 545} \right)$$       = $$\left( {\frac{x}{{100}} \times 3000} \right)$$
$$\eqalign{ \Rightarrow 30x = 1849.2 + 228.9 \cr \Rightarrow 30x = 2078.1 \cr \Rightarrow x = \frac{{2078.1}}{{30}} \cr \Rightarrow x = 69.27 \cr} $$

Solution:
Let the distance be x km and the original time time taken be y hours.
Then,
Original speed = $$\frac{x}{y}$$ km/hr
New speed :
$$\eqalign{ = \left( {\frac{x}{{80\% {\text{ of }}y}}} \right){\text{ km/hr}} \cr = \left( {\frac{5}{4}.\frac{x}{y}} \right){\text{ km/hr}} \cr} $$
Increase in speed :
$$\eqalign{ {\text{ = }}\left( {\frac{{5x}}{{4y}} - \frac{x}{y}} \right) \cr {\text{ = }}\frac{x}{{4y}} \cr} $$
∴ Increase %
$$\eqalign{ {\text{ = }}\left( {\frac{x}{{4y}} \times \frac{y}{x} \times 100} \right)\% \cr = 25\% \cr} $$

Solution:
Lets 85% of 485.5 = 50% of x
Then,
$$\eqalign{ \Rightarrow \left( {\frac{{85}}{{100}} \times 485.5} \right) = \frac{{50x}}{{100}} \cr \Rightarrow \frac{x}{2} = 412.675 \cr \Rightarrow x = 825.35 \cr} $$