Can anyone help me with how to solve the above problem?

Question 2

As we want make values as low as possible we are looking at extreme cases .

For baseline scores suppose there are 100 men and 100 women, then we will have 45 homeowners, thus eliminate all options greater than this.

Now, lets take a extreme case say we gave 195 men and 5 women, so we will have a total of 39 men + 1.25 women = 40.25 homeowners (lets try different values to eliminate the decimal)

180 men and 20 women( 36+5 = 41 ) , thus ans is E

Where did the denominator ‘9’ of the first and second term go?

I feel the algebra is wrong there… It should be

X/9 + (9XY - X) /45

Now, make the denominator common , your left most term will become 5x

So the common term would be ignored as it’s not present in the options?

For the first problem

Charges for the first 1/9 mile = x

Charges for the rest of distance = (total number of 1/9 miles - 1) * \frac{x}{5}

= (9y - 1)* \frac{x}{5}

= \frac{9xy - x}{5}

Total charges = x + \frac{9xy - x}{5}

@mittalswathi08 is correct

\frac{5x + 9xy - x}{45} = \frac{x}{9} + \frac{9xy - x}{45}

The reason this is not the answer is because this is wrong:

The cost of taxicab for the first 1/9 miles is simply x, not x/9

Yeah, you are correct!!

For the second question, I took a different approach than Vishu’s.

Basically, I agree on using extreme case but I will set the extreme case in this way:

The question asks us to find the minimum number. So how about we ignore 0.25, and use 200 as the total number of men. In this way, we get 40. But we know 40 cannot stand as the final answer because we also need to count that 0.25 portion in. Hence, imagine we decrease the number of men and allocate some number to women. However, no need to allocate too many and therefore 41 is a reasonable answer.

It should be x+ \frac{x}{5}\times\left(9y-1\right)