AP ICET 26th April 2019 Shift-2

Let the length  and breadth of the rectangle is a and b.

Area of the rectangle $$=ab$$

As per the condition given in the question, new length of the rectangle = 1.15a and 0.8b

Hence, new area of the rectangle $$=1.15a\times 0.8b=0.92ab$$

So, the required area $$=\dfrac{(ab-0.92ab)\times 100}{ab}=8\%$$

Hence the area of the new rectangle decreased by $$8\%$$.

Let the cost price of one pen = 1Rs.

So, cost price of 10 pens = 10Rs.

Selling price of 8pens = cost price of 10 pens

So, selling price of one pen$$=\dfrac{10}{8}=1.25$$Rs.

Hence profit earned by the trader $$=\dfrac{(1.25-1)\times 100}{1}=25\%$$

Let the cost price of the article is x Rs.

As per the condition, the marked price of article $$=\dfrac{125x}{100}=1.25x$$Rs.

The price of the article after the discount $$=\dfrac{1.25x \times 85}{100}=1.25\times 0.85 \times x$$

The selling price of the article $$1.25\times 0.85 \times x=9350$$Rs.

$$x=\dfrac{9350}{1.25\times 0.85}=8800$$Rs.

Amount invested by A=3.5 lakhs

Amount invested by B=3 lakh

Ratio in the profit = 2:1

Let B started after x months.

So, $$3.5\times 12 : 3\times x= 2:1$$

$$x=\dfrac{3.5\times 12}{3\times 2}$$

$$x=7 months$$

Total amount invested by $$P=400000\times 4+440000\times 8=5120000$$Rs.

Total amount invested by $$Q=600000\times 4+640000\times 8=7520000$$Rs.

Hence the ratio in the investment $$= \dfrac{5120000}{7520000}=\dfrac{512}{752}=\dfrac{32}{47}$$

Hence, the share of P in the investment $$=\dfrac{32\times 112180}{79}=45440$$Rs.

The share of Q $$=\dfrac{47\times 112180}{79}=66740$$Rs.

Hence the required difference $$=66740-45440=21300Rs.$$

As per the question,

A can empty tank in 8 hour and B can empty the tank in 12 hour.

If A and B both open at a time, then tank will be fill in one hour$$\dfrac{1}{8}+\dfrac{1}{12}=\dfrac{3+2}{24}=\dfrac{5}{24}$$

Hence tank will be filled, together by A and B $$\dfrac{24}{5}hour$$

Let C can empty the tank in in x hour.

Hence, it will be filled $$\dfrac{5}{24}-\dfrac{1}{x}=\dfrac{1}{6}$$

$$\dfrac{1}{x}=\dfrac{5-4}{24}=\dfrac{1}{24}$$

Hence, leak can empty the tank in 24 hour.

As per the question,

A can fill the tank in 6 hour.

So, A can fill the tank in one hour $$= \dfrac{1}{6}$$

Tap is open for 3 hour, so tank filled in 3 hour $$= \dfrac{3}{6}=\dfrac{1}{2}$$

Tap B can  empty the tank in 4 hour. So, tap B can empty the tank in one hour $$=\dfrac{1}{4}$$

Hence, tank will be empty in one hour $$=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{3-2}{12}=\dfrac{1}{12}$$

Hence, it will empty in 12 hour. Hence it will empty in $$\dfrac{12}{2}=6 hour$$

Time by which tank will be empty is 3:00PM

Let the usual speed of the man is v, the distance between the home and office is d and time taken by the man is t min.

$$t=\dfrac{d}{v}-----(i)$$

As per the condition given in the question,

As per the given condition,
Let the length of the train =x
Let the speed of the train =v m
Hence, $$t=\dfrac{x}{v}$$
$$\Rightarrow \dfrac{x}{v}=30--------(i)$$
Train crosses the platform in$$50=\dfrac{x+100}{v}--------------(ii)$$
From equation (i) and (ii)
$$\Rightarrow \dfrac{30v+100}{v}=50$$
$$\Rightarrow 50v-30v=100$$
$$\Rightarrow 20v=100$$
$$\Rightarrow v=5m/sec$$
So, length of the train $$x=30\times v=30\times 5=150m$$

Let B alone can finish the work in x days.

A alone can finish the work in 24 days.

Hence A can finish the work in one day $$=\dfrac{1}{24}$$

So, A can finish the work in 4 days $$=\dfrac{4}{24}=\dfrac{1}{6}$$

The remaining work $$=\dfrac{5}{6}$$

A and B together can finish the work in one days $$\dfrac{1}{24}+\dfrac{1}{x}=\dfrac{5}{8\times 6}$$

$$\Rightarrow \dfrac{1}{x}=\dfrac{5}{48}-\dfrac{1}{24}=\dfrac{3}{48}$$

So, $$x=16$$days

Hence, B alone can finish the work in 16 days.