RPF 11 Jan 2019

Let total distance = $$2d$$ km

Using, time = distance/speed

=> $$\frac{d}{20}+\frac{d}{30}=7.5$$

=> $$\frac{30d+20d}{600}=7.5$$

=> $$5d=7.5\times60$$

=> $$d=7.5\times12=90$$

$$\therefore$$ Total distance of the entire journey = $$2\times90=180$$ km

=> Ans - (D)

Let ideal time taken = $$t$$ hours

Also, speed is inversely proportional to time.

=> $$\frac{20}{24}=\frac{t-\frac{7.5}{60}}{t+\frac{7.5}{60}}$$

=> $$5t+\frac{7.5}{12}=6t-\frac{7.5}{10}$$

=> $$6t-5t=\frac{7.5}{12}+\frac{7.5}{10}$$

=> $$t=\frac{37.5+45}{60}=\frac{82.5}{60}$$

$$\therefore$$ Distance = speed $$\times$$ time

= $$20\times(\frac{82.5}{60}+\frac{7.5}{60})$$

= $$\frac{90}{3}=30$$ km

=> Ans - (B)

Selling price = Rs. 1596 and profit % = 14%

=> Cost price = $$\frac{1596}{(100+14)}\times100$$

= $$14\times100=Rs.$$ $$1400$$

$$\therefore$$ Amount which is equal to half of the cost price of the product = $$\frac{1400}{2}=Rs.$$ $$700$$

=> Ans - (C)

Let number of employees in the firm initially = 100 and wages per head = Rs. 100

=> Total wages = $$100\times100=Rs.$$ $$10,000$$

Number of employees after increase of 15% = $$100+\frac{15}{100}\times100=115$$

Similarly, new wages per head = $$Rs.$$ $$85$$

=> New wages = $$85\times115=Rs.$$ $$9775$$

$$\therefore$$ Decrease in total wages = $$\frac{10000-9775}{10000}\times100=2.25\%$$

=> Ans - (A)

Quantity of milk in 175 litres mixture = $$\frac{3}{(3+4)}\times175=75$$ litres

=> Quantity of water = $$175-75=100$$ litres

Let $$x$$ litres of water is to be added.

=> $$\frac{75}{102+x}=\frac{3}{5}$$

=> $$375=300+3x$$

=> $$3x=375-300=75$$

=> $$x=\frac{75}{3}=25$$

$$\therefore$$ Quantity of water to be further added is 25 litres.

=> Ans - (A)

Numbers which are divisible by 24, must first be divisible by 3 and also by 8. Checking divisibility by 8.

109644, 190446 and 446190 are not divisible by 8, so only first number is divisible by 8 and 3.

Thus, 109464 is divisible by 24.

=> Ans - (A)