Question 1: The speed of an slow train is $x$ km/hr and that of an fast train is $(x+25) \ km/hr$.

• Find the time taken by each train to cover 300 km.
• If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.

Time taken by slow train $= \frac{300}{x} h\ r$

Time taken by fast train $= \frac{300}{x+25} \ hr$

Given $\frac{300}{x} = \frac{300}{x+25} + 2$ $300(x+25) = x[300+2(x+25)]$ $2x^2+50x-7500 \Rightarrow x = 50 \ or \ -75 (ruled \ out)$

The speed of the slow train = 50 km/hr and that of the fast train is 75 km/hr. $\\$

Question 2: If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. find the speed of the car.

Let the speed of the car be $x \ km/hr$

Therefore $\frac{36}{x} = \frac{36}{x+10} + \frac{18}{60}$ $\frac{360}{x} = \frac{360}{x+10} +3$ $360x+3600=3x^2+390x$ $3x^2+30x-3600=0 \Rightarrow x = 30 \ or \ -40 (ruled \ out)$

Therefore speed of the car is 30 km/hr. $\\$

Question 3: If the speed of an airplane is reduced by 40 km per hr, it takes 20 minutes more to cover 1200 km. find the speed of the airplane.

Let the speed of the airplane be $x \ km/hr$

Therefore $\frac{1200}{x} + \frac{20}{60} = \frac{1200}{x-40}$ $\frac{1200}{x} + \frac{1}{3} = \frac{1200}{x-40}$ $(3600+x)(x-40) = 3600x$ $x^2-40x-144000 =0 \Rightarrow x = 400 \ or \ -360 (ruled \ out)$

Therefore speed of the airplane is 400 km/hr. $\\$

Question 4: A car covers a distance of 400 km at certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.    

Let the speed of the car be $x \ km/hr$

Therefore $\frac{400}{x} = \frac{400}{x+12} + \frac{100}{60}$ $\frac{2400}{x} = \frac{2400}{x+12} +10$ $2400x+28800=10x^2+2520x$ $10x^2-120x-28800=0 \Rightarrow x = -60\ or \ 48 (ruled \ out)$

Therefore speed of the car is 48 km/hr. $\\$

Question 5: A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of $x \ km/hr$. and the remaining distance at a speed of $(x+2) \ km/hr$. if she takes 2 hrs 30 minutes to cover the whole distance, find $x$. $\frac{6}{x}+\frac{6}{x+2} = \frac{150}{60}$ $6(x+2)+6x=\frac{5}{2}x(x+2)$ $24x+24= 5x^2+10x$ $5x^2-14x-24=0 \Rightarrow x = 4 or -1.2 \ (ruled \ out)$

Her speed is 4 km/hr. $\\$

Question 6: A car made a run of 390 km in $x$ hours. If the speed had been 4 km/hr more, it would have taken 2 hr less for the journey. Find $x$.

Given $\frac{390}{x}=\frac{390}{x+4} + 2$ $390(x+4) = x[390+2(x+4)]$ $390x+1560=2x^2+398x$ $2x^2+8x-1560 = 0 \Rightarrow x = 26 or -30 \ (ruled \ out)$

Hence the speed of the car is 26 km/hr. $\\$

Question 7: A goods train leaves a station at 6 pm followed by an express train which leaves at 8 pm and travels 20 km/hr faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speed of both the train remains constant between the two stations; calculate their speeds.

Given $\frac{1040}{x}=\frac{1040}{x+20}+\frac{156}{60}$ $\frac{10400}{x}=\frac{10400}{x+20}+26$ $10400(x+20) = x[10400+26(x+20)]$ $26x^2+520-208000=0 \Rightarrow x = 80 \ or \ -100 (ruled \ out)$

Hence the speed of the two trains are 80 km/hr and 100 km/hr. $\\$

Question 8: A man bought an article for $Rs. \ x$ and sold it for Rs. 16. If his loss was $x$ percent, find the cost price of the article.

Cost price $= x \ Rs.$

Selling Price $= 16 \ Rs.$ $Loss = x\% \ of \ Cost \ Price = \frac{x}{100} \times x = \frac{x^2}{100}$

Therefore $x-\frac{x^2}{100} = 16$ $x^2-100x+1600 = 0 \Rightarrow x = 80 \ Rs. or 20 \ Rs.$ $\\$

Question 9: A trader bought an article for Rs. $x$ and sold to for Rs. 52, thereby making a profit of $(x-10)$ percent on his outlay. Calculate the cost price.

Cost price $= x \ Rs.$

Selling Price $= 52 \ Rs.$ $Profit = (x-10)\% \ of \ Cost \ Price = (\frac{x-10}{100} ) \times x$

Therefore $52- x = \frac{x-10}{100} x$ $x^2+90x-5200 = 0 \Rightarrow x = 40 \ Rs. or -130 \ (ruled \ out) \ Rs.$

Therefore the cost price is 40 Rs. $\\$

Question 10: By selling a chair for Rs. 75, a person gained as much percent as its cost. Calculate the cost of the chair.

Cost price $= x \ Rs.$

Selling Price $= 75 \ Rs.$ $Profit = (x)\% \ of \ Cost \ Price = (\frac{x}{100} ) \times x =\frac{x^2}{100}$

Therefore $75- x = \frac{x^2}{100} x$ $x^2+100x-7500 = 0 \Rightarrow x = 50 \ Rs. or -150 \ (ruled \ out) \ Rs.$

Therefore the cost price is 50 Rs. $\\$