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

• 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. $\displaystyle \text{Time taken by slow train } = \frac{300}{x} \text{ hr }$ $\displaystyle \text{Time taken by fast train } = \frac{300}{x+25} \text{ hr }$ $\displaystyle \text{Given } \frac{300}{x} = \frac{300}{x+25} + 2$ $\displaystyle 300(x+25) = x[300+2(x+25)]$ $\displaystyle 2x^2+50x-7500 \Rightarrow x = 50 \text{ or } -75$ (ruled out)

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

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. $\displaystyle \text{Let the speed of the car be } x \text{ km/hr }$ $\displaystyle \text{Therefore } \frac{36}{x} = \frac{36}{x+10} + \frac{18}{60}$ $\displaystyle \frac{360}{x} = \frac{360}{x+10} +3$ $\displaystyle 360x+3600=3x^2+390x$ $\displaystyle 3x^2+30x-3600=0 \Rightarrow x = 30 \text{ or } -40$ (ruled out)

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

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. $\displaystyle \text{Let the speed of the airplane be} x \text{ km/hr }$ $\displaystyle \text{Therefore } \frac{1200}{x} + \frac{20}{60} = \frac{1200}{x-40}$ $\displaystyle \frac{1200}{x} + \frac{1}{3} = \frac{1200}{x-40}$ $\displaystyle (3600+x)(x-40) = 3600x$ $\displaystyle x^2-40x-144000 =0 \Rightarrow x = 400 \text{ or } -360$ (ruled out)

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

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. $\displaystyle \text{Let the speed of the car be} x \text{ km/hr }$ $\displaystyle \text{Therefore } \frac{400}{x} = \frac{400}{x+12} + \frac{100}{60}$ $\displaystyle \frac{2400}{x} = \frac{2400}{x+12} +10$ $\displaystyle 2400x+28800=10x^2+2520x$ $\displaystyle 10x^2-120x-28800=0 \Rightarrow x = -60\text{ or } 48$ (ruled out)

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

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 $\displaystyle x \text{ km/hr }$. and the remaining distance at a speed of $\displaystyle (x+2) \text{ km/hr }$. if she takes 2 hrs 30 minutes to cover the whole distance, find $\displaystyle x$. $\displaystyle \frac{6}{x}+\frac{6}{x+2} = \frac{150}{60}$ $\displaystyle 6(x+2)+6x=\frac{5}{2}x(x+2)$ $\displaystyle 24x+24= 5x^2+10x$ $\displaystyle 5x^2-14x-24=0 \Rightarrow x = 4 \text{ or } -1.2$ (ruled out)

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

Question 6: A car made a run of 390 km in $\displaystyle x$ hours. If the speed had been 4 km/hr more, it would have taken 2 hr less for the journey. Find $\displaystyle x$. $\displaystyle \text{Given } \frac{390}{x}=\frac{390}{x+4} + 2$ $\displaystyle 390(x+4) = x[390+2(x+4)]$ $\displaystyle 390x+1560=2x^2+398x$ $\displaystyle 2x^2+8x-1560 = 0 \Rightarrow x = 26 \text{ or } -30$ (ruled out)

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

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

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

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

Cost price $\displaystyle = x \text{ Rs. }$

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

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

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

Cost price $\displaystyle = x \text{ Rs. }$

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

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

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

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 $\displaystyle = x \text{ Rs. }$

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

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

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