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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.

 Answer:

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.

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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.

Answer:

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.

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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.

 Answer:

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.

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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.

 Answer:

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 60 km/hr.

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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  .

Answer:

\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.

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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 .

 Answer:

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.

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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.

Answer:

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.

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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.

 Answer:

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. 

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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.

Answer:

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.

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

Answer:

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.

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