Question 1: The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A care travels at a speed of $x \ km/hr$ and the train travels at a speed which is 16 km/hr faster than the car. Calculate;

• The time taken by the car to reach town B from A, in terms of $x$;
• The time taken by the train, to reach town B from A, in terms of $x$.
• If the train takes 2 hours less than the car to reach town B, obtain an equation in $x$, and solve it.
• Hence, find the speed of train. [1998]

Time taken by the car to reach town B from A $= \frac{216}{x}$

The time taken by the train, to reach town B from A  $= \frac{208}{x+16}$

Given that the train takes 2 hours less than the car to reach town B

$\frac{216}{x} - \frac{208}{x+16} = 2$

$216x+3456-208x=2x^2+32x$

$2x^2+24x-3456=0 \Rightarrow x = 36\ or -48 \ (not \ possible)$

Therefore speed of train $= (36+16) = 52 \ km/hr$

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Question 2: A trader buys $x$ articles for a total cost of Rs.600.

• Write down the cost of one article in terms of $x$. If the cost per article were Rs.5 more, the number of articles that can be bought for Rs.600 would be four less.
• Write down the equation in terms of x for the above situation and solve it for $x$. [1999]

Cost of one article $= \frac{600}{x}$

New Cost $= \frac{600}{x} + 5$

Therefore

$\frac{600}{(\frac{600}{x}+5)} = x-4$

$600 = (x-4)(\frac{600}{x} + 5)$

$600x = 5x^2+600x-20x-2400$

$5x^2-20x-2400 = 0 \Rightarrow x = 24 \ or -20 \ (not \ possible)$

Therefore the number of articles bought is 24.

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Question 3: A hotel bill for a number of people for overnight stay is Rs.4,800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs.200. find the number of people staying overnight. [2000]

Let the number of people $= x$

Therefore

$\frac{4800}{x} -\frac{4800}{x+4} = 200$

$4800x+19200 - 4800x = 200x^2+800x$

$200x^2+800x-19200=0 \Rightarrow x = 8 \ or -12 \ (not \ possible)$

Hence the number of people staying is 8.

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Question 4: An Airplane traveled a distance of 400 km at an average speed of x km/hr. on the return journey, the speed was increased by 40 km/hr. write down an expression for the time taken for:

• The onward journey
• The return journey
• If the airplane takes 2 hours less in returning, calculate the speed of the airplane. [2002]

The time taken for onward journey $= \frac{400}{x}$

The time taken for return journeys $= \frac{400}{x+40}$

$\frac{400}{x} - \frac{400}{x+40} = 2$

$400x + 16000 - 400x = 2x^2+80x$

$2x^2+80x-16000=0 \Rightarrow x = 71.65 \ or -111.65 \ (not \ possible)$

Hence the speed of the airplane is 71.65 km/hr

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Question 5: 6,500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs.30 less. Find the original number of persons.

Let the number of person $= x$

Therefore

$\frac{6500}{x} - \frac{6500}{x+15} = 30$

$6500x+97500 -6500x= 30(x^2+15x)$

$30x^2+450x-97500=0 \Rightarrow x = 50 \ or -65 \ (not \ possible)$

Hence the number of people is 50.

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Question 6: A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed find its usual speed.

Let the speed of the plane $= x$

Therefore

$\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$

$1500x+375000 -1500x= \frac{1}{2}(x^2+250x)$

$x^2+250-750000=0 \Rightarrow x = 750 \ or -1000 \ (not \ possible)$

Hence the original speed of the plan is 750 km/hr.

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Question 7: Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.

Let the speed of the second train $= x \ km/hr$

Speed of the first train $= (x+5) \ km/hr$

Given

$[2(x+50]^2+[2x]^2 = 50^2$

$4(x^2+25+10x)+4x^2=2500$

$8x^2+40x-2400=0 \Rightarrow x = 15 \ or -20 \ (not \ possible)$

Speed of the first train is 20 km/hr. Speed of the second train is 15 km/hr.

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Question 8: The sum S of first n even natural numbers is given by the relation S=n(n+1). Find n, if the sum is 420.

$n(n+1) = 420$

$n^2+n - 420 = 0 \Rightarrow n = 20 \ or -21 \ (not \ possible)$

Hence n = 20

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Question 9: The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.

Let the age of father $= x \ years$

Therefore Age of son $= (45-x) \ years$

Age of father 5 years ago $= (x-5) \ years$

Age of son 5 years ago $= (40-x) \ years$

Given

$(x-5) \times (40-x) = 124$

$40x -200 -x^2+5x = 124$

$x^2-45x+324 = 0 \Rightarrow x = 36 \ or 9$

Hence father’s age is 36 years and son’s age is 9 years.

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Question 10: In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find:

• The number of rows in the original arrangement
• The number of seats in the auditorium after re-arrangement. [2003]

Let the number of rows $= x$

Therefore the number of seats in a row $= x$

Given

$(2x) \times (x-10) - x \times x = 300$

$2x^2-20x-x^2 = 300$

$x^2-20x-300=0 \Rightarrow x = 30 \ or -10 \ (not \ possible)$

Therefore the number of rows are 30 and each row has 30 seats.

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Question 11: Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, in how many days will Mohan alone complete the work?

No of days that Manoj takes to complete the work$= x$

Therefore the number of days Mohan takes to complete work$= (x-16)$

Given$\frac{1}{x} + \frac{1}{x-16} = \frac{1}{15}$

$15(x-16+x) = x^2-16x$

$30x - 240 = x^2-16x$

$x^2-46x+240 = 0 \Rightarrow x = 40 \ or 6 \ (not \ possible)$

Therefore Mohan will take 24 days to complete the work.

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Question 12: Two years ago, a man’s age was three times the square of his son’s age. In three years of time, his age will be four times his son’s age. Find their present ages.

Let man’s age now $= x \ years$

Let son’s age now  $= y \ years$

Given  $(x-2) = 3 (y-2)^2$

$x+3=4(y+3)$

Solving

$3(y^2+4-4y)+5 = 4y+12$

$3y^2+12-12y+5 = 4y+12$

$3y^2-16y+5= 0 \Rightarrow y = 5 \ or 0.33 \ (not \ possible)$

Hence the fathers age is 29 years.

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Question 13: In certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by 1/14. Find the fraction.

Let the fraction $= \frac{x}{x+3}$

Given

$\frac{x}{x+3} - \frac{x-1}{x+2} = \frac{1}{14}$

$x(x+2)-(x-1)(x+3) = \frac{1}{14} (x+3)(x+2)$

$x^2+2x-(x^2+2x-3) = \frac{1}{14}(x^2+5x+6)$

$42 = x^2+5x+6$

$x^2+5x-36=0 \Rightarrow x = 4, \ or -9 \ (not \ possible)$

Therefore the fraction is $\frac{4}{7}$

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Question 14: In a two-digit number, the ten’s digit is bigger. The product of the digit is 27 and the difference between two digits is 6. Find the number

Let the number be $yx$

$yx= 27$

$y-x=6$

$x(x+6) = 27$

$x^2+6x-27=0 \Rightarrow x = 3 \ or -9 \ (not \ possible)$

Therefore the number is 93

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Question 15: Some school children went on an excursion by a bus to a picnic spot at a distance of 300 km. while returning, it was raining and the bus had to reduce its speed by 5 km/hr and it took two hours longer for returning. Find the time taken to return.

Let the speed of the bus is $x \ km/hr$

Given

$\frac{300}{x-5} -\frac{300}{x} = 2$

$300x - 300(x-5) = 2x(x-5)$

$2x^2-10x-1500=0 \Rightarrow x = 30 \ or -25 \ (not \ possible)$

Hence the speed of the bus was 30 km/hr  &s=1\$

Therefore the time taken to return $= \frac{300}{25} = 12 \ hrs$

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Question 16: 480 is divided equally among $x$ children. If the number of children were 20 more then each would have got Rs.12 less. Find $x$ [2011]

Let the number of children $= x$

Therefore

$\frac{480}{x} - \frac{480}{x+20} = 12$

$480x+9600-480x= 12 (x^2+20x)$

$12x^2+240x-9600=0 \Rightarrow x = 20 \ or -40 \ (not \ possible)$

Hence the number of children is 20

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