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: 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 6: 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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Question 7: 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.    [2012]

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 8: A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number. [2010]

Let the two parts be $x \ and \ y$

Given $x^2+y^2 = 20$

Also $y^2 = 8x$

Substituting it back $x^2+8x-20 = 0$

$\Rightarrow x = 2 or -10$   (ignore this as the number is positive)

Therefore the larger part is $y^ = 16 \Rightarrow y = 4$

Hence the number is $2+4 = 6$

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Question 9: By increasing the speed of the car by 10 km/hr, the time of the journey for a distance of 72 km is reduced by 36 minutes. Find the original speed of the car.    [2005]

Let the original speed of the car $= x$ km/hr

Then the time taken to cover the distance $= \frac{72}{x}$ hrs

The new speed of the car $= (x+10)$ km/hr

Therefore the time taken to cover the distance $= \frac{72}{x+10}$ hrs

Hence given

$\frac{72}{x} - \frac{72}{x+10} = \frac{1}{2}$

$\Rightarrow x = 30$ km/hr

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Question 10: A two digit number is such that the products of the digits is 6. When 9 is added to the number, the digits interchange their places. Find the number.    [2014]

Let the number by $xy$

Given $x \times y = 6$ … … … … … … i)

Also $xy+9 = yx \Rightarrow 10x+y+9 = 10y +x$

$\Rightarrow y = x+1$ … … … … … … ii)

Solving i) and ii)

$x(x+1) = 6$

$x^2+x-6 = 0$

$\Rightarrow x = -3 (not \ possible) \ or \ 2$

Hence $x = 2 \Rightarrow y = 3$

Hence the number is $23$

Question 11: Five years ago, a woman’s age was square of her son’s age. 10 year hence her age will be twice that of her son. Find i) the age of her son five year ago ii) the present age of the woman.    [2007]

Let the age of the son 5 years ago $= x$ years

Therefore the woman’s age 5 years ago $= x^2$ years

The present age of the son $= (x+5)$ years

The present age of woman $= (x^2+5)$ years

10 years hence the son’s age $= (x+15)$ years

10 year hence the woman’s age $= (x^2+15)$ year

Given that $x^2+15 = x(x+15)$

$x^2-2x-15 = 0$

solving $x = 5 or -3 (not \ possible)$

Hence $x = 5$

Therefore

The age of son 5 years ago $= 5$ years

The present age of the woman $= (x^2+5) = 25+5 = 30$ years.

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Question 12: A shop keeper buys a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that he could have bought for Rs. 960 would be 4 more. Taking the original cost of each book to be x Rs. write an equation in x and solve for it.     [2013]

Let the original cost of the book $= x$  Rs.

No of books bought   $= \frac{960}{x}$

If the cost of the book was $= (x-8)$ Rs.

The the number of books bought $= \frac{960}{x-8}$

Given

$\frac{960}{x-8} - \frac{960}{x} = 4$

simplifying $x^2 - 8x -1920 = 0$

$(x-48)(x+40) = 0$

$\Rightarrow x = 48 \ or \ -40 (not \ possible)$ .

Hence the cost of the book is 48 Rs.

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Question 13: Some students planned a picnic. The budget for the food was Rs. 480. As eight of the students failed to join the party, the cost of the food for each member increased by Rs. 10. Find how many students went for the picnic.   [2008]

Let the number of students going to the picnic $= x$

Planned Food budget $= 480$ Rs.

Planned Budget per student $= \frac{480}{x}$

Number of student who actually went for the party $= (x-8)$

Actual food budget per student $= \frac{480}{x-8}$

Given

$\frac{480}{x-8} - \frac{480}{x} = 10$

Simplifying

$x^2-8x-384 = 0$

$(x-24)(x+16) = 0 \Rightarrow x = 24.$

Hence the number of students who actually went for the picnic $= 24-8 = 16$

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Solve each of the following equations for $x$

Question 14: $x^2-3x-9=0$     [2007]

Comparing $x^2-3x-9=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -3 \ and \ c =-9$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}$

Solving we get $x = 4.85, -1.85$

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Question 15: $x^2-5x-10=0$     [2013]

Comparing $x^2-5x-10=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -5 \ and \ c =-10$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}$

Solving we get $x = 6.53, -1.53$

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Question 16: $2x-\frac{1}{x}=7$     [2006]

$2x-\frac{1}{x}=7$

$2x^2-7x-1=0$

Comparing $2x^2-7x-1=0$ with $ax^2+bx+c=0$, we get $a = 2, b = -7 \ and \ c =-1$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-1)}}{2(2)}$

Solving we get $x = 3.64, -0.14$

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Question 17: $5x^2-3x-4=0$      [2012]

Comparing $5x^2-3x-4=0$ with $ax^2+bx+c=0$, we get $a = 5, b = -3 \ and \ c =-4$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-3) \pm \sqrt{(-3)^2-4(5)(-4)}}{2(5)}$

Solving we get $x = 1.243, -0.643$

If we want only three significant figures than $x = 1.24, -0.643$

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Question 18: $(x-1)^2-3x+4=0$      [2014]

$(x-1)^2-3x+4=0$

$x^2+1-2x-3x+4=0$

$x^2-5x+5=0$

Comparing $5x^2-3x-4=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -5 \ and \ c =5$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(5)}}{2(1)}$

Solving we get $x = 3.618, 1.382$

If we want only two significant figures than $x = 3.6, 1.3$

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Question 19: $x^2-5x-10=0$    [2005]

Comparing $x^2-5x-10=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -5 \ and \ c =-10$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}$

Solving we get $x = 6.53, -1.53$

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Question 20: $3x^2-x-7=0$   [2004]

Comparing $3x^2-x-7=0$ with $ax^2+bx+c=0$, we get $a = 3, b = -1 \ and \ c =-7$

Since $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Therefore $x = \frac{-(-1) \pm \sqrt{(-1)^2-4(3)(-7)}}{2(3)}$

Solving we get $x = 1.703, -1.3699$

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Question 21: $px^2-4x+3=0$  [2010]

Comparing $px^2-4x+3=0$ with $ax^2+bx+c=0$, we get $a = p, b = -4 \ and \ c =3$

For roots to be equal, we should have $b^2-4ac = 0$

$(-4)^2-4(p)(3)=0$

$16-12p=0$

$p=\frac{4}{3}$

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Question 22: $x^2+2(m-1)x+(m+5)=0$  [2012]

Comparing $x^2+2(m-1)x+(m+5)=0$ with $ax^2+bx+c=0$, we get $a = 1, b = 2(m-1) \ and \ c =(m+5)$

For roots to be equal, we should have $b^2-4ac = 0$

$(2(m-1))^2-4(1)(m+5)=0$

$4(m^2+1-2m)-4(m+5)=0$

$4m^2+4-8m-4m-20=0$

$4m^2-12m-16=0$

$m^2-3m-4=0$

$m^2-4m+m-4=0$

$m(m-4)+(m-4)=0$

$(m-4)(m+1)=0 \Rightarrow m = 4, -1$