Which of the equations among the following is/are quadratic equation(s) ? $q_1 : x^2 + x = (x+1)^2$, $q_2 : x-1= x^2 -1$, $q_3 : x^4 = x^2$, $q_4 : \sqrt{x} = x^2\sqrt{x} +1$
Which of the following equations is/are not quadratic ? $q_1: (x-1)^2 = x^2$, $q_2: (x-1)^3 = x^3$ $q_3: (x+1)^3 = 3x^2$, $q_4: (\sqrt{x}+1)^2 = 2\sqrt{x}$
Two water taps together can fill a tank in $3\frac{1}{3}$ hours. The tap of larger diameter takes $5$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
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Let the time taken by the tap of smaller diameter to fill the tank separately be '$x$' hours and the time taken by the tap of larger diameter to fill the tank separately be $(x - 5)$ hours. A.T.Q. $\frac{1}{x} + \frac{1}{x-5} = \frac{3}{10}$ $\Rightarrow 3x^2 - 35x + 50 = 0$ $\Rightarrow (x - 10) (3x - 5) = 0$ $x = 10$ or $x = \frac{5}{3}$ But $x = \frac{5}{3}$ is not possible, so $x = 10$ $\therefore$ time taken by the tap of smaller diameter to fill the tank separately is $10$ hours and time taken by the tap of larger diameter to fill the tank separately is $10-5=5$ hours
Sum of the areas of two squares is $468$ m$^2$. If the difference of their perimeters is $24$ m, find the lengths of the sides of the two squares.
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Let the lengths of the sides of the two squares be '$x$' m and '$y$' m s.t. $x > y$ A.T.Q. $x^2 + y^2 = 468$ -----(1) (1/2 Mark) $4x - 4y = 24$ $\Rightarrow x-y=6$ -----(2) (1/2 Mark) From (1) and (2), we get $y^2 + 6y - 216 = 0$ $\Rightarrow y = 12$ and $y = -18$ (1 Mark) But side of a square is always positive, So, $y = 12$ and $x = 18$ (1 Mark) Hence, the lengths of the sides of two squares are $12$ m and $18$ m.
In a $2$-digit number, the digit at the unit's place is $5$ less than the digit at the ten's place. The product of the digits is $36$. Find the number.
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Let digit at ten's place be $x$ then digit at unit's place $= x - 5$ $x(x - 5) = 36$ $\Rightarrow x^2 - 5x - 36 = 0$ $(x-9)(x + 4) = 0$ $x \ne -4$ so, $x = 9$ $\therefore$ Required number is $94$
Three consecutive integers are such that sum of the square of second and product of other two is $161$. Find the three integers.
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Let the three numbers be $x, x+1$ and $x+2$ ($\frac{1}{2}$) $(x + 1)^2 + x(x + 2) = 161$ $\Rightarrow x^2+2x-80 = 0$ (1) $\Rightarrow (x+10)(x-8)=0$ $\therefore x= 8$ or $-10$ (1) So, the numbers are $8, 9, 10$ or $-10, -9, -8$ ($\frac{1}{2}$)
A dealer sells an article for ₹75 and gains as much percent as the cost price of the article. Find the cost price of the article.
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Let the cost price of the article be $x$ $\therefore$ Gain % $= x$ $x = \frac{75-x}{x} \times 100$ $\Rightarrow x^2+ 100x - 7500 = 0$ $\Rightarrow (x - 50)(x + 150) = 0$ $x \ne -150 \therefore x = 50$ So, the cost price of the article is ₹50
Find two consecutive negative integers, sum of whose squares is $481$.
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Let two consecutive negative integers be $x$ and $(x + 1)$ According to given statement, $x^2 + (x + 1)^2 = 481$ (1 Mark) $\Rightarrow x^2 + x - 240 = 0$ (1/2 Mark) $\Rightarrow (x + 16) (x - 15) = 0$ (1 Mark) $x = -16$ or $15$ $x = 15$ does not satisfy the given condition. So, required integers are $-16$ and $-15$. (1/2 Mark)
A rectangular field is $16$ m long and $10$ m wide. There is a path of equal width all around it, having an area of $120$ sq.m. Find the width of the path.
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Let the width of the path be $x$ m. A. T. Q. $$\begin{aligned}& (16 + 2x)(10 + 2x) - 16 \times 10 = 120 \\ & \Rightarrow 4x^2 + 52x - 120 = 0 \text{ or } x^2 + 13x - 30 = 0 \\ & \Rightarrow (x - 2)(x + 15) = 0 \\ & \Rightarrow x = 2 \text{ (Rejecting } x = -15) \\ & \therefore \text{Width of the path is } 2 \text{ m.}\end{aligned}$$
While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by $x$ units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide. Based on the above information, answer the following questions: (I) Write an algebraic equation depicting the above information. (II) Write the corresponding quadratic equation in standard form. (III) What should be the new dimensions of the enlarged photo? OR Can any rational value of $x$ make the new area equal to $220\text{cm}^2$
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(i) $(18 + x) (12 + x) = 2(18\times12)$ (ii) $x^2 + 30x - 216 = 0$ (iii) Solving : $x^2 + 30x – 216 = 0$ $\Rightarrow (x + 36) (x – 6) = 0$ $x \neq -36 \therefore x = 6$. new dimensions are $24$ cm $\times 18$ cm OR (iii) If $(18+ x) (12 + x) = 220$ then $x^2 + 30x - 4=0$ Here $D = 900 + 16 = 916$ which is not a perfect square. Thus we can't have any such rational value of $x$.
A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway. The total area of the lawn and the walkway is $360$ square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are $12$ metres by $10$ metres. Based on the information given above, answer the following questions: (i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway $= x \operatorname{m}$. (ii) (a) Solve the quadratic equation to find the width of the walkway 'x'. OR (b) If the cost of paving the walkway at the rate of ₹50 per square metre is ₹12,000, calculate the area of the walkway. (iii) Find the perimeter of the lawn.
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(i) $(12 + 2x)(10 + 2x) = 360$ $4x^2 + 44x - 240 = 0$ or $x^2 + 11x - 60 = 0$ (ii)(a) $(x + 15)(x - 4) = 0$ $x = 4$ $\therefore$ width of the walkway $= 4 \operatorname{m}$ OR (ii)(b) Area of the walkway $= \frac{12000}{50}$ $= 240 \operatorname{m}^2$ (iii) Perimeter of the lawn $= 2(12 + 10) = 44 \operatorname{m}$
A train travels at a certain average speed for a distance of $54$ km and then travels a distance of $63$ km at an average speed of $6$ km/h more than the first speed. If it takes $3$ hours to complete the journey, what was its first average speed ?
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Let first average speed of the train be $x$ km/hr. $\frac{54}{x} + \frac{63}{x + 6} = 3$ $\Rightarrow 54(x + 6) + 63x = 3x^2 + 18x$ $\Rightarrow 3x^2 - 99x - 324 = 0$ or $x^2 - 33x - 108 = 0$ $\Rightarrow (x - 36) (x + 3) = 0$ $\Rightarrow x = 36, -3$ (rejected) Therefore, first average speed of the train was $36$ km/hr.
Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes $2$ hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.
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Let the time taken by smaller diameter tap be $x$ hrs. Time taken by larger diameter tap is $(x - 2)$ hrs. Therefore $\frac{1}{x-2} + \frac{1}{x} = \frac{8}{15}$ $\Rightarrow 15(2x - 2) = 8x(x - 2)$ $\Rightarrow 8x^2 - 46x + 30 = 0$ $\Rightarrow 4x^2 - 23x + 15 = 0$ $\Rightarrow (4x - 3)(x - 5) = 0$ $\Rightarrow x = \frac{3}{4}$ as $x-2 < 0$ (rejected) or $x = 5$ Smaller diameter tap fills in $5$ hrs. Larger diameter tap fills in $3$ hrs.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $\frac{16}{21}$, find the fraction.
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Let numerator be $x$, then denominator be $(2x + 1)$ Fraction = $\frac{x}{2x+1}$ A.T.Q. $\frac{x}{2x+1} + \frac{2x+1}{x} = \frac{58}{21}$ $\Rightarrow 11x^2 - 26x - 21 = 0$ $\Rightarrow (x - 3)(11x + 7) = 0$ $x \neq -\frac{7}{11}$ So, $x = 3$ $\therefore$ Fraction = $\frac{3}{7}$
A train travels a distance of $90$ km at a constant speed. Had the speed been $15$ km/h more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
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Let the original speed be $x$ km/h New speed $= (x + 15)$ km/h A.T.Q. $$\begin{aligned}& \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \\ & \Rightarrow x^2 + 15x - 2700 = 0 \\ & \Rightarrow(x + 60) (x - 45) = 0 \\ & x \neq -60 \text{ , } x = 45 \\ & \text{The original speed of the train } = 45\text{km/h}\end{aligned}$$
At present, Sourav's age is $3$ years more than the square of his son Ravi's age. When Ravi grows to his father's present age, Sourav's age would be $6$ years less than $13$ times the present age of Ravi. Find present ages of Ravi and Sourav.
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Let the present age of Ravi be 'r' years and the present age of Sourav be 's' years Therefore, $s = 3 + r^2$ --- (1) Ravi grows to father's present age in $(s - r)$ years. $\therefore$ father's age after $(s - r)$ years = $(2s - r)$ years and Ravi's age after $(s - r)$ years = 's' years Therefore, $2s - r = 13r - 6$ or $s = 7r - 3$ --- (2) Using (1) and (2), $r^2 - 7r + 6 = 0$ $\Rightarrow (r - 6)(r - 1) = 0$ $\Rightarrow r = 6, 1$ Ignoring $r = 1$ as $s \neq 4$ r = $6$ Hence $s = 39$
The sum of the areas of two squares is $52$ cm$^2$ and difference of their perimeters is $8$ cm. Find the lengths of the sides of the two squares.
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Let the lengths of the sides of two squares be 'x' cm and 'y' cm such that $x > y$. ATQ $x^2 + y^2 = 52$ ----- (1) $4x - 4y = 8$ or $x - y = 2$ ----- (2) From (1) and (2), we have $y^2 + 2y - 24 = 0$ $\Rightarrow (y + 6) (y - 4) = 0$ $\therefore y = 4$ So, $x = 2 + 4 = 6$ $\therefore$ Lengths of the sides of two squares are $6$ cm and $4$ cm respectively.
The numerator of a fraction is $3$ less than its denominator. If $2$ is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.
A train travelling at a uniform speed for $360$ km would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Find the original speed of the train.
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Let the original speed of train be 'x' km/h ATQ $\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60}$ ($2$) $\Rightarrow x^2 + 5x - 2250 = 0$ ($1$) $\Rightarrow (x + 50)(x - 45) = 0$ ($1$) So, $x = 45$ $\therefore$ Original speed of the train is $45$ km/h. ($1$)
A faster train takes one hour less than a slower train for a journey of $200$ km. If the speed of the slower train is $10$ km/hr less than that of the faster train, find the speeds of the two trains.
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Let the speed of faster train be $x$ km/h $\therefore$ speed of slower train $= (x – 10)$ km/h According to the question, $\frac{200}{x-10} - \frac{200}{x} = 1$ (I) (2 Marks) $\Rightarrow x^2 - 10x - 2000 = 0$ (II) (1 Mark) $\Rightarrow (x - 50)(x + 40) = 0$ (III) (1 Mark) $\therefore x = 50$ $x = -40$ (Rejected) (IV) (1/2 Mark) Hence, speed of faster train $= 50$ km/h and speed of slower train $= 40$ km/h (V) (1/2 Mark)
The sum of the areas of two squares is $640$ m$^2$. If the difference in their perimeters is $64$ m, find the sides of the two squares.
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Let the sides of the two squares be $x$ m and $y$ m $(x > y)$ $x^2 + y^2 = 640$ (I) (1 Mark) and $4x – 4y = 64 \Rightarrow y = x - 16$ (II) (1 Mark) $\therefore x^2 + (x – 16)^2 = 640$ (III) (1 Mark) $\Rightarrow x^2 - 16x – 192 = 0$ (IV) (1 Mark) $\Rightarrow (x – 24)(x + 8) = 0$ $\therefore x = 24$ (V) (1/2 Mark) $x = -8$ (Rejected) $\Rightarrow y = 24 – 16 = 8$ (VI) (1/2 Mark) Hence the sides of the two squares are $24$ m and $8$ m
Venkat can row a boat in still water at the speed of 12 km/h. He ferries tourists 15 km upstream and 18 km downstream in 3 hours. Find the speed of the stream.
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Let speed of the stream be $x$ km/h $\frac{15}{12 - x} + \frac{18}{12 + x} = 3$ (I) (2) $\Rightarrow x^2 - x - 12 = 0$ (II) (1) $\Rightarrow (x - 4)(x + 3) = 0$ (III) (1) $\Rightarrow x = 4, -3$ x = -3 (rejected) $\therefore x = 4$ (IV) (1) The speed of the stream = 4 km/h
By selling an article for ₹$48$, a trader loses as much percent as half of the cost price of the article. Calculate the cost price and loss amount of the article.
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Let the cost price (CP)of the article be $x$ Loss = $\frac{x}{2}\%$ of $x = \frac{x^2}{200}$ (I Mark) Selling price = $48 = x - \frac{x^2}{200}$ (II Mark) $\Rightarrow x^2 - 200x + 9600 = 0$ (III Mark) $\Rightarrow (x - 120)(x - 80) = 0$ (IV Mark) $\Rightarrow x = 120,80$ (V Mark) When CP = ₹$120$, Loss = $120 - 48 = 72$ (VI Mark) When CP = ₹$80$, Loss = $80 - 48 = 32$ (VII Mark)
Two water taps together can fill a tank in $8\frac{8}{3}$ hours. The tap of larger diameter takes $4$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
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Let tap of smaller diameter takes $x$ hours to fill the tank. $\therefore$ tap of larger diameter takes $x - 4$ hours to fill the tank. $\frac{1}{x} + \frac{1}{x-4} = \frac{9}{80}$ (2 Marks) $\Rightarrow 9x^2-196x + 320 = 0$ (1 Mark) $\Rightarrow (x-20)(9x – 16) = 0$ (1 Mark) $\Rightarrow x = 20, \frac{16}{9}$ $x = \frac{16}{9}$ (rejected) (1/2 Mark) $\therefore x = 20$ Hence, tap of smaller diameter and larger diameter takes $20$ hours & $16$ hours respectively, to fill the tank. (1/2 Mark)
A rectangular floor area can be completely tiled with $200$ square tiles. If the side length of each tile is increased by $1$ unit, it would take only $128$ tiles to cover the floor. (i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information. (ii) Write the corresponding quadratic equation in standard form. (iii) (a) Find the value of $x$, the length of side of a tile by factorisation. OR (b) Solve the quadratic equation for $x$, using quadratic formula.
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p+1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.
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For real roots, $D \geq 0$. $[-2(p+1)]^2 - 4p^2 \geq 0 \Rightarrow p \geq -\frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2} + 1$ marks). $\therefore$ smallest value of $p = -\frac{1}{2}$ ($\frac{1}{2}$ mark). At $p = -\frac{1}{2}$ given equation becomes $x^2 - 2(-\frac{1}{2} + 1)x + (-\frac{1}{2})^2 = 0$ ($\frac{1}{2}$ mark). $x^2 - x + \frac{1}{4} = 0$ or $4x^2 - 4x + 1 = 0$ (1 mark). $(2x-1)(2x-1) = 0$. $\therefore$ roots are $\frac{1}{2}, \frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2}$ marks).
Express $x - \frac{1}{x} = 3$ as a quadratic equation in standard form and hence find its roots. Also, find the value of 'a' for which the equation $x + \frac{1}{x} = a$, when expressed as a quadratic equation, has real and equal roots.