Quadratic Equations — Class 10 Maths PYQs

(B) $(x+\sqrt{x})^2=2x\sqrt{x}$

(A) $5$

(A) $0$

(C) $1$

(A) $0$

(c) $q_2$ only

(d) $q_1$ and $q_3$

(a) $q_1, q_3$ and $q_4$

(c) $0.1$

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

(a) $x^2 - 6x + 7 = 0$

(D) $a = 0, b = - 6$

(D) $pq$

(B) $-11$

(D) $-14$

(C) $-\frac{3}{2}$

(A) $7x^2 - 50 x + 7 = 0$

(D) $a + b + 3 = 0$

(A) $a = 3$

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

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.

Let one number be $x \Rightarrow \text{another number} = 15 - x$
Therefore, $\frac{1}{x} + \frac{1}{15-x} = \frac{3}{10}$
$\frac{15-x + x}{x(15 - x)} = \frac{3}{10} \Rightarrow 150 = 3x(15 - x)$
$3x^2 - 45x + 150 = 0$
$x^2 - 15x + 50 = 0 \Rightarrow (x - 10)(x - 5) = 0$
$\Rightarrow x = 10, 5$
Numbers are $10, 5$ or $5, 10$

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 $94$

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 $-10, -9, -8$ ($\frac{1}{2}$)

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

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)

Let the two consecutive even numbers be $x$ and $x + 2$.
A.T.Q.
$x^2 + (x + 2)^2 = 452$
$\Rightarrow x^2 + 2x - 224 = 0$
$\Rightarrow (x + 16)(x - 14) = 0$
$\Rightarrow x = 14$
Required numbers are $14$ and $16$.

Let the width of the path be $x$ m.
A. T. Q. $$\begin{aligned}& (16 + 2x)(10 + 2x) - 16 \times 10 = 120 \\ & \Rightarrow 4x^2 + 52x - 120 = 0 \text{ or } x^2 + 13x - 30 = 0 \\ & \Rightarrow (x - 2)(x + 15) = 0 \\ & \Rightarrow x = 2 \text{ (Rejecting } x = -15) \\ & \therefore \text{Width of the path is } 2 \text{ m.}\end{aligned}$$

Let two consecutive odd numbers be x and $x + 2$
A.T.Q.
$x^2 + (x + 2)^2 = 514$
$\Rightarrow 2x^2 + 4x - 510 = 0$ or $x^2 + 2x - 255 = 0$
$\Rightarrow (x + 17) (x - 15) = 0$
$\Rightarrow x = 15$
Required numbers are 15 and 17

Let length and breadth of the park be $a$ metres and $b$ metres respectively.
$2(a + b) = 100 \dots(i)$ ($\frac{1}{2}$ mark)
$ab = 600 \dots(ii)$ ($\frac{1}{2}$ mark)
using (i) & (ii) we get $a^2 - 50a + 600 = 0$ (1 mark)
$\Rightarrow a = 30 \text{ or } 20$ ($\frac{1}{2}$ mark)
and $b = 20 \text{ or } 30$ ($\frac{1}{2}$ mark)
$\therefore \text{length} = 30 \text{ m, breadth} = 20 \text{ m or vice versa}$

Let the number be $x$. $x + \frac{1}{x} = \frac{13}{6} \Rightarrow 6x^2 - 13x + 6 = 0 \Rightarrow (2x-3)(3x-2) = 0 \Rightarrow x = \frac{3}{2}$ or $\frac{2}{3}$

If time taken by $B$ be $x$ days, then $A$ takes $(x - 6)$ days
A.T.Q. $\frac{1}{x-6} + \frac{1}{x} = \frac{1}{4}$
$\implies x^2 - 14x + 24 = 0$
$\implies (x-12)(x-2) = 0$
$\implies x = 12$
$x = 2$ (rejected)
$\therefore B$ will take $12$ days to finish the work

Let no. of rows be $x = $ no. of seats in each row.
Total seats in beginning $= x \times x = x^2$.
New number of rows $= 2x$.
New number of seats in each row $= x-10$.
New total seats $= 2x(x-10)$.
According to the problem, new total seats = original total seats + $300$.
(a) $2x(x - 10) = x^2 + 300$
$2x^2 - 20x = x^2 + 300$
$x^2 - 20x - 300 = 0$
(b) (i) To find $x$ (number of rows in original arrangement):
$x^2 - 20x - 300 = 0$
$(x - 30)(x + 10) = 0$
$x = 30$ or $x = -10$.
Since number of rows cannot be negative, $x = 30$.
So, there are $30$ rows in the original arrangement.
OR
(ii) Number of seats in the auditorium in the beginning $= x^2 = 30^2 = 900$.
(c) Number of seats after re-arrangement $= x^2 + 300 = 900 + 300 = 1200$.

(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 $x$.

(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 $= 2(12 + 10) = 44 \operatorname{m}$

Let the original duration of the tour be $x$ days
Original daily expenses = ₹ $\frac{4200}{x}$
New daily expenses = ₹ $\frac{4200}{x+3}$
$\frac{4200}{x} - \frac{4200}{x+3} = 70$ (2 Marks)
$\Rightarrow x^2 + 3x - 180 = 0$ (1 1/2 Marks)
$\Rightarrow (x + 15)(x – 12) = 0$ (1 Mark)
$\Rightarrow x = -15, x = 12$
$x = -15$ (rejected)
$x = 12$ (1/2 Mark)
$\therefore$ Original duration of the tour = $12$ days

Let the altitude of triangle be $x$ cm
then Base of triangle = $(x + 10)$ cm
Area of triangle = $600$ cm$^2$
$\frac{1}{2} \times x \times (x + 10) = 600$ (1 1/2 Marks)
$\Rightarrow x^2+10x-1200 = 0$ (1 Mark)
$\Rightarrow (x + 40)(x – 30) = 0$ (1 Mark)
$\Rightarrow x = -40, x = 30$
$x = -40$ (rejected)
$x = 30$ (1/2 Mark)
$\therefore$ Altitude = $30$ cm
Base = $40$ cm (1/2 Mark)
Hypotenuse = $\sqrt{(30)^2 + (40)^2}= 50$ cm (1/2 Mark)

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.

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.

Let the present age of Nidhi be $x$ years.
According to question, $(x - 7)^2 = 5x + 1$
$x^2- 19x + 48 = 0$
$(x - 16)(x - 3) = 0$
$x = 16,3$
$x \neq 3$
$\therefore x = 16$
Hence, the present age of Nidhi $= 16$ years

Let the number of books bought initially be $x$
According to question,
$\frac{1800}{x} - \frac{1800}{x+15} = 20$
$x^2 + 15x - 1350 = 0$
$(x + 45)(x - 30) = 0$
$x = -45$
$\therefore x = 30$
So, the number of books bought initially $= 30$

Let number of students who attended picnic be $x$.
A.T.Q.
$\frac{500}{x} - \frac{500}{x+5} = 5$
$\Rightarrow x^2 + 5x - 500 = 0$
$\Rightarrow (x + 25) (x - 20) = 0$
$x = -25, x = 20$
But number of students can't be negative.
Hence, $x = 20$
Therefore, number of students who attended picnic is $20$.

Let original speed of aircraft be $x \text{ km/hr}$.
A.T.Q.
$\frac{2800}{x-100} - \frac{2800}{x} = \frac{1}{2}$
$\Rightarrow x^2 - 100x - 560000 = 0$
$\Rightarrow (x-800)(x + 700) = 0$
$x \neq -700$ So, $x = 800$
Original Duration $\frac{2800}{800} = \frac{7}{2}$ hrs or $3$ hrs $30$ min.

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 = $\frac{3}{7}$

Let the original speed be $x$ km/h
New speed $= (x + 15)$ km/h
A.T.Q.
$$\begin{aligned}& \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \\ & \Rightarrow x^2 + 15x - 2700 = 0 \\ & \Rightarrow(x + 60) (x - 45) = 0 \\ & x \neq -60 \text{ , } x = 45 \\ & \text{The original speed of the train } = 45\text{km/h}\end{aligned}$$

Let the required no. be $10x + y$
Here $xy = 18$ ---- (i)
$(10x + y) - 63 = 10y + x$
or $x - y = 7$ ---- (ii)
Solving (i) and (ii) to get
$x=9$ and $y=2$
Hence required number is $92$.

Let present age of son = $x$ years
and present age of man = $2x^2$ years
A.T.Q.
$$\begin{aligned}& 3(x + 8) + 4 = 2x^2 + 8 \\ & \Rightarrow 2x^2 - 3x - 20 = 0 \\ & \Rightarrow (2x+5) (x - 4) = 0 \\ & x \neq -\frac{5}{2}\end{aligned}$$ So, $x = 4$
Present age of son = $4$ years
Present age of man = $32$ years

Let the two digit number be $10x + y$
$xy = 14$ ..... (i) (1 Mark)
$10x + y + 45 = 10y + x$
$y-x = 5$ .....(ii) (1 Mark)
From (i) and (ii)
$x(x+5) = 14 \Rightarrow x^2 + 5x - 14 = 0$ (1/2 Mark)
$(x+7)(x-2) = 0$ (1/2 Mark)
$x = 2$ (as $x \neq -7$) (1 Mark)
Number = $27$ (1 Mark)

Let the side of first square = $x$ cm (1/2 Mark)
$\therefore$ Side of second square = $(x + 4)$ cm (1/2 Mark)
$x^2 + (x + 4)^2 = 400$ (1 Mark)
$x^2 + x^2 + 8x + 16 = 400 \Rightarrow 2x^2 + 8x - 384 = 0 \Rightarrow x^2 + 4x - 192 = 0$ (1 Mark)
$(x + 16)(x - 12) = 0$ (1/2 Mark)
$x = 12$ (as $x \neq -16$) (1/2 Mark)
Side of squares = $12$cm and $16$cm (1 Mark)

Let two numbers be $x$ and $$\begin{aligned}& (18 - x) \\ & A.T.Q. \\ & \frac{1}{x} + \frac{1}{18-x} = \frac{1}{4} \\ & \Rightarrow x^2 - 18x + 72 = 0 \\ & \Rightarrow (x - 12)(x - 6) = 0 \\ & \Rightarrow x = 12, x = 6\ \therefore\end{aligned}$$ two numbers are $12$ and $6$.

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 $s = 39$

Let sides be $x, y$. $x + y + 25 = 60 \implies y = 35 - x$. $x^2 + y^2 = 25^2 \implies x^2 + (35 - x)^2 = 625 \implies x^2 - 35x + 300 = 0 \implies (x - 20)(x - 15) = 0 \implies x = 20, 15$. Sides are $15$ cm and $20$ cm.

Let speed be $x$ km/h. $\frac{480}{x - 8} - \frac{480}{x} = 3 \implies x^2 - 8x - 1280 = 0 \implies (x - 40)(x + 32) = 0 \implies x = 40$. Speed $= 40$ km/h.

Let unit digit be $y$
and ten's digit = $x$
hence, the two digit number = $10x + y$, (1/2)
ATQ
$xy = 12$ ..............(i) (1)
$10x + y + 36 = 10y + x$
$x - y + 4 = 0$ ....... (ii) (1)
From (i) and (ii)
$x^2 + 4x - 12 = 0$ (1/2)
$(x + 6)(x - 2) = 0$ (1/2)
Hence, $x = 2$
and $y = 6$ (1/2)
$\therefore$ Number = $26$ (1/2)

Let marks scored in Mathematics be $x$
and marks scored in Science be $y$
ATQ,
$x + y = 32$ ..............(i) (1)
and $(x + 4)(y - 2) = 253$ ...... (ii) (1)
from (i) and (ii)
$x^2 - 26x + 133 = 0$ (1)
$(x - 19)(x - 7) = 0$ (1)
$x = 19, 7$ (1/2)
$x = 19 \Rightarrow y = 13$
$x = 7 \Rightarrow y = 25$ (1/2)
Hence, marks in Mathematics and Science are $19, 13$ or $7, 25$

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.

Let the speed in upward direction be 'x' km/h
and the speed in downward direction = $(x + 10)$ km/h
ATQ
$\frac{150}{x} - \frac{150}{x+10} = \frac{5}{2}$
$\Rightarrow x^2 + 10 x - 600 = 0$
$\Rightarrow (x+30)(x - 20) = 0$
$\therefore x = 20$
and $x + 10 = 20 + 10 = 30$
Therefore, speeds in upward and downward direction are $20$ km/h and $30$ km/h respectively.

Let denominator be $x$
$\therefore$ Numerator $= (x - 3)$
Therefore, fraction $= \frac{x-3}{x}$ ($1$)
ATQ
$\frac{x-3}{x} + \frac{x-3+2}{x+2} = \frac{29}{20}$ ($1$)
$\Rightarrow 11x^2 - 98x - 120 = 0$ ($1$)
$\Rightarrow (x-10)(11x + 12) = 0$ ($1/2$)
So, $x = 10$ ($1/2$)
$\therefore$ Fraction $= \frac{7}{10}$ ($1$)

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. ($1$)

Let the length of shortest side be $x$ m
$\therefore$ length of longest side = $(x + 4)$ m
and length of third side = $(x + 2)$ m
Now, $(x + 4)^2 = x^2 + (x + 2)^2$
$\Rightarrow x^2 - 4x - 12 = 0$
$\Rightarrow (x – 6)(x + 2) = 0$
$\Rightarrow x = 6$
$\therefore$ sides are $6$ m, $8$ m and $10$ m
Area = $\frac{1}{2} \times 6 \times 8 = 24$ m$^2$
Perimeter = $6+8+10 = 24$ m
Difference = $0$

Let digit at unit place be $x$ and digit at tens place be $y$
$\therefore$ number $= 10y + x$
$ATQ$
$10y + x = 7(x + y)$
$\implies 3y = 6x$ or $y = 2x$ --- (1)
Also, $10y + x = 5xy + 2$ --- (2)
from (1) and (2), we get $10x^2 - 21x + 2 = 0$
$\implies (x - 2)(10x - 1) = 0$
$\therefore x = 2$
So, $y = 4$
$\therefore$ Required number is 42.

Let the consecutive odd numbers be $x$ and $x + 2 \implies x^2 + (x + 2)^2 = 650 \implies 2x^2 + 4x - 646 = 0$ or $x^2 + 2x - 323 = 0 \implies (x + 19)(x - 17) = 0 \implies x = 17 \therefore$ Odd numbers are $17, 19$

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)

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

Let the average speed of the aircraft be $x$ km/h
Time taken by aircraft $= \frac{600}{x}$
Time taken by aircraft when flight is delayed by $30$ min $= \frac{600}{x-200}$
$\frac{600}{x-200}$ - $\frac{600}{x}$ = ½ (2 Marks)
x²-200x - 240000 = 0 (1 Mark)
⇒ (x-600)(x + 400) = 0 (1 Mark)
x = 600, x = - 400
x = - 400 (rejected) (1/2 Mark)
∴ x = 600$\\therefore$ Scheduled duration of the flight $= \frac{600}{600} = 1$ hour (1/2 Mark)

Let time taken to fill the pool by larger diameter pipe alone be $x$ hours
Time taken to fill the pool by smaller diameter pipe alone = $(x + 10)$ hours
$\frac{4}{x}$ + $\frac{9}{x + 10}$ = ½ (2 Marks)
x² - 16x - 80 = 0 (1 Mark)
⇒ (x + 4)(x - 20) = 0 (1 Mark)
⇒ x = -4, x = 20
x = -4 (rejected) (1/2 Mark)
∴ x = 20
Time taken to fill the pool by larger diameter pipe alone = $20$ hours
Time taken to fill the pool by smaller diameter pipe alone = $30$ hours (1/2 Mark)

Let the marks obtained in Maths be $x$
Then the marks obtained in Science = $30 - x$
$(x + 2)(30 - x - 3) = 210$ (2 Marks)
$\Rightarrow x^2 - 25x + 156 = 0$ (1 Mark)
$\Rightarrow (x-13)(x - 12) = 0$ (1/2 Mark)
$\therefore x = 13, x = 12$ (1/2 Mark)
Either marks in Maths and Science are 13 and 17 respectively (1/2 Mark)
or marks in Maths and Science are 12 and 18 respectively (1/2 Mark)

Let the shortest side be $x$ cm
Then hypotenuse = $(2x + 6)$ cm
and the third side = $(3x - 6)$ cm
$x^2 + (3x - 6)^2 = (2x + 6)^2$ (2 Marks)
$\Rightarrow 6x^2 - 60x = 0$ (1 Mark)
$\Rightarrow 6x(x - 10) = 0$ (1/2 Mark)
$\Rightarrow x = 0, x = 10$
x = 0 (rejected) (1/2 Mark)
$\therefore x = 10$
and hypotenuse = 26 cm (1/2 Mark)
and third side = 24 cm (1/2 Mark)

Let original duration of the tour be $x$ days.
$\therefore$ Daily expense is $\frac{5400}{x}$
According to the question,
$\frac{5400}{x} - \frac{5400}{x+5} = 180$ (I) (2 Marks)
$\Rightarrow x^2 + 5x - 150 = 0$ (II) (1 Mark)
$\Rightarrow (x + 15) (x - 10) = 0$ (III) (1 Mark)
$\Rightarrow x = -15, 10$
$\therefore x \neq -15$
$\therefore x = 10$ (IV) ($\frac{1}{2}$ Mark)
$\therefore$ Original duration of tour is $10$ days and daily expense is ₹$540$. (V) ($\frac{1}{2}$ Mark)

Let the original length of cloth be $x$ metre.
$\therefore$ Original cost is $\frac{2100}{x}$ per metre.
According to the question,
$\frac{2100}{x} - \frac{2100}{x+2} = 120$ (I) (2 Marks)
$\Rightarrow x^2 + 2x - 35 = 0$ (II) (1 Mark)
$\Rightarrow (x + 7) (x - 5) = 0$ (III) (1 Mark)
$\Rightarrow x = -7,5$
$\therefore x \neq -7$
$\therefore x = 5$ (IV) ($\frac{1}{2}$ Mark)
$\therefore$ The original price of cloth per metre is ₹$420$ and original length is $5$ m. (V) ($\frac{1}{2}$ Mark)

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

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)

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)

Let the consecutive integers be $x, (x + 1)$ and $(x + 2)$ (1 Mark)
$x^2 + (x + 1)(x + 2) = 67$ (1 Mark)
$\Rightarrow 2x^2 + 3x - 65 = 0$ (1 Mark)
$\Rightarrow (x - 5)(2x + 13) = 0$ (1 Mark)
$\Rightarrow x = 5$ or $x = -\frac{13}{2}$ (1 Mark)
neglecting $x = -\frac{13}{2}$, as $x$ is a positive integer. ($\frac{1}{2}$ Mark)
So, $x = 5$
Thus, integers are $5, 6$ and $7$ ($\frac{1}{2}$ Mark)

(a) Let the sides of squares be $x$ and $y$
A.T.Q.
$x^2 + y^2 = 2650$ --------(i) [1 mark]
$4x + 4y = 280 \Rightarrow x + y = 70$ ---(ii) [1 mark]
getting $2x^2 - 140x + 2250 = 0$ or $x^2 - 70x + 1125 = 0$ [1 mark]
$\Rightarrow (x - 25)(x - 45) = 0$
$\Rightarrow x = 25$ and $x = 45$ [1 mark]
$\therefore y = 45$ and $y = 25$
sides of square are $25 \text{ cm}$ and $45 \text{ cm}$. [1 mark]

$\frac{x - 2 - x}{x (x - 2)} = 3$ [1 mark]
$\Rightarrow 3x^2 - 6x + 2 = 0$ [1 mark]
$\text{Discriminant} = 36 - 24 = 12$ [1 mark]
$\text{Roots are } \frac{6 + \sqrt{12}}{6} \text{ and } \frac{6 - \sqrt{12}}{6}$
$\text{or } 1 + \frac{\sqrt{3}}{3} \text{ and } 1 - \frac{\sqrt{3}}{3}$ [1+1 marks]

Let the two consecutive odd integers be $x$ and $x + 2$ [$\frac{1}{2}$ mark]
$x^2 + (x + 2)^2 = 290$ [$1\frac{1}{2}$ marks]
$2x^2 + 4x - 286 = 0$ or $x^2 + 2x - 143 = 0$ [$1\frac{1}{2}$ marks]
$(x - 11)(x + 13) = 0$
$x = 11$ [$1$ mark]
Required odd integers are $11$ and $13$ [$\frac{1}{2}$ mark]

Let width be $x\text{ m}$ and length be $(2x + 1)\text{ m}$ [$\frac{1}{2}$ mark]
A.T.Q. $(2x + 1)x = 300$ [$1\frac{1}{2}$ marks]
$2x^2 + x - 300 = 0$ [$1\frac{1}{2}$ marks]
$(x - 12)(2x + 25) = 0$
$x = 12$ [$1$ mark]
(Rejecting $x = -\frac{25}{2}$)
$\text{length} = 25\text{ m and width} = 12\text{ m}$ [$\frac{1}{2}$ mark]

Let the numbers be $x, x + 2$
$x^2 + (x + 2)^2 = 100$ (1½ Mark)
simplifying we get
$2x^2 + 4x - 96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x - 6) = 0$ (1/2 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are 6, 8 (1/2 Mark)

Let speed of stream be $x$ km/h (0.5 Mark)
Speed of boat upstream= $(20 – x)$ km/h
Speed of boat downstream = $(20 + x)$ km/h (0.5 Mark)
$\therefore \frac{48}{20-x} - \frac{48}{20+x} = 1$ (1.5 Mark)
$x^2 + 96x – 400 = 0$
$(x + 100) (x - 4) = 0$ (1 Mark)
$x = 4, x = -100$ (rejected)
Hence, speed of stream is $4$ km/h. (1.5 Mark)

(a) $2,-5$

(D) $p, -(p + 1)$

(A) $\pm 6$

(C) $5, -3$

(B) $\frac{a-1}{2}, -\frac{a+1}{2}$

(B) irrational and distinct

Answer (B) not real

Discriminant $= 16$
$x = \frac{-10 \pm \sqrt{16}}{2\sqrt{3}}$
$x = -\frac{3}{\sqrt{3}}, -\frac{7}{\sqrt{3}}$ or $x = -\sqrt{3}, -\frac{7}{3}\sqrt{3}$

Discriminant $= (-4a^2)^2 - 4 \times 4 \times (a^4 - b^4) = 16b^4$
Since, Discriminant $> 0$
Therefore, the given equation has real and distinct roots

$\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$
$\Rightarrow x^2 – 3x + 2 = 0$
$\Rightarrow (x - 2)(x – 1) = 0$
$\Rightarrow x = 1, 2$

$$\begin{aligned}& 4 (2x + 3) - 5x = 3x (2x + 3) \\ & \Rightarrow 6x^2 + 6x - 12 = 0\end{aligned}$$ or $$\begin{aligned}& x^2 + x - 2 = 0 \\ & \Rightarrow (x - 1) (x + 2) = 0 \\ & \Rightarrow x = 1, x = -2\end{aligned}$$

$\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$
$\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)} = \frac{10}{3}$
Simplifying, we get $2 x^2 - 19x + 42 = 0$
$\Rightarrow (x – 6)(2x – 7) = 0$
$\Rightarrow x = 6$ or $x = \frac{7}{2}$

$\frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \implies x[x - 1 + 2(x - 2)] = 6(x - 1)(x - 2) \implies 3x^2 - 13x + 12 = 0 \implies (3x - 4)(x - 3) = 0 \implies x = 3, x = \frac{4}{3}$

(i) Discriminant $= (p^2 - q^2)^2 + 4p^2q^2 = (p^2 + q^2)^2$ [$1 + 1$ mark]
(ii) $\therefore x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2p^2}$ [$1$ mark]
$= \frac{q^2}{p^2}, -1$ [$1 + 1$ mark]

Let the three consecutive positive integers be $x, x + 1, x + 2$ [$1$ mark]
A.T.Q. $x^2 + (x + 1) (x + 2) = 67$ [$1$ mark]
$\Rightarrow 2x^2 + 3x - 65 = 0$ [$1$ mark]
$\Rightarrow (2x + 13) (x - 5) = 0$ [$1$ mark]
$\Rightarrow x = 5, x = -\frac{13}{2}$ (rejected)
So the three consecutive positive integers are 5, 6 and 7 [$1$ mark]

(A) Given equation can be written as
$24(18 + x) – 24(18 – x) = 324 – x^2$ (1 Mark)
i.e., $x^2 + 48x – 324 = 0$ (1 Mark)
$D = 48^2 – 4(-324) = 3600$ (1 Mark)
Roots are $\frac{-48 \pm 60}{2}$ (1 Mark)
i.e., $6, -54$ (1 Mark)
OR
(B) Let the numbers be $x, x + 2$ (1/2 Mark)
$x^2 + (x + 2)^2 = 100$ (1 1/2 Mark)
simplifying we get
$2x^2 + 4x-96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x – 6) = 0$ (1 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are $6, 8$ (1/2 Mark)

Given equation can be written as
$24(18 + x) - 24(18 - x) = 324 - x^2$ (1 Mark)
i.e., $x^2 + 48x - 324 = 0$ (1 Mark)
$D = 48^2 - 4(-324) = 3600$ (1 Mark)
Roots are $\frac{-48 \pm 60}{2}$ (1 Mark)
i.e., 6, -54 (1/2 Mark)

$\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$
$\Rightarrow 3[(x - 2) (x – 5) + (x – 3) (x − 4)] = 10 (x – 3) (x - 5)$ (1.5 Mark)
$\Rightarrow 4x^2 - 38x + 84 = 0$ or $2x^2 - 19x + 42 = 0$ (1.5 Mark)
$\Rightarrow (x – 6) (2x – 7) = 0$ (1 Mark)
$\Rightarrow x = 6, x = \frac{7}{2}$ (1 Mark)

(C) $-1$

(D) $a^2 - 4bc$

(B) $17$

(C) 49

(B) $49$

(b) $1-4a^2$

$4x^2-5 = 0$
$a = 4, b = 0, c=-5$
Discriminant = $b^2 - 4ac = 0-4 (4) (-5) = 80 > 0$
$\Rightarrow$ roots are real and distinct.

(i) $200 x^2 = 128 (x + 1)^2$
(ii) $25x^2 = 16x^2 + 32x + 16$
$\Rightarrow 9x^2 - 32x - 16 = 0$
(iii) (a) $9x^2 - 32x - 16 = 0$
$\Rightarrow (9x + 4) (x - 4) = 0$
$x \neq -\frac{4}{9}$ so, $x = 4$
OR
(iii) (b) $x = \frac{32\pm\sqrt{1024+576}}{18} = \frac{32\pm 40}{18}$
$x \neq -\frac{4}{9}$ so, $x = 4$

(c) $6$

(b) $- x^2 + 4x + 4 = 0$

(D) $\frac{b^2}{4a}$

(B) $-\frac{3}{2}$

(b) $(x - \sqrt{3})^2$

(A) $4$

(B) $ax^2 - a^2x + 1 = 0$

(A) $l = n$

$2x^2 - 9x + 4 = 0$
$a = 2, b = -9, c = 4$
Let $\alpha, \beta$ be roots of $2x^2 – 9x + 4 = 0$
Sum = $\alpha + \beta = -\frac{b}{a} = -\frac{-9}{2} = \frac{9}{2}$
Product of roots = $\alpha\beta = \frac{c}{a} = \frac{4}{2} = 2$

Roots of $3x^2-2x-1=0$ are $-\frac{1}{3}, 1$ (1+1 Mark)
Roots of the quadratic equation $ax^2+bx+c=0$ are $\frac{5}{3}$ and $3$ (1/2+1/2 Mark)

Roots of $2x^2 - 5x + 2 = 0$ are $2$ and $\frac{1}{2}$ (1 + 1 Mark)
Roots of quadratic equation $ax^2 + bx + c = 0$ are $4$ and $1$ (1/2 + 1/2 Mark)

Roots of $5x^2 - 2x - 3 = 0$ are $1, -\frac{3}{5}$ (1+1 Mark)
Roots of equation $ax^2 + bx + c = 0$ are $0$ and $-\frac{8}{5}$ (1/2+1/2 Mark)

(b) $+ 12$

(a) $\frac{25}{2}$

(c) $\pm 2\sqrt{2}$

(B) $2$

(c) $ac=\frac{b^2}{4}$

(D) not-real

(C) not real

(C) $\frac{b^2}{4a}$

(D) Assertion (A) is false, but Reason (R) is true.

(D) $0, 8$

(d) $-\frac{1}{2}$

(D) -3

(A) $(x+1)^2 = 2x+1$

(A) $x^2 + 2x = 0$

(C) $x^2 + 1 = 0$

(A) $\pm\sqrt{24}$

(A) $3$ OR (B) $-3$

(B) $\sqrt{\frac{8}{3}}$

(C) $-\frac{9}{4}$

(B) $-5$ (1 Mark)

(B) $\frac{49}{12}$ (1 Mark)

(B) $\frac{9}{4}$

(B) $\frac{9}{4}$

Answer (C) $2$

(D) Assertion (A) is false, but Reason (R) is True.

(A) $0$

(C) $6$ only

(A) $2$

(B) atmost two

(B) $p^2 = -4qr$

(A) $b^2 = ac$

(D) $1$

(A) The roots of the given equation are reciprocal of each other.

(A) $-1$

$0$ or $-4$

(A) $\frac{3}{2}$

(C) $\frac{1}{5}$

For real and equal roots, $D = 0$
$k^2 - 16 = 0$
$k = \pm 4$

Discriminant $(D) = (q - r)^2 - 4 (p - q) (r - p)$ (I) ($\frac{1}{2}$ Mark)
$= (q + r - 2p)^2$ (II) ($\frac{1}{2}$ Mark)
Substituting, $q + r = 2p$
$\Rightarrow D = (2p - 2p)^2 = 0$ (III) ($\frac{1}{2}$ Mark)
$\therefore$ Roots of the given equation are equal. (IV) ($\frac{1}{2}$ Mark)

Discriminant $= 33$
$\Rightarrow x = \frac{9 \pm \sqrt{33}}{8}$

Discriminant $= (-4\sqrt{3})^2 - 4 \times 4 \times 3 = 0$
$\Rightarrow$ The given equation has real and equal roots

$px(x - 2) + 6 = 0 \Rightarrow px^2 – 2px + 6 = 0$
$a = p, b = -2p, c = 6$
Quadratic equation has equal roots, $\therefore D = 0$
$b^2-4ac = 0 \Rightarrow 4p^2-24p = 0$
$4p (p – 6) = 0$
$p = 0, p = 6$
$p = 0$ rejected $\therefore p = 6$

Let roots of the quadratic equation be $$\begin{aligned}& \alpha, 6\alpha \\ & px^2 - 14x + 8 = 0 \\ & therefore \alpha + 6\alpha = -\frac{-14}{p} \Rightarrow 7\alpha = \frac{14}{p} \Rightarrow \alpha = \frac{2}{p}\end{aligned}$$and $$\begin{aligned}& \alpha \cdot 6\alpha = \frac{8}{p} \Rightarrow 6\alpha^2 = \frac{8}{p} \Rightarrow 6 \cdot (\frac{2}{p})^2 = \frac{8}{p} \\ & \Rightarrow 6 \cdot \frac{4}{p^2} = \frac{8}{p} \\ & \Rightarrow \frac{24}{p^2} = \frac{8}{p} \\ & \Rightarrow 24p = 8p^2 \Rightarrow 8p^2 - 24p = 0 \Rightarrow 8p(p-3) = 0 \\ & \Rightarrow p = 0 \text{ or } p = 3 \\ & \text{Since } p \neq 0 \text{ (coefficient of } x^2 \text{ cannot be zero)} \\ & \Rightarrow p = 3\end{aligned}$$

For equal roots, $D = 0$
$[-(7p + 2)^2] - 4(2p + 1)(7p - 3) = 0 \implies 7p^2 - 24p - 16 = 0 \implies (7p + 4)(p - 4) = 0 \implies p = 4, p = -\frac{4}{7}$
For $p = 4$, the equation is $9x^2 - 30x + 25 = 0$ whose roots are $\frac{5}{3}, \frac{5}{3}$
For $p = -\frac{4}{7}$, the equation is $x^2 - 14x + 49 = 0$ whose roots are $7, 7$

Given equation has real and equal roots if
$\{-(7p + 2)\}^2 - 4(2p + 1)(7p - 3) = 0$
$\Rightarrow 7p^2 - 24p - 16 = 0$
$\Rightarrow (7p + 4) (p - 4) = 0$
$\Rightarrow p = -\frac{4}{7}, p = 4$

For real and equal roots.
$[-2(3k + 1)]^2 - 4(k + 1)(8k + 1) = 0$
$\Rightarrow k^2 - 3k = 0$
$\therefore k = 0, k = 3$

For real and equal roots, $D = b^2 - 4ac = 0$
$36 (k + 1)^2 - 4 (k + 1)\times 3 (k + 9) = 0$
$\Rightarrow k^2 - 2k - 3 = 0$
$\Rightarrow (k - 3) (k + 1) = 0$
$k \neq - 1$ So, $k = 3$

For real and equal roots, $D = 0$
$\therefore [-(p + 1)]^2 - 4(p + 4) = 0$
$\implies p^2 - 2p - 15 = 0$
$\implies (p - 5)(p + 3) = 0$
$\therefore p = 5, -3$
For $p = 5, 9x^2 - 6x + 1 = 0 \implies (3x - 1)(3x - 1) = 0 \therefore x = \frac{1}{3}, \frac{1}{3}$
For $p = -3, x^2 + 2x + 1 = 0 \implies (x + 1)(x + 1) = 0 \therefore x = -1, -1$
Hence roots are $\frac{1}{3}, \frac{1}{3}$ and $-1, -1$ for $p = 5$ and $p = -3$ respectively.

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).

(a) Let the smaller number be $y$ and greater number be $x$.
A.T.Q.
$x^2 - y^2 = 180$
$y^2 = 8x$
$\Rightarrow x^2 - 8x = 180$
$x^2 - 8x - 180 = 0$
$(x - 18) (x + 10) = 0$
$x = 18, x = -10$ (rejected)
$\therefore$ The numbers are $18$ and $12$

For equal roots; $b^2 - 4ac = 0$
$k^2 - 24 = 0$
$\Rightarrow k = \pm 2\sqrt{6}$
Equations are
$2x^2 + 2\sqrt{6}x + 3 = 0$; $2x^2 - 2\sqrt{6}x + 3 = 0$
Roots are $x = -\sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}}$; $x = \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}$

$x - \frac{1}{x} = 3 \implies x^2 - 1 = 3x \implies x^2 - 3x - 1 = 0$ (1 Mark)
$D = (-3)^2 - 4(1)(-1) = 9 + 4 = 13$ (1 Mark)
Roots are $\frac{3 + \sqrt{13}}{2}$, $\frac{3 - \sqrt{13}}{2}$ (1 Mark)
Now, $x + \frac{1}{x} = a \implies x^2 - ax + 1 = 0$ (1/2 Mark)
Since roots are real and equal, $D = 0$
$(-a)^2 - 4(1)(1) = 0 \implies a^2 - 4 = 0$ (1/2 Mark)
$a^2 = 4 \implies a = 2$ or $a = -2$ (1/2+1/2 Mark)