Question 515 Marks
Find the value of a and b for which $\text{x}=-\frac{3}4{}$ and x = -2 are the root of the equation $ax^2 + bx - 6 = 0.$
AnswerMultiplying (2) by 4 adding the result from (1), we get
11a = 44
⇒ a = 4
Putting a = 4 in (1), we get
3 × 4 + 4b = 32
⇒ 4b = 32 - 12 = 20
$\Rightarrow\text{b}=\frac{20}{4}=5$
$\therefore$ a = 4 and b = 5
View full question & answer→Question 525 Marks
Some students planned a picnic. The total budget for food was Rs. 2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by Rs. 20. How many students attended the picnic and how much did each student pay for the food?
AnswerLet the number of students be x.
Then, cost of food for each student $=\text{Rs. }\frac{2000}{\text{x}}$
If the number of students decreased by 5, then
New cost of food for each students $=\text{Rs. }\frac{2000}{\text{x}-5}$
It is given that:
$\frac{2000}{\text{x}-5}-\frac{2000}{\text{x}}=20$
$\Rightarrow\frac{\text{2000x}-\text{2000x}+10000}{\text{x}^2-\text{5x}}=20$
$\Rightarrow 10000 = 20x^2 - 100x$
$\Rightarrow 20x^2 - 100x - 10000 = 0$
$\Rightarrow x^2 - 5x - 500 = 0$
$\Rightarrow x^2 - 25x + 20x - 500 = 0$
$\Rightarrow x(x - 25) + 20(x - 25) = 0$
$\Rightarrow (x - 25)(x + 20) = 0$
$\Rightarrow x - 25 = 0 or x + 20 = 0$
$\Rightarrow x = 25 or x = -20$
Since number of students cannot be negative, $\text{x}\neq-20$
⇒ x = 25
⇒ x - 5 = 20
$\Rightarrow\frac{2000}{\text{x}-5}=\frac{2000}{20}=100$
Hence, the number of students who attended the picnic is 20 and the cost of food dor each student is Rs. 100.
View full question & answer→Question 535 Marks
Two natural numbers differ by 3 and their product is 504. Find the numbers.
AnswerLet the required number be x and (x - 3).
Then, we have
$x(x - 3) = 504$
$\Rightarrow x^2 - 3x - 504 = 0$
$\Rightarrow\text{x}=\frac{3\pm\sqrt{9+2016}}{2}$
$=\frac{3\pm\sqrt{2025}}{2}$
$=\frac{(3\pm45)}{2}$
$\Rightarrow\text{x}=\frac{3+45}{2}=24$ or $\text{x}=\frac{(3-45)}{2}=\frac{-42}{2}=-21$
Hence, the required numbers are (24, 21) or (-21 and - 24).
View full question & answer→Question 545 Marks
The speed of a boat in still water is 15km/hr. It goes 30km upstream and return back at the same point in 4 hours 30 minutes. Find the speed of the stream.
AnswerLet the speed of the stream be x km/h.It is given that the speed of a boat in still water is 15km/h.
Now,
Speed of the boat upstream = Speed of the boat in still water - Speed of the stream = (15 - x)km/h
Speed of the boat downstream = Speed of the boat in still water + Speed of the stream = (15 + x)km/h
We know that,
$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
Time is taken for upstream journey + Time taken for the downstream journey = 4h 30min
$\therefore\frac{30}{15-\text{x}}+\frac{30}{15+\text{x}}=4\frac{1}{2}$
$\Rightarrow30\Big[\frac{(15+\text{x})+(15-\text{x})}{(15+\text{x})(15-\text{x})}\Big]=\frac{9}{2}$
$\Rightarrow30\Big(\frac{30}{225-\text{x}^2}\Big)=\frac{9}{2}$
$\Rightarrow225-\text{x}^2=\frac{30\times30\times2}{9}$
$\Rightarrow225-\text{x}^2=200$
$\Rightarrow\text{x}^2=225-200=25$
$\Rightarrow\text{x}=\pm5$
Since speed cannot be negative, therefore, x = 5.
Thus, the speed of the stream is 5km/h.
View full question & answer→Question 555 Marks
Two taps running together can fill a tank in $3\frac{1}{13}\ \text{hours.}$ If one pipe takes 3 hours more than the other to fill the tank then how much time will each tap take to fill the tank
AnswerLet the faster tap takes x minutes to fill the tank.
Then, the other tap takes (x + 3) minute
$\therefore\frac{1}{\text{x}}+\frac{1}{(\text{x}+3)}=\frac{13}{40}$
$\Rightarrow\frac{(\text{x}+3)+\text{x}}{\text{x}(\text{x}+3)}=\frac{13}{40}$
$\Rightarrow 40(2x + 3) = 13(x^2 + 3x)$
$\Rightarrow 13x^2 - 41x - 120 = 0$
$\Rightarrow 13x^2 - 65x + 24x - 120 = 0$
$\Rightarrow 13x(x - 5) + 24(x - 5) = 0$
$\Rightarrow (13x + 24)(x - 5) = 0$
$\Rightarrow x - 5 = 0 or 13x + 24 = 0$
⇒ x = 5 or $\text{x}=-\frac{24}{3}$
Since time cannot be negative, $\text{x}\neq-\frac{24}{3}$
⇒ x = 5
The faster tap takes 5 minutes to fill the tank.
Then, the other tap takes (5 + 3) minutes = 8 minutes.
View full question & answer→Question 565 Marks
The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}.$ Find the numbers.
AnswerLet the required numbers be x and (x - 5).
Then, we have
x > x - 5
$\Rightarrow\frac{1}{\text{x}}<\frac{1}{\text{x}-5}$
$\therefore\frac{1}{\text{x}-5}-\frac{1}{\text{x}}=\frac{5}{14}$
$\Rightarrow\frac{\text{x}-\text{x}+5}{\text{x}(\text{x}-5)}=\frac{5}{14}$
$\Rightarrow 70 = 5x^2 - 25x$
$\Rightarrow 5x^2 - 25x - 70 = 0$
$\Rightarrow x^2 - 5x - 14 = 0$
$\Rightarrow x^2 - 7x + 2x - 14 = 0$
$\Rightarrow x(x - 7) + 2(x - 7) = 0$
$\Rightarrow (x - 7)(x + 2) = 0$
$\Rightarrow x - 7 = 0 or x + 2 = 0$
$\Rightarrow x = 7 or x = -2$
Since x is a natural number, x ≠ -2
⇒ x = 7 and x - 5 = 2
Hence, the required numbers are 7 and 2.
View full question & answer→Question 575 Marks
If the roots of the equations $ax^2 + 2bx + c = 0$ and $\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0$ are simultaneously real then prove that $b^2 = ac.$
AnswerIt is given that the roots of the equation $ax^2 + 2bx + c = 0 $are real.
$\therefore\text{D}_1=(\text{2b})^2-4\times\text{a}\times\text{c}\ge0$
$\Rightarrow4(\text{b}^2-\text{ac})\ge0$
$\Rightarrow\text{b}^2-\text{ac}\ge0\ \dots(1)$
Also, the roots of the equation $\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0$ are real.
$\therefore\text{D}_2=\big(-2\sqrt{\text{ac}}\big)^2-4\times\text{b}\times\text{b}\ge0$
$\Rightarrow4(\text{ac}-\text{b}^2)\ge0$
$\Rightarrow-4(\text{b}^2-\text{ac})\ge0$
$\Rightarrow\text{b}^2-\text{ac}\le0\ \dots(2)$
The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if:
$b^2- ac = 0$
$\Rightarrow b^2 = ac$
View full question & answer→Question 585 Marks
A train covers a distance of 480km at a uniform speed. If the speed had been 8km/h less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.
AnswerLet the usual speed of the train be x km/hr
Time taken to cover 480km $=\frac{480}{\text{x}}\ \text{hours.}$
Time taken to cover 480km when the speed is decreased by 8km/hr $=\frac{480}{\text{x}-8}\ \text{hours}.$
$\therefore\frac{480}{\text{x}}=\frac{480}{\text{x}-8}-3$
$\Rightarrow\frac{1500}{\text{x}}-\frac{1500}{\text{x}+100}=\frac{1}{2}$
$\Rightarrow\frac{480\text{x}-3840-480\text{x}}{\text{x}^2-\text{8x}}=-3$
$\Rightarrow -3840 = -3x^2 + 24x$
$\Rightarrow 3x^2 - 24x - 3840 = 0$
$\Rightarrow x^2 - 8x - 1280 = 0$
$\Rightarrow x^2 - 40x + 32x - 1280 = 0$
$\Rightarrow x(x - 40) + 32(x - 40) = 0$
$\Rightarrow (x - 40)(x + 32) = 0$
$\Rightarrow x - 40 = 0 or x + 32 = 0$
$\Rightarrow x = 40 or x = -32$
Since speed cannot be negative, $\text{x}\neq-32.$
⇒ x = 40
Hence, the usual speed of the plane is 40km/hr.
View full question & answer→Question 595 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$15x^2 - 28 = x$
AnswerGiven,
$15x^2 - 28 = x$
$15x^2 - x - 28 = 0$
$On comparing it with ax^2 + bx + c = 0, we get:$
$a = 15, b = -1 and c = -28$
Discriminant D is given by:
$D = (b^2- 4ac)$
$= (-1)^2 - 4 \times 15 \times (-28)$
$= 1 - (-1680)$
$= 1 + 1680$
$= 1681$
$= 1681 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-1)+\sqrt{1681}}{2\times15}$
$=\frac{1+41}{30}$
$=\frac{42}{30}$
$=\frac{7}{5}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-1)-\sqrt{1681}}{2\times15}$
$=\frac{1-41}{30}$
$=\frac{-40}{30}$
$=\frac{-4}{3}$
Thus, the roots of the equation are $\frac{7}{5}$ and $\frac{-4}{3}.$
View full question & answer→Question 605 Marks
The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.
AnswerLet x, y be the two natural numbers and x > y.
Then,
$\therefore$ $x^2 - y^2 = 45 ...(1)$
Also, square of smaller number = 4 larger number
$\Rightarrow y^2 = 4x ...(2)$
Putting value of $y^2$ from (1), we get
$x^2 - 4x = 45$
$\Rightarrow x^2 - 4x - 45 = 0$
$\Rightarrow x^2 - 9x + 5x - 45 = 0 or x(x - 9) + 5(x - 9) = 0$
$\Rightarrow (x - 9)(x - 5) = 0$
$\Rightarrow x = 9, -5$
But $\text{x}\neq-5\ \ \therefore\text{x}=9$
From (2), $y^2 = 4x = 4 \times 9 = 36$
$\therefore$ y = 6
Thus, the two required numbers are 9 and 6
View full question & answer→Question 615 Marks
A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.
AnswerLet the speed of the stream be x km/h. The speed of boat in still stream = 18km/h. Speed of boat up to the stream = 18 - x km/h $\therefore$ Time taken by boat to go up the stream 24km $=\frac{24}{18-\text{x}}\text{h}$
$\therefore$ Time taken by boat to go down the stream
$=\frac{24}{18+\text{x}}\text{h}$
Time taken by the boat to go up the stream is 1 hour more than the time taken down the stream.
$\therefore\frac{24}{18-\text{x}}-\frac{24}{18+\text{x}}=1$
$\Rightarrow\frac{1}{18-\text{x}}-\frac{1}{18+\text{x}}=\frac{1}{24}$
$\Rightarrow\frac{18+\text{x}-(18-\text{x})}{(18+\text{x})(18-\text{x})}=\frac{1}{24}$
$\Rightarrow\frac{2\text{x}}{324-\text{x}^2}=\frac{1}{24}$
$\Rightarrow 324 - x^2 = 48x$
$\Rightarrow x^2 + 48x - 324 = 0$
$\Rightarrow x^2 + 54x - 6x - 324 = 0$
$\Rightarrow x(x + 54) - 6(x + 54) = 0$
$\Rightarrow (x + 54)(x - 6) = 0$
$\Rightarrow x + 54 = 0 or x - 6 = 0$
$\Rightarrow x = -54 or x = 6$
Since the speed cannot be negative,
$\text{x}\neq-54.$
$\therefore$ x = 6
Hence, the speed of the stream = 6km/h.
View full question & answer→Question 625 Marks
The sum of two natural numbers is 15 and the sum of their reciprocals is $\frac{3}{10}.$ Find the numbers.
AnswerLet the required numbers be x and (15 - x).
Then, we have
$\frac{1}{\text{x}}+\frac{1}{15-\text{x}}=\frac{3}{10}$
$\Rightarrow\frac{15-\text{x}+\text{x}}{\text{x}(15-\text{x})}=\frac{3}{10}$
$\Rightarrow 150 = 45x - 3x^2$
$\Rightarrow 3x^2 - 45x + 150 = 0$
$\Rightarrow x^2 - 15x + 50 = 0$
$\Rightarrow x^2 - 10x - 5x + 50 = 0$
$\Rightarrow x(x - 10) - 5(x - 10) = 0$
$\Rightarrow (x - 10)(x - 5) = 0$
$\Rightarrow x - 10 = 0 or x - 5 = 0$
$\Rightarrow x = 10 or x = 5$
Hence, the required numbers are 5 and 10.
View full question & answer→Question 635 Marks
A man busy a number of pens for Rs. 180. If he had bought 3 more pens for the same amount, each pen would have cost him Rs. 3 less. How many pens did he buy?
AnswerLet the number of pens bought be x.
Then, cost of x pens = Rs. 180
⇒ cost of pen pen $=\text{Rs. }\frac{180}{\text{x}}$
If number of pens bought is (x + 3), then
Cost of one pen $=\text{Rs. }\frac{180}{\text{x}+3}$
It is given that
$\frac{180}{\text{x}}-\frac{180}{\text{x}+3}=3$
$\Rightarrow\frac{\text{180x}+\text{540x}-\text{180x}}{\text{x}^2+\text{3x}}=3$
$\Rightarrow 540x = 3x^2 + 9x$
$\Rightarrow 3x^2 + 9x - 540 = 0$
$\Rightarrow x^2 + 9x - 108 = 0$
$\Rightarrow x^2 + 15x - 12x - 108 = 0$
$\Rightarrow x(x + 15) - 12(x + 15) = 0$
$\Rightarrow (x + 15)(x - 12) = 0$
$\Rightarrow x + 15 = 0 or x - 12 = 0$
$\Rightarrow x = -15 or x = -12$
Since number of pens cannot be negative, $\text{x}\neq-15$
⇒ x = 12
Hence, he bought 12 pens.
View full question & answer→Question 645 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\text{x}^2-2\sqrt2\text{x}+1=0$
AnswerThe given equation is $2\text{x}^2-2\sqrt2\text{x}+1=0$.Comparing it with $ax^2 + bx + c = 0,$ we get
$\text{a}=2,\ \text{b}=-2\sqrt2$ and c = 1
$\therefore$ Discriminant, $D = b^2 - 4ac$
$=(-2\sqrt2)^2-4\times2\times1$
$=8-8=0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=0$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(-2\sqrt2\big)+\sqrt{0}}{2\times2}$
$=\frac{2\sqrt2}{4}$
$=\frac{\sqrt2}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(-2\sqrt2\big)-\sqrt0}{2\times2}$
$=\frac{2\sqrt2}{4}$
$=\frac{\sqrt2}{2}$
Hence, $\frac{\sqrt2}{2}$is the repeated root of the given equation.
View full question & answer→Question 655 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$x^2 - kx + 9 = 0$
AnswerThe given equation is $x^2-k x+9=0$
$\therefore D=(-k)^2-4 \times 1 \times 9$
$D=k^2-36$
The given equation has real and distinct roots if $D>0$.
$\therefore k^2-36>0$
$\Rightarrow(k-6)(k+6)>0$
$\Rightarrow k<-6 \text { or } k>6$
View full question & answer→Question 665 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$x^2 + x + 2 = 0$
AnswerThe given equation is $x ^2+ x +2=0$ Comparing it with $ax ^2+ bx + c =0$, we get
$a=1, b=1 \text { and } c=2$
$\therefore \text { Discriminant, } D=b^2-4 a c$
$=1^2-4 \times 1 \times 2$
$=1-8$
$=-7<0$
Hence, the given equation has no real roots (or real roots does not exist).
View full question & answer→Question 675 Marks
Solve the following equations by using the method of completing the square:
$x^2 + 8x - 2 = 0$
Answer$x^2 + 8x - 2 = 0$
$\Rightarrow x^2 + 8x = 2$
$\Rightarrow x^2 + 2 \times x \times 4 + 4^2 = 2 + 4^2 (Adding 4^2 on both sides)$
$\Rightarrow (x + 4)^2= 2 + 16 = 18$
$\Rightarrow\text{x}+4=\pm\sqrt{18}=\pm3\sqrt2$ (Taking square root on both sides)
$\Rightarrow\text{x}+4=3\sqrt2$ or $\text{x}+4=-3\sqrt2$
$\Rightarrow\text{x}=-4+3\sqrt2$ or $\text{x}=-4-3\sqrt2$
Hence, $\big(-4+3\sqrt2\big)$ and $\big(-4-3\sqrt2\big)$ are the roots of the given equation.
View full question & answer→Question 685 Marks
The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.
AnswerLet the age of a boy be x years. Then,His brother's age = (25 - x) years
It is given that
$x(25 - x) = 126$
$\Rightarrow 25x - x^2 = 126$
$\Rightarrow x^2 - 25x + 126$
$\Rightarrow x^2 - 18x - 7x + 126 = 0$
$\Rightarrow x(x - 18) - 7(x - 18) = 0$
$\Rightarrow (x - 18)(x - 7) = 0$
$\Rightarrow x - 18 = 0 or x - 7 = 0$
$\Rightarrow x = 18 or x = 7$
Hence, their ages are 18 years and 7 years.
View full question & answer→Question 695 Marks
Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.
AnswerSuppose two years ago, son's age be $x$ years. Then, man's age two years ago $=3 x^2$ years Present age of son $=(x+2)$ years And, present age of father $=\left(3 x^2+2\right)$ years Three year's hence, we have Son's age $=(x+2+3)=(x+5)$ years Father's age $=\left(3 x^2+2+3\right)=\left(3 x^2+5\right)$ years It is given that
$3x^2 + 5 = 4(x + 5)$
$\Rightarrow 3x^2 + 5 = 4x + 20$
$\Rightarrow 3x^2 - 4x - 15 = 0$
$\Rightarrow 3x^2 - 9x + 5x - 15 = 0$
$\Rightarrow 3x(x - 3) + 5(x - 3) = 0$
$\Rightarrow (x - 3)(3x + 5) = 0$
$\Rightarrow x - 3 = 0 or 3x + 5 = 0$
⇒ x = 3 or $\text{x}=\frac{-5}{3}$ Since, age cannot be in fraction,
$\text{x}\neq\frac{-5}{3}.$
$\Rightarrow x = 3$
Thus, present age of son $=(x+2)=5$ years.
And, present age of father $=\left(3 x^2+2\right)=3(3)^2+2=29$ years.
View full question & answer→Question 705 Marks
The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.
AnswerLet the base of triangle be x meter. Then, altitude of triangle = (x + 7)meter. $\therefore$ Area of triangle
$=\frac{1}{2}\times\text{x}\times(\text{x}\times\text{7})\text{cm}^2$
$\therefore\frac{1}{2}\times(\text{x}^2+\text{7x})=165$
$\Rightarrow x^2 + 7x - 330 = 0$
$\Rightarrow x2 + 22x - 15x - 330 = 0$
$\Rightarrow x(x + 22) - 15(x + 22) = 0$
$\Rightarrow x = -22 or x = 15 \Rightarrow x = 15$
$[\because$ base cannot be negative$]$Thus, the base of the triangle = 15m
And the altitude of triangle = (15 + 7) = 22m.
View full question & answer→Question 715 Marks
The sum of the areas of two squares is $640m^2$. If the difference in their perimeters be 64m, find the sides of the two squares.
AnswerLet x and y be the lenght of the two square fields.
$\therefore$ $x^2 + y^2 = 640 ...(1)$
4x - 4y = 64
$\therefore$ x - y = 16 ...(2)
From (2),
x = y + 16,
Putting value of x in (1)
$\Rightarrow (y + 16)^2+ y^2 = 640$
$\Rightarrow y^2 + 32y + 256 + y^2 = 640$
$\Rightarrow 2y^2 + 32y + 256 - 640 = 0$
$\Rightarrow 2y^2 + 32y - 384 = 0$
$\Rightarrow y^2 + 16y - 192 = 0$
$\Rightarrow y^2 + 24y - 8y - 192 = 0$
$\Rightarrow y(y + 24) - 8(y + 24) = 0$
$\Rightarrow (y + 24)(y - 8) = 0$
$\Rightarrow y + 24 = 0 or y - 8 = 0$
$\Rightarrow y = -24 or y = 8$
Putting y = 8 in (2)
x - 8 = 16
$\therefore$ x = 16 + 8 = 24
$\therefore$ Sides of two squares are 24m and 8m respectively.
View full question & answer→Question 725 Marks
The length of a rectangle is thrice as long as the side of a square. The side of the square is 4cm more than width of a the rectangle. Their areas being equal, find their dimensions.
AnswerLet the side of square be x cm.
Then, length of the rectangle = 3x cm
Breadth of the rectangle = (x - 4)cm
$\therefore$ Area of rectangle = Area of square x
$\therefore$ $3x(x - 4) = x^2$
$\Rightarrow 3x^2 - 12x = x^2$
$\Rightarrow 2x^2 - 12x = 0$
$\Rightarrow 2x(x - 6) = 0$
$\Rightarrow 2x = 0 or x - 6 = 0$
$\Rightarrow x = 0 or x = 6$
⇒ x = 6 (Side of the square is never 0)
Thus, side of the square = 6cm
And length of the rectangle = (3 × 6) = 18cm
Then, breadth of the rectangle = (6 - 4)cm = 2cm
View full question & answer→Question 735 Marks
A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.
AnswerLet the length = x meter. Area = length × breadth = 180m$^2$ Breadth $=\Big(\frac{180}{\text{x}}\Big)\text{m}$ $\therefore\text{x}=\frac{180}{\text{x}}+\frac{180}{\text{x}}=39$
$\Rightarrow x^2 - 39x + 360 = 0$
$\Rightarrow x^2 - 24x - 15x + 360 = 0$
$\Rightarrow x(x - 24) - 15(x - 24) = 0$
$\Rightarrow (x - 24)(x - 15) = 0$
$\Rightarrow x = 24 or x = 15$
If length of the rectangle = 15m Then, Breadth $=\Big(\frac{180}{15}\Big)\text{m}=12\text{m}$ Also, if length of rectangle = 24m $\therefore$ Breadth $=\frac{181}{24}\text{m}=\frac{15}2{}\text{m}=7.5\text{m}$
View full question & answer→Question 745 Marks
Solve the following equations by using the method of completing the square:
$\sqrt3\text{x}^2+10\text{x}+7\sqrt3=0$
Answer$\sqrt3\text{x}^2+10\text{x}+7\sqrt3=0$
$3\text{x}^2+10\sqrt3\text{x}+21=0$ $\big($Multiplying both sides by $\sqrt3\big)$
$\Rightarrow\text{3x}^2-10\sqrt3\text{x}=-21$
$\Rightarrow\big(\sqrt3\text{x}\big)^2-2\times\sqrt3\text{x}\times5+5^2\\=-21+5^2$ [Adding $5^2$ on both sides]
$\Rightarrow\big(\sqrt3\text{x}+5\big)^2=-21+25$
$=4=2^2$
$\Rightarrow\sqrt3\text{x}+5=\pm2$ (Taking square root on both sides)
$\Rightarrow\sqrt3\text{x}+5=2$ or $\sqrt3\text{x}+5=-2$
$\Rightarrow\sqrt3\text{x}=-3$ or $\sqrt3\text{x}=-7$
$\Rightarrow\text{x}=-\frac{3}{\sqrt3}=-\sqrt3$ or $\text{x}=-\frac{7}{\sqrt3}=-\frac{7\sqrt3}{3}$
Hence, $-\sqrt3$ and $-\frac{7\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 755 Marks
The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.
AnswerLet the required consecutive positive odd numbers be x and (x + 2).
Then, we have
$x^2 + (x + 2)^2 = 514$
$\Rightarrow x^2 + x^2 + 4x + 4 = 514$
$\Rightarrow 2x^2 + 4x - 510 = 0$
$\Rightarrow x^2 + 2x - 255 = 0$
$\Rightarrow x^2 - 15x + 17x - 255 = 0$
$\Rightarrow x(x - 15) + 17(x - 15) = 0$
$\Rightarrow (x - 15)(x + 17) = 0$
$\Rightarrow x - 15 = 0 or x + 17 = 0$
$\Rightarrow x = 15 or x = -17$
Since x is a positive number, x ≠ -17
⇒ x = 15
⇒ x + 2 = 15 + 2 = 147
Hence, the required positive odd numbers are 15 and 17.
View full question & answer→Question 765 Marks
Find two consecutive positive even integers whose product is 288.
AnswerLet the required consecutive positive even integers be x and (x + 2).
Then, we have
$x \times (x + 2) = 288$
$\Rightarrow x^2 + 2x - 288 = 0$
$\Rightarrow x^2 + 18x - 16x - 288 = 0$
$\Rightarrow x(x + 18) - 16(x + 18) = 0$
$\Rightarrow (x + 18)(x - 16) = 0$
$\Rightarrow x + 18 = 0 or x - 16 = 0$
$\Rightarrow x = -18 or x = 16$
Since x is a positive integer, x ≠ -18
⇒ x = 16
⇒ x + 2 = 16 + 2 = 18
Hence, the required consecutive positive even integers are 16 and 18.
View full question & answer→Question 775 Marks
The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.
AnswerLet the breadth of hall = x meters.
Then, lenght of the hall = (x + 3) meters.
$\therefore$ Area = lenght × breadth $= 238m^2$
$\Rightarrow x \times (x + 3) = 238$
$\Rightarrow x^2 + 3x - 238 = 0$
$\Rightarrow x(x + 17) - 14(x + 17) = 0$
$\Rightarrow (x + 17)(x - 14) = 0$
$\Rightarrow x + 17 = 0 or x - 14 = 0$
$\Rightarrow x = -17 or x = 14$
⇒ x = 14 $[\because$ breadth cannot be negative$]$
Thus, the breadth of hall is 14m
And lenght of the hall is (14 + 3)m = 17m.
View full question & answer→Question 785 Marks
If the roots of the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal prove that $\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}.$
AnswerIt is given that the roots of the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal.
$\therefore$ D = 0
$\Rightarrow [-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$
$\Rightarrow 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$
$\Rightarrow 4(a^2c^2 + b^2d^2 + 2abcd) - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2) = 0$
$\Rightarrow (-a^2d^2 + 2abcd - b^2c^2) = 0$
$\Rightarrow -(a^2d^2 - 2abcd + b^2c^2) = 0$
$\Rightarrow (ad - bc)^2 = 0$
$\Rightarrow ad - bc = 0$
$\Rightarrow ad = bc$
$\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}.$
Hence Proved.
View full question & answer→Question 795 Marks
A rectangular field is 16m long and 10m wide. There is a path of uniform width all around it, having an area of $120m^2$. Find the width of the path.
AnswerLet the width of the path be x m Length of the field including the path = 16 + x + x = 16 + 2x
Breadth of the field including the path = 10 + x + x = 10 + 2x
Now, (Area of the field including path) - (Area of the field excluding path)= Area of the path
$\Rightarrow (16 + 2x)(10 + 2x) - (16 \times 10) = 120$
$\Rightarrow 160 + 32x + 20x + 4x^2 - 160 = 120$
$\Rightarrow 4x^2 + 52x - 120 = 0$
$\Rightarrow x^2 + 13x - 30 = 0$
$\Rightarrow x^2 + (15 - 2)x + 30 = 0$
$\Rightarrow x^2 + 15x - 2x + 30 = 0$
$\Rightarrow x(x + 15) - 2(x + 15) = 0$
$\Rightarrow (x + 15)(x - 2) = 0$
$\Rightarrow x + 15 = 0 or x - 2 = 0$
$\Rightarrow x = -15 or x = 2$
$\Rightarrow x = 2 $$[\because$ width cannot be negative$]$Thus, the width of the path is 2m.
View full question & answer→Question 805 Marks
The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2cm and exceeds twice the length of the altitude by 1cm. Find the length of each side of the triangle.
AnswerLet the base of the triangle be x Then, hypotenuse = (x + 2)cm $\therefore$ (x + 2) - (2 altitude) = 1
⇒ altitude $=\frac{1}{2}(\text{x}+1)$ By applying pythagoras theorem we have, $\therefore(\text{x}+1)^2=\text{x}^2+\frac{1}{4}(\text{x}+1)^2$ $\Rightarrow\text{x}^2+4+\text{4x}$ $=\text{x}^2+\frac{\text{x}^2}{4}+\frac{1}{4}+\frac{1}{2}\text{x}$
$\Rightarrow4+\text{4x}=\frac{\text{x}^2}{4}+\frac{1}{4}+\frac{\text{x}}2{}$+
$\Rightarrow-\frac{\text{x}^2}{4}+\frac{7}{2}\text{x}+\frac{15}{4}=0$
$\Rightarrow -x^2 + 15 + 14x = 0$
$\Rightarrow x^2 - 14x - 15 = 0$
$\Rightarrow x^2 - 15x + x - 15 = 0$
$\Rightarrow x(x - 15) + 1(x - 15) = 0$
$\Rightarrow (x - 15)(x + 1) = 0$
$\Rightarrow x = 15 or x = -1$
$\Rightarrow x = 15$
$[\because$ base cannot be negative$]$Thus, the base of the triangle be = 15m
Then, hypotenuse of triangle = (15 + 2) = 17cm
And altituted of triangle $=\frac{1}{2}(15+1)=8\text{cm}$
View full question & answer→Question 815 Marks
A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.
AnswerLet the tens digit be x and units digit be y.
then, 10x + y = 4(x + y) and 10x + y = 2xy
$\Rightarrow y = 2x and 10x + y = 2xy$
$Putting y = 2x in 10x + y = 2xy$
$\Rightarrow 10x + 2x = 2x.2x$
$\Rightarrow 4x^2 - 12x = 0$
$\Rightarrow 4x(x - 3) = 0$
$\Rightarrow x - 3 = 0 or x = 3$
Hence, the tens digit is 3 and units digits is (2, 3) is
Hence the required number is 36.
View full question & answer→Question 825 Marks
Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.
AnswerLet the required number be x and y.
Then,
$x^2 + y^2 = 25(x + y) ...(1)$
$x^2 + y^2 = 50(x + y) ...(2)$
$\Rightarrow 25(x + y) = 50(x - y)$
$\Rightarrow x + y = 2(x - y)$
$\Rightarrow x = 3y$
Putting x = 3y in (1), we get
$9y^2 + y^2 = 100y$
$\Rightarrow 10y^2 - 100y = 0$
$\Rightarrow 10y(y - 10) = 0$
$\Rightarrow y = 10$
Hence, x = 30 and y = 10
View full question & answer→Question 835 Marks
Solve the following quadratic equation:
$2^{2x} - 3.2^{(x+2)} + 32= 0$
Answer$2^{2x} - 3.2^{(x+2)} + 32= 0 2^{2x} - 3.2^x.2^2 + 32 = 0 y^2 - 12y + 32 = 0$
$where 2^x = y \Rightarrow y^2 - 8y - 4y + 32 = 0$
$\Rightarrow y(y - 8) - 4(y - 8) = 0$
$\Rightarrow (y - 8)(y - 4) = 0$
$\Rightarrow y - 8 = 0 or y - 4 = 0$
$\Rightarrow y = 8 or y = 4$
$\Rightarrow 2^x = 8$
$\Rightarrow 2^x = (2)^3$
$\Rightarrow x = 3$
$\Rightarrow 2^x = 4$
$\Rightarrow 2^x = (2)^2$
$\Rightarrow x = 2$
Hence, 3 and 2 are the roots of given equation.
View full question & answer→Question 845 Marks
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500km away, in time, the pilot increased the speed by 100km/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?
AnswerLet the original speed of the train be x km/hr Then, Time taken to cover 1500km with the original speed $=\frac{1500}{\text{x}}\text{hrs}$ Time taken to cover 1500km with the speed of (x + 100)km/hr
$=\frac{1500}{\text{x}+100}\text{hrs}$
$\therefore\frac{1500}{\text{x}}=\frac{1500}{\text{x}+100}+\frac{1}{2}$
$\Rightarrow\frac{1500}{\text{x}}-\frac{1500}{\text{x}+100}=\frac{1}{2}$
$\Rightarrow\frac{1500\text{x}+150000-1500\text{x}}{\text{x}^2+\text{100x}}=\frac{1}{2}$
$\Rightarrow 300000 = x^2 + 100x$
$\Rightarrow x^2 + 100x - 300000 = 0$
$\Rightarrow x^2 + 600x - 500x - 300000 = 0$
$\Rightarrow x(x + 600) - 500(x + 600) = 0$
$\Rightarrow (x + 600)(x - 500) = 0$
$\Rightarrow x + 600 = 0 or x - 500 = 0$
$\Rightarrow x = -600 or x = 500$
Since speed cannot be negative,
$\text{x}\neq-600.$
⇒ x = 500 Hence, the original speed of the plane is 500km/hr.Yes, we apperciate the value shown by the pilot.
View full question & answer→Question 855 Marks
A train covers a distance of 300km at a uniform speed. If the speed of the train is increased by 5 km/hr, it take 2 hours less in the journey. Find the original speed of the train.
AnswerDistance travelled = 300km
Let the original speed of train be x kmph
Recall that, time taken $=\frac{\text{Distance}}{\text{Speed}}$
Hence time taken $=\frac{300}{\text{x}}$
Given that the speed of the train is increased by 5km an hour.
Hence the new speed is (x + 5)kmph
Time taken to cover 300km $=\frac{300}{(\text{x}+5)}$
Given that the time taken 2 hours less when compared to the previous time.
$\Rightarrow\frac{300}{\text{x}}-\frac{300}{\text{x}+5}=2$
$\Rightarrow\frac{300(\text{x}+5)-300\text{x}}{\text{x}(\text{x+5})}=2$
$\Rightarrow 300x + 1500 - 300x = 2x^2 + 10x$
$\Rightarrow 2x^2 + 10x - 1500 = 0$
$\Rightarrow x^2 + 5x - 750 = 0$
$\Rightarrow x^2 + 30x - 25x - 750 = 0$
$\Rightarrow x(x + 30) - 25(x + 30) = 0$
$\Rightarrow (x + 30)(x - 25) = 0$
$\Rightarrow x + 30 = 0 or x - 25 = 0$
$\therefore$ $x = -30 or x = 25$
Since speed cannot be negative, hence the original speed of train is 25kmph.
View full question & answer→Question 865 Marks
The sum of a natural number and its reciprocal is $\frac{65}{8}.$ Find the number.
AnswerLet the required numbers be x.
Then, its reciprocal $=\frac{1}{\text{x}}$
Thus, we have
$\text{x}+\frac{1}{\text{x}}=\frac{65}{8}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{65}8{}$
$\Rightarrow 8x^2+ 8 = 65x$
$\Rightarrow 8x^2 - 65x + 8 = 0$
$\Rightarrow 8x^2 - 64x - x + 8 = 0$
$\Rightarrow 8x(x - 8) - 1(x - 8) = 0$
$\Rightarrow (x - 8)(8x - 1) = 0$
$\Rightarrow x - 8 = 0 or 8x - 1 = 0$
⇒ x = 8 or $\text{x}=\frac{1}{8}$
Since x is a natural number, $\text{x}\neq\frac{1}{8}.$
⇒ x = 8
Hence, the required number is 8.
View full question & answer→Question 875 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
AnswerGiven: $\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$ On comparing it with $ax^2 + bx + c = 0,$ we get:
$\text{x}=\sqrt3,\ \text{b}=-2\sqrt2$ and $\text{c}=-2\sqrt3$
$\therefore$ Discriminant D is given by: $D = (b^2 - 4ac)$
$=\big(-2\sqrt2\big)^2-4\times\sqrt3\times\big(-2\sqrt3\big)$
$=8+24$ $=32>0$ So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{32}=4\sqrt2$ $\therefore\ \alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$ $=\frac{-\big(-2\sqrt2\big)+4\sqrt{2}}{2\times\sqrt3}$
$=\frac{6\sqrt2}{2\sqrt3}$
$=\sqrt6$ $\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big(-2\sqrt2\big)-4\sqrt{2}}{2\times\sqrt3}$
$=\frac{-2\sqrt2}{2\sqrt3}$
$=-\frac{\sqrt6}{3}$
Hence, $\sqrt6$ and $-\frac{\sqrt6}{3}$ are the roots of the given equation.
View full question & answer→Question 885 Marks
Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
AnswerLet the smaller part and larger part be x, 16 - x.
Then,
$\Rightarrow 2x^2- (16 - x)^2 = 164$
$\Rightarrow 2x^2 - (256 + x^2 - 32x) = 164$
$\Rightarrow 2x^2 - 256 - x^2 + 32x = 164$
$\Rightarrow x^2 + 32x - 256 - 164 = 0$
$\Rightarrow x^2 + 32x - 420 = 0$
$\Rightarrow x^2 + 42x - 10x - 420 = 0$
$\Rightarrow x(x + 42) - 10(x + 42)$
$\Rightarrow x + 42 = 0 or x - 10 = 0$
$\Rightarrow x = -42 or x = 10$
-42 is not a positive part
Hence, the larger and smaller parts are 10, 6 respectively.
View full question & answer→Question 895 Marks
The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.
AnswerLet the breadth of a rectangle = x meter
Then, lenght of the rectangle = 3x meter
$\therefore$ Area = lenght × breadth $= 147cm^2$
$\Rightarrow x \times 3x = 147$
$\Rightarrow 3x^2 = 147$
$\Rightarrow x^2 = 49$
$\Rightarrow\text{x}=\sqrt{49}$
$\Rightarrow\text{x}=\pm7$
$\Rightarrow\text{x}=7$ or $ \text{x}=-7$
$\Rightarrow\text{x}=7$ $[\because$ breadth cannot be negative$]$
Thus, breadth of rectangle = 7m
And lenght of rectangle = (3 × 7)m = 21m.
View full question & answer→Question 905 Marks
The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2\frac{9}{10}.$ Find the fraction.
AnswerLet the numerator and denominator be x. x+ 3
Then,
$\frac{\text{x}}{(\text{x}+3)}+\frac{(\text{x}+3)}{\text{x}}=2\frac{9}{10}$
$\Rightarrow\frac{\text{x}^2+(\text{x}+3)^2}{\text{x}(\text{x}+3)}=\frac{29}{10}$
$\Rightarrow\frac{\text{x}^2+\text{x}^2+9+\text{6x}}{\text{x}^2+\text{3x}}=\frac{{29}}{10}$
$\Rightarrow 20x^2 + 90 + 60x = 29x^2 + 87x$
$\Rightarrow 20x^2 - 29x^2 + 60x - 87x + 90 = 0$
$\Rightarrow -9x^2- 27x + 90 = 0$
$\Rightarrow x^2 + 3x - 10 = 0$
$\Rightarrow x^2 + (5x - 2x) - 10 = 0$
$\Rightarrow x(x + 5) - 2(x + 5) = 0$
$\Rightarrow (x + 5)(x - 2) = 0$
$\Rightarrow x + 5 = 0 or x - 2 = 0$
$\Rightarrow x = -5 or x = 2$
Hence, numerator and denominator are 2 and 5 respectively and fraction is $\frac{2}{5}.$
View full question & answer→Question 915 Marks
The sum of a number and its reciprocal is $2\frac{1}{30}.$ Find the number.
AnswerLet the required number be x.
Then, its reciprocal $=\frac{1}{\text{x}}$
Thus, we have
$\text{x}+\frac{1}{\text{x}}=2\frac{1}{30}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{61}{30}$
$\Rightarrow 30x^2 + 30 = 61x$
$\Rightarrow 30x^2 - 61x + 30 = 0$
$\Rightarrow 30x^2- 36x - 25x + 30 = 0$
$\Rightarrow 6x(5x - 6) - 5(5x - 6) = 0$
$\Rightarrow (5x - 6)(6x - 5) = 0$
$\Rightarrow 5x - 6 = 0 or 6x - 5 = 0$
$\Rightarrow\text{x}=\frac{6}{5}$ or $\text{x}=\frac{5}{6}$
Hence, the required number is $\frac{6}{5}$ or $\frac{5}{6}.$
View full question & answer→Question 925 Marks
A train travels at a certain average speed for a distance of 54km and then travels a distance of 63km at an average of 6km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
AnswerLet x be the first speed of the train.
We know that $\frac{\text{Distance}}{\text{Speed}}=\text{Time}$
Thus, we have
$\frac{54}{\text{x}}+\frac{63}{\text{x}+6}=3\ \text{hours}.$
$\Rightarrow\frac{54(\text{x}+6)+63\text{x}}{\text{x}(\text{x}+6)}=3$
$\Rightarrow 54(x + 6) + 63x = 3x(x + 6)$
$\Rightarrow 54x + 324 + 63x = 3x^2 + 18x$
$\Rightarrow 117x + 324 = 3x^2 + 18x$
$\Rightarrow 3x^2 - 117x - 324 + 18x = 0$
$\Rightarrow 3x^2 - 99x - 324 = 0$
$\Rightarrow x^2 - 33x - 108 = 0$
$\Rightarrow x^2 - 36x + 3x - 108 = 0$
$\Rightarrow x(x - 36) + 3(x - 36) = 0$
$\Rightarrow (x - 36)(x + 3) = 0$
$\Rightarrow x - 36 = 0 or x + 3 = 0$
$\Rightarrow x = 36 or x = -3$
Since speed cannot be negative
Hence, the intial speed of the train is 36km/hr.
View full question & answer→Question 935 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$5x^2 - kx + 1 = 0$
AnswerThe given equation is $5x^2 - kx + 1 = 0$
$\therefore$$ D = (-k)^2 - 4 \times 5 \times 1$
$D = k^2 - 20$
The given equation has real and distinct roots if D > 0.
$\therefore\text{k}^2-20>0$
$\Rightarrow\text{k}^2-\big(2\sqrt5\big)^2>0$
$\Rightarrow\big(\text{k}-2\sqrt5\big)\big(\text{k}+2\sqrt5\big)>0$
$\Rightarrow\text{k}<-2\sqrt5$ or $\text{k}>2\sqrt5$
View full question & answer→Question 945 Marks
A motorboat whose speed is 9km/hr in still water, goes 15km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.
AnswerLet the speed of the stream be = x km/h.Speed of the boat in still water = 9km/h
Speed of the boat down stream = 9 + x km/h
$\therefore$ time taken by boat to go 15km downstream $=\frac{15}{9+\text{x}}\text{hrs.}$
Speed of boat upstream = 9 - x
$\therefore$ time taken by boat to go 15km up stream $=\frac{15}{9-\text{x}}\text{hrs.}$
Total time $=\frac{15}{9+\text{x}}+\frac{15}{9-\text{x}}=3\frac{45}{60}=3\frac{3}{4}=\frac{15}{4}$
Dividing by 15
$\frac{1}{9+\text{x}}+\frac{1}{9-\text{x}}=\frac{1}{4}$
$\Rightarrow\frac{9+\text{x}+9-\text{x}}{(9+\text{x})(9-\text{x})}=\frac{1}{4}$
$\Rightarrow\frac{18}{81-\text{x}^2}=\frac{1}{4}$
$\Rightarrow81-\text{x}^2=72$
$\therefore\text{x}^2=81-72=9$
$\therefore\text{x}=\pm3$
But $\text{x}\neq-3$
$\therefore$ Speed of the stream = 3km/h.
View full question & answer→Question 955 Marks
The perimeter of a rectangular plot is 62m and its area is 228 sq metres. Find the dimensions of the plot.
AnswerPerimeter of a rectangle = 62m.
Let the lenght of the rectangle be x m. Then, breadth of the rectangle
$=\frac{\text{Perimeter}}{2}-\text{Length}$
$=\frac{62}{2}-\text{x}=(31-\text{x})\text{m}$Now, Area $= 228m^2$
Area = lenght × breadth $= 228m^2$
$\Rightarrow x(31 - x) = 228$
$\Rightarrow 31x - x^2 = 228$
$\Rightarrow x^2 - 31x + 228 = 0$
$\Rightarrow x^2 + 3x - 238 = 0$
$\Rightarrow x^2 - 19x - 12x + 228 = 0$
$\Rightarrow x(x - 19) - 12(x - 19) = 0$
$\Rightarrow (x - 19)(x - 12) = 0$
$\Rightarrow x - 19 = 0 or x - 12 = 0$
$\Rightarrow x = 19 or x = -12$
$\Rightarrow x = 19$
$[\because$ breadth cannot be negative$]$
Hence, the lenght of a rectangle is 19m and the breadth of a rectangle is 12m.
View full question & answer→Question 965 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}^2-(\sqrt3+1)\text{x}+\sqrt3=0$
AnswerThe given equation is:$\text{x}^2-(\sqrt3+1)\text{x}+\sqrt3=0$
Comparing it with $ax^2 + bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-(\sqrt3+1)$ and $\text{c}=\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=[-(\sqrt3+1)]^2-4\times1\times\sqrt3$
$=3+1+2\sqrt3-4\sqrt3$
$=3-2\sqrt3+1$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{\big(\sqrt3-1\big)^2}=\sqrt3-1$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\big(\sqrt3-1\big)}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\big(\sqrt3-1\big)}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3+1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and 1 are the roots of the given equation.
View full question & answer→Question 975 Marks
If a and b are real and a ≠ b then show that the roots of the equation $(a - b)x^2 + 5(a + b)x - 2(a - b) = 0$ are real and unequal.
AnswerThe given equation is $(a-b) x^2+5(a+b) x-2(a-b)=0$
$\therefore D=[5(a+b)]^2-4 \times(a-b) \times[-2(a-b)]$
$D=25(a+b)^2+8(a-b)^2$
Since $a$ and $b$ are real and $a \neq b$, so $(a-b)^2>0$ and $(a+b)^2>0$.
$\therefore 8(a-b)^2>0 \ldots(1)$ (Product of two positive numbers is always positive)
Also, $25(a+b)^2>0 \ldots(2)$ (Product of two positive numbers is always positive)
Adding (1) and (2), we get
$25(a+b)^2+8(a-b)^2>0$ (Sum of two positive numbers is always positive)
$\Rightarrow D>0$
Hence, the roots of the given equation are real and unequal.
View full question & answer→Question 985 Marks
One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.
AnswerLet the age of the son be x and age of man be y. 1 year ago
$\therefore$ (y - 1) = 8(x - 1)
$\Rightarrow y - 1 = 8x - 8$
$\Rightarrow y - 8x = -7$
$\Rightarrow y = -7 + 8x Also, y = x^2$
$\Rightarrow (-7 + 8x) = x^{2$
$} \Rightarrow 8x - 7 = x^2$
$\Rightarrow x^2 - 8x + 7 = 0$
$\Rightarrow x^2 - 7x - x + 7 = 0$
$\Rightarrow x(x - 7) - 1(x - 7) = 0$
$\Rightarrow (x - 7)(x - 1) = 0$
$\Rightarrow x - 7 = 0 or x - 1 = 0$
$\Rightarrow x = 7 or x = 1$
⇒ x = 7 $[\text{x}\neq1]$ Age of the son = 7 years.
Age of man = (-7 + 8 × 7) = -7 + 56 = 49 years.
View full question & answer→Question 995 Marks
The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}.$ Find the numbers.
AnswerLet the required numbers be x and (9 - x).
Then, we have
$\frac{1}{\text{x}}+\frac{1}{9-\text{x}}=\frac{1}{2}$
$\Rightarrow\frac{9-\text{x}+\text{x}}{\text{x}(9-\text{x})}=\frac{1}{2}$
$\Rightarrow 18 = 9x - x^2$
$\Rightarrow x^2 - 9x + 18 = 0$
$\Rightarrow x^2 - 6x - 3x + 18 = 0$
$\Rightarrow x(x - 6) - 3(x - 6) = 0$
$\Rightarrow (x - 6)(x - 3) = 0$
$\Rightarrow x - 6 = 0 or x - 3 = 0$
$\Rightarrow x = 6 or x = 3$
Hence, the required numbers are 3 and 6.
View full question & answer→Question 1005 Marks
Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
AnswerLet the time taken by the smaller pipe to fill the tank be x hours.
Then, the time taken by the larger pipe = (x - 9) hours.
Part of tank filled by smaller pipe in 1 hour $=\frac{1}{\text{x}}$
Part of tank filled by smaller pipe in 1 hour $=\frac{1}{\text{x}-9}$
It is given that the tank can be filled in 6 hours by both the pipes together.
$\therefore\frac{1}{\text{x}}+\frac{1}{\text{x}-9}=\frac{1}{6}$
$\Rightarrow\frac{100}{\text{x}}+\frac{100}{\text{x}+5}=9$
$\Rightarrow\frac{\text{x}-9+\text{x}}{\text{x}^2-\text{9x}}=\frac{1}{6}$
$\Rightarrow 6(2x - 9) = x^2 - 9x$
$\Rightarrow 12x - 54 = x^2- 9x$
$\Rightarrow x^2 - 21x + 54 = 0$
$\Rightarrow x^2 - 18x + 3x + 54 = 0$
$\Rightarrow x(x - 18) - 3(x - 18) = 0$
$\Rightarrow (x - 18)(x - 3) = 0$
$\Rightarrow x - 18 = 0 or x - 3 = 0$
$\Rightarrow x = 18 or x = 3$
Since time cannot be less than 9, $\text{x}\neq3$
⇒ x = 18
⇒ x - 9 = 18 - 9 = 9
Hence, the smaller pipe taken 18 hours to fills the tank and the larger pipe takes 9 hours to fills the tank.
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