Question 14 Marks
The sum of the reciprocals of Meena's ages (in years) $3$ years ago and $5$ years hence is $\frac{1}{3}.$ Find her present age.
AnswerLet the present age of Meena be $x$ years.
Then,
$3$ years ago, Meena's age $= (x - 3)$ years.
$5$ years hence, Meena's age $= (x + 5)$ years
It is given that
$\frac{1}{\text{x}-3}+\frac{1}{\text{x}+5}=\frac{1}{3}$
$\Rightarrow\frac{\text{x}+5+\text{x}-3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{3}$
$\Rightarrow\frac{\text{2x}+2}{\text{x}^2+\text{2x}-15}=\frac{1}{3}$
$\Rightarrow 6x + 6 = x^2 + 2x - 15$
$\Rightarrow x^2 - 4x - 21 = 0$
$\Rightarrow x^2 - 7x + 3x - 21 = 0$
$\Rightarrow x(x - 7) + 3(x - 7) = 0$
$\Rightarrow (x - 7)(x + 3) = 0$
$\Rightarrow x - 7 = 0 or x + 3 = 0$
$\Rightarrow x = 7$ or $x = -3$
Since age cannot be negative, $\text{x}\neq-3.$
$\Rightarrow x = 7$
Hence, Meena's present age is $7$ years.
View full question & answer→Question 24 Marks
A teacher on attempting to arrange the students for mass drill in the form of a solid square found that $24$ students were left. When he increased the size of the square by one student, he found that he was short of $25$ students. Find the number of students.
AnswerLet there be x rows and number of student in each row be x.
Then, total number of students $= (x^2 + 24)$
$\Rightarrow x^2 + 24 = (x + 1)^2 - 25$
$\Rightarrow x^2 + 24 + x^2 + 1 + 2x - 25$
$\Rightarrow 2x = 48$
$\Rightarrow x = 24$
Hence total number of student
$= [(24)^2 + 24] = 567 + 24 = 600$
Total number of students is $600$.
View full question & answer→Question 34 Marks
Solve the following equations by using the method of completing the square:
$3x^2 - 2x - 1 = 0$
Answer$3x^2 - 2x - 1 = 0$
$\Rightarrow 9x^2 - 6x - 3 = 0$ (Multiplying both sides by $3$)
$\Rightarrow 9x^2 - 6x = 3$
$\Rightarrow (3x)^2 - 2 \times 3x \times 1 + 1^2 = 3 + 1^2$ [Adding $1^2$ on both sides]
$\Rightarrow (3x - 1)^2 = 3 + 1 = 4 = (2)^2$
$\Rightarrow\text{3x}-1=\pm2$ (Taking square root on both sides)
$\Rightarrow 3x - 1 = 2$ or $3x - 1 = -2$
$\Rightarrow 3x = 3$ or $3x = -1$
$\Rightarrow x = 1$ or $\text{x}=-\frac{1}{3}$
Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.
View full question & answer→Question 44 Marks
Solve the following equations by using the method of completing the square:
$2x^2 + 5x - 3 = 0$
Answer$2x^2 + 5x - 3 = 0$
$\Rightarrow 4x^2 + 10x - 6 = 0$ (Multiplying both sides by $2$)
$\Rightarrow 4x^2 + 10x = 6$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\frac{5}{2}+\Big(\frac{5}{2}\Big)^2\\=6+\Big(\frac{5}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{5}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}+\frac{5}{2}\Big)^2$
$=6+\frac{25}{4}$
$=\frac{24+25}{4}=\frac{49}{7}=\Big(\frac{7}{2}\Big)^2$
$\Rightarrow\text{2x}+\frac{5}{2}=\pm\frac{7}{2}$ (Taking square root on both sides)
$\Rightarrow\text{2x}+\frac{5}{2}=\frac{7}{2}$ or $\text{2x}+\frac{5}{2}=-\frac{7}{2}$
$\Rightarrow\text{2x}=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1$ or $\text{2x}=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6$
$\Rightarrow\text{x}=\frac{1}{2}$ or $x = -3$
Hence $\frac{1}{2}$ and -3 are the roots of the given equation.
View full question & answer→Question 54 Marks
The product of Tanvy's age (in years) $5$ years ago and her age $8$ years later is $30$. Find her present age.
AnswerLet the present age of Tanvy be x years.
Then,$(x - 5)(x + 8) = 30$
$\Rightarrow x^2 + 8x - 5x - 40 = 30 $
$\Rightarrow x^2 + 3x - 40 - 30 = 0 $
$\Rightarrow x^2 + 3x - 70 = 0 $
$\Rightarrow x^2 + 10x - 7x - 70 = 0$
$ \Rightarrow x(x + 10) - 7(x + 10) = 0$
$ \Rightarrow (x + 10)(x - 7) = 0 $
$\Rightarrow x + 10 = 0$ or $x - 7 = 0 $
$\Rightarrow x = -10$ or $x = 7$
$\Rightarrow x = $7 $(\because$ age cannot be negative$)$
Hence, the present age of Tanvy is $7$ years.
View full question & answer→Question 64 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
AnswerThe given equation is $\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$.Comparing it with $ax^2 + bx + c = 0$, we get
$\text{a}=\sqrt2,\ \text{b}=7$ and $\text{x}=5\sqrt2$
$\therefore$ Discriminant, $D = b^2 - 4ac$
$=(7)^2-4\times\sqrt2\times5\sqrt2$
$=49-40 = 9>0$
So, the given equation has real roots.
$\sqrt{\text{D}}=\sqrt9=3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7+3}{2\times\sqrt2}$
$=\frac{-4}{2\sqrt2}$
$=-\sqrt2$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7-3}{2\times\sqrt2}$
$=\frac{-10}{2\sqrt2}$
$=-\frac{5\sqrt2}{2}$
Hence, $-\sqrt2$ and $=-\frac{5\sqrt2}{2}$ are the roots of the given equation.
View full question & answer→Question 74 Marks
A person on tour has $Rs. 10800$ for his expenses. If he extends his tour by $4$ days, he has to cut down his daily expenses by $Rs. 90$. Find the original duration of the tour.
AnswerLet the original duration of a tour be $x$ days.
Total expenditure on tour $= Rs. 10800$
$\therefore$ Expenditure per day $=\text{Rs. }\frac{10800}{\text{x}}$
Duration of extended tour $= (x + 4)$days
$\therefore$ Expenditure per day according to new schedule $=\text{Rs. }\frac{10800}{\text{x}+4}$
It is given that:
$\frac{10800}{\text{x}}-\frac{10800}{\text{x}+4}=90$
$\Rightarrow\frac{\text{10800x}+43200-\text{10800x}}{\text{x}^2+\text{4}}=90$
$\Rightarrow 43200 = 90x^2 + 360x$
$\Rightarrow 90x^2 + 360x - 43200 = 0$
$\Rightarrow x^2 + 4x - 480 = 0$
$\Rightarrow x^2 + 24x - 20x - 480 = 0$
$\Rightarrow x(x + 24) - 20(x + 24) = 0$
$\Rightarrow (x + 24)(x - 20) = 0$
$\Rightarrow x + 24 = 0$ or $x - 20 = 0$
$\Rightarrow x = -24$ or $x = 20$
Since number of days cannot be negative, $\text{x}\neq-24$
$\Rightarrow x = 20$
Hence, the original duration of the tour is $20$ days.
View full question & answer→Question 84 Marks
The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.
AnswerLet the altitude of triangle be x meter.
Hence, base = 3x meter
$\therefore$ Area of triangle $=\frac{1}{2}\times(3\text{x}\times\text{x})\text{cm}^2$
$=\frac{1}{2}\times\text{3x}^2=96$
$\Rightarrow\text{x}^2=\frac{96\times2}{3}$
$\Rightarrow\text{x}^2=64$
$\Rightarrow\text{x}=\sqrt{64}$
$\Rightarrow\text{x}=\pm8$
$\therefore\text{x}=8\ [\because$ lenght of altitude can never be negative$]$
Hence, altitude of triangle is 8cm.
And base of triangle = 3x = (3 × 8)cm = 24cm.
View full question & answer→Question 94 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2\text{x}^2+6\sqrt3\text{x}-60=0$
AnswerThe given equation is: $2\text{x}^2+6\sqrt3\text{x}-60=0$
On comparing it with $ax^2 + bx + c = 0$,
we get: $\text{x}=2,\ \text{b}=6\sqrt3$ and $\text{c}=-60$
$\therefore$ Discriminant $D$ is given by: $D = (b^2 - 4ac)$ $=\big(6\sqrt3\big)^2-4\times2\times(-60)$
$=108+480$ $=588>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{588}=14\sqrt3$
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-6\sqrt3+14\sqrt{3}}{2\times2}$
$=\frac{8\sqrt3}{4}$ $=2\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-6\sqrt3-14\sqrt{3}}{2\times2}$
$=\frac{-20\sqrt3}{4}$ $=-5\sqrt3$
Hence, $2\sqrt3$ and $-5\sqrt3$ are the roots of the given equation.
View full question & answer→Question 104 Marks
A takes $10$ days than the time taken by B to finish a piece of work. If both A and B together can finish the work in $12$ days, find the time taken by B to finish the work.
AnswerSuppose $B$ alone takes $x$ days to finish the work
Then, $A$ alone can finish it in $(x - 10)$ days.
Now, (A's one day's work) + (B's one day work) $=\frac{1}{\text{x}-10}+\frac{1}{\text{x}}$
And, $(A + B)$'s one day's work $=\frac{1}{12}$
$\therefore\frac{1}{\text{x}-10}+\frac{1}{\text{x}}=\frac{1}{12}$
$\Rightarrow\frac{\text{x}+\text{x}-10}{\text{x}(\text{x}-10)}=\frac{1}{12}$
$\Rightarrow 12(2x - 10) = x(x - 10)$
$\Rightarrow 24x - 120 = x^2 - 10x$
$\Rightarrow x^2 - 34x + 120 = 0$
$\Rightarrow x^2 - 30x - 4x + 120 = 0$
$\Rightarrow x(x - 30) - 4(x - 30) = 0$
$\Rightarrow (x - 30)(x - 4) = 0$
$\Rightarrow x - 30 = 0$ or $x - 4 = 0$
$\Rightarrow x = 30$ or $x = 4$
Since number cannot be less that $10$, $\text{x}\neq4$
$\Rightarrow x = 30$
Hence, $B$ alone can finish the work in $30$ days.
View full question & answer→Question 114 Marks
In a class test, the sum of the marks obtained by P in mathematics and science is $28$.
Had he got $3$ more marks in mathematics and $4$ marks less in science, the product of marks obtained in the two subjects would have been $180$.
Find the marks obtained him in the two subjects separately.
AnswerLet the marks obtained by $P$ in mathematics $= x$ Then,
marks obtained by him in science $= 28 - x$
It is given that $(x + 3)(28 - x - 4) = 180$
$\Rightarrow (x + 3)(24 - x) = 180 $
$\Rightarrow 24x - x^2 + 72 - 3x = 180 $
$\Rightarrow 21x - x^2 = 108$
$\Rightarrow x^2 - 12x - 9x + 108 = 0$
$\Rightarrow x(x - 12) - 9(x - 12) = 0$
$\Rightarrow (x - 12)(x - 9) = 0$
$\Rightarrow x - 12 = 0$ or $x - 9 = 0$
$\Rightarrow x = 12$ or $x = 9$
When $x = 12, 28 - x = 28 - 12 = 16$
When $x = 9 28 - x = 28 - 9 = 19$
Hence,
Marks in mathematics$ = 12$ and marks in science $= 16$
or
Marks in mathematics $= 9$ and marks in science $= 19$
View full question & answer→Question 124 Marks
Solve the following quadratic equation: $x^2 - 4ax - b^2 + 4a^2 = 0$
Answer$x^2 - 4ax - b^2 + 4a^2 = 0$
$\Rightarrow x^2 - 4ax + (4a^2 - b^2) = 0$
$\Rightarrow x^2 - 4ax + (2a + b)(2a - b) = 0$
$\Rightarrow x^2 - (2a + b)x - (2a - b)x + (2a + b)(2a - b) = 0$
$\Rightarrow x[x - (2a + b)] - (2a - b)[x + (2a + b)] = 0$
$\Rightarrow [x - (2a + b)][x - (2a - b)] = 0$
$\Rightarrow x - (2a + b) = 0$ or $x - (2a - b) = 0$
$\Rightarrow x = 2a + b$ or $x = 2a - b$
View full question & answer→Question 134 Marks
The sum of the squares of two consecutive multiples of $7$ is $1225$. Find the multiples.
AnswerLet the required numbers be $x$ and $(x + 7)$.
Then, we have
$x^2+ (x + 7)^2 = 1225$
$\Rightarrow x^2 + x^2 + 14x + 49 = 1225$
$\Rightarrow 2x^2+ 14x - 1176 = 0$
$\Rightarrow x^2 + 7x - 588 = 0$
$\Rightarrow x^2 + 28x - 21x - 588 = 0$
$\Rightarrow x(x + 28) - 21(x + 28) = 0$
$\Rightarrow (x + 28)(x - 21) = 0$
$\Rightarrow x + 28 = 0$ or $x - 21 = 0$
$\Rightarrow x = -28$ or $x = 21$
When $x = -28$
$x + 7 = -28 + 7 = -21$
When $x = 21$
$x + 7 = 21 + 7 = 28$
Hence, the required numbers are $21, 28$ or $-21, -28.$
View full question & answer→Question 144 Marks
The distance between Mumbai and Pune is $192\ km$. Travelling by the Deccan Queen, it takes $48$ minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by $20\ km/hr$.
AnswerLet the speed of the Deccan Queen $= x\ kmph$.
The speed of other train $= (x - 20)\ kmph.$
Then, time taken by Deccan Queen $=\Big(\frac{192}{\text{x}}\Big)\text{h}$
Time taken by other train $=\Big(\frac{192}{\text{x}-20}\Big)\text{h}$
Difference of time taken by two trains is $\frac{48}{60}\neq\frac{4}{5}\text{h}$
$\therefore\frac{192}{\text{x}-20}-\frac{192}{\text{x}}=\frac{4}{5}$
$\Rightarrow\frac{1}{\text{x}-20}-\frac{1}{\text{x}}=\frac{1}{240}$
$\Rightarrow\frac{\text{x}-\text{x}+20}{\text{x}^2-20\text{x}}=\frac{1}{240}$
$\Rightarrow x^2 - 20x - 4800 = 0$
$\Rightarrow x^2 - 80x + 60x - 4800 = 0$
$\Rightarrow x(x - 80) + 60(x - 80) = 0$
$\Rightarrow (x - 80)(x + 60) = 0$
$\Rightarrow x - 80 = 0$ or $x + 60 = 0$
$\Rightarrow x = -80$ or $x = -60$
Since the speed cannot be negative, $\text{x}\neq-60.$
$\Rightarrow x = 80$
Hence, the speed of Deccan Queen $= 80\ km/h$.
View full question & answer→Question 154 Marks
Solve the following quadratic equation:
$abx^2 +(b^2 - ac)x - bc = 0$
Answer$abx^2 +(b^2 - ac)x - bc = 0$
$\Rightarrow abx^2 + b^2x - acx - bc = 0$
$\Rightarrow bx(ax + b) - c(ax + b) = 0$
$\Rightarrow (ax + b)(bx + c) = 0$
$\Rightarrow (ax + b) = 0$ or $(bx - c) = 0$
$\Rightarrow\text{x}=\frac{-\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
Hence, $\frac{-\text{b}}{\text{a}}$ and $\frac{\text{c}}{\text{b}}$ are the roots of the given equation.
View full question & answer→Question 164 Marks
Solve the following equations by using the method of completing the square:
$\text{4x}^2+4\sqrt3+3=0$
Answer$\text{4x}^2+4\sqrt3+3=0$
$\Rightarrow\text{4x}^2+4\sqrt3\text{x}=-3$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\sqrt3+\big(\sqrt3\big)^2\\=-3+\big(\sqrt3\big)^2$ $[$ Adding $\big(\sqrt3\big)^2$ on Both sides$]$
$\Rightarrow\big(\text{2x}+\sqrt3\big)^2=-3+3=0$
$\Rightarrow\text{2x}+\sqrt3=0$
$\Rightarrow\text{x}=-\frac{\sqrt3}{2}$
Hence, $-\frac{\sqrt3}2{}$ is the required root of the given equation.
View full question & answer→Question 174 Marks
A dealer sells an article for $Rs.\ 75$ and gains as much percent as the cost price of the article. Find the cost price of the article.
AnswerLet the cost price of an article be $Rs. x$. Then, gain $= x \%$ of $x$
⇒ Gain $=\text{Rs. }\Big(\text{x}\times\frac{\text{x}}{100}\Big)=\text{Rs}.\Big(\frac{\text{x}^2}{100}\Big)$
$\therefore$ S.P. = C.P. + Gain $=\text{x}+\frac{\text{x}^2}{100}$
But, S.P. $= Rs.\ 75$
$\Rightarrow\text{x}+\frac{\text{x}^2}{100}=75$
$\Rightarrow 100x + x^2 = 7500 $
$\Rightarrow x^2 + 100x - 7500 = 0 $
$\Rightarrow x^2 + 150x - 50x - 7500 = 0$
$ \Rightarrow x(x + 150) - 50(x + 150) = 0$
$ \Rightarrow (x + 150)(x - 50) = 0 $
$\Rightarrow x + 150 = 0$ or $x - 50 = 0 $
$\Rightarrow x = -150$ or $x = 50$
Since the price cannot be negative,
$\text{x}\neq-150$ $\Rightarrow x = 50$
Thus, the cost price of an article is $Rs.\ 50$.
View full question & answer→Question 184 Marks
Find two consecutive multiples of $3$ whose product is $648$.
AnswerLet the required consecutive multiples of $3$ be $3x$ and $3(x + 1)$.
Then, we have
$3x \times 3(x + 1) = 648$
$\Rightarrow 9x^2 + 9x - 648 = 0$
$\Rightarrow x^2 + x - 72 = 0$
$\Rightarrow x^2 + 9x - 8x - 72 = 0$
$\Rightarrow x(x + 9) - 8(x + 9) = 0$
$\Rightarrow (x + 9)(x - 8) = 0$
$\Rightarrow x + 9 = 0$ or $x - 8 = 0$
$\Rightarrow x = 9$ or $x = 8$
Since x is a positive integer, $x ≠ -9$
$\Rightarrow x = 8$
$\Rightarrow 3x = 3 \times 8 = 24$ and $3(x + 1) = 3(9) = 27$
Hence, the required consecutive multiples of $3$ are $24$ and $27$.
View full question & answer→Question 194 Marks
The hypotenuse of a right-angled triangle is $1$ metre less than twice the shortest side.
If the third is $1$ metre more than the shortest side, find the sides of the triangle.
AnswerLet the shorter side of triangle be $x$ meter.
Then, its hypotenuse $= (2x + 1)$ meter
And let the altitude $= (x + 1)$ meter.
Then, $(2x - 1)^2 = x^2 + (x + 1)^2$
$\Rightarrow 4x^2 + 1 - 4x = x^2 + x^2 + 1 + 2x$
$\Rightarrow 2x^2- 6x = 0$
$\Rightarrow 2x(x - 3) = 0$
$\Rightarrow (x - 3) = 0$ or $2x = 0$
$\Rightarrow x = 3$ or $x = 0$
$\Rightarrow x = 3$ $[\because$ base cannot be zero$]$
thus, Base $= 3m$
Hypotenuse $= (2 \times 3 - 1)m = 5m$
Altitude $= (3 + 1)m = 4m.$
View full question & answer→Question 204 Marks
The hypotenuse of a right-angled triangle is $20$ metres. If the difference between the length of the other sides be $4$ metres, find the other sides.
AnswerLet the other side of triangle be $x$ and $(x - 4)$ meters.
By Pythagoras theorem, we have
$\Rightarrow x^2 + (x - 4)^2 = 330$
$\Rightarrow x^2 + x^2 + 16 - 8x = 400$
$\Rightarrow 2x^2 - 8x - 384 = 0$
$\Rightarrow x^2 - 4x - 192 = 0$
$\Rightarrow x^2 - 16x + 12x - 192 = 0$
$ \Rightarrow x(x - 16) + 12(x - 16) = 0$
$\Rightarrow (x - 16)(x + 12) = 0 $
$\Rightarrow x = 16$ or $x = -12$
$\Rightarrow x = 16$ $[\because$ height cannot be negative$]$
Thus, the height of the triangle be $= 16m$
And the base of the triangle $= (16 - 4) = 12m.$
View full question & answer→Question 214 Marks
Two pipes running together can fill a tank in $11\frac{1}{9}$ minutes. If one pipe takes $5$ minutes more than the other to fill the tank separately,
find the time in which each pipe would fill the tank separately.
AnswerSuppose the faster pipe takes $x$ minutes to fill the tank.
Then, the slower pipe will take $(x + 5)$ minutes to fill the tank.
$\therefore$ Protion of the tank filled by the faster pipe in one minute $=\frac{1}{\text{x}}$
⇒ Protion of the tank filled by the faster pipe in $\frac{100}{9}$ minutes.
$=\frac{1}{\text{x}}\times\frac{100}{9}$
$=\frac{100}{\text{9x}}$
Similarly, Protion of the tank filled by the slower pipe in $\frac{100}{9}$ minutes.
$=\frac{1}{\text{x}+5}\times\frac{100}{9}$
$=\frac{100}{9(\text{x}+5)}$
It is given that the tank is filled in $\frac{100}{9}$ minutes.
$\therefore\frac{100}{\text{9x}}+\frac{100}{9(\text{x}+5)}=1$
$\Rightarrow\frac{100}{\text{x}}+\frac{100}{\text{x}+5}=9$
$\Rightarrow\frac{100\text{x}+500+\text{100x}}{\text{x}^2+\text{5x}}=9$
$\Rightarrow 200x + 500 = 9x^2 + 45x$
$\Rightarrow 9x^2 - 155x - 500 = 0$
$\Rightarrow 9x^2 - 180x + 25x - 500 = 0$
$\Rightarrow 9x(x - 20) + 25(x - 20) = 0$
$\Rightarrow (x - 20)(9x + 25) = 0$
$\Rightarrow x - 20 = 0$ or $9x + 25 = 0$
$\Rightarrow x = 20$ or $\text{x}=-\frac{25}{9}$
Since time cannot be negative, $\text{x}\neq-\frac{25}{9}$
$\Rightarrow x = 20$
$\Rightarrow x + 5 = 20 + 5 = 25$
Hence, the faster pipe fills the tank in $20$ minutes and the slower pipe fills the tank in $25$ minutes.
View full question & answer→Question 224 Marks
Divide $57$ into two parts whose product is $680$.
AnswerLet the one part be $x$.
Then, the other part will be $(57 - x)$.
Thus, we have
$x(57 - x) = 680$
$\Rightarrow 57x - x^2= 680$
$\Rightarrow x^2 - 57x + 680 = 0$
$\Rightarrow x^2 - 17x - 40x + 680 = 0$
$\Rightarrow x(x - 17) - 40(x - 17) = 0$
$\Rightarrow (x - 17)(x - 40) = 0$
$\Rightarrow x - 17 = 0$ or $x - 40 = 0$
$\Rightarrow x = 17$ or $x = 40$
Hence, the two parts are $17$ and $40$.
View full question & answer→Question 234 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 2ax + (a^2 - b^2) = 0$
AnswerGiven, $x^2 - 2ax + (a^2 - b^2) = 0$ On comparing it with $Ax^2 + bx + c = 0,$
we get: $A = 1, B = -2a$ and $C = (a^2 - b^2)$ Discriminant $D$ is given by:
$D = B^2 - 4AC = (-2a)^2 - 4 \times 1 \times (a^2 - b^2) = 4a^2 - 4a^2 + 4b^2 = 4b^2 > 0$
Hence, the roots of the equation are real.Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-2\text{a})+\sqrt{4\text{b}^2}}{2\times1}$
$=\frac{2\text{a}+2\text{b}}{2}$
$=\frac{2(\text{a}+\text{b})}{2}$
$=(\text{a}+\text{b})$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{2\text{a}-2\text{b}}{2}$
$=\frac{2(\text{a}-\text{b})}{2}$
$=(\text{a}-\text{b})$
Hence, the roots of the equation are $(a + b)$ and $(a - b)$.
View full question & answer→Question 244 Marks
Divide $27$ into two parts such that the sum of their reciprocals is $\frac{3}{20}.$
AnswerLet the one part be $x$.
Then, the other part will be $(27 - x)$.
Thus, we have
$\frac{1}{\text{x}}+\frac{1}{27-\text{x}}=\frac{3}{20}$
$\Rightarrow\frac{27-\text{x}+\text{x}}{\text{x}(27-\text{x})}=\frac{3}{20}$
$\Rightarrow\frac{27}{\text{27x}-\text{x}^2}=\frac{3}{20}$
$\Rightarrow 27 \times 20 = 3(27x) - 3x^2$
$\Rightarrow 540 = 81x - 3x^2$
$\Rightarrow 3x^2- 81x + 540 = 0$
$\Rightarrow x^2 - 27x + 180 = 0$
$\Rightarrow x^2 - 15x - 12x + 180 = 0$
$\Rightarrow x(x - 15) - 12(x - 15) = 0$
$\Rightarrow (x - 15)(x - 12) = 0$
$\Rightarrow x - 15 = 0$ or $x - 12 = 0$
$\Rightarrow x = 15$ or $x = 12$
Hence, the two parts are $15$ and $12$.
View full question & answer→Question 254 Marks
The sum of the squares of two consecutive positive integers is $365$. Find the integers.
AnswerLet the required consecutive positive integers be $x$ and $(x + 1)$.
Then, we have
$x^2 + (x + 1)^2 = 365$
$\Rightarrow x^2 + x^2 + 2x + 1 = 365$
$\Rightarrow 2x^2 + 2x - 364 = 0$
$\Rightarrow x^2 + x + 182 = 0$
$\Rightarrow x^2 + 14x - 13x - 182 = 0$
$\Rightarrow x(x + 14) - 13(x + 14) = 0$
$\Rightarrow (x + 14)(x - 13) = 0$
$⇒ x + 14 = 0$ or $x - 13 = 0$
$\Rightarrow x = -14$ or $x = 13$
Since x is a positive integer, $x ≠ -14$
$\Rightarrow x = 13$
$\Rightarrow x + 1 = 13 + 1 = 14$
Hence, the required positive integers are $13$ and $14$.
View full question & answer→Question 264 Marks
Solve the following quadratic equation: $4x^2 - 4a^2x + (a^4 - b^4) = 0$
Answer$4x^2 - 4a^2x + (a^4 - b^4) = 0$
$\Rightarrow 4x^2 - 2(a^2 + b^2)x - 2(a^2 - b^2)x + (a^4 - b^4) = 0$
$\Rightarrow 2x[2x - (a^2 + b^2)] - (a^2 - b^2)[2x + (a^2+ b^2)] = 0$
$\Rightarrow [2x - (a^2 + b^2)][2x - (a^2 - b^2)] = 0$
$\Rightarrow 2x - (a^2 + b^2) = 0$ or $2x - (a^2 - b^2) = 0$
$\Rightarrow\text{x}=\frac{\text{a}^2+\text{b}^2}{2}$ or $\text{x}=\frac{\text{a}^2-\text{b}^2}{2}$
View full question & answer→Question 274 Marks
If $ad \neq bc$ then prove that the equation $(a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0$ has no real roots.
AnswerCompare the given quadratic equation with $Ax^2 + Bx + C = 0$
Here $A = a^2 + b^2, B = 2(ac + bd)$ and $C = c^2 + d^2$
Consider, $B^2 - 4AC = [2(ac + bd)]^2 - 4(a^2 + b^2) \times (c^2 + d^2)$
$= 4[a^2c^2+ 2abcd + b^2d^2] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2]$
$= 4a^2c^2 + 8abcd + 4b^2d^2 - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2$
$= 8abcd - 4a^2d^2 - 4b^2c^2$
$= -4[4a^2d^2 + 4bc^2 - 2abcd]$
$= -4[ad - bc]^2$
Hence the given equation has no real roots unless $ad \neq bc$
View full question & answer→Question 284 Marks
The sum of two natural numbers is $28$ and their product is $192$. Find the numbers.
AnswerLet the required natural numbers be $x$ and $(28 - x)$.
Then, we have
$x \times (28 - x) = 192$
$\Rightarrow 28x - x^2 = 192$
$\Rightarrow x^2 - 28x + 192 = 0$
$\Rightarrow x^2 - 16x - 12x + 192 = 0$
$\Rightarrow x(x - 16) - 12(x - 16) = 0$
$\Rightarrow (x - 16)(x - 12) = 0$
$\Rightarrow x - 16 = 0$ or $x - 12 = 0$
$\Rightarrow x = 16$ or $x = 12$
Hence, the required numbers are $16$ and $12$.
View full question & answer→Question 294 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2x^2 + ax - a^2 = 0$
AnswerThe given equation is: $2x^2 + ax - a^2 = 0$
Comparing it with $Ax^2 + Bx + C = 0$
$A = 2, B = a$ and $C = -a^2$
Discriminant, $D = B^2 - 4AC$
$= a^2 - 4 \times 2 \times -a^2$
$= a^2 + 8a^2$
$= 9a^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{9\text{a}^2}=\text{3a}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-\text{a}+\text{3a}}{2\times2}$
$=\frac{\text{2a}}{4}$
$=\frac{\text{a}}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-\text{a}-\text{3a}}{2\times2}$
$=\frac{-\text{4a}}{4}$
$=-\text{a}$
View full question & answer→Question 304 Marks
In a class test, the sum of Kamal's marks in Mathematics and English is $40$. Had he got $3$ marks more in Mathematics and $4$ marks less in English, the product of the marks would have been $360$. Find his marks in two subjects separately.
AnswerLet the marks obtained by kamal in mathematice and english be $x$ and $y$ .
$\therefore x+y=40 \ldots(1)$
$\text { and }(x+3)(y-4)=360$
$\text { From (1) } y=40-x$
Putting value of $y$ in (2)
$(x+3)(40-x-4)=360$
$\Rightarrow(x+3)(36-x)=360$
or $36 x-x^2+108-3 x=360$
$\Rightarrow-x^2+33 x-252=0 \text { or } x^2-33 x+252=0$
$\Rightarrow x^2-21 x-12 x+252=0$
$\text { or } x(x-21)-12(x-21)=0$
$\Rightarrow(x-21)(x-12)=0$
$\therefore$ when $x-21=0, x=21$
when $x-12=0, x=12$
for $x=21,21+y=40 \therefore y=19$
$\text { for } x=12,12+y=40 \therefore y=28$
The marks obtained by kamal in mathematics and english respectively are $(21,19)$ or $(12,28)$.
View full question & answer→Question 314 Marks
A train travels $180\ km$ at a uniform speed. If the speed had been 9km/hr more, it would have taken $1$ hour less for the same journey. Find the speed of the train.
AnswerLet the uniform speed of the train be x km/hr.
Time taken to cover 180km $=\frac{180}{\text{x}}\ \text{hours}$
Time taken to cover 180km when the speed is increased by 9km/hr $=\frac{180}{\text{x}+9}\ \text{hours}$
$\therefore\frac{180}{\text{x}}-\frac{180}{\text{x}+9}=1$
$\Rightarrow\frac{\text{180x}+1620-\text{180x}}{\text{x}^2+\text{9x}}=1$
$\Rightarrow 1620 = x^2 + 9x$
$\Rightarrow x^2 + 9x - 1620 = 0$
$\Rightarrow x^2 + 45x - 36x - 1620 = 0$
$\Rightarrow x(x + 45) - 36(x + 45) = 0$
$\Rightarrow (x + 45)(x - 36) = 0$
$\Rightarrow x + 45 = 0$ or $x - 36 = 0$
$\Rightarrow x = -45$ or $x = 36$
Since the speed cannot be negative, $\text{x}\neq-45.$
$\Rightarrow x = 36$
Hence, the uniform speed of the train is $36\ km/hr$.
View full question & answer→Question 324 Marks
The numerator of a fraction is $3$ less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}.$ Find the fraction.
AnswerLet the denominator of the fraction be $x$.
Then, the numerator of the fraction will be (x - 3).
$\therefore$ Fraction $=\frac{\text{x}-3}{\text{x}}$
$\frac{\text{x}-3}{\text{x}+1}=\frac{\text{x}-3}{\text{x}}-\frac{1}{15}$
$\Rightarrow\frac{\text{x}-3}{\text{x}}-\frac{\text{x}-3}{\text{x}+1}=\frac{1}{15}$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)-\text{x}(\text{x}-3)}{\text{x}(\text{x}+1)}=\frac{1}{15}$ when
$\Rightarrow\frac{(\text{x}^2-\text{2x}-3)-(\text{x}^2-\text{3x})}{\text{x}^2+\text{x}}=\frac{1}{15}$
$\Rightarrow\frac{\text{x}-3}{\text{x}^2+\text{x}}=\frac{1}{15}$
$\Rightarrow $ 15x - 45 = x^2 + x$
$\Rightarrow x^2 - 14x + 45 = 0$
$\Rightarrow x^2- 9x - 5x + 45 = 0$
$\Rightarrow x(x - 9) - 5(x - 9) = 0$
$\Rightarrow (x - 9)(x - 5) = 0$
$\Rightarrow x - 9 = 0 or x - 5 = 0$
$\Rightarrow x = 9$ or $x = 5$
When $x = 9,$
$x - 3 = 9 - 3 = 6$
$\Rightarrow $ Fraction $=\frac{6}{9}=\frac{2}{3}$
When $x = 5,$
$x - 3 = 5 - 3 = 2$
⇒ Fraction $=\frac{2}{5}$
Since, numerator is $3$ less than the denominator, requierd fraction is $\frac{2}{5}.$
View full question & answer→Question 334 Marks
Find two consecutive positive odd integers whose product is $483$.
AnswerLet the required consecutive positive odd integers be $x$ and $(x + 2)$.
Then, we have
$x \times (x + 2) = 483$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x + 23)(x - 21) = 0$
$\Rightarrow x + 23 = 0$ or $x - 21 = 0$
$\Rightarrow x = -23$ or $x = 21$
Since x is a positive integer,$x ≠ -23$
$\Rightarrow x = 21$
$\Rightarrow x + 2 = 21 + 2 = 23$
Hence, the required consecutive positive odd integers are $21$ and $23$.
View full question & answer→Question 344 Marks
The sum of a natural number and its positive square root is $132.$ Find the number.
AnswerLet the natural number be x.
Then, its positive square root will be $\sqrt{\text{x}}$
Accroding to given information, we have
$\text{x}+\sqrt{\text{x}}=132$
$\Rightarrow y^2 + y = 132$ where, $\sqrt{\text{x}}=\text{y}$
$\Rightarrow y^2 + y - 132 = 0$
$\Rightarrow y^2 + 12y - 11y - 132 = 0$
$\Rightarrow y(y + 12) - 11(y + 12) = 0$
$\Rightarrow (y + 12)(y - 11) = 0$
$\Rightarrow y + 12 = 0$ or $y - 11 = 0$
$\Rightarrow y = -12$ or $y = 11$
Since square root of a number cannot be negative, $\text{y}\neq-12$
Hence, $y = 11$
$\Rightarrow\sqrt{\text{x}}=11$
$\Rightarrow\text{x}=121$
Thus, the required natural numbr is $121.$
View full question & answer→Question 354 Marks
The sum of a natural number and its square is $156.$ Find the number.
AnswerLet the natural number be $x$.
Then, its square will be $x^2$
Accroding to given information, we have
$\Rightarrow x + x^2 = 156$
$\Rightarrow x^2 + x - 156 = 0$
$\Rightarrow x^2 + 13x - 12x - 156 = 0$
$\Rightarrow x(x + 13) - 12(x + 13) = 0$
$\Rightarrow (x + 13)(x - 12) = 0$
$\Rightarrow x + 13 = 0$ or $x - 12 = 0$
$\Rightarrow x = -13$ or $x = 12$
Since x is a natural number, $x ≠ -13$
Hence, the required natural number is $12$.
View full question & answer→Question 364 Marks
The length of a rectangle is twice its breadth and its area is $288 cm^2$. Find the dimensions of the rectangle.
AnswerLet the breadth of a rectangle $=x cm$
Then, lenght of the rectangle $=2 x cm$
$\therefore$ Area $=$ lenght $\times$ breadth $=288 cm^2$
$\Rightarrow 2 x \times x =288$
$\Rightarrow 2 x ^2=288$
$\Rightarrow x ^2=144$
$\Rightarrow x =\sqrt{144}$
$\Rightarrow x = \pm 12$
$\Rightarrow x =12[\because$ breadth cannot be negative]
Thus, breadth of rectangle $=12 cm$
And lenght of rectangle $=(2 \times 12)=24 cm$.
View full question & answer→Question 374 Marks
A man is $3\frac{1}{2}$ times as old as his son. If the sum of the squares of their ages is 1325, find the ages of the father and the son.
Answer$\Rightarrow\text{x}^2+\Big(\frac{7}{2}\text{x}\Big)^2=1325$ $\Rightarrow\frac{\text{x}^2}{1}+\frac{\text{49x}^2}{4}=1325$ $\Rightarrow\frac{\text{4x}^2+\text{49x}^2}{4}=1325$ $\Rightarrow\text{53x}^2=1325\times4$ $\Rightarrow\text{x}^2=\frac{1325\times4}{53}=100$ $\Rightarrow\text{x}^2=\sqrt{100}=10$ Son's ages $\Rightarrow\text{x}=10$ Father's age$\Rightarrow\frac{7}{2}\times10=35\text{yrs.}$
View full question & answer→Question 384 Marks
A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
AnswerLet the onglnal speed of the truck = s km/hr New speed of the truck = (s + 20)km/hr Time taken for 150km + Time token for 200km = 5 $\frac{150}{\text{x}}+\frac{200}{(\text{s}+20)}=5$
$\Rightarrow\frac{150\text{x}+3000+\text{200s}}{\text{s}(\text{s}+20)}=5$
$\Rightarrow\frac{350\text{s}+3000}{\text{s}(\text{s}+20)}=5$
$\Rightarrow\frac{50(\text{7s}+60)}{\text{s}(\text{s}+20)}=5$10(7s + 60) = s(s + 20)
$\Rightarrow 70s + 600 = s^2 + 20s$
$\Rightarrow s^2 - 50s - 600 = 0$
$\Rightarrow s^2 - 60s - 10s - 600 = 0$
$\Rightarrow s(s - 60) - 10(s - 60) = 0$
$\Rightarrow (s - 60)(s - 10) = 0$
$\Rightarrow s - 60 = 0$ or $s - 10 = 0$
$\Rightarrow s = 60$ or $s = 10$
$\Rightarrow s = 10$ [Not possible] $\therefore$ First speed of the truck = 60km/hr.
View full question & answer→Question 394 Marks
The area of a right-triangle is $600cm^2.$ If the base of the triangle exceeds the altitude by $10\ cm$, find the dimensions of the triangle.
AnswerLet the altitude of triangle be $x\ cm$.
Then, base of triangle is $(x + 10)cm$
$\therefore$ Area $= 600cm^2$
$\Rightarrow\frac{1}{2}\times$ base $\times $ altitude $= 600cm^2$
$\Rightarrow\frac{1}{2}\times(\text{x}+10)\times\text{x}=600$
$\Rightarrow x^2 +$ $10x = 1200$
$\Rightarrow x^2 + 10x - 1200 = 0$
$\Rightarrow x^2 + 40x - 30x - 1200 = 0$
$\Rightarrow x(x + 40) - 30(x + 40) = 0$
$\Rightarrow (x + 40)(x - 30) = 0$
$\Rightarrow x = -40$ or $x = 30$
$\therefore$ $x = 30$ $[\because$ length of altitude cannot be negative$]$
Hence, altitude of triangle is $30\ cm$ and base of triangle 40cm.
$(\text { Hypotenuse })^2=(30)^2+(40)^2$
$=900+1600=2500$
$\therefore \text { Hypotenuse }=50$
View full question & answer→Question 404 Marks
The sum of the squares of two consecutive positive even numbers is $452$. Find the numbers.
AnswerLet the required consecutive positive even numbers be $x$ and $(x + 2)$.
Then, we have
$x^2 + (x + 2)^2 = 452$
$\Rightarrow x^2 + x^2 + 4x + 4 = 452$
$\Rightarrow 2x^2 + 4x - 448 = 0$
$\Rightarrow x^2 + 2x - 224 = 0$
$\Rightarrow x^2 - 14x + 16x - 224 = 0$
$\Rightarrow x(x - 14) + 16(x - 14) = 0$
$\Rightarrow (x - 14)(x + 16) = 0$
$\Rightarrow x - 14 = 0$ or $x + 16 = 0$
$\Rightarrow x = 14$ or $x = -16$
Since x is a positive number, $x \neq -16$
$\Rightarrow x = 14$
$\Rightarrow x + 2 = 14 + 2 = 16$
Hence, the required positive even numbers are $14$ and $16$.
View full question & answer→Question 414 Marks
A passenger train takes $2$ hours less for a journey of $300\ km$ if its speed is increased by $5\ km/hr$ from its usual speed. Find its usual speed.
AnswerLet the usual speed of the passenger train be x km/hr.
Time taken to cover $300km$ $=\frac{300}{\text{x}}\ \text{hours}$
Time taken to cover $300km$ when the speed is increased by $5\ km\ /hr$ $=\frac{300}{(\text{x}+5)}\ \text{hours}$
It is given that the time taken to cover 300km is decreased by $2$ hours.
$\therefore\frac{300}{\text{x}}-\frac{300}{\text{x}+5}=2$
$\Rightarrow\frac{300\text{x}+1500-300\text{x}}{\text{x}^2+5\text{x}}=2$
$\Rightarrow 1500 = 2x^2 + 10x$
$\Rightarrow 2x^2 + 10x - 1500 = 0$
$\Rightarrow x^2 + 5x - 750 = 0$
$\Rightarrow x^2 + 30x - 25x - 750 = 0$
$\Rightarrow x(x + 30) - 25(x + 30) = 0$
$\Rightarrow (x + 30)(x - 25) = 0$
$\Rightarrow x + 30 = 0$ or $x - 25 = 0$
$\Rightarrow x = -30$ or $x = 25$
Since the speed cannot be negative, $\text{x}\neq30.$
$\Rightarrow x = 25$
Hence the original speed of train is $25\ km\ ph$.
View full question & answer→Question 424 Marks
The product of two consecutive positive integers is $306$. Find the integers.
AnswerLet the required consecutive positive integers be $x$ and $(x + 1)$.
Then, we have
$x \times (x + 1) = 306$
$\Rightarrow x^2 + x - 306 = 0$
$\Rightarrow x^2 + 18x - 17x - 306 = 0$
$\Rightarrow x(x + 18) - 17(x + 18) = 0$
$\Rightarrow (x + 18)(x - 17) = 0$
$\Rightarrow x + 18 = 0$ or $x - 17 = 0$
$\Rightarrow x = -18$ or $x = 17$
Since x is a positive integer, $x \neq -18$
$\Rightarrow x = 17$
$\Rightarrow x + 1 = 17 + 1 = 18$
Hence, the required positive integers are $17$ and $18$.
View full question & answer→Question 434 Marks
$300$ apples are distributed equally among a certain number of students. Had there been $10$ more students, each would have received one apple less. Find the number of students.
AnswerLet the number of students be x, then
$\frac{300}{\text{x}}-\frac{300}{(\text{x}+10)}=1$
$\Rightarrow\frac{1}{\text{x}}-\frac{1}{(\text{x}+10)}=\frac{1}{300}$
$\Rightarrow\frac{\text{x}+10-\text{x}}{\text{x}(\text{x}+10)}=\frac{1}{300}$
$\Rightarrow x(x + 10) = 3000$
$\Rightarrow x^2+ 10x - 3000 = 0$
$\Rightarrow x^2 + 60x - 50x - 3000 = 0$
$\Rightarrow x(x + 60) - 50(x + 60) = 0$
$\Rightarrow (x + 60)(x - 50) = 0$
$\Rightarrow x + 60 = 0$ or $x - 50 = 0$
$\Rightarrow x = -60$ or $x = 50$
$\Rightarrow x = 50$ $(\because$ number of students cannot be negative$)$
Hence the number of students is $50$.
View full question & answer→Question 444 Marks
Solve the following equations by using the method of completing the square:
$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
Answer$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-\big(\sqrt2+1\big)\text{x}=-\sqrt2$
$\Rightarrow\text{x}^2-2\times\text{x}\times\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2+1}{2}\Big)^2\\=-\sqrt2+\Big(\frac{\sqrt2+1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{\sqrt2+1}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big[\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)\Big]^2=\frac{-4\sqrt2+2+1+2\sqrt2}{4}$
$=\frac{2-2\sqrt2+1}{4}=\Big(\frac{\sqrt2-1}{2}\Big)^2$
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\pm\Big(\frac{\sqrt2-1}{2}\Big)$ (Taking square root on both sides)
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\frac{2\sqrt2}{2}=\sqrt2$ or $\text{x}=\frac{2}2{}=1$
Hence, $\sqrt2$ and 1 are the roots of the given equation.
View full question & answer→Question 454 Marks
A two-digit number is such that the product of its digits is $14$. If $45$ is added to the number, the digits interchange their places. Find the number.
AnswerLet the tens digit and units digit of required number be x and y respectively.
$\therefore\ \text{xy}=14$
$\Rightarrow\text{y}=\frac{14}{\text{x}}\ \dots(1)$
the number =$ (10x + y)$
$(10x + y) + 45 = (10x + y)$
$9x - 9y = -45$
$x - y = -5 ...(2)$
Putting $\text{y}=\frac{14}{\text{x}}$ from (1) in (2), we get
$\text{x}-\frac{14}{\text{x}}=-5$
$\Rightarrow x^2 + 5x - 14 = 0$
$\Rightarrow x^2 + 7x - 2x - 14 = 0$
$\Rightarrow x(x + 7) - 2(x + 7) = 0$
$\Rightarrow (x + 7)(x - 2) = 0$
$\Rightarrow x + 7 = 0$ or $x - 2 = 0$
$\Rightarrow x = - 7$ or $x = 2$
$\therefore$ x = 2 $[\because$ a digit cannot be negative$]$
Putting $x = 2$ in (1), we get $y = 8$
The ten digit is 2 and unit digit is $7$
Hence, the required number is $27.$
View full question & answer→Question 464 Marks
If the price of a book is reduced by $Rs. 5$, a person can buy $4$ more books for $Rs. 600$. Find the original price of the book.
AnswerLet the original price of the book $= Rs. x.$
$\therefore$ Number of books bought for $Rs. 600$ $=\frac{600}{\text{x}}$
Reduced price of the book $= Rs. (x - 5)$
$\therefore$ Number of books bought for $Rs. 600$ $=\frac{600}{\text{x}-5}$
It is given that:
$\frac{600}{\text{x}-5}-\frac{600}{\text{x}}=4$
$\Rightarrow\frac{\text{600x}-\text{600x}+3000}{\text{x}^2-\text{5x}}=4$
$\Rightarrow 3000 = 4x^2 - 20x$
$\Rightarrow 4x^2 - 20x - 3000 = 0$
$\Rightarrow x^2 - 5x - 750 = 0$
$\Rightarrow x^2 - 30x + 25x - 750 = 0$
$\Rightarrow x(x - 30) + 25(x - 30) = 0$
$\Rightarrow (x - 30)(x + 25) = 0$
$\Rightarrow x - 30 = 0$ or $x + 25 = 0$
$\Rightarrow x = 30$ or $x = -25$
Since price of the book cannot be negative, $\text{x}\neq-25$
$\Rightarrow x = 30$
Hence, the original price of a book is $Rs. 30$
View full question & answer→Question 474 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
AnswerThe given equation is:$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
$\Rightarrow\frac{\text{m}^2\text{x}^2+\text{n}^2}{\text{mn}}=1-2\text{x}$
$\Rightarrow\text{m}^2\text{x}^2+\text{n}^2=\text{mn}-2\text{mnx}$
$\Rightarrow\text{m}^2\text{x}^2+2\text{mnx}+\text{n}^2-\text{mn=0}$
This equation is the form $ax^2+ bx + c = 0,$ where $a = m^2, b = 2mn$ and $c = n^2 - mn$.
$\therefore$ Discriminant $\text{D}=\text{b}^2-4\text{ac}$
$=(2\text{mn})^2-4\times\text{m}^2\times(\text{n}^2-\text{mn})$
$=4\text{m}^2\text{n}^2-4\text{m}^2\text{n}^2+4\text{m}^3\text{n}$
$=4\text{m}^3\text{n}>0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{4\text{m}^3\text{n}}=2\text{m}\sqrt{\text{mn}}$
$\therefore\alpha =\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-2\text{mn}+2\text{m}\sqrt{\text{mn}}}{2\times\text{m}^2}$
$=\frac{2\text{m}\big(-\text{n}+\sqrt{\text{mn}}\big)}{2\text{m}^2}$
$=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-2\text{mn}-2\text{m}\sqrt{\text{mn}}}{2\times\text{m}^2}$
$=\frac{-2\text{m}\big(\text{n}+\sqrt{\text{mn}}\big)}{2\text{m}^2}$
$=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$
Hence, $\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$ and $\frac{-\text{n}-\sqrt{\text{mn}}}{\text{m}}$ are the roots of the given equation.
View full question & answer→Question 484 Marks
The difference of two natural numbers is $3$ and the difference of their reciprocals is $\frac{3}{28}.$ Find the numbers.
AnswerLet the required numbers be x and (x - 3).
Then, we have
$x > x - 3$
$\Rightarrow\frac{1}{\text{x}}<\frac{1}{\text{x}-3}$
$\therefore\frac{1}{\text{x}-3}-\frac{1}{\text{x}}=\frac{3}{28}$
$\Rightarrow\frac{\text{x}-\text{x}+3}{\text{x}(\text{x}-3)}=\frac{3}{28}$
$\Rightarrow 84 = 3x^2 - 9x$
$\Rightarrow 3x^2 - 9x - 84 = 0$
$\Rightarrow x^2 - 3x - 28 = 0$
$\Rightarrow x^2 - 7x + 4x - 28 = 0$
$\Rightarrow x(x - 7) + 4(x - 7) = 0$
$\Rightarrow (x - 7)(x + 4) = 0$
$\Rightarrow x - 7 = 0$ or $x + 4 = 0$
$⇒ x = 7$ or $x = -4$
$$Since x is a natural number, $x \neq -4$
$\Rightarrow x = 7$ and $x - 3 = 4$
Hence, the required numbers are $7$ and $4$.
View full question & answer→Question 494 Marks
Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is $46$. Find the integers.
AnswerLet the three consecutive numbers be x, x + 1, x + 2.
Sum of square of first and product of the other two
$\Rightarrow x^2 + (x + 1)(x + 2) = 46$
$\Rightarrow x^2 + (x^2 + 3x + 2) = 4$6 or $2x^2 + 3x - 44 = 0$
$\Rightarrow 2x^2 + 11x - 8x - 44 = 0$ or $x(2x + 11) - 4(2x - 11) = 0$
$\Rightarrow (x - 4)(2x + 11) = 0$
$\therefore\ \text{x}=4$ or $\frac{-11}{2}$
But $\text{x}\neq\frac{-11}{2}\ \ \therefore\text{x}=4$
$\therefore$ required numbers are $4, 5$ and $6$
View full question & answer→Question 504 Marks
By using the method of completing the square, show that the equation $2x^2 + x + 4 = 0$ has no real roots.
Answer$2x^2 + x + 4 = 0$
$\Rightarrow 4x^2 + 2x + 8 = 0$ (Multiplying both sides by $2$)
$\Rightarrow 4x^2 + 2x = -8$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=-8+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}+\frac{1}{2}\Big)^2=-8+\frac{1}{4}$
$=-\frac{31}{4}<0$
But $\Big(\text{2x}+\frac{1}{2}\Big)^2$ cannot be negative for any real value of $x$.
So, there is no real value of x satisfying the given equation.
Hence, the given equation has no real roots.
View full question & answer→Question 514 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
$11a = 44$
$\Rightarrow a = 4$
Putting $a = 4$ in (1), we get
$3 \times 4 + 4b = 32$
$\Rightarrow 4b = 32 - 12 = 20$
$\Rightarrow\text{b}=\frac{20}{4}=5$
$\therefore$ a = 4 and b = 5
View full question & answer→Question 524 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, $\text{x}\neq-20$
$\Rightarrow x = 25$
$\Rightarrow x - 5 = 20$
$\Rightarrow\frac{2000}{\text{x}-5}=\frac{2000}{20}=100$
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 534 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 544 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 554 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 $13x + 24 = 0$
$\Rightarrow x = 5$ or $\text{x}=-\frac{24}{3}$
Since time cannot be negative, $\text{x}\neq-\frac{24}{3}$
$\Rightarrow x = 5$
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 564 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,$x \neq -2$
$\Rightarrow x = 7$ and $x - 5 = 2$
Hence, the required numbers are $7$ and $2$.
View full question & answer→Question 574 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 584 Marks
A train covers a distance of $480\ km$ 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 $480\ km$ $=\frac{480}{\text{x}}\ \text{hours.}$
Time taken to cover $480\ km$ 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, $\text{x}\neq-32.$
$\Rightarrow x = 40$
Hence, the usual speed of the plane is 40km/hr.
View full question & answer→Question 594 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 604 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 614 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 624 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 634 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, $\text{x}\neq-15$
$\Rightarrow x = 12$
Hence, he bought $12$ pens.
View full question & answer→Question 644 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 654 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 $x^2 - kx + 9 = 0$
$\therefore$ $D = (-k)^2- 4 \times 1 \times 9$
$D = k^2 - 36$
The given equation has real and distinct roots if $D > 0$.
$\therefore$ $k^2- 36 > 0$
$\Rightarrow (k - 6)(k + 6) > 0$
$\Rightarrow k < -6 or k > 6$
View full question & answer→Question 664 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
$a = 1, b = 1$ and $c = 2$
$\therefore$ Discriminant, $D = b^2- 4ac$
$= 1^2- 4 \times 1 \times 2$
$= 1 - 8$
$= -7 < 0$
Hence, the given equation has no real roots (or real roots does not exist).
View full question & answer→Question 674 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 684 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
$x(25-x)=126$
$\Rightarrow 25 x-x^2=126$
$\Rightarrow x^2-25 x+126$
$\Rightarrow x^2-18 x-7 x+126=0$
$\Rightarrow x(x-18)-7(x-18)=0$
$\Rightarrow(x-18)(x-7)=0$
$\Rightarrow x-18=0 \text { or } x-7=0$
$\Rightarrow x=18 \text { or } x=7$
Hence, their ages are $18$ years and $7$ years.
View full question & answer→Question 694 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 $= 3x^2 $
years Present age of son = (x + 2) years And,
present age of father $= (3x^2 + 2)$ years Three year's hence,
we have Son's age = (x + 2 + 3) = (x + 5) years
Father's age $= (3x^2 + 2 + 3) = (3x^2 + 5) $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 $= (3x^2+ 2) = 3(3)^2 + 2 = 29 $years.
View full question & answer→Question 704 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 714 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 $24\ m$ and $8\ m$ respectively.
View full question & answer→Question 724 Marks
The length of a rectangle is thrice as long as the side of a square. The side of the square is $4\ cm$ 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 $x = 6$
$\Rightarrow x = 6$ (Side of the square is never $0$)
Thus, side of the square $= 6\ cm$
And length of the rectangle $= (3 \times 6) = 18\ cm$
Then, breadth of the rectangle =$ (6 - 4)cm = 2\ cm$
View full question & answer→Question 734 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.
Question 744 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 754 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, $x \neq -17$
$\Rightarrow x = 15$
$\Rightarrow x + 2 = 15 + 2 = 147$
Hence, the required positive odd numbers are $15$ and $17$.
View full question & answer→Question 764 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, $x \neq -18$
$\Rightarrow x = 16$
$\Rightarrow x + 2 = 16 + 2 = 18$
Hence, the required consecutive positive even integers are $16$ and $18$.
View full question & answer→Question 774 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 $x = 14$
$\Rightarrow x = 14$ $[\because$ breadth cannot be negative$]$
Thus, the breadth of hall is $14\ m$
And lenght of the hall is $(14 + 3)m$ = $17m.$
View full question & answer→Question 784 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 794 Marks
A rectangular field is $16\ m$ long and $10\ m$ wide. There is a path of uniform width all around it, having an area of $120\ m^2$. 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 804 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
$\Rightarrow$ 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 814 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 824 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 834 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 844 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 $1500\ km$ with the original speed
$=\frac{1500}{\text{x}}\text{hrs}$ Time taken to cover $1500\ km$ 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,
$\text{x}\neq-600.$
$\Rightarrow x = 500$
Hence, the original speed of the plane is $500\ km/hr$.
Yes, we apperciate the value shown by the pilot.
View full question & answer→Question 854 Marks
A train covers a distance of $300\ km$ 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 $= 300\ km$
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 $300\ km$ $=\frac{300}{(\text{x}+5)}$
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 $25\ kmph$.
View full question & answer→Question 864 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 $8x - 1 = 0$
$\Rightarrow x = 8$ or $\text{x}=\frac{1}{8}$
Since $x$ is a natural number, $\text{x}\neq\frac{1}{8}.$
$\Rightarrow x = 8$
Hence, the required number is $8$.
View full question & answer→Question 874 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 884 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 894 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 $\times $ 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 $= (3 \times 7)m = 21m.$
View full question & answer→Question 904 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 914 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 924 Marks
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 of $6\ km/hr$ 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 934 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 944 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 954 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 964 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 974 Marks
If a and b are real and $a \neq b$ 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
$25(a+b)^2+8(a-b)^2>0 \text { (Sum of two positive numbers is always positive) }$
$\Rightarrow D>0$
Hence, the roots of the given equation are real and unequal.
View full question & answer→Question 984 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 $
$\Rightarrow x = 7$
$[\text{x}\neq1]$ Age of the son $= 7$ years.
Age of man $= (-7 + 8 × 7) = -7 + 56 = 49$ years.
View full question & answer→Question 994 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 1004 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 $=\frac{1}{x}$
Part of tank filled by smaller pipe in $1$ hour $=\frac{1}{x-9}$
It is given that the tank can be filled in $6$ hours by both the pipes together.
$\therefore \frac{1}{x}+\frac{1}{x-9}=\frac{1}{6}$
$\Rightarrow \frac{100}{x}+\frac{100}{x+5}=9$
$\Rightarrow \frac{x-9+x}{x^2-9 x}=\frac{1}{6}$
$\Rightarrow 6(2 x-9)=x^2-9 x$
$\Rightarrow 12 x-54=x^2-9 x$
$\Rightarrow x^2-21 x+54=0$
$\Rightarrow x^2-18 x+3 x+54=0$
$\Rightarrow x(x-18)-3(x-18)=0$
$\Rightarrow(x-18)(x-3)=0$
$\Rightarrow x-18=0 \text { or } x-3=0$
$\Rightarrow x=18 \text { or } x=3$
Since time cannot be less than $9, x \neq 3$
$\Rightarrow x=18$
$\Rightarrow x-9=18-9=9$
Hence, the smaller pipe taken $18$ hours to fills the tank and the larger pipe takes $9$ hours to fills the tank.
View full question & answer→Question 1014 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}-\frac{1}{\text{x}}=3,\ \text{x}\neq0$
View full question & answer→