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→