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 6 x+6=x^2+2 x-15 $
$ \Rightarrow x^2-4 x-21=0 $
$ \Rightarrow x^2-7 x+3 x-21=0 $
$ \Rightarrow x(x-7)+3(x-7)=0 $
$ \Rightarrow(x-7)(x+3)=0 $
$ \Rightarrow x-7=0 \text { or } x+3=0 $
$ \Rightarrow x=7 \text { 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 $=\left(x^2+24\right)$
$\Rightarrow x^2+24=(x+1)^2-25 $
$ \Rightarrow x^2+24+x^2+1+2 x-25 $
$ \Rightarrow 2 x=48 $
$ \Rightarrow x=24$
Hence total number of student
$=\left[(24)^2+24\right]=567+24=600$
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:
$ 3 x^2-2 x-1=0 $
Answer$ 3 x^2-2 x-1=0 $
$ \Rightarrow 9 x^2-6 x-3=0 \text { (Multiplying both sides by } 3 \text { ) } $
$ \Rightarrow 9 x^2-6 x=3 $
$ \Rightarrow(3 x)^2-2 \times 3 x \times 1+1^2=3+1^2 \text { [Adding } 1^2 \text { on both sides] } $
$ \Rightarrow(3 x-1)^2=3+1=4=(2)^2$
$\Rightarrow\text{3x}-1=\pm2$ (Taking square root on both sides)
$⇒ 3x - 1 = 2$ or $3x - 1 = -2$
$⇒ 3x = 3$ or $3x = -1$
$⇒ x = 1$ 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$ 2 x^2+5 x-3=0$
$ \Rightarrow 4 x^2+10 x-6=0(\text { Multiplying both sides by } 2)$
$ \Rightarrow 4 x^2+10 x=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+8 x-5 x-40=30 $
$ \Rightarrow x^2+3 x-40-30=0 $
$ \Rightarrow x^2+3 x-70=0 $
$ \Rightarrow x^2+10 x-7 x-70=0 $
$ \Rightarrow x(x+10)-7(x+10)=0 $
$ \Rightarrow(x+10)(x-7)=0 $
$ \Rightarrow x+10=0 \text { or } x-7=0 $
$ \Rightarrow x=-10 \text { 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=90 x^2+360 x $
$ \Rightarrow 90 x^2+360 x-43200=0 $
$ \Rightarrow x^2+4 x-480=0 $
$ \Rightarrow x^2+24 x-20 x-480=0 $
$ \Rightarrow x(x+24)-20(x+24)=0 $
$ \Rightarrow(x+24)(x-20)=0 $
$ \Rightarrow x+24=0 \text { or } x-20=0 $
$ \Rightarrow x=-24 \text { 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) $=\frac{1}{x-10}+\frac{1}{x}$
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(2 x-10)=x(x-10) $
$ \Rightarrow 24 x-120=x^2-10 x $
$ \Rightarrow x^2-34 x+120=0 $
$ \Rightarrow x^2-30 x-4 x+120=0 $
$ \Rightarrow x(x-30)-4(x-30)=0 $
$ \Rightarrow(x-30)(x-4)=0 $
$ \Rightarrow x-30=0 \text { or } x-4=0 $
$ \Rightarrow x=30 \text { 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 24 x-x^2+72-3 x=180 $
$ \Rightarrow 21 x-x^2=108 $
$ \Rightarrow x^2-12 x-9 x+108=0 $
$ \Rightarrow x(x-12)-9(x-12)=0 $
$ \Rightarrow(x-12)(x-9)=0 $
$ \Rightarrow x-12=0 \text { or } x-9=0 $
$ \Rightarrow x=12 \text { 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:
$x^2-4 a x-b^2+4 a^2=0$
Answer$x^2-4 a x-b^2+4 a^2=0$
$\Rightarrow x^2-4 a x+\left(4 a^2-b^2\right)=0$
$\Rightarrow x^2-4 a x+(2 a+b)(2 a-b)=0$
$\Rightarrow x^2-(2 a+b) x-(2 a-b) x+(2 a+b)(2 a-b)=0$
$\Rightarrow x[x-(2 a+b)]-(2 a-b)[x+(2 a+b)]=0$
$\Rightarrow[x-(2 a+b)][x-(2 a-b)]=0$
$\Rightarrow x-(2 a+b)=0 \text { or } x-(2 a-b)=0$
$\Rightarrow x=2 a+b \text { or } x=2 a-b$
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+14 x+49=1225$
$ \Rightarrow 2 x^2+14 x-1176=0$
$ \Rightarrow x^2+7 x-588=0$
$ \Rightarrow x^2+28 x-21 x-588=0$
$ \Rightarrow x(x+28)-21(x+28)=0$
$ \Rightarrow(x+28)(x-21)=0$
$ \Rightarrow x+28=0 \text { or } x-21=0$
$ \Rightarrow x=-28 \text { 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^2-20 x-4800=0 $
$ \Rightarrow x^2-80 x+60 x-4800=0 $
$ \Rightarrow x(x-80)+60(x-80)=0 $
$ \Rightarrow(x-80)(x+60)=0 $
$ \Rightarrow x-80=0 \text { or } x+60=0 $
$ \Rightarrow x=-80 \text { 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:
$ a b x^2+\left(b^2-a c\right) x-b c=0 $
Answer$ a b x^2+\left(b^2-a c\right) x-b c=0 $
$ \Rightarrow a b x^2+b^2 x-a c x-b c=0 $
$ \Rightarrow b x(a x+b)-c(a x+b)=0 $
$ \Rightarrow(a x+b)(b x+c)=0 $
$ \Rightarrow(a x+b)=0 \text { or }(b x-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 $x$
$\Rightarrow$ 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 100 x+x^2=7500 $
$ \Rightarrow x^2+100 x-7500=0 $
$ \Rightarrow x^2+150 x-50 x-7500=0 $
$ \Rightarrow x(x+150)-50(x+150)=0 $
$ \Rightarrow(x+150)(x-50)=0 $
$ \Rightarrow x+150=0 \text { or } x-50=0 $
$ \Rightarrow x=-150 \text { 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
$3 x \times 3(x+1)=648 $
$\Rightarrow 9 x^2+9 x-648=0 $
$\Rightarrow x^2+x-72=0 $
$\Rightarrow x^2+9 x-8 x-72=0 $
$\Rightarrow x(x+9)-8(x+9)=0 $
$\Rightarrow(x+9)(x-8)=0 $
$\Rightarrow x+9=0 \text { or } x-8=0 $
$\Rightarrow x=9 \text { or } x=8$
Since x is a positive integer,$ x \neq -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.
$ \text { Then, }(2 x-1)^2=x^2+(x+1)^2 $
$ \Rightarrow 4 x^2+1-4 x=x^2+x^2+1+2 x $
$ \Rightarrow 2 x^2-6 x=0 $
$ \Rightarrow 2 x(x-3)=0 $
$ \Rightarrow(x-3)=0 \text { or } 2 x=0 $
$ \Rightarrow x=3 \text { 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+16-8 x=400$
$ \Rightarrow 2 x^2-8 x-384=0$
$ \Rightarrow x^2-4 x-192=0$
$ \Rightarrow x^2-16 x+12 x-192=0$
$ \Rightarrow x(x-16)+12(x-16)=0$
$ \Rightarrow(x-16)(x+12)=0$
$ \Rightarrow x=16 \text { 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 $=\frac{1}{\text{x}}$
\Rightarrow 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 200 x+500=9 x^2+45 x $
$ \Rightarrow 9 x^2-155 x-500=0 $
$ \Rightarrow 9 x^2-180 x+25 x-500=0 $
$ \Rightarrow 9 x(x-20)+25(x-20)=0 $
$ \Rightarrow(x-20)(9 x+25)=0 $
$ \Rightarrow x-20=0 \text { or } 9 x+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 57 x-x^2=680 $
$ \Rightarrow x^2-57 x+680=0 $
$ \Rightarrow x^2-17 x-40 x+680=0 $
$ \Rightarrow x(x-17)-40(x-17)=0 $
$ \Rightarrow(x-17)(x-40)=0 $
$ \Rightarrow x-17=0 \text { or } x-40=0 $
$ \Rightarrow x=17 \text { 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:
$x^2-2 a x+\left(a^2-b^2\right)=0$
AnswerGiven,
$x^2-2 a x+\left(a^2-b^2\right)=0$
On comparing it with $\mathrm{Ax}^2+\mathrm{bx}+\mathrm{c}=0$, we get:
$A=1, B=-2 a \text { and } C=\left(a^2-b^2\right)$
Discriminant $D$ is given by:
$ D=B^2-4 A C $
$ =(-2 a)^2-4 \times 1 \times\left(a^2-b^2\right) $
$ =4 a^2-4 a^2+4 b^2 $
$ =4 b^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(27 x)-3 x^2 $
$ \Rightarrow 540=81 x-3 x^2 $
$ \Rightarrow 3 x^2-81 x+540=0 $
$ \Rightarrow x^2-27 x+180=0 $
$ \Rightarrow x^2-15 x-12 x+180=0 $
$ \Rightarrow x(x-15)-12(x-15)=0 $
$ \Rightarrow(x-15)(x-12)=0 $
$ \Rightarrow x-15=0 \text { or } x-12=0 $
$ \Rightarrow x=15 \text { or } x=12$
$\text { Hence, the two parts are } 15 \text { and } 12 .$
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+2 x+1=365 $
$ \Rightarrow 2 x^2+2 x-364=0 $
$ \Rightarrow x^2+x+182=0 $
$ \Rightarrow x^2+14 x-13 x-182=0 $
$ \Rightarrow x(x+14)-13(x+14)=0 $
$ \Rightarrow(x+14)(x-13)=0 $
$ \Rightarrow x+14=0 \text { or } x-13=0 $
$ \Rightarrow x=-14 \text { or } x=13$
$\text { Since } x \text { is a positive integer, } x \neq-14$
$ \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:
$ 4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
Answer$ 4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
$\Rightarrow 4 x^2-2\left(a^2+b^2\right) x-2\left(a^2-b^2\right) x+\left(a^4-b^4\right)=0$
$\Rightarrow 2 x\left[2 x-\left(a^2+b^2\right)\right]-\left(a^2-b^2\right)\left[2 x+\left(a^2+b^2\right)\right]=0$
$\Rightarrow\left[2 x-\left(a^2+b^2\right)\right]\left[2 x-\left(a^2-b^2\right)\right]=0$
$\Rightarrow 2 x-\left(a^2+b^2\right)=0 \text { or } 2 x-\left(a^2-b^2\right)=0$
$\Rightarrow x=\frac{a^2+b^2}{2} \text { or } x=\frac{a^2-b^2}{2}$
$\Rightarrow\text{x}=\frac{\text{a}^2+\text{b}^2}{2}$ or $\text{x}=\frac{\text{a}^2-\text{b}^2}{2}$
View full question & answer→Question 274 Marks
If $a d \neq b c$ then prove that the equation $\left(a^2+b^2\right) x^2+2(a c+b d) x+\left(c^2+d^2\right)=0$ has no real roots.
AnswerCompare the given quadratic equation with $\mathrm{Ax}^2+\mathrm{Bx}+\mathrm{C}=0$
Here $A=a^2+b^2, B=2(a c+b d)$ and $C=c^2+d^2$
Consider, $B^2-4 A C=[2(a c+b d)]^2-4\left(a^2+b^2\right) \times\left(c^2+d^2\right)$
$=4\left[a^2 c^2+2 a b c d+b^2 d^2\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right]$
$=4 a^2 c^2+8 a b c d+4 b^2 d^2-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2$
$=8 a b c d-4 a^2 d^2-4 b^2 c^2$
$=-4\left[4 a^2 d^2+4 b c^2-2 a b c d\right]$
$=-4[a d-b c]^2$
Hence the given equation has no real roots unless $a d \neq b c$
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$
$\Rightarrow 28 x-x^2=192$
$\Rightarrow x^2-28 x+192=0$
$\Rightarrow x^2-16 x-12 x+192=0$
$\Rightarrow x(x-16)-12(x-16)=0$
$\Rightarrow(x-16)(x-12)=0$
$\Rightarrow x-16=0 \text { or } x-12=0$
$\Rightarrow x=16 \text { or } x=12$
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:
$2 x^2+a x-a^2=0$
AnswerThe given equation is: $2 x^2+a x-a^2=0$
Comparing it with $\mathrm{Ax}^2+\mathrm{Bx}+\mathrm{C}=0$
$A=2, B=a$ and $C=-a^2$ Discriminant, $D=B^2-4 A C$
$=a^2-4 \times 2 \times-a^2 $
$ =a^2+8 a^2 $
$ =9 a^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 $y.$
$\therefore x + y = 40 ...(1)$
and $(x + 3)(y - 4) = 360 ...(2)$
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 $
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 $x = 21, 21 + y = 40 \therefore y = 19$
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 $9\ km/hr$ 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+9 x $
$ \Rightarrow x^2+9 x-1620=0 $
$ \Rightarrow x^2+45 x-36 x-1620=0 $
$ \Rightarrow x(x+45)-36(x+45)=0 $
$ \Rightarrow(x+45)(x-36)=0 $
$ \Rightarrow x+45=0 \text { or } x-36=0 $
$ \Rightarrow x=-45 \text { 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 15 x-45=x^2+x$
$ \Rightarrow x^2-14 x+45=0$
$ \Rightarrow x^2-9 x-5 x+45=0$
$ \Rightarrow x(x-9)-5(x-9)=0$
$ \Rightarrow(x-9)(x-5)=0$
$ \Rightarrow x-9=0 \text { or } x-5=0$
$ \Rightarrow x=9 \text { or } x=5$
When $x = 9,$
$x - 3 = 9 - 3 = 6$
⇒ 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
$x \times(x+2)=483$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x+23)(x-21)=0$
$\Rightarrow x+23=0 \text { or } x-21=0$
$\Rightarrow x=-23 \text { or } x=21$
$\text { Since } x \text { is a positive integer, } x \neq-23$
$\Rightarrow x=21$
$\Rightarrow x+2=21+2=23$
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
$\text{x}+\sqrt{\text{x}}=132$
$\Rightarrow y^2+y=132 \text { where, } \sqrt{x}=y$
$\Rightarrow y^2+y-132=0$
$\Rightarrow y^2+12 y-11 y-132=0$
$\Rightarrow y(y+12)-11(y+12)=0$
$\Rightarrow(y+12)(y-11)=0$
$\Rightarrow y+12=0 \text { or } y-11=0$
$\Rightarrow y=-12 \text { or } y=11$
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+13 x-12 x-156=0$
$ \Rightarrow x(x+13)-12(x+13)=0$
$ \Rightarrow(x+13)(x-12)=0$
$ \Rightarrow x+13=0 \text { or } x-12=0$
$ \Rightarrow x=-13 \text { or } x=1$
Since $x$ is a natural number, $x \neq-13$
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 $288cm^2$. Find the dimensions of the rectangle.
AnswerLet the breadth of a rectangle $= x\ cm$
Then, lenght of the rectangle $= 2x\ cm$
$ \therefore \text { Area }=\text { lenght } \times \text { breadth }=288 \mathrm{~cm}^2 $
$ \Rightarrow 2 x \times x=288 $
$ \Rightarrow 2 x^2=288 $
$ \Rightarrow x^2=144$
$\Rightarrow\text{x}=\sqrt{144}$
$\Rightarrow\text{x}=\pm12$
$\Rightarrow\text{x}=12$ $[\because$ breadth cannot be negative$]$
Thus, breadth of rectangle $= 12cm$
And lenght of rectangle $= (2 × 12) = 24cm.$
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 $150\ km +$ Time token for $200\ km = 5$
$\frac{150}{\text{x}}+\frac{200}{(\text{s}+20)}=5$
$\Rightarrow\frac{150\text{x}+3000+\text{200s}}{\text{s}(\text{s}+20)}=5$
$\Rightarrow\frac{350\text{s}+3000}{\text{s}(\text{s}+20)}=5$
$\Rightarrow\frac{50(\text{7s}+60)}{\text{s}(\text{s}+20)}=5$
$ 10(7 s+60)=s(s+20) $
$ \Rightarrow 70 s+600=s^2+20 s $
$ \Rightarrow s^2-50 s-600=0 $
$ \Rightarrow s^2-60 s-10 s-600=0 $
$ \Rightarrow s(s-60)-10(s-60)=0 $
$ \Rightarrow(s-60)(s-10)=0 $
$ \Rightarrow s-60=0 \text { or } s-10=0 $
$ \Rightarrow s=60 \text { or } s=10 $
$ \Rightarrow s=10[\text { Not possible }]$
$\therefore \text { First speed of the truck }=60 \mathrm{~km} / \mathrm{hr} \text {. }$
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 $(x+10) \mathrm{cm}$
$ \therefore \text { Area }=600 \mathrm{~cm}^2 $
$ \Rightarrow \frac{1}{2} \times \text { base } \times \text { altitude }=600 \mathrm{~cm}^2 $
$ \Rightarrow \frac{1}{2} \times(\mathrm{x}+10) \times \mathrm{x}=600 $
$ \Rightarrow \mathrm{x}^2+10 \mathrm{x}=1200 $
$ \Rightarrow \mathrm{x}^2+10 \mathrm{x}-1200=0 $
$ \Rightarrow \mathrm{x}^2+40 \mathrm{x}-30 \mathrm{x}-1200=0 $
$ \Rightarrow \mathrm{x}(\mathrm{x}+40)-30(\mathrm{x}+40)=0 $
$ \Rightarrow(\mathrm{x}+40)(\mathrm{x}-30)=0 $
$ \Rightarrow \mathrm{x}=-40 \text { or } \mathrm{x}=30 $
$ \therefore \mathrm{x}=30[\because \text { length of altitude cannot be negative }]$
Hence, altitude of triangle is 30 cm and base of triangle 40 cm .
$(\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+4 x+4=452 $
$ \Rightarrow 2 x^2+4 x-448=0 $
$ \Rightarrow x^2+2 x-224=0 $
$ \Rightarrow x^2-14 x+16 x-224=0 $
$ \Rightarrow x(x-14)+16(x-14)=0 $
$ \Rightarrow(x-14)(x+16)=0 $
$ \Rightarrow x-14=0 \text { or } x+16=0 $
$ \Rightarrow x=14 \text { 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 5km/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 $300\ km$ $=\frac{300}{\text{x}}\ \text{hours}$
Time taken to cover $300\ km$ when the speed is increased by 5km/hr $=\frac{300}{(\text{x}+5)}\ \text{hours}$
It is given that the time taken to cover $300\ km$ 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=2 x^2+10 x $
$ \Rightarrow 2 x^2+10 x-1500=0 $
$ \Rightarrow x^2+5 x-750=0 $
$ \Rightarrow x^2+30 x-25 x-750=0 $
$ \Rightarrow x(x+30)-25(x+30)=0 $
$ \Rightarrow(x+30)(x-25)=0 $
$ \Rightarrow x+30=0 \text { or } x-25=0 $
$ \Rightarrow x=-30 \text { or } x=25$
Since the speed cannot be negative, $\text{x}\neq30.$
$⇒ x = 25$
Hence the original speed of train is $25\ kmph.$
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+18 x-17 x-306=0 $
$ \Rightarrow x(x+18)-17(x+18)=0 $
$ \Rightarrow(x+18)(x-17)=0 $
$ \Rightarrow x+18=0 \text { or } x-17=0 $
$ \Rightarrow x=-18 \text { 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+10 x-3000=0 $
$ \Rightarrow x^2+60 x-50 x-3000=0 $
$ \Rightarrow x(x+60)-50(x+60)=0 $
$ \Rightarrow(x+60)(x-50)=0 $
$ \Rightarrow x+60=0 \text { or } x-50=0 $
$ \Rightarrow x=-60 \text { or } x=50 $
$ \Rightarrow x=50(\because \text { 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.
$\therefore\ \text{xy}=14$
$\Rightarrow\text{y}=\frac{14}{\text{x}}\ \dots(1)$
$ \text { the number }=(10 x+y) $
$ (10 x+y)+45=(10 x+y) $
$ 9 x-9 y=-45 $
$ x-y=-5 \ldots(2)$
Putting $\mathrm{y}=\frac{14}{\mathrm{x}}$ from $(1)$ in $(2)$, we get
$ \mathrm{x}-\frac{14}{\mathrm{x}}=-5 $
$ \Rightarrow \mathrm{x}^2+5 \mathrm{x}-14=0 $
$ \Rightarrow \mathrm{x}^2+7 \mathrm{x}-2 \mathrm{x}-14=0 $
$ \Rightarrow \mathrm{x}(\mathrm{x}+7)-2(\mathrm{x}+7)=0 $
$ \Rightarrow(\mathrm{x}+7)(\mathrm{x}-2)=0 $
$ \Rightarrow \mathrm{x}+7=0 \text { or } \mathrm{x}-2=0 $
$ \Rightarrow \mathrm{x}=-7 \text { or } \mathrm{x}=2$
$\therefore \mathrm{x}=2[\because$ 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=4 x^2-20 x $
$ \Rightarrow 4 x^2-20 x-3000=0 $
$ \Rightarrow x^2-5 x-750=0 $
$ \Rightarrow x^2-30 x+25 x-750=0 $
$ \Rightarrow x(x-30)+25(x-30)=0 $
$ \Rightarrow(x-30)(x+25)=0 $
$ \Rightarrow x-30=0 \text { or } x+25=0 $
$ \Rightarrow x=30 \text { 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 $a x^2+b x+c=0$, where
$\mathrm{a}=\mathrm{m}^2, \mathrm{~b}=2 \mathrm{mn}$ and $\mathrm{c}=\mathrm{n}^2-\mathrm{mn}$.
$\therefore$ 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=3 x^2-9 x $
$ \Rightarrow 3 x^2-9 x-84=0 $
$ \Rightarrow x^2-3 x-28=0 $
$ \Rightarrow x^2-7 x+4 x-28=0 $
$ \Rightarrow x(x-7)+4(x-7)=0 $
$ \Rightarrow(x-7)(x+4)=0 $
$ \Rightarrow x-7=0 \text { or } x+4=0 $
$ \Rightarrow x=7 \text { or } x=-4$
Since x is a natural number, $x ≠ -4$
$⇒ x = 7$ 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+\left(x^2+3 x+2\right)=46 \text { or } 2 x^2+3 x-44=0 $
$ \Rightarrow 2 x^2+11 x-8 x-44=0 \text { or } x(2 x+11)-4(2 x-11)=0 $
$ \Rightarrow(x-4)(2 x+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$ 2 x^2+x+4=0 $
$ \left.\Rightarrow 4 x^2+2 x+8=0 \text { (Multiplying both sides by } 2\right) $
$ \Rightarrow 4 x^2+2 x=-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.
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