Question 13 Marks
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by$\frac{1}{14}$ Find the fraction.
AnswerLet the numerator of a fraction $=x$
then denominator $=x+3$
then fraction $=\frac{1}{14}$
Now according to the condition,
new fraction $\frac{x-1}{x+31}=\frac{x}{x+3}-\frac{1}{14}$
$\Rightarrow \frac{x-1}{x+2}=\frac{14 x-x-3}{14(x+3)}$
$\Rightarrow \frac{x-1}{x+2}=\frac{13 x-3}{14 x+42}$
$\Rightarrow( x -1)(14 x +42)=(13 x -3)( x +2)$
$\Rightarrow 14 x ^2+42 x -14 x -42=13 x ^2+26 x -3 x 6$
$\Rightarrow 14 x ^2+28 x 42-13 x ^2-23 x +6=0$
$\Rightarrow x ^2+5 x -36=0$
$\Rightarrow x ^2+9 x -4 x -36=0$
$x(x+9)-4(x+9)=0$
$\Rightarrow( x +9)( x -4)=0$
Either $x+9=0$,
then $x=-9$,
but it is not possible as the fraction is positive.
or
$x-4=0$
then $x=4$
$\therefore$ Fraction $=\frac{x}{x+3}=\frac{4}{4+3}=\frac{4}{7}$.
View full question & answer→Question 23 Marks
Find three consecutive odd integers, the sum of whose squares is $83$.
AnswerLet the three numbers be $x, x + 2, x + 4$
According to statements,
$(x)^2 + (x + 2)^2 + (x + 4)^2 = 83$
$\Rightarrow x^2 + x^2 + 4x + 4 + x^2 + 8x + 16 = 83$
$\Rightarrow 3x^2 + 12x + 20 = 83$
$\Rightarrow 3x^2 + 12x + 20 - 83 = 0$
$\Rightarrow 3x^2 + 12x - 63 = 0$
$\Rightarrow x^2 + 4x - 21 = 0$
$\Rightarrow x^2 + 7x - 3x - 21 = 0$
$\Rightarrow x(x + 7) -3(x + 7) = 0$
$\Rightarrow (x - 3)(x + 7) = 0$
Either $x - 3 = 0$,
then $x = 3$
or
$x + 7 = 0,$
then $x = -7$
∴ Number will be $3, 3 + 2, 3 + 4 = 3, 5, 7$
or
Numbers will be $-7, -7 + 2, -7 + 4 = -7, -5. -3$.
View full question & answer→Question 33 Marks
Find three successive even natural numbers, the sum of whose squares is $308$.
AnswerLet first even number $= 2x$
second even number $= 2x + 2$
third even number $= 2x + 4$
Now according to the condition,
$(2x)^2 + (2x + 2)^2 + (2x + 4)^2 = 308$
$\Rightarrow 4x^2 + 4x^2 + 8x + 4 + 4x^2 + 16x + 16 = 308$
$\Rightarrow 12x^2+ 24x + 20 - 308 = 0$
$\Rightarrow 12x^2 + 24x - 288 = 0$
$\Rightarrow x^2 + 2x - 24 = 0$ ...(Dividing y 12)
$\Rightarrow x^2 + 6x - 4x - 24 = 0$
$\Rightarrow x(x + 6) -4(x + 6) = 0$
$\Rightarrow (x + 6)(x - 4) = 0$
Either $x + 6 = 0$,
then $x = -6$
But i is not a natural number, hence not possible.
or
$x - 4 = 0,$
then $x = 4$
∴ First even natural number = 2x = 2 x 4 = 8
second number = 8 + 2 = 10
and the third number = 10 + 2 = 12.
View full question & answer→Question 43 Marks
There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?
AnswerLet the first integer $=x$
then second integer $=x+1$
and third integer $=x+2$
Now according to the condition,
$
\begin{aligned}
& x^2+(x+1)(x+2)=154 \\
& \Rightarrow x^2+x^2+3 x+2-154=0 \\
& \Rightarrow 2 x^2+3 x-152=0 \\
& \Rightarrow 2 x^2+19 x-16 x-152=0 \\
& \Rightarrow x (2 x +19)-8(2 x+19)=0 \\
& \Rightarrow(2 x+19)(x-8)=0
\end{aligned}
$
Either $2 x+19=0$,
then $2 x=-19$
$
\Rightarrow x =-\frac{19}{2}
$
But it is not possible as it is not an positive integer. or
$
x-8=0
$
then $x=8$
$\therefore$ Numbers are $8,(8+1)-9$ and $(8+2)=10$.
View full question & answer→Question 53 Marks
The difference between the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.
AnswerLet the larger number $=x$
then smaller number $=y$
Now according to the condition,
$
x ^2- y ^2=45.....(1)
$
and $y^2=4 x....(2)$
Substituting the value of $y^2$ from (2) in (1)
$
\begin{aligned}
& x^2-4 x=45 \\
& \Rightarrow x^2-4 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
\end{aligned}
$
Either $x-9=0$,
then $x=9$
or $x+5=0$,
then $x=-5$
(i) When $x=9$, the larger number $=9$
and smaller number
$
\begin{aligned}
& y=\sqrt{4 x} \\
& =\sqrt{4 \times 9} \\
& =\sqrt{36} \\
& \therefore y=6
\end{aligned}
$
(ii) When $x=-5$, then larger number $=-5$
$
\begin{aligned}
& y=\sqrt{4 x} \\
& =\sqrt{4 \times 5} \\
& =\sqrt{-20}
\end{aligned}
$
which is not possible.
Hence numbers are 6,9 .
View full question & answer→Question 63 Marks
The sum of the ages of Vivek and his younger brother Amit is $47$ years. The product of their ages in years is $550$. Find their ages.
AnswerLet Vivek’s present age be x years.
His brother’s age = (47 – x) years
According to question,
$x(47 – x) = 550$
$\Rightarrow 47x – x^2 = 550$
$\Rightarrow x^2 – 47x + 550 = 0$
$\Rightarrow x^2 – 25x – 22x + 550 = 0$
$\Rightarrow x(x – 25) – 22(x – 25) – 0$
$\Rightarrow (x - 25)(x - 22) = 0$
$\Rightarrow x - 25 = 0$ or $x - 22 = 0$
$\Rightarrow x = 25$ or $x = 22$
When $x = 25$, then $47 - x = 47 - 25 = 22$
When $x = 22$, then $47 - x = 47 - 22 = 25$ ...(does not satisfy the given condition)
∴ Vivek's age = $x = 25$ years.
His younger brother's age = 22 years.
View full question & answer→Question 73 Marks
A trader buys x articles for a total cost of Rs 600.
(i) Write down the cost of one article in terms of x.
If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it to find x.
AnswerNumber of articles $=x$
The total cost of articles $=\operatorname{Rs} 600$
1) Cost of one article $=\operatorname{Rs} \frac{600}{x}$
2) From the given information we have
$
\begin{aligned}
& \frac{600}{x-4}-\frac{600}{x}=5 \\
& \frac{600 x-600 x+2400}{x(x-4)}=5 \\
& \frac{480}{x(x-4)}=1 \\
& x^2-4 x-480=0 \\
& x^2-24 x+20 x-480=0 \\
& (x-24)+20(x-24)=0 \\
& (x-24)(x+20)=0 \\
& x=24,-20
\end{aligned}
$
Since number of article cannnot be negative So $x=24$.
View full question & answer→Question 83 Marks
2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x – 4), find x.
AnswerCost of $2 x$ articles $=5 x+54$
Cost of 1 article $=\frac{5 x+54}{2 x}.....(1)$
Again cost of $x+2$ articles $=10 x-4$
$\therefore$ Cost of 1 article $=\frac{10 x-4}{x+2}.....(2)$
From (1) and (2),
$
\begin{aligned}
& \frac{5 x+54}{2 x}=\frac{10 x-4}{x+2} \\
& \Rightarrow(5 x+54)(x+2)=2 x(10 x-4) \\
& \Rightarrow 5 x^2+10 x+54 x+108-20 x^2-8 x \\
& \Rightarrow 5 x^2+10 x+54 x+108-20 x^2+8 x=0 \\
& \Rightarrow-15 x^2+72 x+108=0 \\
& \Rightarrow 5 x^2-24 x-36=0 \ldots(\text { Dividing by }-3) \\
& \Rightarrow 5 x^2-30 x+6 x-36=0 \\
& \Rightarrow 5 x(x-6)+6(x-6)=0 \\
& \Rightarrow(x-6)(5 x+6)=0
\end{aligned}
$
Either $x-6=0$,
then $x=6$
or
$
5 x+6=0 \text {, }
$
then $5 x=-6$
$
\Rightarrow x =\frac{-6}{5} \text {. }
$
but it is not possible as it is in negative.
$
\therefore x=6 \text {. }
$
View full question & answer→Question 93 Marks
Rs. 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got Rs. 12 less. Find ‘x’.
AnswerShare of each child $=\text { ₹ } \frac{480}{x}$
Now, number of children
$
= x +20
$
$\because$ Share of each child
$
=\text { ₹ } \frac{480}{x+20}
$Now, According to the question
$
\begin{aligned}
& \frac{480}{x}-\frac{480}{x+20}=12 \\
& \Rightarrow \frac{480 x+9,600-480 x}{x(x+20)}=12 \\
& \Rightarrow 9,600=12 x(x+20) \\
& \Rightarrow 800=x^2+20 x \\
& \Rightarrow x^2+20 x-800=0 \\
& \Rightarrow x^2+40 x-20 x-800=0 \\
& \Rightarrow x(x+40)-20(x+40)=0 \\
& \Rightarrow(x-20)(x+40)=0 \\
& \Rightarrow x=20
\end{aligned}
$
or
$\Rightarrow x=-40 \ldots$ (not possible)
$
\therefore x=20
$
View full question & answer→Question 103 Marks
The sum of two numbers is $9$ and the sum of their squares is $41$. Taking one number as x, form ail equation in x and solve it to find the numbers.
AnswerSum of two numbers = 9
Let first number = x
then second number = 9 – x
Now according to the condition,
$(x)^2 + (9 - x)^2 = 41$
$\Rightarrow x^2 + 81 - 18x + x^2 - 41 = 0$
$\Rightarrow 2x^2 - 18x + 40 = 0$
$\Rightarrow x^2 - 9x + 20 = 0 ..$.(Dividing by $2)$
$\Rightarrow x^2 - 4x - 5x + 20 = 0$
$\Rightarrow (x - 4) -5(x - 4) = 0$
$\Rightarrow (x - 4) (x - 5) = 0$
Either x - 4 = 0,
then x = 4
or
x - 5 = 0,
then x = 5
(i) If x = 4, then first number = 4
and second number = 9 - 4 = 5
(ii) If x = 5, then first number = 5
ans second number = 9 - 5 = 4
Hence numbers are 4 and 5.
View full question & answer→Question 113 Marks
The distance by road between two towns A and B is 216 km, and by rail, it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(1) the time is taken by the car to reach town B from A, in terms of x;
(2) the time is taken by the train to reach town B from A, in terms of x.
(3) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.
(4) Hence, find the speed of the train.
AnswerSpeed of car $=x km / hr$
Speed of train $=(x+16) km / hr$
1) we know Time $=\frac{\text { Distance }}{\text { Speed }}$
Time is taken by the car to reach town B From $A=\frac{216}{ x }$ hrs
2) Time taken by the train to reach town $B$ from $A=\frac{208}{x+16}$ hrs
3) From the given information
$
\begin{aligned}
& \frac{216}{x}-\frac{208}{x+16}=2 \\
& \frac{216 x+3456-208 x}{x(x+16)}=2 \\
& \frac{8 x+3456}{x(x+16)}=2 \\
& 4 x+1728=x^2+16 x \\
& x^2+12 x-1728=0 \\
& x^2+48 x-36 x-1728=0 \\
& x(x+48)-36(x+48)=0 \\
& (x+48)(x-36)=0 \\
& x=-48,36
\end{aligned}
$
But speed cannnot be negative So $x=36$
4) Speed of train $=(36+16) km / hr =52 km / hr$
View full question & answer→Question 123 Marks
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/hr more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
AnswerLet the original speed of the car be $x$
$km / hr$,
so, Time taken by car $=\frac{400}{x}$ hrs.
Again, Speed $=(x+12) km / hr$
Time taken by car $=\frac{400}{x+12}$
so, $\frac{400}{x}-\frac{400}{x+12}=1 hr +\frac{40}{60}$
$400 \frac{(x+12-x)}{x(x+12)}=\frac{5}{3}$
$\frac{4800}{x^2+12 x}=\frac{5}{3}$
$\Rightarrow 5\left(x^2+12 x\right)=14,400$
$\Rightarrow x ^2+12 x -2,880=0$
$\Rightarrow x ^2+60 x -48 x -2,880=0$
$\Rightarrow x ( x +60)-48( x +60)=0$
$\Rightarrow(x+60)(x-48)=0$
Either, $x+60=0$
$x=-60 \ldots$ (Neglect, Speed can't be negative)
or
$x-48=0$
$x =48$
$\Rightarrow$ Original speed of the car is $48 km / hr$.
View full question & answer→Question 133 Marks
The speed of an express train is x km/hr arid the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.
AnswerLet the speed of express train is $x$
$Km / hr$. Speed of ordinary train is $( x -12) km / hr$.
Time require to cover for each train is $\frac{240}{x}$ and $\frac{240}{x-12}$ respectively.
According to question
$
\begin{aligned}
& \frac{240}{x-12}-\frac{240}{x}=1 \\
& \frac{240 x-240(x-12)}{(x-12)(x)}=1 \\
& 240 x-240(x-12)=x(x-12) \\
& x^2-12 x-2880=0 \\
& (x-60)(x+48)=0 \\
& \therefore x=60 km / hr .
\end{aligned}
$
Speed of the express train is $60 km / hr$.
View full question & answer→Question 143 Marks
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10km/h and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate 'x'.
AnswerTime taken by bus to cover total distance with speed $km / h =\frac{240}{ x }$ hrs
Time taken by bus to cover total distance with speed $(x-10) km / h =\left(\frac{240}{ x -10}\right)$ hrs
According to the given condition, we have
$
\begin{aligned}
& \frac{240}{x-10}-\frac{240}{x}=2 \\
& \Rightarrow 240\left(\frac{1}{x-10}-\frac{1}{x}\right)=2 \\
& \Rightarrow \frac{1}{x-10}-\frac{1}{x}=\frac{1}{120} \\
& \Rightarrow \frac{x-x+10}{x(x-10)}=\frac{1}{120} \\
& \Rightarrow \frac{10}{x^2-10 x}=\frac{1}{120} \\
& \Rightarrow x 2-10 x=1200 \\
& \Rightarrow x 2-10 x-1200=0 \\
& \Rightarrow(x-40)(x+30)=0 \\
& \Rightarrow x-40=0 \text { or } x+30=0 \\
& \Rightarrow x=40 \text { or } x=-30
\end{aligned}
$
Since the speed cannot be negative, we have $x=40 km / h$
View full question & answer→Question 153 Marks
In an auditorium, the number of rows are equal to the number of seats in each row.If the number of rows is doubled and number of seats in each row is reduced by $5$, then the total number of seats is increased by $375$. How many rows were there?
AnswerLet the number of rows = x
then no. of seats in each row = x
and total number of seats = $x \times x = x^2$
According to the condition,
$2x x (x - 5) = x^2 + 375$
$\Rightarrow 2x^2- 10x = x^2 + 375$
$\Rightarrow 2x^2 - 10x - x^2 - 375 = 0$
$\Rightarrow x^2 - 10x - 375 = 0$
$\Rightarrow x^2 - 25x + 15x - 375 = 0$
$\Rightarrow x(x - 25) + 15(x - 25) = 0$
$\Rightarrow (x - 25)(x + 15) = 0$
Either $x - 25 = 0$,
then $x = 25$
or
$x + 15 = 0$,
then $x = -15$,
but it is not possible as it is negative.
∴ Number of rows = 25.
View full question & answer→Question 163 Marks
Find two consecutive even natural numbers such that the sum of their squares is $340$.
AnswerLet first even natural number $= 2x$
Then second number $= 2x + 2$
According to the condition,
$(2x)^2 + (2x + 2)^2 = 340$
$4x^2 + 4x + 8x + 4 = 340$
$\Rightarrow 8x^2+ 8x + 4 - 340 = 0$
$\Rightarrow 8x^2 + 8x - 336 = 0$
$\Rightarrow x^2 + x - 42 = 0$ ...(Dividing by 8)
$\Rightarrow x^2 + 7x - 6x - 42 = 0$
$\Rightarrow x(x + 7) -6(x + 7) = 0$
$\Rightarrow (x + 7)(x - 6) = 0$
EIther $x + 7 = 0$,
then $x = -7$
or
$x - 6 = 0$,
then $x = 6$
∴ First even natural number $= 2x$
$= 2 \times 6$
$= 12$
and second
$= 12 + 2$
$= 14$
$\therefore$ Numbers are $12, 14$.
View full question & answer→Question 173 Marks
In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.
AnswerTotal number of students $=480$
Let the number of students in each row $=x$
then the number of rows $=\frac{480}{x}$
According to the conditions, $x =\frac{480}{x}+4$
$
\begin{aligned}
& \Rightarrow x ^2=480+4 \\
& \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
\end{aligned}
$
Either $x-24=0$
or
$
\begin{aligned}
& x+20=0 \\
& \Rightarrow x=24
\end{aligned}
$
or
$
x=-20
$
which is not possible as it is negative
$\therefore$ Number of students in each row $=24$.
View full question & answer→Question 183 Marks
If the product of two consecutive even integers is $224$, find the integers.
AnswerLet first even integer = $2x$
then second even integer = $2x + 2$
According to the condition,
$2x x (2x + 2) = 244$
$\Rightarrow 4x^2 + 4x - 244 = 0$
$\Rightarrow x^2 + x - 56 = 0$
$\Rightarrow x^2 + 8x - 7x - 56 = 0$
$\Rightarrow x(x + 8) -7(x + 8) = 0$
$\Rightarrow (x + 8)(x - 7) = 0$
Either $x + 8 = 0$,
then $x = -8$
∴ First even integer = $2 x (-8) = -16$
and second even integer = $-16 + 2 = -14$
or
$x - 7 = 0$,
then $x = 7$
∴ First even integer = $2x = 2 x 7 = 14$
and second even integer = $14 + 2 = 16$.
View full question & answer→Question 193 Marks
Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is $2$ cm longer than the base and $4$ cm longer than the shortest side, find the lengths of the rods.
AnswerLet the length of hypotenuse = x cm
then base = (x – 2)cm
and shortest side = x – 4
According to the condition,
$(x)^2 = (x - 2)^2 + (x - 4)^2$
$\Rightarrow x^2= x^2 - 4x + 4 + x^2- 8x + 16$
$\Rightarrow x^2 = 2x^2 - 12x + 20$
$\Rightarrow 2x^2 - 12x + 20 - x2 = 0$
$\Rightarrow x^2 - 12x + 20 = 0$
$\Rightarrow x^2- 10x - 2x + 20 = 0$
$\Rightarrow x(x - 10) -2(x - 10) = 0$
$\Rightarrow (x - 10)(x - 2) = 0$
Either $x - 10 = 0$,
then $x = 10$
or
$x - 2 = 0$,
then $x = 2$,
but it is not possible as the hypotenuse is the longest side.
∴ Hypotenuse = 10cm
Base = 10 - 2 = 8cm
and shortest side = 10 - 4 = 6cm.
View full question & answer→Question 203 Marks
If the product of two positive consecutive even integers is $288$, find the integers.
AnswerLet first positive even integer = $2x$
then second even integer = $2x + 2$
According to the condition,
$2x \times (2x + 2) = 288$
$\Rightarrow 4x^2 + 4x – 288 = 0$
$\Rightarrow x^2+ x – 72 = 0$ ...(Dividing by 4)
$\Rightarrow x^2 - 9x - 8x - 72 = 0$
$\Rightarrow x(x + 9) -8(x + 9) = 0$
$\Rightarrow (x + 9)(x - 8) = 0$
EIther $x + 9 = 0$,
then $x = -9$
But it is not possible as it is negative
or
$x - 8 = 0$,
then $x = 8$
∴ First even integer $= 2x$
$= 2 x 8$
$= 16$
and second even integer
$= 16 + 2$
$= 18$.
View full question & answer→Question 213 Marks
The hypotenuse of grassy land in the shape of a right triangle is $1$ metre more than twice the shortest side. If the third side is $7$ metres more than the shortest side, find the sides of the grassy land.
AnswerLet the shortest side = x
Hypotenuse $= 2x + 1$
and third side $= x + 7$
According to the condition,
$(2x + 1)^2 = x^2 + (x + 7)2$
$\Rightarrow 4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49$
$\Rightarrow 4x^2 + 4x + 1 - 2x^2- 14x - 49 = 0$
$\Rightarrow 2x^2 - 10x - 48 = 0$
$\Rightarrow x^2- 5x - 24 = 0$...(Dividing by 2)
$\Rightarrow x^2 - 8x + 3x - 24 = 0$
$\Rightarrow x(x - 8) + 3(x - 8) = 0$
$\Rightarrow (x - 8)(x + 3) = 0$
EIther $x - 8 = 0$,
then $x = 8$
or
$x + 3 = 0$,
then $x = -3$,
but it is not possible as it is negative.
∴ Shortest side = 8m
Third side
$= x + 7$
$= 8 + 7$
$= 15m$
and hypotenuse
$= 2x + 1$
$= 8 \times 2 + 1$
$= 16 + 1$
$= 17m$.
View full question & answer→Question 223 Marks
If the perimeter of a rectangular plot is $68$ m and the length of its diagonal is $26$ m, find its area.
AnswerPerimeter = 68 m and diagonal = 26m
Length + breadth = = 34m
Let length = xm
then breadth = (34 – x)m
According to the condition,
$l^2 + b^2 = h^2$
$(x)^2 + (34 - x)^2 = (26)^2$
$\Rightarrow x^2 + 1156 + x^2 - 68x = 676$
$\Rightarrow 2x^2 - 68x + 1156 - 676 = 0$
$\Rightarrow 2x^2 - 68x + 480 = 0$
$\Rightarrow x^2- 34x + 240 = 0$...(Dividing by 2)
$\Rightarrow x^2 - 24x - 10x + 240 = 0$
$\Rightarrow x(x - 24) -10(x - 24) = 0$
$\Rightarrow (x - 24)(x - 10) = 0$
Either $x - 24 = 0$,
then $x = 24$
or
$x - 10 = 0$,
then $x = 10$
∵ Length is greater than breadth
∴ Length = 24m
and breadth $= (34 - 24) = 10m$
and Area $= l \times b = 24 \times 10 = 240m^2$.
View full question & answer→Question 233 Marks
The lengths of the parallel sides of a trapezium are $(x + 9)$ cm and $(2x – 3)$ cm and the distance between them is $(x + 4)$ cm. If its area is $540 cm^2$, find $x$.
AnswerArea of a trapezium $=\frac{1}{2}$
(sum of parallel sides) $x$ height
Lengths of parallel sides are $(x+9)$ and $(2 x-3)$
and height $=(x+4)$
According to the condition,
$
\begin{aligned}
& \frac{1}{2}(x+9+x-3) \times(x+4)=540 \\
& \Rightarrow(3 x+6)(x+4)=540 \times 2 \\
& \Rightarrow 3 x^2+12 x+6 x+24-1080=0 \\
& \Rightarrow 3 x^2+18 x-1056=0 \\
& \Rightarrow x^2+6 x-352=0 \quad \ldots(\text { Dividing by } 3) \\
& \Rightarrow x^2+22 x-16 x-352=0 \\
& \Rightarrow x ( x +22)-16( x +22)=0 \\
& \Rightarrow( x +22)( x -16)=0
\end{aligned}
$
Either $x+22=0$,
then $x=-22$
But it is not possible as it is negative.
or
$
x-16=0
$
then $x=16$.
View full question & answer→Question 243 Marks
The perimeter of a rectangular plot is $180 m$ and its area is $1800 m^2$. Take the length of the plot as x m. Use the perimeter $180 m$ to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.
AnswerThe perimeter of a rectangular field $=180 m$
and area $=1800 m ^2$
Let length $= xm$
But length + breadth $=\frac{180}{2}=90 m$
$\therefore$ breadth $=(90- x ) m$
According to the condition,
$
\begin{aligned}
& x(90-x)=1800 \\
& \Rightarrow 90 x-x^2-1800=0 \\
& \Rightarrow x^2-90 x+1800=0 \\
& \Rightarrow x^2-60 x-30 x+1800=0 \\
& \Rightarrow x(x-60)-30(x-60)=0 \\
& \Rightarrow(x-60)(x-30)=0
\end{aligned}
$
Elther $x-60=0$,
then $x=60$
or
$
x-30=0
$
then $x=30$
$\because$ Length is greater than its breadth
$\therefore$ Length $=60 m$
and breadth $=90-60=30 m$.
View full question & answer→Question 253 Marks
A rectangular garden $10$ m by $16$ m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is $120$ square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.
AnswerLength of garden $= 16 m$
and width $= 10 m$
Let the width of walk $= x m$
Outer length $= 16 + 2x$
and outer width $= 10 + 2x$
Now according to the condition,
$(16 + 2x)(10 + 2x) -16 \times 10 = 120$
$\Rightarrow 160 + 32x + 20x + 4x^2- 160 = 120$
$\Rightarrow 4x^2 + 52x - 120 = 0$
$\Rightarrow x^2 + 13x - 30 = 0$ ...(Dividing by 4)
$\Rightarrow x^2 + 15 x - 2x - 30 = 0$
$\Rightarrow x(x + 15) -2(x + 15) = 0$
$\Rightarrow (x + 15)(x - 2) = 0$
EIther $x + 15 = 0$,
then $x = -15$
But it is not possible.
or
$x - 2 = 0$,
then $x = 2$.
View full question & answer→Question 263 Marks
A rectangle of area $105 cm²$ has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is $44 cm$, write down an equation in x and solve it to determine the dimensions of the rectangle.
AnswerPerimeter of rectangle $= 44 cm$
length + breadth $= 22 cm$
Let length $= x$
then breadth $= 22 – x$
According to the condition,
$x(22 - x) = 105$
$\Rightarrow 22x - x^2 = 105$
$\Rightarrow x^2- 22x + 105 = 0$
$\Rightarrow x^2 - 15x - 7x + 105 = 0$
$\Rightarrow x(x - 15) -7(x - 15) = 0$
$\Rightarrow (x - 15)(x - 7) = 0$
Either $x - 15 = 0$,
then $x = 15$
or
$x - 7 = 0$,
then $x = 7$
As length > breadth,
x = 7 is not admissible.
∴ Length = 15cm
and breadth
$= 22 - 15$
$= 7cm$.
View full question & answer→Question 273 Marks
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
AnswerLet 2-digit number $=x y=10 x+y$
Reversed digits $=y x=10 y+x$
According to question,
$
\begin{aligned}
& x y=6 \\
& y=\frac{6}{x}....(1)
\end{aligned}
$
and
$
\begin{aligned}
& 10 x+y+9=10 y+x \\
& \Rightarrow 10 x+\frac{6}{x}+9=10 \times \frac{6}{x}+x \ldots\left(\text { From }(1) y=\frac{6}{x}\right) \\
& \Rightarrow 10 x^2+6+9 x=60+x^2 \\
& \Rightarrow 10 x^2-x^2+9 x+6-60=0 \\
& \Rightarrow 9 x^2+9 x-54=0 \\
& \Rightarrow x^2+x-6=0 \\
& \Rightarrow x^2+3 x-2 x-6=0 \\
& \Rightarrow x(x+3)-2(x+3)=0 \\
& \Rightarrow x=2 \text { or }-3 \ldots(\text { rejacting }-3)
\end{aligned}
$
putting the value of $x$ in (i)
$
\begin{aligned}
& y=\frac{6}{2}=3 \\
& \therefore 2 \text {-digit } \\
& =10 x+y \\
& =10 \times 2+3 \\
& =23 .
\end{aligned}
$
View full question & answer→Question 283 Marks
Find two consecutive integers such that the sum of their squares is $61$
AnswerLet first integer $= x$
Then second integer $= x + 1$
According to the condition,
$(x)^2 + (x + 1)^2 = 61$
$\Rightarrow x^2 + x^2 + 2x + 1 = 61$
$\Rightarrow 2x^2 + 2x + 1 - 61 = 0$
$\Rightarrow 2x^2+ 2x - 60 = 0$
$\Rightarrow x^2 + x - 30 = 0$ ...(Dividing by 2)
$\Rightarrow x^2 + 6x - 5x - 30 = 0$
$\Rightarrow x(x + 6) -5(x + 6) = 0$
$\Rightarrow (x + 6)(x - 5) = 0$
Either $x + 6 = 0$,
then $x = -6$
or
$x - 5 = 0$,
then $x = 5$
(i) If $x = -6$, then
First integer $= -6$
and second $= -6 + 1 = -5$
(ii) If $x = 5$, then
First integer $= 5$
and second $= 5 + 1 = 6$
∴ Required integers are $(-6, -5),(5, 6)$.
View full question & answer→Question 293 Marks
A two digit number contains the bigger at ten’s place. The product of the digits is $27$ and the difference between two digits is $6$. Find the number.
AnswerLet unit’s digit $= x$
then tens digit $= x + 6$
Number $= x + 10(x + 6)$
$= x + 10x + 60$
$= 11x + 60$
According to the condition,
$x(x + 6) = 27$
$\Rightarrow x^2 + 6x - 27 = 0$
$\Rightarrow x^2 + 9x - 3x - 27 = 0$
$\Rightarrow x(x + 9) -3(x + 9) = 0$
$\Rightarrow (x + 9)(x - 3) = 0$
Either $x + 9 = 0$,
then $x = -9$,
but it is not possible as it is negative.
or
$x - 3 = 0$,
then $x = 3$
∴ Number
$= 11x + 60$
$= 11 \times 3 + 60$
$= 33 + 60$
$= 93$.
View full question & answer→Question 303 Marks
Find two consecutive natural numbers such that the sum of their squares is $61.$
AnswerLet the first natural number $= x$
then second natural number $= x + 1$
According to the condition, $(x)^2 + (x + 1)^2 = 61$
$\Rightarrow x^2 + x^2+ 2x + 1 - 61 = 0$
$\Rightarrow 2x^2 + 2x - 60 = 0$
$\Rightarrow x^2+ x - 30 = 0$
$\Rightarrow x^2 + 6x - 5x - 30 = 0$
$\Rightarrow x(x + 6) -5(x + 6) = 0$
$\Rightarrow (x + 6)(x - 5) = 0$
EIther x + 6 = 0,
then x = -6
or
x - 5 = 0,
then x = 5
∴ The number are positive.
∴ x = -6 is not possible
Hence the first natural number = 5
and secod natural number = 5 + 1 = 6.
View full question & answer→Question 313 Marks
If $– 5$ is a root of the quadratic equation $2x^2 + px – 15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots, find the value of k.
Answer-5 is a root of the quadratic equation
$
\begin{aligned}
& 2 x^2+p x-15=0, \text { then } \\
& \Rightarrow 2(5)^2-p(-5)-15=0 \\
& \Rightarrow 50-5 p-15=0 \\
& \Rightarrow 35-5 p=0 \\
& \Rightarrow 5 p=35=0 \\
& \Rightarrow p=\frac{35}{5}=7 \\
& p\left(x^2+x\right)+k=0 \text { has equal roots } \\
& \Rightarrow p x^2+p x+k=0 \\
& \Rightarrow 7 x^2+7 x+k=0
\end{aligned}
$
Here, $a=7, b=7, c=k$
$
\begin{aligned}
& b^2-4 a c \\
& =(7)^2-4 \times 7 \times k \\
& =49-28 k
\end{aligned}
$
$\because$ Roots are equal
$
\begin{aligned}
& \therefore b^2-4 a c=0 \\
& \Rightarrow 49-28 k=0 \\
& \Rightarrow 28 k=49 \\
& \Rightarrow k=\frac{49}{28}=\frac{7}{4} \\
& \therefore k=\frac{7}{4} .
\end{aligned}
$
View full question & answer→Question 323 Marks
Find the value(s) of p for which the quadratic equation $(2p + 1)x^2 – (7p + 2)x + (7p – 3) = 0$ has equal roots. Also find these roots.
AnswerThe quadratic equation given is $(2 p+1) x^2-(7 p+2) x+(7 p-3)=0$
Comparing with $ax ^2+ bx + c =0$, we have
$
\begin{aligned}
& a=2 p+1, b=-(7 p+2), c=(7 p-3) \\
& D=b^2-4 a c \\
& \Rightarrow 0=[-(7 p+2)]^2-4(2 p+1)(7 p-3) \\
& 0=49 p^2+4+28 p-4\left(14 p^2-6 p+7 p-3\right) \\
& 0=49 p^2+4+28 p-56 p^2-4 p+12 \\
& 0=-7 p^2+24 p+16 \\
& 0=-7 p^2+28 p-4 p+16 \\
& 0=-7 p(p-4)-4(p-4) \\
& 0=(-7 p-4)(p-4) \\
& \Rightarrow-7 p-4=0 \text { or } p-4=0
\end{aligned}
$
Hence, the value of $p=\frac{-4}{7}$ or $p=4$.
View full question & answer→Question 333 Marks
Find the value(s) of m for which each of the following quadratic equation has real and equal roots: $x^2 + 2(m – 1) x + (m + 5) = 0$
Answer$x^2 + 2(m – 1)x + (m + 5) = 0$
Equating with $ax^2 + bx + c = 0$
$a = 1, b = 2(m – 1), c = (m + 5)$
Since equation has real and equal roots.
So, $D = 0$
$\Rightarrow b^2 – 4ac = 0$
$[2(m – 1)^2 – 4 \times 1 \times (m + 5) = 0$
$\Rightarrow 4(m – 1)^2 – 4(m + 5) = 0$
$\Rightarrow 4 [(m – 1)^2 – (m + 5)] = 0$
$\Rightarrow 4 [m^2 – 2m + 1 – m – 5] = 0$
$\Rightarrow m^2 – 3m – 4 = 0$
$\Rightarrow (m + 1)(m – 4) = 0$
Either $m + 1 = 0$
$m = - 1$
or
$m – 4 = 0$
$m = 4$
$m = -1, 4.$
View full question & answer→Question 343 Marks
Find the value(s) of m for which each of the following quadratic equation has real and equal roots: $(3m + 1)x^2 + 2(m + 1)x + m = 0$
Answer$(3 m+1) x^2+2(m+1) x+m=0$
$\text { Here } a=3 m+1, b=2(m+1), c=m$
$D=b^2-4 a c$
$=[2(m+1)]^2-4 x(3 m+1)(m)$
$=4\left(m^2+2 m+1\right)-12 m^2-4 m$
$=4 m^2+8 m+4-12 m^2-4 m$
$=-8 m^2+4 m+4$
$\therefore \text { Roots are equal. }$
$\therefore D=0$
$\Rightarrow-8 m^2+4 m+4=0$
$\Rightarrow 2 m^2-m-1=0 \quad \ldots(\text { Dividing by } 4)$
$\Rightarrow 2 m^2-2 m+m-1=0$
$\Rightarrow 2 m(m-1)+1(m-1)=0$
$\Rightarrow(m-1)(2 m+1)=0$
$\text { Either } m-1=0,$
$\text { then } m=1$
$\text { or }$
$2 m+1=0$
$\text { then } 2 m=-1$
$\Rightarrow m=-\frac{1}{2} .$
View full question & answer→Question 353 Marks
Find the value (s) of k for which each of the following quadratic equation has equal roots : $(k – 4) x^2 + 2(k – 4) x + 4 = 0$
Answer$(k – 4) x^2 + 2(k – 4) x + 4 = 0$
Here $a = k - 4, b = 2(k - 4), c = 4$
$D = b^2 - 4ac$
$= [2(k - 4)]^2 - 4 \times (k - 4) x 4$
$= 4(k^2 + 16 - 8k) - 16(k - 4)$
$= 4(k^2 - 8k + 16) - 16(k - 4)$
$= 4[k^2 - 8k + 16 - 4k + 16]$
$= 4(k^2 - 12k + 32)$
$\because$ Roots are equal
$\therefore D = 0$
$\Rightarrow 4(k^2 - 12k + 32) = 0$
$\Rightarrow k^2 - 12k + 32 = 0$
$\Rightarrow k^2 - 8k - 4k + 32 = 0$
$\Rightarrow k(k - 8) -4(k - 8) = 0$
$\Rightarrow (k - 8)(k - 4) = 0$
Either $k - 8 = 0$,
then $k = 4$
or
$k - 4 = 0$,
then $k = 4$
But $k - 4 ≠ 0$
$k ≠ 4$
$k = 8$.
View full question & answer→Question 363 Marks
Solve the following equation by using formula :
$
\frac{1}{x}+\frac{1}{x-2}=3, x \neq 0,2
$
Answer$\begin{aligned} & \frac{1}{x}+\frac{1}{x-2} \\ & \frac{x-2+x}{x(x-2)}=3 \\ & \Rightarrow \frac{2 x-2}{x^2-2 x}=3 \\ & \Rightarrow 3 x^2-6 x =2 x -2 \\ & \Rightarrow 3 x ^2-6 x -2 x +2=0 \\ & \Rightarrow 3 x ^2-8 x +2=0 \\ & \text { Here } a =3, b =-8, c =2 \\ & b 2-4 ac \\ & =(-8)^2-4 \times 3 \times 2 \\ & =64-24 \\ & =40 \\ & x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ & =\frac{-(-8) \pm \sqrt{40}}{2 \times 3} \\ & =\frac{8 \pm 2 \sqrt{10}}{6} \\ & =\frac{4 \pm \sqrt{10}}{3} \\ & \therefore x=\frac{4+\sqrt{10}}{3} \text { and } \frac{4-\sqrt{10}}{3}\end{aligned}$
View full question & answer→Question 373 Marks
Solve the following equation by using formula :$x-\frac{1}{x}=3, x \neq 0$
Answer$
\begin{aligned}
& x-\frac{1}{x}=3 \\
& x^2-1=3 x \\
& \Rightarrow x^2-3 x-1=0
\end{aligned}
$
Here $a=1, b=-3, c=-1$
$
\begin{aligned}
\therefore & b^2-4 a c \\
= & (-3)^2-4 \times 1 \times(-1) \\
= & 9+4 \\
= & 13
\end{aligned}
$
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-(-3) \pm \sqrt{13}}{2 \times 1} \\
& =\frac{3 \pm \sqrt{13}}{2}
\end{aligned}
$
$\therefore x=\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.
View full question & answer→Question 383 Marks
Solve the following equation by using formula :
$4x^2 – 4ax + (a^2 – b^2) = 0$
Answer$
4 x^2-4 a x+\left(a^2-b^2\right)=0
$
Here $a=4, b=-4 a, c=a^2-b^2$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(-4 a)^2-4 \times 4\left(a^2-b^2\right) \\
& =16 a^2-16\left(a^2-b^2\right) \\
& =16 a^2-16 a^2+16 b^2 \\
& D=16 b^2
\end{aligned}
$
$
\begin{aligned}
& \because \times=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(-4 a) \pm \sqrt{16 b^2}}{2 \times 4} \\
& =\frac{4 a \pm 4 b}{8} \\
& =\frac{a+b}{2}
\end{aligned}
$
$\therefore x_1=\frac{a+b}{2}, x_2=\frac{a-b}{2}$
Hence $x=\frac{a+b}{2}, \frac{a-b}{2}$.
View full question & answer→Question 393 Marks
Solve the following equation by using formula :
$a (x^2 + 1) = (a^2+ 1) x , a \neq 0$
Answer$
\begin{aligned}
& a\left(x^2+1\right)=\left(a^2+1\right) x \\
& a x^2-\left(a^2+1\right) x+a=0 \\
& \text { Here } a=a, b=-\left(a^2+1\right), c=a \\
& D=b^2-4 a c \\
& =\left[-\left(a^2+1\right)\right]^2-4 \times a x a \\
& =a^4+2 a^2+1-4 a^2 \\
& =a^4-2 a+1 \\
& =\left(a^2-1\right)^2 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{\left(a^2+1\right) \pm \sqrt{\left(a^2-1\right)^2}}{2 a} \\
& =\frac{\left(a^2+1\right)+\left(a^2-1\right)}{2 a} \\
& \therefore x_1=\frac{a^2+1+a^2-1}{2 a} \\
& =\frac{2 a^2}{2 a} \\
& =a \\
& x_2=\frac{a^2+1-a^2+1}{2 a} \\
& =\frac{2}{2 a} \\
& =\frac{1}{a}
\end{aligned}
$
Hence $x =a, \frac{1}{a}$.
View full question & answer→Question 403 Marks
Solve the following equation by using formula :$\frac{x+1}{x+3}=\frac{3 x+2}{2 x+3}$
Answer$
\begin{aligned}
& \frac{x+1}{x+3}=\frac{3 x+2}{2 x+3} \\
& (x+1)(2 x+3)=(3 x+2)(x+3) \\
& \Rightarrow 2 x^2+3 x+2 x+3=3 x^2+9 x+2 x+6 \\
& \Rightarrow 2 x^2+5 x+3-3 x^2-11 x-6=0 \\
& \Rightarrow-x^2-6 x-3=0 \\
& \Rightarrow x^2+6 x+3=0
\end{aligned}
$
Here $a=1, b=6, c=3$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(6)^2-4 \times 1 \times 3 \\
& =36-12 \\
& =24
\end{aligned}
$
$
\begin{aligned}
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-6 \pm \sqrt{24}}{2 \times 1} \\
& =\frac{6 \pm \sqrt{4 \times 6}}{2} \\
& =\frac{-6 \pm 2 \sqrt{6}}{2} \\
& =-3 \pm \sqrt{6} \\
& \therefore x_1=-3+\sqrt{6}, x_2=-3-\sqrt{6}
\end{aligned}
$
Hence $x=-3+\sqrt{6},-3-\sqrt{6}$.
View full question & answer→Question 413 Marks
Solve the following equation by using formula :$\frac{x-2}{x+2}+\frac{x+2}{x-2}=4$
Answer$
\begin{aligned}
& \frac{x-2}{x+2}+\frac{x+2}{x-2}=4 \\
& \Rightarrow \frac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4 \\
& \Rightarrow \frac{x^2-4 x+4+x^2+4 x+4}{x^2-4} \\
& \Rightarrow 2 x^2+8=4 x^2-16 \\
& \Rightarrow 2 x^2+8-4 x^2+16=0 \\
& \Rightarrow-2 x^2+24=0 \\
& \Rightarrow x^2-12=0
\end{aligned}
$
Here $a=1, b=0, c=-12$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(0)^2-4 \times 1(-12) \\
& =0+48 \\
& =48
\end{aligned}
$
$
\begin{aligned}
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{0 \pm \sqrt{48}}{2 \times 1} \\
& =\frac{ \pm \sqrt{48}}{2} \\
& =\frac{ \pm \sqrt{16 \times 3}}{2} \\
& = \pm \frac{4 \sqrt{3}}{2} \\
& = \pm 2 \sqrt{3}
\end{aligned}
$
Hence roots are $2 \sqrt{3},-2 \sqrt{3}$.
View full question & answer→Question 423 Marks
Solve the following equation by using formula $\sqrt{3} x^2+10 x-8 \sqrt{3}=0$
Answer$
\begin{aligned}
& \sqrt{3} x^2+10 x-8 \sqrt{3}=0 \\
& \text { Here } a=\sqrt{3}, b=10, c=-8 \sqrt{3} \\
& D=b^2-4 a c \\
& =(10)^2-4 \times \sqrt{3} \times(-8 \sqrt{3}) \\
& =100+96 \\
& =196 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-10 \pm \sqrt{196}}{2 \times \sqrt{3}} \\
& =\frac{-10 \pm 14}{2 \sqrt{3}} \\
& \therefore x_1=\frac{-10+14}{2 \sqrt{3}} \\
& =\frac{4}{2 \sqrt{3}} \\
& =\frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{2 \sqrt{3}}{3} \\
& x_2=\frac{-10-14}{2 \sqrt{3}} \\
& =\frac{-24}{2 \sqrt{3}} \\
& =\frac{-12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-12 \sqrt{3}}{3} \\
& =-4 \sqrt{3}
\end{aligned}
$
Hence $x =\frac{2 \sqrt{3}}{3},-4 \sqrt{3}$.
View full question & answer→Question 433 Marks
Solve the following equation by using formula :$2 x^2+\sqrt{5}-5=0$
Answer$
\begin{aligned}
& 2 x^2+\sqrt{5}-5=0 \\
& \text { Here } a=2, b=\sqrt{5}, c=-5 \\
& D=b^2-4 a c \\
& =(\sqrt{5})^2-4 \times 2 \times(-5) \\
& =5+40 \\
& =45 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-\sqrt{5} \pm \sqrt{45}}{2 \times 2} \\
& =\frac{-\sqrt{5} \pm \sqrt{9 \times 5}}{4} \\
& =\frac{-\sqrt{5} \pm 3 \sqrt{5}}{4} \\
& \therefore x_1=\frac{-\sqrt{5}+3 \sqrt{5}}{4} \\
& =\frac{2 \sqrt{5}}{4} \\
& =\frac{\sqrt{5}}{2} \\
& x_2=\frac{-\sqrt{5}-3 \sqrt{5}}{4} \\
& =\frac{-4 \sqrt{5}}{4} \\
& =-\sqrt{5}
\end{aligned}
$
Hence $x =\frac{\sqrt{5}}{2},-\sqrt{5}$.
View full question & answer→Question 443 Marks
Solve the following equation by using formula :
$25x^2 + 30x + 7 = 0$
Answer$\begin{aligned} & 25 x^2+30 x+7=0 \\ & \text { Here } a=25, b=30, c=7 \\ & D=b^2-4 a c \\ & =(30)^2-4 \times 25 \times 7 \\ & =900-700 \\ & =200 \\ & \because x=\frac{-b \pm \sqrt{D}}{2 a} \\ & =\frac{-30 \pm \sqrt{200}}{2 \times 25} \\ & =\frac{-30 \pm \sqrt{100 \times 2}}{50} \\ & =\frac{-30 \pm 10 \sqrt{2}}{50} \\ & =\frac{-3 \pm \sqrt{2}}{5} \\ & \therefore x_1=\frac{-3+\sqrt{2}}{5} \text { and } x_2=\frac{-3-\sqrt{2}}{5}\end{aligned}$
Hence $x=\frac{-3+\sqrt{2}}{5}, \frac{-3-\sqrt{2}}{5}$.
View full question & answer→Question 453 Marks
Solve the following equation by using formula :
(2x + 3)(3x – 2) + 2 = 0
Answer$
\begin{aligned}
& (2 x+3)(3 x-2)+2=0 \\
& 6 x^2-4 x+9 x-6+2=0 \\
& 6 x^2+5 x-4=0 \\
& \text { Here } a=6, b=5, c=-4 \\
& D=b^2-4 a c \\
& =(5)^2-4 x 6 x(-4) \\
& =25+96 \\
& =121 \\
& \because x=\frac{-b+\sqrt{D}}{2 a} \\
& =\frac{-5 \pm \sqrt{121}}{5 \times 6} \\
& =\frac{-5 \pm 11}{12} \\
& \therefore x_1=\frac{-5+11}{12} \\
& =\frac{6}{12} \\
& =\frac{1}{2} \\
& \therefore x_2=\frac{-5-11}{12} \\
& =\frac{-16}{12} \\
& =\frac{-4}{3}
\end{aligned}
$
Hence $x =\frac{1}{2}, \frac{-4}{3}$.
View full question & answer→Question 463 Marks
Solve the following equation by using formula :
$x^2 + 7x – 7 = 0$
Answer$
x^2+7 x-7=0
$
Here $a =1, b =7, c =-7$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(7)^2-4 \times 1(-7) \\
& =49+28 \\
& =77
\end{aligned}
$
$
\begin{aligned}
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-7 \pm \sqrt{77}}{2 \times 1} \\
& =\frac{-7 \pm \sqrt{77}}{2}
\end{aligned}
$
$\therefore x_1=\frac{-7+\sqrt{77}}{2}$ and $x_2=\frac{-7-\sqrt{77}}{2}$
Hence $x=\frac{-7+\sqrt{77}}{2}, \frac{-7-\sqrt{77}}{2}$.
View full question & answer→Question 473 Marks
Solve the following equation by using formula :
$2x^2 – 6x + 3 = 0$
Answer$
\begin{aligned}
& 2 x^2-6 x+3=0 \\
& \text { Here } a=2, b=-6, c=3 \\
& \text { then } D=b^2-4 a c \\
& =(-6)^2-4 \times 2 \times 3 \\
& =36-24 \\
& =12
\end{aligned}
$
Now
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{(-6) \pm \sqrt{12}}{2 \times 2} \\
& =\frac{6 \pm 2 \sqrt{3}}{4} \\
& \therefore x_1=\frac{6+2 \sqrt{3}}{4} \\
& =\frac{2(3+\sqrt{3})}{4} \\
& =\frac{3+\sqrt{3}}{2} \\
& x_2=\frac{6-2 \sqrt{3}}{4} \\
& =\frac{2(3-\sqrt{3})}{4} \\
& =\frac{3-\sqrt{3}}{2}
\end{aligned}
$
Hence $x=\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$.
View full question & answer→Question 483 Marks
Solve the following equation by using quadratic formula and give your answer correct to 2 decimal places : $2 x-\frac{1}{x}=1$
Answer$
\begin{aligned}
& 2 x-\frac{1}{x}=1 \\
& \Rightarrow 2 x^2-1=7 x \\
& \Rightarrow 2 x^2-7 x-1=0.....(1)
\end{aligned}
$
Comparing (i) with $a x^2+b x+c$, we get,
$
\begin{aligned}
& a=2, b =-7, c =-1 \\
& \because x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& \Rightarrow x =\frac{-(-7) \pm \sqrt{(-7)^2-4(2) \times(-1)}}{2 \times 2} \\
& \Rightarrow \frac{7 \pm \sqrt{49+8}}{4} \\
& \Rightarrow \frac{7 \pm \sqrt{57}}{4} \\
& \Rightarrow x=\frac{7+\sqrt{57}}{4} \text { or } x=\frac{7-\sqrt{57}}{4} \\
& \Rightarrow x=\frac{7+7.55}{4} \text { or } x=\frac{7-7.55}{4} \\
& \Rightarrow x=\frac{14.55}{4} \text { or } x=\frac{-0.55}{4} \\
& \Rightarrow x =3.64 \text { or } x =-0.14 .
\end{aligned}
$
View full question & answer→Question 493 Marks
Solve the following equation by using quadratic formula and give your answer correct to 2 decimal places :$ 4x^2 – 5x – 3 = 0$
AnswerGiven equation $4 x^2-5 x-3=0$
Comparing with $ax ^2+ bx + c =0$, we have
$
\begin{aligned}
& a=4, b=-5, c=-3 \\
& \because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-(-5) \pm \sqrt{(-5)^2-4 \times 4 \times(-3)}}{2 \times 4} \\
& =\frac{5 \pm \sqrt{25+48}}{8} \\
& =\frac{5 \pm \sqrt{73}}{8} \\
& =\frac{5 \pm 8.544}{8} \\
& =\frac{5+8.544}{8} \text { or } \frac{5-8.544}{8} \\
& =\frac{13.544}{8} \text { or } \frac{-3.544}{8} \\
& =1.693 \text { or }-0.443 \\
& =1.69 \text { or }-0.44 . . . \text { (correct to } 2 \text { demical places) }
\end{aligned}
$
View full question & answer→Question 503 Marks
Solve the following equation by using formula :
$2x^2 – 7x + 6 = 0$
Answer$
2 x^2-7 x+6=0
$
Here $a=2, b=-7, c=6$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(-7)^2-4 \times 2 \times 6 \\
& =49-48 \\
& =1
\end{aligned}
$
$
\begin{aligned}
& \therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(-7) \pm \sqrt{1}}{2 \times 2} \\
& =\frac{7+1}{4} \\
& \therefore x_1=\frac{7+1}{4} \\
& =\frac{8}{4} \\
& =2
\end{aligned}
$
and
$
\begin{aligned}
& x_2=\frac{7-1}{4} \\
& =\frac{6}{4} \\
& =\frac{3}{2}
\end{aligned}
$
$
\therefore x=2, \frac{3}{2} \text {. }
$
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