# CBSE 10th Mathematics | Applications of Quadratic Equations

### Solution:

Let the required numbers be x and 15 – x. Then,

According to question, we have

$\displaystyle \frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}$

Þ    $\displaystyle \frac{15-x+x}{x(15-x)}=\frac{3}{10}$

Þ    $\displaystyle \frac{15}{x(15-x)}=\frac{3}{10}$

Þ    $\displaystyle 150=3x(15-x)$

Þ    150 = $\displaystyle 45x-3{{x}^{2}}$

Þ    $\displaystyle 3{{x}^{2}}-45x+150=0$

or    $\displaystyle {{x}^{2}}-15x+50=0$

Þ    $\displaystyle {{x}^{2}}-10x-5x+50=0$

Þ    $\displaystyle x(x-10)-5(x-10)=0$

Þ    $\displaystyle (x-10)(x-5)=0$

Þ    $\displaystyle x-10=0$ or $\displaystyle x-5=0$

Þ    x = 10 or, x = 5

Hence, the two numbers are 10 and 5.

### Solution:

Let the tens digit be x then the units digit $\displaystyle =\frac{14}{x}$    [$\displaystyle \because$ product of two digit is 14]

Then, the two digit number is $\displaystyle \left( 10x+\frac{14}{x} \right)$.

Reverse number = $\displaystyle \left( \frac{140}{x}+x \right)$

According to question, we have

$\displaystyle 10x+\frac{14}{x}+45=\frac{140}{x}+x$

$\displaystyle 10{{x}^{2}}+14+45x=140+{{x}^{2}}$

$\displaystyle 10{{x}^{2}}-{{x}^{2}}+45x+14-140=0$

$\displaystyle 9{{x}^{2}}+45x-126=0$

or     $\displaystyle {{x}^{2}}+5x-14=0$

$\displaystyle {{x}^{2}}+7x-2x-14=0$

$\displaystyle x(x+7)-2(x+7)=0$

$\displaystyle (x+7)(x-2)=0$

$\displaystyle x=2$;    $\displaystyle x=-7$ is rejected as digit can not be negative

\    $\displaystyle x=2$ and other digit $\displaystyle =\frac{14}{2}=7$

Hence, the required number is 10 × 2 + 7 i.e. 27.

### Solution:

Let the tens digit be x and the units digit be y.

Since, the two digit number is 4 times the sum of its digits, we have

$\displaystyle 10x+y=4(x+y)$

Þ    $\displaystyle 10x+y=4x+4y$

Þ    $\displaystyle 6x=3y$

Þ    $\displaystyle x=\frac{y}{2}$                    …(i)

Since, the two-digit number is twice the product of its digits, we have

$\displaystyle (10x+y)=2(xy)$

Þ    $\displaystyle 10x+y-2xy=0$            …(ii)

Substituting $\displaystyle x$= $\displaystyle \frac{y}{2}$ from equation (i) in equation (ii), we have

$\displaystyle 10\times \frac{y}{2}+y-2\times \frac{y}{2}\times y=0$

Þ    $\displaystyle -{{y}^{2}}+5y+y=0$

or    $\displaystyle {{y}^{2}}-6y=$0     Þ    $\displaystyle y(y-6)=0$

Þ    Either y = 0 [rejected]        or    $\displaystyle y-6=0$

Þ    $\displaystyle y=6$

\    $\displaystyle x=\frac{6}{2}=3$        [from (i)]

The required number is (10 × 3 + 6) = 36

### Solution:

Let P be the required location of the pole. Let the distance of the pole from the gate R be x m, i.e. RP = x m. Now the difference of the distances of the pole from the two gates = QP – RP (or, RP – QP) = 7 m. Therefore, QP = (x + 7) m.

Now, QR = 13m, and since QR is a diameter.

$\displaystyle \angle QPR=90{}^\circ$        (Angle in a semicircle is right angle)

Therefore,    $\displaystyle Q{{P}^{2}}+P{{R}^{2}}=Q{{R}^{2}}$    (By Pythagoras theorem)

i.e.,         $\displaystyle {{(x+7)}^{2}}+{{x}^{2}}={{13}^{2}}$

i.e.,        $\displaystyle {{x}^{2}}+14x+49+{{x}^{2}}=169$

i.e.,        $\displaystyle 2{{x}^{2}}+14x-120=0$

So, the distance x of the pole from gate R satisfies the equation

$\displaystyle {{x}^{2}}+7x-60=0$

We will find its discriminant and find whether it would be possible

$\displaystyle {{b}^{2}}-4ac={{7}^{2}}-4\times 1\times (-60)=289$> 0

So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.

Solving the quadratic equation$\displaystyle {{x}^{2}}+7x-60=0$, by quadratic formula, we have

$\displaystyle x=\frac{-b\pm \sqrt{D}}{2a}$

$\displaystyle x=\frac{-7\pm \sqrt{289}}{2}=\frac{-7\pm 17}{2}$

Therefore, x = 5 or –12.

Since x is the distance between the pole and the gate R, it must be positive. Therefore, x = –12 is rejected. So, x = 5.

Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate R and 12 m from the gate Q.

### Solution:

Let the number of pottery articles, produced on a particular day be x.

Cost of production of each article = Rs.(2x + 3)

\    Total cost of production of x articles = Rs.(2x + 3) × x    … (i)

It is given that the total cost of production = Rs.90        …(ii)

\    $\displaystyle (2x+3)x=90$

Þ    $\displaystyle 2{{x}^{2}}+3x-90=0$

Þ    $\displaystyle 2{{x}^{2}}-12x+15x-90=0$

Þ    $\displaystyle 2x(x-6)+15(x-6)=0$

Þ    $\displaystyle (x-6)(2x+15)=0$

Þ    x = 6     or     $\displaystyle x=-\frac{15}{2}$ (rejected as the cost cannot be negative)

\    $\displaystyle x=6$

Hence number of articles produced in a day = 6

Cost of production of each article         = Rs. (2 ´ 6 + 3)

= Rs. 15

### Solution:

According to question, we have

$\displaystyle {{(r+2)}^{2}}+{{(r+9)}^{2}}={{17}^{2}}$    [By Pythagoras theorem]

Þ    $\displaystyle {{r}^{2}}+4r+4+{{r}^{2}}+18r+81=289$

Þ    $\displaystyle 2{{r}^{2}}+22r-204=0$

or    $\displaystyle {{r}^{2}}+11r-102=0$        [Required equation]

Þ    $\displaystyle {{r}^{2}}+17r-6r-102=0$

Þ    $\displaystyle r(r+17)-6(r+17)=0$

Þ    $\displaystyle (r+17)(r-6)=0$

Þ    $\displaystyle r=-17$ or r = 6 cm

As r can not be negative \ r = 17 is rejected

Hence r = 6 cm

### Solution:

Let length of one side of rectangle = (x) m

Total length of wire = 30 m

Then, the length of adjacent side = (30 – 2x) m

\    Area = x (30 – 2x) = 30x – 2×2

By the given condition,

$\displaystyle 30x-2{{x}^{2}}=100$

$\displaystyle 2{{x}^{2}}-30x+100=0$

or    $\displaystyle {{x}^{2}}-15x+50=0$

Þ    $\displaystyle {{x}^{2}}-10x-5x+50=0$

Þ    $\displaystyle x(x-10)-5(x-10)=0$

Þ    $\displaystyle (x-5)(x-10)=0$

Þ    $\displaystyle x-5=0$

or    $\displaystyle x-10=0$

Þ    $\displaystyle x=5$

or    x = 10

Case 1: If one side = 10 m, then 2nd side = 30 – 2 × 10 = 30 – 20 = 10 m

Case 2: If one side = 5 m, then 2nd side = 30 – 2 × 5 = 30 – 10 = 20 m

So, the dimensions of garden are either 10 m × 10 m or 20 m × 5 m.

### Solution:

Let the length of the piece be x metres.

 Cost of the cloth Rs.200 Rs.200 Length of the piece x metres (x + 5) metres Rate per metre Rs.$\displaystyle \frac{200}{x}$ Rs.$\displaystyle \frac{200}{x+5}$

If the piece is 5 m longer then rate per metre is Rs.2 less.

\    $\displaystyle \frac{200}{x}-\frac{200}{x+5}=2$

Þ    $\displaystyle 200\left[ \frac{x+5-x}{x(x+5)} \right]=2$

Þ    $\displaystyle 200\left[ \frac{5}{{{x}^{2}}+5x} \right]=2$

Þ    $\displaystyle {{x}^{2}}+5x=500$

Þ    $\displaystyle {{x}^{2}}+5x-500=0$

Þ    $\displaystyle {{x}^{2}}+25x-20x-500=0$

Þ    $\displaystyle x(x+25)-20(x+25)=0$

Þ    $\displaystyle (x+25)(x-20)=0$

Þ    $\displaystyle x+25=0$    or    $\displaystyle x-20=0$

Þ    $\displaystyle x=-25$ (rejected, as length cannot be negative) or    x = 20

Rate per metre = Rs. $\displaystyle \frac{200}{x}=Rs.\,\frac{200}{20}=$Rs.10

Hence, the length of the piece of cloth is 20 metres and the rate per metre is Rs.10

### Solution:

Let x be the smaller number. Then the square of the larger number will be 18x. Since the sum of the squares of these two positive integers is 208. We have

$\displaystyle {{x}^{2}}+18x=208$

Þ    $\displaystyle {{x}^{2}}+26x-8x-208=0$

Þ    $\displaystyle x(x+26)-8(x+26)=0$

Þ    $\displaystyle (x-8)(x+26)$ = 0

Þ    $\displaystyle x-8=0$    or     $\displaystyle x+26=0$

Þ    $\displaystyle x=8$ or –26

Since the numbers are positive, x = –26 is rejected

\    x = 8 and square of the larger positive integer = 18 × 8 = 144

\    Larger number = $\displaystyle \sqrt{144}=12$

Hence, the larger number is 12 and the smaller number is 8.

### Solution:

Let the numerator of the fraction be x.

Then, the denominator = 2x + 1. Fraction = $\displaystyle \frac{x}{2x+1}$

According to question, we have

$\displaystyle \frac{x}{2x+1}+\frac{2x+1}{x}=\frac{29}{10}$

Þ    $\displaystyle \frac{{{x}^{2}}+{{(2x+1)}^{2}}}{x(2x+1)}=\frac{29}{10}$

Þ    $\displaystyle \frac{{{x}^{2}}+4{{x}^{2}}+4x+1}{2{{x}^{2}}+x}=\frac{29}{10}$

Þ    $\displaystyle \frac{5{{x}^{2}}+4x+1}{2{{x}^{2}}+x}=\frac{29}{10}$

Þ    $\displaystyle 50{{x}^{2}}+40x+10=58{{x}^{2}}+29x$

Þ    $\displaystyle 8{{x}^{2}}-11x-10=0$

Þ    $\displaystyle x=\frac{11\pm \sqrt{121+320}}{16}=\frac{11\pm \sqrt{441}}{16}=\frac{11\pm 21}{6}$

Þ    $\displaystyle x=\frac{32}{16}$ or $\displaystyle -\frac{10}{16}$

Þ    $\displaystyle x=2$ or $\displaystyle \frac{-5}{8}$        [Rejected]

Denominator = 2x + 1 = 2(2) + 1 = 5

Hence, the required fraction is $\displaystyle \frac{2}{5}$.

### Solution:

Let the total number of swans be x. Then,

Number of swans playing on the shore of the tank = $\displaystyle \frac{7}{2}\sqrt{x}$

It is given that there are two remaining swans.

\    $\displaystyle x=\frac{7}{2}\sqrt{x}+2$

Þ    $\displaystyle x-\frac{7}{2}\sqrt{x}-2=0$

Put $\displaystyle \sqrt{x}=y\,\ \ \ \ \Rightarrow \ \ \ \ x={{y}^{2}}$

Þ    $\displaystyle {{y}^{2}}-\frac{7}{2}y-2=0$

Þ    $\displaystyle 2{{y}^{2}}-7y-4=0$

Þ    $\displaystyle 2{{y}^{2}}-8y+y-4=0$

Þ    $\displaystyle 2y(y-4)+1(y-4)=0$

Þ    $\displaystyle (y-4)(2y+1)=0$

Þ    $\displaystyle y=4$ or, $\displaystyle y=-\frac{1}{2}$

Þ    $\displaystyle y=4$                [Q y = –1/2 is not possible]

Þ    $\displaystyle x={{y}^{2}}={{4}^{2}}=1$6

Hence, the total number of swans is 16.

### Solution:

Let the usual speed of the plane be x km/hr. Then,

Time taken to cover 1500 km with the usual speed = $\displaystyle \frac{1500}{x}$hrs.

Time taken to cover 1500 km with the increased speed of

$\displaystyle (x+250)$km/hr = $\displaystyle \frac{1500}{x+250}$

According to question

\    $\displaystyle \frac{1500}{x}=\frac{1500}{x+250}+\frac{1}{2}$        [30 minutes = $\displaystyle \frac{30}{60}$ hr = $\displaystyle \frac{1}{2}$ hr]

Þ    $\displaystyle \frac{1500}{x}-\frac{1500}{x+250}=\frac{1}{2}$

Þ    $\displaystyle \frac{1500x+1500\times 250-1500x}{x(x+250)}=\frac{1}{2}$

Þ    $\displaystyle \frac{1500\times 250}{{{x}^{2}}+250x}=\frac{1}{2}$

Þ    $\displaystyle 750000={{x}^{2}}+250x$

Þ    $\displaystyle {{x}^{2}}+250x-750000=0$

Þ    $\displaystyle {{x}^{2}}+1000x-750x-750000=0$

Þ    $\displaystyle x(x+1000)-750(x+1000)=0$

Þ    $\displaystyle (x+1000)(x-750)=0$

Þ    x = – 1000     or     $\displaystyle x=750$

Þ    $\displaystyle x=750$            [Q speed cannot be negative]

Hence, the usual speed of the plane is 750 km/hr.

### Solution:

Let the speed of the second train be x km/hr. Then, the speed of the first train is $\displaystyle (x+5)$km/hr.

Let O be the position of the railway station from which the two trains leave.

Distance traveled by the first train in 2 hours = OA = Speed × Time

= 2(x + 5) km

Distance traveled by the second train in 2 hours = OB = 2x km

In DOAB by using Pythagoras theorem, we have

$\displaystyle A{{B}^{2}}=O{{A}^{2}}+O{{B}^{2}}$

Þ    $\displaystyle {{50}^{2}}={{\{2(x+5)\}}^{2}}+{{(2x)}^{2}}$

Þ    2500 = $\displaystyle 4{{(x+5)}^{2}}+4{{x}^{2}}$

Þ    $\displaystyle 2500=4({{x}^{2}}+25+10x)+4{{x}^{2}}$

Þ    $\displaystyle 8{{x}^{2}}+40x-2400=0$

or    $\displaystyle {{x}^{2}}+5x-300=$0

Þ    $\displaystyle {{x}^{2}}+20x-15x-300=0$

Þ    $\displaystyle x(x+20)-15(x+20)=0$

Þ    $\displaystyle (x+20)(x-15)=0$

Þ    $\displaystyle x=-20$ or, x = 15

Þ     x = 15

Hence, the speed of the second train is 15 km/hr and, the speed of the first train is 20km/hr.

### Solution:

Let the smaller side of the right triangle be x cm and the larger side by y cm. Then,

$\displaystyle {{x}^{2}}+{{y}^{2}}={{(3\sqrt{5})}^{2}}$        [Using Pythagoras Theorem]

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}=45$                …(i)

If the smaller side is tripled and the larger side be doubled, the new hypotenuse is 15 cm.

\    $\displaystyle {{(3x)}^{2}}+{{(2y)}^{2}}={{15}^{2}}$

Þ    $\displaystyle 9{{x}^{2}}+4{{y}^{2}}=225$

From equation (i), we get $\displaystyle {{y}^{2}}=45-{{x}^{2}}$

Putting $\displaystyle {{y}^{2}}=45-{{x}^{2}}$ in equation (ii), we get

$\displaystyle 9{{x}^{2}}+4(45-{{x}^{2}})=225$

Þ    $\displaystyle 5{{x}^{2}}+180=225$

Þ    $\displaystyle 5{{x}^{2}}=45$    Þ    $\displaystyle {{x}^{2}}=9$         Þ    x = ±3

But, length of a side cannot be negative. Therefore, x= 3.

Putting x = 3 in (i), we get

$\displaystyle 9+{{y}^{2}}=45$    Þ    $\displaystyle {{y}^{2}}=36$    Þ    y = 6

Hence, the length of the smaller side is 3 cm and the length of the larger side is 6cm.

### Solution:

Suppose B alone takes x days to finish the work. Then, A alone can finish it in (x–6) days.

In x days B finishes = 1 work

\ In one day B will finish $\displaystyle =\frac{1}{x}$ work

In (x 6) days A finishes = 1 work

\ In one day A will finish $\displaystyle =\frac{1}{x-6}$ work

\    (A¢s one day’s work) + (B¢s one day’s work) = $\displaystyle \frac{1}{x}+\frac{1}{x-6}$

According to question $\displaystyle (A+B)\prime s$ one day’s work = $\displaystyle \frac{1}{4}$ \ we have

Þ    $\displaystyle \frac{1}{x}+\frac{1}{x-6}=\frac{1}{4}$

Þ    $\displaystyle \frac{x-6+x}{x(x-6)}=\frac{1}{4}$         Þ    $\displaystyle \frac{2x-6}{{{x}^{2}}-6x}=\frac{1}{4}$

Þ     $\displaystyle 8x-24={{x}^{2}}-6x$

Þ    $\displaystyle {{x}^{2}}-14x+24=0$

Þ    $\displaystyle {{x}^{2}}-12x-2x+24=0$

Þ    $\displaystyle (x-12)(x-2)=0$

Þ    x = 12 or, x = 2

But, x cannot be less than 6. So, x = 12.

Hence, B along can finish the work in 12 days.

### Solution:

Let x be the number of hours required by the second pipe along to fill the pool. Then, the first pipe takes (x + 5) hours. While the third pipe takes (x – 4) hours to fill the pool.

In $\displaystyle (x+5)$ hours first pipe fills = 1 tank

In 1 hour first pipe will fill = $\displaystyle \frac{1}{x+5}$ tank

Similarly in 1 hour second pipe will fill = $\displaystyle \frac{1}{x}$ tank

In 1 hour third pipe will fill = $\displaystyle \frac{1}{x-4}$ tank

According to question, we have

Þ    $\displaystyle \frac{1}{x+5}+\frac{1}{x}=\frac{1}{x-4}$     Þ    $\displaystyle \frac{x+x+5}{{{x}^{2}}+5x}=\frac{1}{x-4}$

Þ    $\displaystyle (2x+5)(x-4)={{x}^{2}}+5x$

Þ    $\displaystyle {{x}^{2}}-8x-20=0$

Þ    $\displaystyle {{x}^{2}}-10x+2x-20=0$

Þ    $\displaystyle x(x-10)+2(x-10)$ = 0

Þ    $\displaystyle (x-10)(x+2)=0$

Þ    x = 10 or, x = –2

But, x cannot be negative. So, x = 10.

Hence, the timings required by first, second and third pipes to fill the pool individually are 15 hours, 10 hours and 6 hours respectively.

### Solution:

Let the width of the gravel path be x metres. Then,

Each side of the square flower bed (PQRS) = (44 – 2x) metres.

Now, area of the square field (ABCD)

= 44 × 44 = 1936 m2

Area of the flower bed = $\displaystyle {{(44-2x)}^{2}}$m2

\    Area of the gravel path

= Area of the field (ABCD) – Area of the flower bed (PQRS)

= 1936 – (44 – 2x)2

= 1936 – (1936 – 176x + 4×2)

= (176x – 4×2) m2

Cost of laying the flower bed = (Area of the flower bed) (Rate per sq. m)

= $\displaystyle {{(44-2x)}^{2}}\times \frac{275}{100}=\frac{11}{4}{{(44-2x)}^{2}}=11\,{{(22-x)}^{2}}$

Cost of gravelling the path     = Area of the path × Rate per sq. m

= $\displaystyle (176x-4{{x}^{2}})\frac{150}{100}=6(44x-{{x}^{2}})$

It is given that the total cost of laying the flower bed and gravelling the path is Rs.4904.

\    $\displaystyle 11{{(22-x)}^{2}}+6(44x-{{x}^{2}})=4904$

Þ    $\displaystyle 11(484-44x+{{x}^{2}})+(264x-6{{x}^{2}})=4904$

Þ    $\displaystyle 5{{x}^{2}}-220x+5324=4904$

Þ    $\displaystyle 5{{x}^{2}}-220x+420=0$

or     $\displaystyle {{x}^{2}}-44x+84=0$

Þ    $\displaystyle {{x}^{2}}-42x-2x+84=0$

Þ    $\displaystyle x(x-42)-2(x-42)=0$

Þ    $\displaystyle (x-2)(x-42)=0$

Þ    x = 2 or, x = 42

But, x ¹ 42, as the side of the square is 44 m. Therefore, x = 2.

Hence, the width of the gravel path is 2 metres.