# Represent the following situations in the form of quadratic equations:(i) The area of a rectangular plot is $528\ m^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.(ii) The product of two consecutive positive integers is 306. We need to find the integers.(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

To do:

We have to represent the given situations in the form of quadratic equations.

Solution:

(i) The area of a rectangular plot$=528\ m^2$.

The length of the plot (in meters) is one meter more than twice its breadth.

Let the breadth of the plot be $x\ m$.

This implies,

Length of the plot$=2x+1\ m$.

We know that,

Area of a rectangle of length $l$ and breadth $b$ is $lb$.

Therefore,

Area of the rectangular plot$=(x)(2x+1)\ m^2$.

According to the question,

$x(2x+1)=528$   (From equation 1)

$2x^2+x=528$

$2x^2+x-528=0$

Solving for $x$ by factorization method, we get,

$2x^2+33x-32x-528=0$

$2x(x-32)+33(x-32)=0$

$(2x+33)(x-32)=0$

$2x+33=0$ or $x-32=0$

$2x=-33$ or $x=32$

Length cannot be negative. Therefore, the value of $x=32$.

$2x+1=2(32)+1=64+1=65\ m$

The breadth of the plot is $32\ m$ and the length of the plot is $65\ m$.

(ii) The product of two consecutive positive integers is $306$.

Let the two consecutive integers be $x$ and $x+1$, where $x$ is the smaller integer.

Therefore,

$x(x + 1) = 306$

$x^2 + x – 306 = 0$

$x^2+18x-17x-306=0$

$x(x+18)-17(x+18)=0$

$(x+18)(x-17)=0$

$x+18=0$ or $x-17=0$

$x=17$ or $x=-18$

This implies,

If $x=17$ then $x+1=17+1=18$

If $x=-18$ then $x+1=-18+1=-17$

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360.

Let the present age of Rohan be $x$.

This implies,

The age of the mother $= x+26$

Age of the Rohan after 3 years $= x+3$

Age of the mother after 3 years $=(x+26)+3=x+29$.

Therefore,

$(x+3)(x+29) = 360$

$x(x+29)+3(x+29) = 360$

$x^2+29x+3x+87 = 360$

$x^2+32x+87-360 = 0$

$x^2+39x-7x-273=0$

$x(x+39)-7(x+39)=0$

$(x+39)(x-7)=0$

$x+39=0$ or $x-7=0$

$x=7$ or $x=-39$ which is not possible

The present age of Rohan is $7$ years.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance.

Let the original speed of the train be $x$ km/hr.

This implies,

Time taken by the train to travel 480 km at original speed$=\frac{480}{x}$ hours

Time taken by the train to travel 480 km when the speed is 8 km/hr less than the original speed$=\frac{480}{x-8}$ hours

According to the question,

$\frac{480}{x-8}-\frac{480}{x}=3$

$480(\frac{(x)-(x-8)}{(x)(x-8)})=3$

$480(\frac{8}{x^2-8x})=3$

$(480)(8)=3(x^2-8x)$   (On cross multiplication)

$160(8)=x^2-8x$

$x^2-8x-1280=0$

Solving for $x$ by factorization method, we get,

$x^2-40x+32x-1280=0$

$x(x-40)+32(x-40)=0$

$(x-40)(x+32)=0$

$x-40=0$ or $x+32=0$

$x=40$ or $x=-32$

Speed cannot be negative. Therefore, the value of $x$ is $40$ km/hr.

The original speed of the train is $40$ km/hr.

Updated on: 10-Oct-2022

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