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# 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.

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