Solving (Simple) Problems Based on Quadratic Equations — Important Questions

Answer: (a) $11,12$.

Let the numbers be $x$ and $x+1$. Then $x(x+1)=132\Rightarrow x^2+x-132=0\Rightarrow(x+12)(x-11)=0$. So $x=11$ (rejecting $x=-12$), giving $11$ and $12$.

Answer: (a) $3$ or $\dfrac{1}{3}$.

$x+\dfrac{1}{x}=\dfrac{10}{3}\Rightarrow 3x^2-10x+3=0\Rightarrow(3x-1)(x-3)=0$, so $x=3$ or $x=\dfrac{1}{3}$; both are positive and valid.

Answer: (a) $x^2+5x-500=0$.

Time difference: $\dfrac{100}{x}-\dfrac{100}{x+5}=1\Rightarrow 100(x+5)-100x=x(x+5)\Rightarrow 500=x^2+5x$, i.e. $x^2+5x-500=0$.

Answer: (a) $34$ cm.

Let breadth $=x$; then $x(x+3)=70\Rightarrow x^2+3x-70=0\Rightarrow(x+10)(x-7)=0$, so $x=7$ cm, length $=10$ cm. Perimeter $=2(7+10)=34$ cm.

Answer: (b) Taking the integers as $x$ and $x+1$: $x^2+(x+1)^2=61\Rightarrow 2x^2+2x-60=0\Rightarrow x^2+x-30=0\Rightarrow(x+6)(x-5)=0$, so $x=5$ and the integers are $5,6$ (A true). R is a true fact used only to set up the equation, but the assertion follows from solving the quadratic, so R is not the correct explanation of A.

Let the smaller number be $x$; the other is $x+5$.

$x(x+5)=84\Rightarrow x^2+5x-84=0\Rightarrow(x+12)(x-7)=0$.

Since the numbers are natural, $x=7$. The numbers are $\mathbf{7}$ and $\mathbf{12}$.

Let the number be $x$. Then $x+x^2=90\Rightarrow x^2+x-90=0$.

$(x+10)(x-9)=0\Rightarrow x=9$ or $x=-10$.

So the number is $\mathbf{9}$ or $\mathbf{-10}$ (both satisfy the condition).

Let the father's present age be $x$ years; the son's is $(45-x)$ years.

Five years ago: father $=(x-5)$, son $=(40-x)$.

Given $(x-5)(40-x)=4(x-5)$.

$(x-5)\big[(40-x)-4\big]=0\Rightarrow(x-5)(36-x)=0$.

$x=5$ is rejected (son would be older than father), so $x=36$.

Father $=\mathbf{36}$ years, son $=45-36=\mathbf{9}$ years.

Let the speed be $x$ km/h.

$\dfrac{360}{x}-\dfrac{360}{x+5}=1\Rightarrow 360(x+5)-360x=x(x+5)$.

$1800=x^2+5x\Rightarrow x^2+5x-1800=0\Rightarrow(x+45)(x-40)=0$.

Speed cannot be negative, so $x=\mathbf{40}$ km/h.

Let the shorter side be $x$ cm; the other is $(x+7)$ cm.

By Pythagoras: $x^2+(x+7)^2=13^2$.

$2x^2+14x+49=169\Rightarrow 2x^2+14x-120=0\Rightarrow x^2+7x-60=0$.

$(x+12)(x-5)=0\Rightarrow x=5$ (rejecting $-12$).

The sides are $\mathbf{5}$ cm and $\mathbf{12}$ cm.

Let the faster pipe take $x$ minutes; the slower takes $(x+3)$ minutes.

Together they fill it in $3\dfrac{1}{13}=\dfrac{40}{13}$ minutes, so in one minute they fill $\dfrac{13}{40}$ of the cistern:

$\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}.$

$40\big[(x+3)+x\big]=13x(x+3)\Rightarrow 40(2x+3)=13x^2+39x$.

$80x+120=13x^2+39x\Rightarrow 13x^2-41x-120=0$.

$x=\dfrac{41\pm\sqrt{41^2+4\cdot13\cdot120}}{2\cdot13}=\dfrac{41\pm\sqrt{7921}}{26}=\dfrac{41\pm89}{26}$.

Taking the positive root, $x=\dfrac{130}{26}=5$.

The pipes take $\mathbf{5}$ minutes and $\mathbf{8}$ minutes.

Let the original cost of each article be ₹$x$.

Number bought $=\dfrac{900}{x}$. At ₹$(x-3)$ each, the number would be $\dfrac{900}{x-3}$, which is $10$ more:

$\frac{900}{x-3}-\frac{900}{x}=10.$

$900\cdot\dfrac{x-(x-3)}{x(x-3)}=10\Rightarrow\dfrac{2700}{x^2-3x}=10$.

$x^2-3x=270\Rightarrow x^2-3x-270=0\Rightarrow(x-18)(x+15)=0$.

Cost cannot be negative, so $x=\mathbf{18}$. Each article originally cost ₹$18$.

(i) Return time exceeds onward time by $1$ hour:
$\frac{300}{x-10}-\frac{300}{x}=1.$
$300\big[x-(x-10)\big]=x(x-10)\Rightarrow 3000=x^2-10x$, giving
$x^2-10x-3000=0.$

(ii) $x^2-10x-3000=0\Rightarrow(x-60)(x+50)=0$. Rejecting $x=-50$, the original speed is $\mathbf{60}$ km/h.

(iii) Return speed $=60-10=50$ km/h, so return time $=\dfrac{300}{50}=\mathbf{6}$ hours.