Solving (Simple) Problems Based on Quadratic Equations — ICSE Class 10 Maths Important Questions

Answer: (a) $11,12$11,12.

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

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

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

Answer: (a) $x^2+5x-500=0$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$100/x-100/x+5=1 100(x+5)-100x=x(x+5) 500=x^2+5x, i.e. $x^2+5x-500=0$x^2+5x-500=0.

Answer: (a) $34$34 cm.

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

Answer: (b) Taking the integers as $x$x and $x+1$x+1: $x^2+(x+1)^2=61\Rightarrow 2x^2+2x-60=0\Rightarrow x^2+x-30=0\Rightarrow(x+6)(x-5)=0$x^2+(x+1)^2=61 2x^2+2x-60=0 x^2+x-30=0(x+6)(x-5)=0, so $x=5$x=5 and the integers are $5,6$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 $x$x; the other is $x+5$x+5.

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

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

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

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

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

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

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

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

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

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

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

Let the speed be $x$x km/h.

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

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

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

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

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

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

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

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

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

Together they fill it in $3\dfrac{1}{13}=\dfrac{40}{13}$31/13=40/13 minutes, so in one minute they fill $\dfrac{13}{40}$13/40 of the cistern:

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

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

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

$x=\dfrac{41\pm\sqrt{41^2+4\cdot13\cdot120}}{2\cdot13}=\dfrac{41\pm\sqrt{7921}}{26}=\dfrac{41\pm89}{26}$x=41±√41^2+4·13·120/2·13=41±√7921/26=41±89/26.

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

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

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

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

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

$900\cdot\dfrac{x-(x-3)}{x(x-3)}=10\Rightarrow\dfrac{2700}{x^2-3x}=10$900·x-(x-3)/x(x-3)=102700/x^2-3x=10.

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

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

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

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

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