Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: } \text{The product of two consecutive integers is }56.$
$\displaystyle \text{Find the integers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two consecutive integers be }x\text{ and }(x+1).$
$\displaystyle \therefore x(x+1)=56$
$\displaystyle x^2+x-56=0$
$\displaystyle (x+8)(x-7)=0$
$\displaystyle \Rightarrow x=7\text{ or }x=-8$
$\displaystyle \text{If }x=7,\text{ the integers are }7\text{ and }8.$
$\displaystyle \text{If }x=-8,\text{ the integers are }-8\text{ and }-7.$
$\displaystyle \therefore \text{The required integers are }(7,8)\text{ or }(-8,-7).$
$\\$

$\displaystyle \textbf{Question 2: } \text{The sum of the squares of two consecutive natural numbers is }41.$
$\displaystyle \text{Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two consecutive natural numbers be }x\text{ and }(x+1).$
$\displaystyle x^2+(x+1)^2=41$
$\displaystyle x^2+x^2+2x+1=41$
$\displaystyle 2x^2+2x-40=0$
$\displaystyle x^2+x-20=0$
$\displaystyle (x+5)(x-4)=0$
$\displaystyle \Rightarrow x=-5\text{ or }x=4$
$\displaystyle \text{Since }x\text{ is a natural number, }x\neq -5.$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{The required natural numbers are }4\text{ and }5.$
$\\$

$\displaystyle \textbf{Question 3: } \text{Find the two natural numbers which differ by }5$
$\displaystyle \text{and the sum of their squares is }97.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two natural numbers be }x\text{ and }(x+5).$
$\displaystyle x^2+(x+5)^2=97$
$\displaystyle x^2+x^2+10x+25=97$
$\displaystyle 2x^2+10x-72=0$
$\displaystyle x^2+5x-36=0$
$\displaystyle (x-4)(x+9)=0$
$\displaystyle \Rightarrow x=4\text{ or }x=-9$
$\displaystyle \text{Since }x\text{ is a natural number, }x\neq -9.$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{The required natural numbers are }4\text{ and }9.$
$\\$

$\displaystyle \textbf{Question 4: } \text{The sum of a number and its reciprocal is }4.25.$
$\displaystyle \text{Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number be }x.\text{ Then its reciprocal is }\frac{1}{x}.$
$\displaystyle \therefore x+\frac{1}{x}=4.25$
$\displaystyle x+\frac{1}{x}=\frac{17}{4}$
$\displaystyle \frac{x^2+1}{x}=\frac{17}{4}$
$\displaystyle 4x^2+4=17x$
$\displaystyle 4x^2-17x+4=0$
$\displaystyle 4x^2-16x-x+4=0$
$\displaystyle 4x(x-4)-1(x-4)=0$
$\displaystyle (x-4)(4x-1)=0$
$\displaystyle \Rightarrow x=4\text{ or }x=\frac{1}{4}$
$\displaystyle \therefore \text{The required number is }4\text{ or }\frac{1}{4}.$
$\\$

$\displaystyle \textbf{Question 5: } \text{Two natural numbers differ by }3.$
$\displaystyle \text{Find the numbers, if the sum of their reciprocals is }\frac{7}{10}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two natural numbers be }x\text{ and }(x+3).$
$\displaystyle \frac{1}{x}+\frac{1}{x+3}=\frac{7}{10}$
$\displaystyle \frac{x+3+x}{x(x+3)}=\frac{7}{10}$
$\displaystyle \frac{2x+3}{x(x+3)}=\frac{7}{10}$
$\displaystyle 10(2x+3)=7x(x+3)$
$\displaystyle 20x+30=7x^2+21x$
$\displaystyle 7x^2+x-30=0$
$\displaystyle 7x^2+15x-14x-30=0$
$\displaystyle x(7x+15)-2(7x+15)=0$
$\displaystyle (x-2)(7x+15)=0$
$\displaystyle \Rightarrow x=2\text{ or }x=-\frac{15}{7}$
$\displaystyle \text{Since }x\text{ is a natural number, }x\neq -\frac{15}{7}.$
$\displaystyle \therefore x=2$
$\displaystyle \therefore \text{The required natural numbers are }2\text{ and }5.$
$\\$

$\displaystyle \textbf{Question 6: } \text{Divide }15\text{ into two parts such that the sum}$
$\displaystyle \text{of their reciprocals is }\frac{3}{10}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two parts be }x\text{ and }(15-x).$
$\displaystyle \frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}$
$\displaystyle \frac{15-x+x}{x(15-x)}=\frac{3}{10}$
$\displaystyle \frac{15}{x(15-x)}=\frac{3}{10}$
$\displaystyle 150=3x(15-x)$
$\displaystyle 50=x(15-x)$
$\displaystyle x^2-15x+50=0$
$\displaystyle (x-5)(x-10)=0$
$\displaystyle \Rightarrow x=5\text{ or }x=10$
$\displaystyle \therefore \text{The two parts are }5\text{ and }10.$
$\\$

$\displaystyle \textbf{Question 7: } \text{The sum of the squares of two positive integers is }208.$
$\displaystyle \text{If the square of the larger number is }18\text{ times the smaller number,}$
$\displaystyle \text{find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller number be }x\text{ and the larger number be }y.$
$\displaystyle \therefore x^2+y^2=208$
$\displaystyle \text{Also, }y^2=18x$
$\displaystyle \therefore x^2+18x=208$
$\displaystyle x^2+18x-208=0$
$\displaystyle (x-8)(x+26)=0$
$\displaystyle \Rightarrow x=8\text{ or }x=-26$
$\displaystyle \text{Since }x\text{ is a positive integer, }x\neq -26.$
$\displaystyle \therefore x=8$
$\displaystyle y^2=18\times8=144$
$\displaystyle \therefore y=12$
$\displaystyle \therefore \text{The required positive integers are }8\text{ and }12.$
$\\$

$\displaystyle \textbf{Question 8: } \text{The sum of the squares of two consecutive positive}$
$\displaystyle \text{even numbers is }52.\text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two consecutive positive even numbers be }x\text{ and }(x+2).$
$\displaystyle \therefore x^2+(x+2)^2=52$
$\displaystyle x^2+x^2+4x+4=52$
$\displaystyle 2x^2+4x-48=0$
$\displaystyle x^2+2x-24=0$
$\displaystyle (x+6)(x-4)=0$
$\displaystyle \Rightarrow x=-6\text{ or }x=4$
$\displaystyle \text{Since }x\text{ is a positive even number, }x\neq -6.$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{The required numbers are }4\text{ and }6.$
$\\$

$\displaystyle \textbf{Question 9: } \text{Find two consecutive positive odd numbers, the sum}$
$\displaystyle \text{of whose squares is }74.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two consecutive positive odd numbers be }(2x+1)\text{ and }(2x+3).$
$\displaystyle \therefore (2x+1)^2+(2x+3)^2=74$
$\displaystyle 4x^2+4x+1+4x^2+12x+9=74$
$\displaystyle 8x^2+16x+10=74$
$\displaystyle 8x^2+16x-64=0$
$\displaystyle x^2+2x-8=0$
$\displaystyle (x+4)(x-2)=0$
$\displaystyle \Rightarrow x=-4\text{ or }x=2$
$\displaystyle \text{Since the numbers are positive odd numbers, }x\neq -4.$
$\displaystyle \therefore x=2$
$\displaystyle 2x+1=5\text{ and }2x+3=7$
$\displaystyle \therefore \text{The required numbers are }5\text{ and }7.$
$\\$

$\displaystyle \textbf{Question 10: } \text{The denominator of a positive fraction is one more}$
$\displaystyle \text{than twice the numerator. If the sum of the fraction and its reciprocal}$
$\displaystyle \text{is }2.9,\text{ find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x.$
$\displaystyle \therefore \text{Denominator}=2x+1$
$\displaystyle \therefore \text{Fraction}=\frac{x}{2x+1}$
$\displaystyle \frac{x}{2x+1}+\frac{2x+1}{x}=2.9$
$\displaystyle \frac{x^2+(2x+1)^2}{x(2x+1)}=\frac{29}{10}$
$\displaystyle 10[x^2+(2x+1)^2]=29x(2x+1)$
$\displaystyle 10[x^2+4x^2+4x+1]=29(2x^2+x)$
$\displaystyle 50x^2+40x+10=58x^2+29x$
$\displaystyle 8x^2-11x-10=0$
$\displaystyle 8x^2-16x+5x-10=0$
$\displaystyle 8x(x-2)+5(x-2)=0$
$\displaystyle (x-2)(8x+5)=0$
$\displaystyle \Rightarrow x=2\text{ or }x=-\frac{5}{8}$
$\displaystyle \text{Since the fraction is positive, }x\neq -\frac{5}{8}.$
$\displaystyle \therefore x=2$
$\displaystyle \therefore \text{Fraction}=\frac{2}{2(2)+1}=\frac{2}{5}$
$\displaystyle \therefore \text{The required fraction is }\frac{2}{5}.$
$\\$

$\displaystyle \textbf{Question 11: } \text{Three positive numbers are in the ratio }\frac{1}{2}:\frac{1}{3}:\frac{1}{4}.$
$\displaystyle \text{Find the numbers if the sum of their squares is }244.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three positive numbers be }\frac{x}{2},\frac{x}{3}\text{ and }\frac{x}{4}.$
$\displaystyle \therefore \left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^2+\left(\frac{x}{4}\right)^2=244$
$\displaystyle \frac{x^2}{4}+\frac{x^2}{9}+\frac{x^2}{16}=244$
$\displaystyle \frac{36x^2+16x^2+9x^2}{144}=244$
$\displaystyle \frac{61x^2}{144}=244$
$\displaystyle x^2=\frac{244\times144}{61}$
$\displaystyle x^2=4\times144=576$
$\displaystyle x=24$
$\displaystyle \therefore \frac{x}{2}=12,\quad \frac{x}{3}=8,\quad \frac{x}{4}=6$
$\displaystyle \therefore \text{The required numbers are }12,8\text{ and }6.$
$\\$

$\displaystyle \textbf{Question 12: } \text{Divide }20\text{ into two parts such that three times}$
$\displaystyle \text{the square of one part exceeds the other part by }10.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two parts be }x\text{ and }(20-x).$
$\displaystyle \text{Given, }3x^2-(20-x)=10$
$\displaystyle 3x^2-20+x=10$
$\displaystyle 3x^2+x-30=0$
$\displaystyle 3x^2+10x-9x-30=0$
$\displaystyle x(3x+10)-3(3x+10)=0$
$\displaystyle (x-3)(3x+10)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=-\frac{10}{3}$
$\displaystyle \text{Since the part is positive, }x\neq -\frac{10}{3}.$
$\displaystyle \therefore x=3$
$\displaystyle \therefore 20-x=17$
$\displaystyle \therefore \text{The two parts are }3\text{ and }17.$
$\\$

$\displaystyle \textbf{Question 13: } \text{Three consecutive natural numbers are such that}$
$\displaystyle \text{the square of the middle number exceeds the difference of the squares}$
$\displaystyle \text{of the other two by }60.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive natural numbers be }x,(x+1)\text{ and }(x+2).$
$\displaystyle \therefore (x+1)^2-[(x+2)^2-x^2]=60$
$\displaystyle x^2+2x+1-[x^2+4x+4-x^2]=60$
$\displaystyle x^2+2x+1-(4x+4)=60$
$\displaystyle x^2-2x-3=60$
$\displaystyle x^2-2x-63=0$
$\displaystyle (x-9)(x+7)=0$
$\displaystyle \Rightarrow x=9\text{ or }x=-7$
$\displaystyle \text{Since }x\text{ is a natural number, }x\neq -7.$
$\displaystyle \therefore x=9$
$\displaystyle \therefore \text{The required numbers are }9,10\text{ and }11.$
$\\$

$\displaystyle \textbf{Question 14: } \text{Out of three consecutive positive integers, the middle}$
$\displaystyle \text{number is }p.\text{ If three times the square of the larger is greater}$
$\displaystyle \text{than the sum of the squares of the other two numbers by }67,$
$\displaystyle \text{calculate the value of }p.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The three consecutive positive integers are }(p-1),p\text{ and }(p+1).$
$\displaystyle \text{Given, }3(p+1)^2-[(p-1)^2+p^2]=67$
$\displaystyle 3(p^2+2p+1)-[p^2-2p+1+p^2]=67$
$\displaystyle 3p^2+6p+3-(2p^2-2p+1)=67$
$\displaystyle p^2+8p+2=67$
$\displaystyle p^2+8p-65=0$
$\displaystyle (p+13)(p-5)=0$
$\displaystyle \Rightarrow p=-13\text{ or }p=5$
$\displaystyle \text{Since }p\text{ is a positive integer, }p\neq -13.$
$\displaystyle \therefore p=5$
$\displaystyle \therefore \text{The value of }p\text{ is }5.$
$\\$

$\displaystyle \textbf{Question 15: } A\text{ can do a piece of work in }x\text{ days and }B$
$\displaystyle \text{can do the same work in }(x+16)\text{ days. If both working together}$
$\displaystyle \text{can do it in }15\text{ days, calculate }x.$
$\displaystyle \text{Answer:}$
$\displaystyle A\text{'s one day work}=\frac{1}{x}$
$\displaystyle B\text{'s one day work}=\frac{1}{x+16}$
$\displaystyle \text{Together, their one day work}=\frac{1}{15}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x+16}=\frac{1}{15}$
$\displaystyle \frac{x+16+x}{x(x+16)}=\frac{1}{15}$
$\displaystyle \frac{2x+16}{x(x+16)}=\frac{1}{15}$
$\displaystyle 15(2x+16)=x(x+16)$
$\displaystyle 30x+240=x^2+16x$
$\displaystyle x^2-14x-240=0$
$\displaystyle (x-24)(x+10)=0$
$\displaystyle \Rightarrow x=24\text{ or }x=-10$
$\displaystyle \text{Since time cannot be negative, }x\neq -10.$
$\displaystyle \therefore x=24$
$\displaystyle \therefore \text{The value of }x\text{ is }24.$
$\\$

$\displaystyle \textbf{Question 16: } \text{One pipe can fill a cistern in }3\text{ hours less than}$
$\displaystyle \text{the other. Two pipes together can fill the cistern in }6\text{ hours }40\text{ minutes.}$
$\displaystyle \text{Find the time that each pipe will take to fill the cistern.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the slower pipe fill the cistern in }x\text{ hours.}$
$\displaystyle \therefore \text{The faster pipe fills it in }(x-3)\text{ hours.}$
$\displaystyle 6\text{ hours }40\text{ minutes}=6\frac{2}{3}\text{ hours}=\frac{20}{3}\text{ hours}$
$\displaystyle \therefore \text{Together, their one hour work}=\frac{3}{20}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-3}=\frac{3}{20}$
$\displaystyle \frac{x-3+x}{x(x-3)}=\frac{3}{20}$
$\displaystyle 20(2x-3)=3x(x-3)$
$\displaystyle 40x-60=3x^2-9x$
$\displaystyle 3x^2-49x+60=0$
$\displaystyle 3x^2-45x-4x+60=0$
$\displaystyle 3x(x-15)-4(x-15)=0$
$\displaystyle (x-15)(3x-4)=0$
$\displaystyle \Rightarrow x=15\text{ or }x=\frac{4}{3}$
$\displaystyle \text{Since }x>3,\ x\neq \frac{4}{3}.$
$\displaystyle \therefore x=15$
$\displaystyle \therefore x-3=12$
$\displaystyle \therefore \text{The two pipes will take }15\text{ hours and }12\text{ hours respectively.}$
$\\$

$\displaystyle \textbf{Question 17: } \text{A positive number is divided into two parts such that}$
$\displaystyle \text{the sum of the squares of the two parts is }20.\text{ The square of the larger}$
$\displaystyle \text{part is }8\text{ times the smaller part. Taking }x\text{ as the smaller part}$
$\displaystyle \text{of the two parts, find the number.}\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller part be }x\text{ and the larger part be }y.$
$\displaystyle \therefore x^2+y^2=20$
$\displaystyle \text{Also, }y^2=8x$
$\displaystyle \therefore x^2+8x=20$
$\displaystyle x^2+8x-20=0$
$\displaystyle x^2+10x-2x-20=0$
$\displaystyle x(x+10)-2(x+10)=0$
$\displaystyle (x-2)(x+10)=0$
$\displaystyle \Rightarrow x=2\text{ or }x=-10$
$\displaystyle \text{Since }x\text{ is positive, }x\neq -10.$
$\displaystyle \therefore x=2$
$\displaystyle y^2=8\times2=16$
$\displaystyle \therefore y=4$
$\displaystyle \therefore \text{The required number is }2+4=6.$
$\\$