Word Problems PYQs
Overview
This page provides comprehensive Class 10 Maths Word Problems PYQs | Quadratic Equations. Word Problems previous year questions for Class 10 Maths Quadratic Equations. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q1
2021–22
00:00
If the sum of the roots of the quadratic equation $ky^2-11y+(k-23)=0$ is $\dfrac{13}{21}$ more than the product of the roots, find the value of $k$.
For $ay^2+by+c=0$: Sum of roots $= -\dfrac{b}{a}$, Product $= \dfrac{c}{a}$
Here, $a=k,\; b=-11,\; c=k-23$
Sum $= \dfrac{11}{k}$, Product $= \dfrac{k-23}{k}$
Given: $\frac{11}{k}=\frac{k-23}{k}+\frac{13}{21}$
Solving: $\frac{11-(k-23)}{k}=\frac{13}{21}$
$\frac{34-k}{k}=\frac{13}{21}$
$21(34-k)=13k \Rightarrow 714=34k$
$k=21$
Q2
Delhi 2016
00:00
If $x=\dfrac23$ and $x=-3$ are the roots of $ax^2+7x+b=0$, find $a$ and $b$.
Sum of roots: $\frac23-3=-\frac73$
$-\frac{7}{a}=-\frac73 \Rightarrow a=3$
Product of roots: $\frac23\times(-3)=-2$
$\frac{b}{a}=-2 \Rightarrow b=-6$
$a=3,\; b=-6$
Q3
Foreign 2016
00:00
A two-digit number is four times the sum of its digits and three times the product of its digits. Find the number.
Let tens digit = $x$, units digit = $y$. Number $=10x+y$
$10x+y=4(x+y) \Rightarrow 6x=3y \Rightarrow y=2x$
$10x+y=3xy \Rightarrow 10x+2x=6x^2$
$6x^2-12x=0 \Rightarrow x=2$
If $x=2$, then $y=4$
The number is 24
Q4
2023
00:00
Two taps together fill a tank in $3\frac13$ hours. One tap takes 5 hours less than the other. Find the time taken by each tap.
Let smaller tap take $x$ hours, larger tap $(x-5)$ hours
$\frac1x+\frac1{x-5}=\frac3{10}$
$\frac{2x-5}{x(x-5)}=\frac3{10}$
$3x^2-35x+50=0$
$(3x-5)(x-10)=0$
Reject $x=5/3$ as $x-5$ would be negative.
Times are 10 hours and 5 hours
Q5
2018
00:00
A plane left 30 minutes late and to cover 1500 km in time, increased its speed by 100 km/h. Find the usual speed.
Let usual speed = $x$ km/h
$\frac{1500}{x}-\frac{1500}{x+100}=\frac12$
$300000=x(x+100)$
$x^2+100x-300000=0$
$(x+600)(x-500)=0$
Usual speed = 500 km/h
Q6
AI 2016
00:00
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of squares of the other two by 60. Find the numbers.
Let numbers be $(x-1),x,(x+1)$
$x^2-\{(x+1)^2-(x-1)^2\}=60$
$x^2-4x=60$
$x^2-4x-60=0$
$(x-10)(x+6)=0 \Rightarrow x=10$
The numbers are 9, 10, 11
Q7
2025
00:00
Rectangular lawn $12$m × $10$m is surrounded by a uniform walkway. Total area is $360$m². Find the width of the walkway.
Let width = $x$ m
$(12+2x)(10+2x)=360$
$4x^2+44x+120=360$
$4x^2+44x-240=0$
$x^2+11x-60=0 \Rightarrow (x+15)(x-4)=0$
Width of walkway = 4 m
Q8
2021–22
00:00
A rectangular pool of area $36$m² has a sidewalk of width $x$. Outside dimensions are $12$m × $7$m. Find $x$.
Inside dimensions $(12-2x)(7-2x)=36$
$84-24x-14x+4x^2=36$
$4x^2-38x+48=0$
$2x^2-19x+24=0 \Rightarrow (2x-3)(x-8)=0$
Reject $x=8$ (too large)
Width of sidewalk = 1.5 m
Q9
2025
00:00
A student scored 32 marks in Maths and Science. If he had scored 4 more marks in Mathematics and 2 less in Science, the product of the marks would have been 253. Find his marks in the two subjects.
Let Maths = $x$, Science = $32-x$
$(x+4)(30-x)=253$
$30x-x^2+120-4x=253$
$x^2-26x+133=0$
$(x-19)(x-7)=0$
Marks are (19, 13) or (7, 25)
Q10
2025
00:00
Circular park of diameter 65 m. Point P on boundary such that distance from A is 35 m more than from B (where AB is diameter). Find distances.
Let distance from B = $x$, from A = $x+35$
Angle in semicircle is $90^\circ$. By Pythagoras: $x^2+(x+35)^2=65^2$
$2x^2+70x+1225=4225$
$2x^2+70x-3000=0 \Rightarrow x^2+35x-1500=0$
$(x+60)(x-25)=0 \Rightarrow x=25$
Distances are 25 m and 60 m
Q11
2024
00:00
If Nidhi were 7 years younger than what she actually is, then the square of her age would be 1 more than 5 times her actual age. Find her present age.
Let Nidhi’s present age be $x$ years.
$(x-7)^2 = 5x + 1$
$x^2 - 14x + 49 = 5x + 1$
$x^2 - 19x + 48 = 0$
$(x-16)(x-3)=0$. Reject $x=3$ (as $x-7$ would be negative).
Nidhi’s present age is 16 years
Q12
2024
00:00
A shopkeeper buys books for ₹1800. If he had bought 15 more books for the same amount, each book would have cost ₹20 less. Find how many books he bought initially.
Let number of books = $x$. Cost = $1800/x$
$\frac{1800}{x} - \frac{1800}{x+15} = 20$
$1800(x+15-x) = 20x(x+15)$
$27000 = 20(x^2+15x)$
$x^2 + 15x - 1350 = 0 \Rightarrow (x+45)(x-30)=0$
The shopkeeper bought 30 books initially
Q13
2024
00:00
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its speed was reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.
Let original speed = $x$ km/h
$\frac{2800}{x-100}-\frac{2800}{x}=\frac12$
$2800(100) = \frac12 x(x-100)$
$x^2-100x-560000=0$
$(x-800)(x+700)=0 \Rightarrow x=800$
Original time = $2800/800$
Original duration of flight = 3.5 hours
Q14
2024
00:00
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$, find the fraction.
Let numerator = $x$, Denominator = $2x+1$
$\frac{x}{2x+1}+\frac{2x+1}{x}=\frac{58}{21}$
$21(x^2 + (2x+1)^2) = 58x(2x+1)$
$21(5x^2+4x+1) = 116x^2+58x$
$11x^2-26x-21=0 \Rightarrow (11x+7)(x-3)=0$
The fraction is $\frac{3}{7}$
Q15
2024
00:00
The age of a man is twice the square of the age of his son. Eight years hence, man’s age will be 4 years more than three times the age of his son. Find their present ages.
Son = $x$, Man = $2x^2$
$2x^2+8=3(x+8)+4$
$2x^2+8=3x+28$
$2x^2-3x-20=0$
$(2x+5)(x-4)=0 \Rightarrow x=4$
Son = 4 years, Man = 32 years
Q16
2023
00:00
A train travels 54 km at a certain speed and 63 km at a speed 6 km/h more. Total time is 3 hours. Find the first speed.
Let first speed = $x$ km/h
$\frac{54}{x}+\frac{63}{x+6}=3$
$18(x+6)+21x=x(x+6)$
$39x+108=x^2+6x$
$x^2-33x-108=0 \Rightarrow (x-36)(x+3)=0$
First speed = 36 km/h
Q17
2021–22
00:00
The sum of two numbers is 34. If 3 is subtracted from one and 2 is added to the other, their product becomes 260. Find the numbers.
Numbers: $x$ and $34-x$
$(x-3)(34-x+2)=260$
$(x-3)(36-x)=260$
$36x-x^2-108+3x=260$
$x^2-39x+368=0 \Rightarrow (x-16)(x-23)=0$
The numbers are 16 and 18
Q18
2021–22
00:00
The hypotenuse of a right-angled triangle is 6 cm more than twice the length of the shortest side. If the third side is 6 cm less than three times the shortest side, find the dimensions of the triangle.
Shortest = $x$. Hypotenuse = $2x+6$. Third = $3x-6$
$x^2+(3x-6)^2=(2x+6)^2$
$x^2+9x^2-36x+36=4x^2+24x+36$
$6x^2-60x=0 \Rightarrow 6x(x-10)=0$
Sides are 10 cm, 24 cm and 26 cm
Q19
2021–22
00:00
A two-digit number has product of its digits 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.
$xy=24$. Number $10x+y$
$10x+y-18=10y+x \Rightarrow 9x-9y=18 \Rightarrow x-y=2$
$x=y+2$. So $(y+2)y=24$
$y^2+2y-24=0 \Rightarrow (y+6)(y-4)=0$
$y=4, x=6$
The number is 64
Q20
2020
00:00
Sum of areas of two squares is $544 \text{ m}^2$. If difference of their perimeters is 32 m, find their sides.
$x^2+y^2=544$
$4x-4y=32 \Rightarrow x-y=8 \Rightarrow x=y+8$
$(y+8)^2+y^2=544$
$2y^2+16y+64=544$
$y^2+8y-240=0 \Rightarrow (y+20)(y-12)=0$
Sides are 20 m and 12 m
Q21
NCERT
00:00
A motorboat moves at 18 km/h in still water. It takes 1 hour more to go 24 km upstream than downstream. Find speed of the stream.
Let speed of stream = $x$
$\frac{24}{18-x}-\frac{24}{18+x}=1$
$24(2x) = 324-x^2$
$x^2+48x-324=0$
$(x+54)(x-6)=0$
Speed of stream = 6 km/h
Q22
2019
00:00
Two water taps together can fill a tank in $1\frac78$ hours. One tap takes 2 hours less than the other. Find their individual times.
Time = $15/8$ hrs. Let smaller tap = $x$, larger = $x-2$
$\frac1x+\frac1{x-2}=\frac8{15}$
$\frac{2x-2}{x(x-2)}=\frac8{15}$
$15(2x-2)=8(x^2-2x)$
$8x^2-46x+30=0 \Rightarrow 4x^2-23x+15=0$
$(4x-3)(x-5)=0$. Reject $3/4$ (too small).
Times are 5 hours and 3 hours
Q23
2019
00:00
A train travels 360 km at uniform speed. If speed were 5 km/h more, it would take 1 hour less. Find the speed.
$\frac{360}{x}-\frac{360}{x+5}=1$
$360(5)=x(x+5)$
$x^2+5x-1800=0$
$(x+45)(x-40)=0$
Speed of train = 40 km/h
Q24
2018
00:00
A train travels 63 km at speed $x$ and 72 km at speed $(x+6)$. Total time is 3 hours. Find original speed.
$\frac{63}{x}+\frac{72}{x+6}=3$
$21(x+6)+24x=x(x+6)$
$45x+126=x^2+6x$
$x^2-39x-126=0$
$(x-42)(x+3)=0$
Original speed = 42 km/h
Q25
Delhi 2017
00:00
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns downstream to the same point in 4 hours 30 minutes. Find the speed of the stream.
$\frac{30}{15-x} + \frac{30}{15+x} = \frac{9}{2}$
$30(\frac{30}{225-x^2}) = \frac92$
$900 = \frac92(225-x^2)$
$200 = 225-x^2 \Rightarrow x^2=25$
Speed of the stream = 5 km/h
Q26
AI 2017
00:00
Two taps together can fill a tank in $3\frac{1}{13}$ hours. If one tap takes 3 hours more than the other to fill the tank separately, find the time taken by each tap.
Time = $40/13$. Let faster = $x$, slower = $x+3$
$\frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}$
$40(2x+3) = 13(x^2+3x)$
$13x^2-41x-120=0$
$(13x+24)(x-5)=0$
The taps take 5 hours and 8 hours
Q27
Delhi 2016
00:00
A plane was delayed by half an hour. To cover 1500 km in time, its speed was increased by 250 km/h. Find the usual speed of the plane.
$\frac{1500}{x} - \frac{1500}{x+250} = \frac12$
$1500(250) = \frac12 x(x+250)$
$x^2 + 250x - 750000 = 0$
$(x+1000)(x-750)=0$
Usual speed = 750 km/h
Q28
Delhi 2016
00:00
Find $x$ in terms of $a, b, c$: $\frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c}, \; x \neq a,b,c$
$\frac{a(x-b)+b(x-a)}{(x-a)(x-b)} = \frac{2c}{x-c}$
$(ax-ab+bx-ab)(x-c) = 2c(x^2-ax-bx+ab)$
Simplifying leads to $x(a+b-2c) = ab+bc+ca-3abc$ (Correction: $ab+ac+bc-2abc$)
Actually: $(a+b)x - 2ab = \frac{2c(x-a)(x-b)}{x-c}$
Solving gives linear equation in $x$.
$x=\frac{ab+ac+bc-2abc}{a+b-2c}$
Q29
Delhi 2016
00:00
Time taken to cover 150 km was 2½ hours more than return journey. Speed while returning was 10 km/h more. Find speed in each direction.
$\frac{150}{x} - \frac{150}{x+10} = \frac52$
$150(10) = \frac52 x(x+10)$
$600 = x^2+10x$
$x^2+10x-600=0 \Rightarrow (x+30)(x-20)=0$
Speeds are 20 km/h and 30 km/h
Q30
AI 2016
00:00
A motor boat whose speed in still water is 24 km/h takes 1 hour more to go 32 km upstream than downstream. Find speed of stream.
$\frac{32}{24-x}-\frac{32}{24+x}=1$
$32(2x) = 576-x^2$
$x^2+64x-576=0$
$(x+72)(x-8)=0$
Speed of stream = 8 km/h
Q31
NCERT 2016
00:00
A rectangular park has breadth 3 m less than length. Its area is 4 m² more than an isosceles triangular park with base equal to breadth and height 12 m. Find dimensions of rectangle.
Length $x$, Breadth $x-3$. Area $x(x-3)$
Triangle Area $\frac12(x-3)(12) = 6(x-3)$
$x(x-3) = 6(x-3) + 4$
$x^2-3x = 6x-18+4$
$x^2-9x+14=0 \Rightarrow (x-7)(x-2)=0$
If $x=2$, breadth is negative. So $x=7$.
Length = 7 m, Breadth = 4 m
Q32
Foreign 2016
00:00
Two taps together can fill a tank in 9 hours 36 minutes. One tap takes 8 hours less than the other. Find individual times.
Time = $9.6$ hrs = $48/5$. Let slower = $x$, faster = $x-8$
$\frac1x+\frac1{x-8}=\frac5{48}$
$48(2x-8) = 5(x^2-8x)$
$5x^2-136x+384=0$
$(x-24)(5x-16)=0$. Reject $16/5$.
The taps take 24 hours and 16 hours
Q33
AI 2016
00:00
Two pipes together can fill a tank in $11\frac19$ minutes. One pipe takes 5 minutes more than the other. Find individual times.
Time = $100/9$. Let faster = $x$, slower = $x+5$
$\frac1x+\frac1{x+5}=\frac9{100}$
$100(2x+5) = 9(x^2+5x)$
$9x^2-155x-500=0$
$(x-20)(9x+25)=0$
Pipes take 20 minutes and 25 minutes
Q34
2023
00:00
While designing a school year book, the length and width of a photograph are increased by $x$ units each to double its area. The original dimensions are 18 cm × 12 cm. Find new dimensions.
Original Area = $216$. New Area = $432$
$(18+x)(12+x) = 432$
$x^2+30x+216=432$
$x^2+30x-216=0$
$(x+36)(x-6)=0 \Rightarrow x=6$
New dimensions: 24 cm × 18 cm
Q35
2023
00:00
A passenger plane had to increase its speed by 250 km/h to reach a destination 1500 km away in time after being delayed by 30 minutes. Find the usual speed of the plane.
Same as Q27.
$\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$
$x^2 + 250x - 750000 = 0$
$(x+1000)(x-750)=0$
The usual speed of the plane is 750 km/h
Q36
2020
00:00
Sum of the areas of two squares is 544 m². If the difference of their perimeters is 32 m, find the sides of the squares.
Same as Q20.
$x^2 + y^2 = 544$
$x - y = 8$
$(y+8)^2 + y^2 = 544$
$y=12, x=20$
Sides of the squares are 20 m and 12 m
Q37
2019
00:00
A motor boat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than downstream. Find the speed of the stream.
Same as Q21.
$\frac{24}{18-x} - \frac{24}{18+x} = 1$
$x^2 + 48x - 324 = 0$
$(x+54)(x-6)=0$
Speed of stream = 6 km/h
Q38
2019
00:00
A student scored a total of 32 marks in Mathematics and Science. If he had scored 4 more marks in Mathematics and 2 less in Science, the product of the marks would have been 253. Find his marks in the two subjects.
Same as Q9.
Maths $x$, Science $32-x$
$(x+4)(30-x) = 253$
$x^2 - 26x + 133 = 0$
$(x-19)(x-7)=0$
Marks are (19, 13) or (7, 25)
If the sum of the roots of the quadratic equation $ky^2-11y+(k-23)=0$ is $\dfrac{13}{21}$ more than the product of the roots, find the value of $k$.
For $ay^2+by+c=0$: Sum of roots $= -\dfrac{b}{a}$, Product $= \dfrac{c}{a}$
Here, $a=k,\; b=-11,\; c=k-23$
Sum $= \dfrac{11}{k}$, Product $= \dfrac{k-23}{k}$
Given: $\frac{11}{k}=\frac{k-23}{k}+\frac{13}{21}$
Solving: $\frac{11-(k-23)}{k}=\frac{13}{21}$
$\frac{34-k}{k}=\frac{13}{21}$
$21(34-k)=13k \Rightarrow 714=34k$
$k=21$
Q2
Delhi 2016
00:00
If $x=\dfrac23$ and $x=-3$ are the roots of $ax^2+7x+b=0$, find $a$ and $b$.
Sum of roots: $\frac23-3=-\frac73$
$-\frac{7}{a}=-\frac73 \Rightarrow a=3$
Product of roots: $\frac23\times(-3)=-2$
$\frac{b}{a}=-2 \Rightarrow b=-6$
$a=3,\; b=-6$
Q3
Foreign 2016
00:00
A two-digit number is four times the sum of its digits and three times the product of its digits. Find the number.
Let tens digit = $x$, units digit = $y$. Number $=10x+y$
$10x+y=4(x+y) \Rightarrow 6x=3y \Rightarrow y=2x$
$10x+y=3xy \Rightarrow 10x+2x=6x^2$
$6x^2-12x=0 \Rightarrow x=2$
If $x=2$, then $y=4$
The number is 24
Q4
2023
00:00
Two taps together fill a tank in $3\frac13$ hours. One tap takes 5 hours less than the other. Find the time taken by each tap.
Let smaller tap take $x$ hours, larger tap $(x-5)$ hours
$\frac1x+\frac1{x-5}=\frac3{10}$
$\frac{2x-5}{x(x-5)}=\frac3{10}$
$3x^2-35x+50=0$
$(3x-5)(x-10)=0$
Reject $x=5/3$ as $x-5$ would be negative.
Times are 10 hours and 5 hours
Q5
2018
00:00
A plane left 30 minutes late and to cover 1500 km in time, increased its speed by 100 km/h. Find the usual speed.
Let usual speed = $x$ km/h
$\frac{1500}{x}-\frac{1500}{x+100}=\frac12$
$300000=x(x+100)$
$x^2+100x-300000=0$
$(x+600)(x-500)=0$
Usual speed = 500 km/h
Q6
AI 2016
00:00
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of squares of the other two by 60. Find the numbers.
Let numbers be $(x-1),x,(x+1)$
$x^2-\{(x+1)^2-(x-1)^2\}=60$
$x^2-4x=60$
$x^2-4x-60=0$
$(x-10)(x+6)=0 \Rightarrow x=10$
The numbers are 9, 10, 11
Q7
2025
00:00
Rectangular lawn $12$m × $10$m is surrounded by a uniform walkway. Total area is $360$m². Find the width of the walkway.
Let width = $x$ m
$(12+2x)(10+2x)=360$
$4x^2+44x+120=360$
$4x^2+44x-240=0$
$x^2+11x-60=0 \Rightarrow (x+15)(x-4)=0$
Width of walkway = 4 m
Q8
2021–22
00:00
A rectangular pool of area $36$m² has a sidewalk of width $x$. Outside dimensions are $12$m × $7$m. Find $x$.
Inside dimensions $(12-2x)(7-2x)=36$
$84-24x-14x+4x^2=36$
$4x^2-38x+48=0$
$2x^2-19x+24=0 \Rightarrow (2x-3)(x-8)=0$
Reject $x=8$ (too large)
Width of sidewalk = 1.5 m
Q9
2025
00:00
A student scored 32 marks in Maths and Science. If he had scored 4 more marks in Mathematics and 2 less in Science, the product of the marks would have been 253. Find his marks in the two subjects.
Let Maths = $x$, Science = $32-x$
$(x+4)(30-x)=253$
$30x-x^2+120-4x=253$
$x^2-26x+133=0$
$(x-19)(x-7)=0$
Marks are (19, 13) or (7, 25)
Q10
2025
00:00
Circular park of diameter 65 m. Point P on boundary such that distance from A is 35 m more than from B (where AB is diameter). Find distances.
Let distance from B = $x$, from A = $x+35$
Angle in semicircle is $90^\circ$. By Pythagoras: $x^2+(x+35)^2=65^2$
$2x^2+70x+1225=4225$
$2x^2+70x-3000=0 \Rightarrow x^2+35x-1500=0$
$(x+60)(x-25)=0 \Rightarrow x=25$
Distances are 25 m and 60 m
Q11
2024
00:00
If Nidhi were 7 years younger than what she actually is, then the square of her age would be 1 more than 5 times her actual age. Find her present age.
Let Nidhi’s present age be $x$ years.
$(x-7)^2 = 5x + 1$
$x^2 - 14x + 49 = 5x + 1$
$x^2 - 19x + 48 = 0$
$(x-16)(x-3)=0$. Reject $x=3$ (as $x-7$ would be negative).
Nidhi’s present age is 16 years
Q12
2024
00:00
A shopkeeper buys books for ₹1800. If he had bought 15 more books for the same amount, each book would have cost ₹20 less. Find how many books he bought initially.
Let number of books = $x$. Cost = $1800/x$
$\frac{1800}{x} - \frac{1800}{x+15} = 20$
$1800(x+15-x) = 20x(x+15)$
$27000 = 20(x^2+15x)$
$x^2 + 15x - 1350 = 0 \Rightarrow (x+45)(x-30)=0$
The shopkeeper bought 30 books initially
Q13
2024
00:00
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its speed was reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.
Let original speed = $x$ km/h
$\frac{2800}{x-100}-\frac{2800}{x}=\frac12$
$2800(100) = \frac12 x(x-100)$
$x^2-100x-560000=0$
$(x-800)(x+700)=0 \Rightarrow x=800$
Original time = $2800/800$
Original duration of flight = 3.5 hours
Q14
2024
00:00
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$, find the fraction.
Let numerator = $x$, Denominator = $2x+1$
$\frac{x}{2x+1}+\frac{2x+1}{x}=\frac{58}{21}$
$21(x^2 + (2x+1)^2) = 58x(2x+1)$
$21(5x^2+4x+1) = 116x^2+58x$
$11x^2-26x-21=0 \Rightarrow (11x+7)(x-3)=0$
The fraction is $\frac{3}{7}$
Q15
2024
00:00
The age of a man is twice the square of the age of his son. Eight years hence, man’s age will be 4 years more than three times the age of his son. Find their present ages.
Son = $x$, Man = $2x^2$
$2x^2+8=3(x+8)+4$
$2x^2+8=3x+28$
$2x^2-3x-20=0$
$(2x+5)(x-4)=0 \Rightarrow x=4$
Son = 4 years, Man = 32 years
Q16
2023
00:00
A train travels 54 km at a certain speed and 63 km at a speed 6 km/h more. Total time is 3 hours. Find the first speed.
Let first speed = $x$ km/h
$\frac{54}{x}+\frac{63}{x+6}=3$
$18(x+6)+21x=x(x+6)$
$39x+108=x^2+6x$
$x^2-33x-108=0 \Rightarrow (x-36)(x+3)=0$
First speed = 36 km/h
Q17
2021–22
00:00
The sum of two numbers is 34. If 3 is subtracted from one and 2 is added to the other, their product becomes 260. Find the numbers.
Numbers: $x$ and $34-x$
$(x-3)(34-x+2)=260$
$(x-3)(36-x)=260$
$36x-x^2-108+3x=260$
$x^2-39x+368=0 \Rightarrow (x-16)(x-23)=0$
The numbers are 16 and 18
Q18
2021–22
00:00
The hypotenuse of a right-angled triangle is 6 cm more than twice the length of the shortest side. If the third side is 6 cm less than three times the shortest side, find the dimensions of the triangle.
Shortest = $x$. Hypotenuse = $2x+6$. Third = $3x-6$
$x^2+(3x-6)^2=(2x+6)^2$
$x^2+9x^2-36x+36=4x^2+24x+36$
$6x^2-60x=0 \Rightarrow 6x(x-10)=0$
Sides are 10 cm, 24 cm and 26 cm
Q19
2021–22
00:00
A two-digit number has product of its digits 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.
$xy=24$. Number $10x+y$
$10x+y-18=10y+x \Rightarrow 9x-9y=18 \Rightarrow x-y=2$
$x=y+2$. So $(y+2)y=24$
$y^2+2y-24=0 \Rightarrow (y+6)(y-4)=0$
$y=4, x=6$
The number is 64
Q20
2020
00:00
Sum of areas of two squares is $544 \text{ m}^2$. If difference of their perimeters is 32 m, find their sides.
$x^2+y^2=544$
$4x-4y=32 \Rightarrow x-y=8 \Rightarrow x=y+8$
$(y+8)^2+y^2=544$
$2y^2+16y+64=544$
$y^2+8y-240=0 \Rightarrow (y+20)(y-12)=0$
Sides are 20 m and 12 m
Q21
NCERT
00:00
A motorboat moves at 18 km/h in still water. It takes 1 hour more to go 24 km upstream than downstream. Find speed of the stream.
Let speed of stream = $x$
$\frac{24}{18-x}-\frac{24}{18+x}=1$
$24(2x) = 324-x^2$
$x^2+48x-324=0$
$(x+54)(x-6)=0$
Speed of stream = 6 km/h
Q22
2019
00:00
Two water taps together can fill a tank in $1\frac78$ hours. One tap takes 2 hours less than the other. Find their individual times.
Time = $15/8$ hrs. Let smaller tap = $x$, larger = $x-2$
$\frac1x+\frac1{x-2}=\frac8{15}$
$\frac{2x-2}{x(x-2)}=\frac8{15}$
$15(2x-2)=8(x^2-2x)$
$8x^2-46x+30=0 \Rightarrow 4x^2-23x+15=0$
$(4x-3)(x-5)=0$. Reject $3/4$ (too small).
Times are 5 hours and 3 hours
Q23
2019
00:00
A train travels 360 km at uniform speed. If speed were 5 km/h more, it would take 1 hour less. Find the speed.
$\frac{360}{x}-\frac{360}{x+5}=1$
$360(5)=x(x+5)$
$x^2+5x-1800=0$
$(x+45)(x-40)=0$
Speed of train = 40 km/h
Q24
2018
00:00
A train travels 63 km at speed $x$ and 72 km at speed $(x+6)$. Total time is 3 hours. Find original speed.
$\frac{63}{x}+\frac{72}{x+6}=3$
$21(x+6)+24x=x(x+6)$
$45x+126=x^2+6x$
$x^2-39x-126=0$
$(x-42)(x+3)=0$
Original speed = 42 km/h
Q25
Delhi 2017
00:00
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns downstream to the same point in 4 hours 30 minutes. Find the speed of the stream.
$\frac{30}{15-x} + \frac{30}{15+x} = \frac{9}{2}$
$30(\frac{30}{225-x^2}) = \frac92$
$900 = \frac92(225-x^2)$
$200 = 225-x^2 \Rightarrow x^2=25$
Speed of the stream = 5 km/h
Q26
AI 2017
00:00
Two taps together can fill a tank in $3\frac{1}{13}$ hours. If one tap takes 3 hours more than the other to fill the tank separately, find the time taken by each tap.
Time = $40/13$. Let faster = $x$, slower = $x+3$
$\frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}$
$40(2x+3) = 13(x^2+3x)$
$13x^2-41x-120=0$
$(13x+24)(x-5)=0$
The taps take 5 hours and 8 hours
Q27
Delhi 2016
00:00
A plane was delayed by half an hour. To cover 1500 km in time, its speed was increased by 250 km/h. Find the usual speed of the plane.
$\frac{1500}{x} - \frac{1500}{x+250} = \frac12$
$1500(250) = \frac12 x(x+250)$
$x^2 + 250x - 750000 = 0$
$(x+1000)(x-750)=0$
Usual speed = 750 km/h
Q28
Delhi 2016
00:00
Find $x$ in terms of $a, b, c$: $\frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c}, \; x \neq a,b,c$
$\frac{a(x-b)+b(x-a)}{(x-a)(x-b)} = \frac{2c}{x-c}$
$(ax-ab+bx-ab)(x-c) = 2c(x^2-ax-bx+ab)$
Simplifying leads to $x(a+b-2c) = ab+bc+ca-3abc$ (Correction: $ab+ac+bc-2abc$)
Actually: $(a+b)x - 2ab = \frac{2c(x-a)(x-b)}{x-c}$
Solving gives linear equation in $x$.
$x=\frac{ab+ac+bc-2abc}{a+b-2c}$
Q29
Delhi 2016
00:00
Time taken to cover 150 km was 2½ hours more than return journey. Speed while returning was 10 km/h more. Find speed in each direction.
$\frac{150}{x} - \frac{150}{x+10} = \frac52$
$150(10) = \frac52 x(x+10)$
$600 = x^2+10x$
$x^2+10x-600=0 \Rightarrow (x+30)(x-20)=0$
Speeds are 20 km/h and 30 km/h
Q30
AI 2016
00:00
A motor boat whose speed in still water is 24 km/h takes 1 hour more to go 32 km upstream than downstream. Find speed of stream.
$\frac{32}{24-x}-\frac{32}{24+x}=1$
$32(2x) = 576-x^2$
$x^2+64x-576=0$
$(x+72)(x-8)=0$
Speed of stream = 8 km/h
Q31
NCERT 2016
00:00
A rectangular park has breadth 3 m less than length. Its area is 4 m² more than an isosceles triangular park with base equal to breadth and height 12 m. Find dimensions of rectangle.
Length $x$, Breadth $x-3$. Area $x(x-3)$
Triangle Area $\frac12(x-3)(12) = 6(x-3)$
$x(x-3) = 6(x-3) + 4$
$x^2-3x = 6x-18+4$
$x^2-9x+14=0 \Rightarrow (x-7)(x-2)=0$
If $x=2$, breadth is negative. So $x=7$.
Length = 7 m, Breadth = 4 m
Q32
Foreign 2016
00:00
Two taps together can fill a tank in 9 hours 36 minutes. One tap takes 8 hours less than the other. Find individual times.
Time = $9.6$ hrs = $48/5$. Let slower = $x$, faster = $x-8$
$\frac1x+\frac1{x-8}=\frac5{48}$
$48(2x-8) = 5(x^2-8x)$
$5x^2-136x+384=0$
$(x-24)(5x-16)=0$. Reject $16/5$.
The taps take 24 hours and 16 hours
Q33
AI 2016
00:00
Two pipes together can fill a tank in $11\frac19$ minutes. One pipe takes 5 minutes more than the other. Find individual times.
Time = $100/9$. Let faster = $x$, slower = $x+5$
$\frac1x+\frac1{x+5}=\frac9{100}$
$100(2x+5) = 9(x^2+5x)$
$9x^2-155x-500=0$
$(x-20)(9x+25)=0$
Pipes take 20 minutes and 25 minutes
Q34
2023
00:00
While designing a school year book, the length and width of a photograph are increased by $x$ units each to double its area. The original dimensions are 18 cm × 12 cm. Find new dimensions.
Original Area = $216$. New Area = $432$
$(18+x)(12+x) = 432$
$x^2+30x+216=432$
$x^2+30x-216=0$
$(x+36)(x-6)=0 \Rightarrow x=6$
New dimensions: 24 cm × 18 cm
Q35
2023
00:00
A passenger plane had to increase its speed by 250 km/h to reach a destination 1500 km away in time after being delayed by 30 minutes. Find the usual speed of the plane.
Same as Q27.
$\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$
$x^2 + 250x - 750000 = 0$
$(x+1000)(x-750)=0$
The usual speed of the plane is 750 km/h
Q36
2020
00:00
Sum of the areas of two squares is 544 m². If the difference of their perimeters is 32 m, find the sides of the squares.
Same as Q20.
$x^2 + y^2 = 544$
$x - y = 8$
$(y+8)^2 + y^2 = 544$
$y=12, x=20$
Sides of the squares are 20 m and 12 m
Q37
2019
00:00
A motor boat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than downstream. Find the speed of the stream.
Same as Q21.
$\frac{24}{18-x} - \frac{24}{18+x} = 1$
$x^2 + 48x - 324 = 0$
$(x+54)(x-6)=0$
Speed of stream = 6 km/h
Q38
2019
00:00
A student scored a total of 32 marks in Mathematics and Science. If he had scored 4 more marks in Mathematics and 2 less in Science, the product of the marks would have been 253. Find his marks in the two subjects.
Same as Q9.
Maths $x$, Science $32-x$
$(x+4)(30-x) = 253$
$x^2 - 26x + 133 = 0$
$(x-19)(x-7)=0$
Marks are (19, 13) or (7, 25)