CBSE Maths › Class 10 PYQs › Quadratic Equations

Quadratic Equations: 5 marks Questions (CBSE Class 10)

62 different 5 marks questions on Quadratic Equations from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (73)2 marks (38)3 marks (23)4 marks (13)5 marks (62)

Q345 marksLong AnswerQuadratic Equations CBSE 2026 · Basic 430/5/1

ABCD is a rectangle of dimensions 80 cm $\times$ 60 cm. Another rectangle PQRS is drawn inside ABCD leaving space of equal width $x$ cm along the edges of ABCD. If area PQRS is half of the area ABCD, then find the value of $x$ .

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Answer:

x = 10

Dimensions of PQRS: $(80 - 2 x)$ cm and $(60 - 2 x)$ cm.
$(80 - 2 x) (60 - 2 x) = \frac{1}{2} \times 80 \times 60 = 2400$
$4800 - 280 x + 4 x^{2} = 2400 \Rightarrow x^{2} - 70 x + 600 = 0$
$(x - 10) (x - 60) = 0$ , so $x = 10$ or $x = 60$
$x = 60$ makes $60 - 2 x$ negative, so $x = 10$ .

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Basic 430/5/1

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the same journey. Find the original speed of the train.

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Answer: 45 km/h

Let the original speed be x km/h.
$\frac{90}{x} - \frac{90}{x + 15} = \frac{1}{2}$
$\frac{90 \times 15}{x ( x + 15 )} = \frac{1}{2} \Rightarrow x^{2} + 15 x - 2700 = 0$
$(x + 60) (x - 45) = 0$
Speed cannot be negative, so x = 45 km/h.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q335 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/1/1

A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.

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Answer: Slower train 40 km/hr, faster train 50 km/hr

Let the speed of the slower train be $x$ km/hr; faster train $(x + 10)$ km/hr.
$\frac{200}{x} - \frac{200}{x + 10} = 1$
$200 (x + 10) - 200 x = x (x + 10)$
$x^{2} + 10 x - 2000 = 0$
$(x + 50) (x - 40) = 0$ , so $x = 40$ (speed cannot be negative).
Slower train: 40 km/hr; faster train: 50 km/hr.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q33 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/1/1

The sum of the areas of two squares is 640 m $^{2}$ . If the difference in their perimeters is 64 m, find the sides of the two squares.

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Answer: 24 m and 8 m

Let the sides be $x$ m and $y$ m with $x > y$ .
$4 x - 4 y = 64$ , so $x = y + 16$ .
$x^{2} + y^{2} = 640$ : $(y + 16)^{2} + y^{2} = 640$
$2 y^{2} + 32 y + 256 - 640 = 0$ , i.e. $y^{2} + 16 y - 192 = 0$
$(y + 24) (y - 8) = 0$ , so $y = 8$ (side cannot be negative).
$x = 24$ . The sides are 24 m and 8 m.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q335 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/1

A person on a tour has ₹ 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. Find the original duration of the tour.

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Answer: 12 days

Let the original duration be x days. Daily expense = $\frac{4200}{x}$ .
$\frac{4200}{x} - \frac{4200}{x + 3} = 70$
$4200 \times 3 = 70 x (x + 3)$
$x^{2} + 3 x - 180 = 0$
$(x + 15) (x - 12) = 0$
x = 12 (x cannot be negative), so the tour was for 12 days.

Q33 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/1

The area of a right-angled triangle is 600 $c m^{2}$ . If the base of the triangle exceeds the altitude by 10 cm, find all the three dimensions of the triangle.

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Answer: Altitude 30 cm, base 40 cm, hypotenuse 50 cm

Let the altitude be x cm; base = (x + 10) cm.
$\frac{1}{2} x (x + 10) = 600$
$x^{2} + 10 x - 1200 = 0$
$(x + 40) (x - 30) = 0$ , so x = 30.
Altitude = 30 cm, base = 40 cm.
Hypotenuse $= 3 0^{2} + 4 0^{2} = 50$ cm.

Q345 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/2

In a flight of 600 km, an aircraft slowed down its speed due to bad weather. Its average speed for the trip reduced by 200 km/h from its usual speed and time of flight increased by 30 minutes. Find the scheduled duration of the flight.

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Answer: 1 hour

Let the usual speed be x km/h.
$\frac{600}{x - 200} - \frac{600}{x} = \frac{1}{2}$
$600 \times 200 \times 2 = x (x - 200)$
$x^{2} - 200 x - 240000 = 0$
$(x - 600) (x + 400) = 0$ , so x = 600 km/h.
Scheduled duration $= \frac{600}{600} = 1$ hour.

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/2

Two pipes are used to fill a swimming pool. If the pipe of the larger diameter is used for 4 hours and the pipe of the smaller diameter for 9 hours, only half of the pool can be filled. Find how long it would take for each pipe to fill the pool, separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.

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Answer: Larger pipe 20 hours, smaller pipe 30 hours

Let the larger pipe take x hours; the smaller takes (x + 10) hours.
$\frac{4}{x} + \frac{9}{x + 10} = \frac{1}{2}$
$8 (x + 10) + 18 x = x (x + 10)$
$x^{2} - 16 x - 80 = 0$
$(x - 20) (x + 4) = 0$ , so x = 20.
Larger pipe: 20 hours; smaller pipe: 30 hours.

Q355 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/3

In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

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Answer: Maths 13, Science 17 or Maths 12, Science 18

Let Maths marks = x; Science marks = 30 - x.
$(x + 2) (30 - x - 3) = 210$
$(x + 2) (27 - x) = 210$
$- x^{2} + 25 x + 54 = 210$
$x^{2} - 25 x + 156 = 0$
$(x - 12) (x - 13) = 0$ , so x = 12 or 13.
Maths 12, Science 18; or Maths 13, Science 17.

Q35 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/2/3

The length of hypotenuse (in cm) of a right-angled triangle is 6 cm more than twice the length of its shortest side. If the length of its third side is 6 cm less than thrice the length of its shortest side, find the dimensions of the triangle.

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Answer: 10 cm, 24 cm and 26 cm

Let the shortest side be x cm; hypotenuse = (2x + 6) cm; third side = (3x - 6) cm.
$x^{2} + (3 x - 6)^{2} = (2 x + 6)^{2}$
$x^{2} + 9 x^{2} - 36 x + 36 = 4 x^{2} + 24 x + 36$
$6 x^{2} - 60 x = 0$ , so x = 10 (x ≠ 0).
Sides: 10 cm, 24 cm and 26 cm.

Q325 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/4/1

A person on tour has ₹ 5,400 for his expenses. If he extends his tour by 5 days, he has to cut down his daily expenses by ₹ 180. Find the original duration of the tour and daily expense.

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Answer: 10 days; ₹ 540 per day

Let the original duration be x days. Daily expense $= \frac{5400}{x}$ .
$\frac{5400}{x} - \frac{5400}{x + 5} = 180$
$5400 \times 5 = 180 x (x + 5) \Rightarrow x^{2} + 5 x - 150 = 0$
$(x + 15) (x - 10) = 0 \Rightarrow x = 10$ (x cannot be negative)
Daily expense $= \frac{5400}{10} = ₹ 540$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/4/1

The total cost of certain piece of cloth was ₹ 2,100. During special sale time, the shopkeeper offered 2 m extra cloth for free thus reducing the price of cloth per metre by ₹ 120. What was the original per metre price of cloth and its length ?

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Answer: Length 5 m; original price ₹ 420 per metre

Let the original length be x m. Original price per metre $= \frac{2100}{x}$ .
$\frac{2100}{x} - \frac{2100}{x + 2} = 120$
$2100 \times 2 = 120 x (x + 2) \Rightarrow x^{2} + 2 x - 35 = 0$
$(x + 7) (x - 5) = 0 \Rightarrow x = 5$
Price per metre $= \frac{2100}{5} = ₹ 420$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q325 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/5/1

Venkat can row a boat in still water at the speed of 12 km/h. He ferries tourists 15 km upstream and 18 km downstream in 3 hours. Find the speed of the stream.

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Answer: 4 km/h

Let the speed of the stream be x km/h. Upstream speed $= 12 - x$ , downstream speed $= 12 + x$ .
$\frac{15}{12 - x} + \frac{18}{12 + x} = 3$
$15 (12 + x) + 18 (12 - x) = 3 (144 - x^{2})$
$396 - 3 x = 432 - 3 x^{2}$
$3 x^{2} - 3 x - 36 = 0$ , i.e. $x^{2} - x - 12 = 0$
$(x - 4) (x + 3) = 0$ , so $x = 4$ (speed cannot be negative).
Speed of the stream = 4 km/h

Q325 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/5/2

By selling an article for ₹ 48, a trader loses as much percent as half of the cost price of the article. Calculate the cost price and loss amount of the article.

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Answer: Cost price ₹ 80 with loss ₹ 32, or cost price ₹ 120 with loss ₹ 72

Let the cost price be ₹ x. Loss percent $= \frac{x}{2}$ .
Loss $= x \times \frac{x /2}{100} = \frac{x ^{2}}{200}$
SP = CP − Loss: $x - \frac{x ^{2}}{200} = 48$
$x^{2} - 200 x + 9600 = 0$
$(x - 80) (x - 120) = 0$ , so $x = 80$ or $x = 120$ .
If CP = ₹ 80: loss = 40%, loss amount = ₹ 32 (80 − 32 = 48).
If CP = ₹ 120: loss = 60%, loss amount = ₹ 72 (120 − 72 = 48).
Both values satisfy the conditions.

Q355 marksLong AnswerQuadratic Equations CBSE 2026 · Standard 30/5/3

Two water taps together can fill a tank in $8 \frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

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Answer: Smaller tap: 20 hours; larger tap: 16 hours

Let the smaller tap take x hours; the larger takes $(x - 4)$ hours. Together: $8 \frac{8}{9} = \frac{80}{9}$ hours.
$\frac{1}{x} + \frac{1}{x - 4} = \frac{9}{80}$
$80 (2 x - 4) = 9 x (x - 4)$
$9 x^{2} - 196 x + 320 = 0$
$x = \frac{196 \pm 38416 - 11520}{18} = \frac{196 \pm 164}{18}$ , so $x = 20$ or $x = \frac{16}{9}$
$x = \frac{16}{9}$ gives $x - 4 < 0$ , so reject it. Hence x = 20.
Smaller tap: 20 hours; larger tap: 16 hours.

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/1/1

The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.

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Answer: 18 and 12

Let the greater number be $x$ and the smaller $y$ . Then $y^{2} = 8 x$ and $x^{2} - y^{2} = 180$ .
So $x^{2} - 8 x - 180 = 0$ .
$(x - 18) (x + 10) = 0$ , so $x = 18$ ( $x > 0$ ).
$y^{2} = 144$ , so $y = 12$ .
The numbers are 18 and 12 (check: $324 - 144 = 180$ ).

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/1/1

Find the value(s) of k for which the equation $2 x^{2} + k x + 3 = 0$ has real and equal roots. Hence, find the roots of the equations so obtained.

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Answer:

k = \pm 26

; for

k = 26

the roots are

- \frac{6}{2}, - \frac{6}{2}

; for

k = - 26

the roots are

\frac{6}{2}, \frac{6}{2}

For equal roots, $D = k^{2} - 4 (2) (3) = 0$ , so $k^{2} = 24$ and $k = \pm 26$ .
Equal roots are $x = - \frac{b}{2 a} = - \frac{k}{4}$ .
For $k = 26$ : $2 x^{2} + 26 x + 3 = (2 x + 3)^{2} = 0$ , so $x = - \frac{6}{2}$ (twice).
For $k = - 26$ : $(2 x - 3)^{2} = 0$ , so $x = \frac{6}{2}$ (twice).

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/2/1

The sum of areas of two squares is 2650 cm $^{2}$ . If the sum of their perimeters is 280 cm, find the sides of the two given squares.

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Answer: 45 cm and 25 cm

Let the sides be x cm and y cm.
$4 x + 4 y = 280 \Rightarrow x + y = 70 \Rightarrow y = 70 - x$ .
$x^{2} + (70 - x)^{2} = 2650 \Rightarrow 2 x^{2} - 140 x + 4900 - 2650 = 0$ .
$2 x^{2} - 140 x + 2250 = 0 \Rightarrow x^{2} - 70 x + 1125 = 0$ .
$(x - 45) (x - 25) = 0 \Rightarrow x = 45$ or $x = 25$ .
If $x = 45$ , $y = 25$ (and vice versa). Check: $4 5^{2} + 2 5^{2} = 2025 + 625 = 2650$ .
The sides are 45 cm and 25 cm.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/2/1

Express the equation $\frac{1}{x} - \frac{1}{x - 2} = 3$ , $(x \neq = 0, 2)$ as a quadratic equation in standard form. Hence, find the roots of the quadratic equation so obtained.

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Answer:

3 x^{2} - 6 x + 2 = 0

; roots

x = \frac{3 + 3}{3}

and

x = \frac{3 - 3}{3}

$\frac{( x - 2 ) - x}{x ( x - 2 )} = 3 \Rightarrow \frac{- 2}{x ^{2} - 2 x} = 3$ .
$- 2 = 3 x^{2} - 6 x \Rightarrow 3 x^{2} - 6 x + 2 = 0$ (standard form).
Here $a = 3$ , $b = - 6$ , $c = 2$ ; $D = 36 - 24 = 12$ .
$x = \frac{6 \pm 12}{6} = \frac{6 \pm 2 3}{6} = \frac{3 \pm 3}{3}$ .
Both roots are different from 0 and 2, so they are valid.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/3/1

Find two consecutive odd integers, sum of whose squares is 290.

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Answer: 11 and 13, or −13 and −11

Let the integers be $x$ and $x + 2$ .
$x^{2} + (x + 2)^{2} = 290 \Rightarrow 2 x^{2} + 4 x + 4 = 290 \Rightarrow x^{2} + 2 x - 143 = 0$ .
$(x + 13) (x - 11) = 0$ , so $x = 11$ or $x = - 13$ .
The integers are 11 and 13, or $- 13$ and $- 11$ .
Check: $121 + 169 = 290$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/3/1

A charity trust decides to build a rectangular hall having an area of 300 $m^{2}$ . The length of the hall is one metre more than twice its width. Find the length and breadth of the hall.

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Answer: Length = 25 m, breadth = 12 m

Let the breadth be $x$ m; then the length is $(2 x + 1)$ m.
$x (2 x + 1) = 300 \Rightarrow 2 x^{2} + x - 300 = 0$ .
$2 x^{2} + 25 x - 24 x - 300 = 0 \Rightarrow (2 x + 25) (x - 12) = 0$ .
$x = 12$ (breadth cannot be negative).
Breadth = 12 m, length = $2 \times 12 + 1 = 25$ m.

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/4/1

It is given that $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$ ; $(p \neq = 0)$
(i) Show that the discriminant (D) of above equation is a perfect square.
(ii) Find the roots of the equation.

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Answer: (i)

D = (p^{2} + q^{2})^{2}

(ii)

x = \frac{q ^{2}}{p ^{2}}

x = - 1

(i) $a = p^{2}$ , $b = p^{2} - q^{2}$ , $c = - q^{2}$ .
$D = (p^{2} - q^{2})^{2} - 4 p^{2} (- q^{2}) = p^{4} - 2 p^{2} q^{2} + q^{4} + 4 p^{2} q^{2} = (p^{2} + q^{2})^{2}$ , a perfect square.
(ii) $x = \frac{- ( p ^{2} - q ^{2} ) \pm ( p ^{2} + q ^{2} )}{2 p ^{2}}$ .
Taking +: $x = \frac{2 q ^{2}}{2 p ^{2}} = \frac{q ^{2}}{p ^{2}}$ .
Taking $-$ : $x = \frac{- 2 p ^{2}}{2 p ^{2}} = - 1$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Basic 430/4/1

Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.

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Answer: 5, 6, 7

Let the integers be $x$ , $x + 1$ , $x + 2$ .
$x^{2} + (x + 1) (x + 2) = 67$
$2 x^{2} + 3 x + 2 = 67$ , i.e. $2 x^{2} + 3 x - 65 = 0$ .
$(2 x + 13) (x - 5) = 0$ , so $x = 5$ or $x = - \frac{13}{2}$ .
$x$ is a positive integer, so $x = 5$ ; the numbers are 5, 6, 7.
Check: $25 + 6 \times 7 = 67$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q345 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/1/1

The perimeter of a right triangle is $60$ cm and its hypotenuse is $25$ cm. Find the lengths of other two sides of the triangle.

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Answer:

15

cm and

20

Let the other two sides be $x$ cm and $(35 - x)$ cm (since $60 - 25 = 35$ ).
By Pythagoras: $x^{2} + (35 - x)^{2} = 2 5^{2}$
$2 x^{2} - 70 x + 1225 = 625$
$x^{2} - 35 x + 300 = 0$
$(x - 15) (x - 20) = 0$ , so $x = 15$ or $x = 20$ .
The other two sides are $15$ cm and $20$ cm.

Also asked in: 2025 Standard 30/1/3

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/1/1

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. Find the speed of the train.

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Answer:

40

km/h

Let the speed be $x$ km/h.
$\frac{480}{x - 8} - \frac{480}{x} = 3$
$\frac{480 \times 8}{x ( x - 8 )} = 3$ , so $x (x - 8) = 1280$
$x^{2} - 8 x - 1280 = 0$
$(x - 40) (x + 32) = 0$ , so $x = 40$ or $x = - 32$ .
Speed cannot be negative, so the speed of the train is $40$ km/h.

Also asked in: 2025 Standard 30/1/3

Q355 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/1/2

A two-digit number is such that the product of its digits is $12$ . When $36$ is added to this number, the digits interchange their places. Find the number.

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Answer:

26

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$(10 y + x) - (10 x + y) = 36$ , so $9 (y - x) = 36$ and $y = x + 4$ .
$x y = 12$ : $x (x + 4) = 12$ , i.e. $x^{2} + 4 x - 12 = 0$ .
$(x + 6) (x - 2) = 0$ , so $x = 2$ (a digit cannot be negative) and $y = 6$ .
The number is $26$ . Check: $2 \times 6 = 12$ and $26 + 36 = 62$ .

Q35 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/1/2

A student scored a total of $32$ marks in class tests in Mathematics and Science. Had he scored $2$ marks less in Science and $4$ marks more in Mathematics, the product of his marks would have been $253$ . Find his marks in the two subjects.

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Answer: Mathematics

19

, Science

13

; or Mathematics

7

, Science

25

Let the marks in Mathematics be $x$ ; then Science $= 32 - x$ .
$(x + 4) (32 - x - 2) = 253$ , i.e. $(x + 4) (30 - x) = 253$ .
$30 x - x^{2} + 120 - 4 x = 253$ , so $x^{2} - 26 x + 133 = 0$ .
$(x - 7) (x - 19) = 0$ , so $x = 7$ or $x = 19$ .
If $x = 19$ : Mathematics $19$ , Science $13$ (check: $23 \times 11 = 253$ ).
If $x = 7$ : Mathematics $7$ , Science $25$ (check: $11 \times 23 = 253$ ).

Q335 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/2/1

The sum of the areas of two squares is 52 cm $^{2}$ and difference of their perimeters is 8 cm. Find the lengths of the sides of the two squares.

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Answer: 6 cm and 4 cm

Let the sides be x cm and y cm with $x > y$
$4 x - 4 y = 8$ , so $x = y + 2$
$(y + 2)^{2} + y^{2} = 52$
$2 y^{2} + 4 y - 48 = 0$ , i.e. $y^{2} + 2 y - 24 = 0$
$(y + 6) (y - 4) = 0$ , so $y = 4$ (side cannot be negative)
$x = 6$ : sides are 6 cm and 4 cm

Also asked in: 2025 Standard 30/2/2

Q33 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/2/1

The time taken by a person to travel an upward distance of 150 km was $2 \frac{1}{2}$ hours more than the time taken in the downward return journey. If he returned at a speed of 10 km/h more than the speed while going up, find the speeds in each direction.

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Answer: Upward speed 20 km/h; downward speed 30 km/h

Let the upward speed be x km/h; downward speed $= x + 10$ km/h
$\frac{150}{x} - \frac{150}{x + 10} = \frac{5}{2}$
$\frac{1500}{x ( x + 10 )} = \frac{5}{2}$ , so $x^{2} + 10 x - 600 = 0$
$(x + 30) (x - 20) = 0$ , so $x = 20$ (speed cannot be negative)
Upward 20 km/h, downward 30 km/h

Also asked in: 2025 Standard 30/2/2

Q355 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/2/3

The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $1 \frac{9}{20}$ . Find the original fraction.

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Answer:

\frac{7}{10}

Let the numerator be x; denominator $= x + 3$
$\frac{x}{x + 3} + \frac{x + 2}{x + 5} = \frac{29}{20}$
$20 [x (x + 5) + (x + 2) (x + 3)] = 29 (x + 3) (x + 5)$
$40 x^{2} + 200 x + 120 = 29 x^{2} + 232 x + 435$
$11 x^{2} - 32 x - 315 = 0$
$x = \frac{32 \pm 1024 + 13860}{22} = \frac{32 \pm 122}{22}$ , so $x = 7$ or $x = - \frac{45}{11}$ (rejected: negative and not an integer)
Original fraction $= \frac{7}{10}$ (check: $\frac{7}{10} + \frac{9}{12} = \frac{29}{20}$ )

Q35 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/2/3

A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

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Answer: 45 km/h

Let the original speed be x km/h
$\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60} = \frac{4}{5}$
$\frac{1800}{x ( x + 5 )} = \frac{4}{5}$ , so $x (x + 5) = 2250$
$x^{2} + 5 x - 2250 = 0$ , i.e. $(x + 50) (x - 45) = 0$
$x = 45$ (speed cannot be negative): original speed 45 km/h

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/4/1

The sides of a right triangle are such that the longest side is 4 m more than the shortest side and the third side is 2 m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle.

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Answer: Sides 6 m, 8 m, 10 m; difference = 0 (area 24, perimeter 24)

Let shortest side $= x$ m; longest $= x + 4$ ; third side $= x + 2$
Pythagoras: $x^{2} + (x + 2)^{2} = (x + 4)^{2}$
$x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0 \Rightarrow x = 6$ ( $x > 0$ )
Sides: 6 m, 8 m, 10 m
Area $= \frac{1}{2} \times 6 \times 8 = 24$ ; perimeter $= 6 + 8 + 10 = 24$
Difference $= 24 - 24 = 0$

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/4/1

Express the equation $\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}$ ; $(x \neq = 3, 5)$ as a quadratic equation in standard form. Hence, find the roots of the equation so formed.

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Answer:

2 x^{2} - 19 x + 42 = 0

; roots

6

and

\frac{7}{2}

$\frac{( x - 2 ) ( x - 5 ) + ( x - 4 ) ( x - 3 )}{( x - 3 ) ( x - 5 )} = \frac{10}{3}$
$\frac{2 x ^{2} - 14 x + 22}{x ^{2} - 8 x + 15} = \frac{10}{3}$
$6 x^{2} - 42 x + 66 = 10 x^{2} - 80 x + 150$
$4 x^{2} - 38 x + 84 = 0 \Rightarrow 2 x^{2} - 19 x + 42 = 0$ (standard form)
$D = 361 - 336 = 25$ ; $x = \frac{19 \pm 5}{4}$
$x = 6$ or $x = \frac{7}{2}$ (both allowed since $x \neq = 3, 5$ )

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/5/1

A 2-digit number is seven times the sum of its digits and two (2) more than 5 times the product of its digits. Find the number.

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Answer: 42

Let tens digit $= a$ , units digit $= b$ ; number $= 10 a + b$ .
$10 a + b = 7 (a + b) \Rightarrow 3 a = 6 b \Rightarrow a = 2 b$
$10 a + b = 5 ab + 2 \Rightarrow 21 b = 10 b^{2} + 2$
$10 b^{2} - 21 b + 2 = 0 \Rightarrow (b - 2) (10 b - 1) = 0$
$b = 2$ ( $b = \frac{1}{10}$ is not a digit), so $a = 4$ .
Number $= 42$ . Check: $7 (4 + 2) = 42$ and $5 (8) + 2 = 42$ .

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/5/1

Find the value(s) of $p$ for which the quadratic equation given as $(p + 4) x^{2} - (p + 1) x + 1 = 0$ has real and equal roots. Also, find the roots of the equation(s) so obtained.

Show answer & solution

Answer:

p = 5

(roots

\frac{1}{3}, \frac{1}{3}

) or

p = - 3

(roots

- 1, - 1

)

Equal roots: $D = (p + 1)^{2} - 4 (p + 4) = 0$
$p^{2} - 2 p - 15 = 0 \Rightarrow (p - 5) (p + 3) = 0 \Rightarrow p = 5$ or $p = - 3$
$p = 5$ : $9 x^{2} - 6 x + 1 = 0 \Rightarrow (3 x - 1)^{2} = 0 \Rightarrow x = \frac{1}{3}, \frac{1}{3}$
$p = - 3$ : $x^{2} + 2 x + 1 = 0 \Rightarrow (x + 1)^{2} = 0 \Rightarrow x = - 1, - 1$

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q325 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/6/1

There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter.
An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.

Diagram for CBSE 2025 Class 10 Maths question 32

Show answer & solution

Answer: PA = 60 m and PB = 25 m

$∠ A P B = 9 0^{\circ}$ (angle in a semicircle).
Let $P B = x$ m, so $P A = x + 35$ m.
$x^{2} + (x + 35)^{2} = 6 5^{2} \Rightarrow 2 x^{2} + 70 x + 1225 = 4225$
$x^{2} + 35 x - 1500 = 0 \Rightarrow (x + 60) (x - 25) = 0$
$x = 25$ (length cannot be negative)
PA = 60 m, PB = 25 m

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2025 · Standard 30/6/1

Find the smallest value of $p$ for which the quadratic equation $x^{2} - 2 (p + 1) x + p^{2} = 0$ has real roots. Hence, find the roots of the equation so obtained.

Show answer & solution

Answer:

p = - \frac{1}{2}

; roots

\frac{1}{2}, \frac{1}{2}

For real roots, $D \geq 0$ : $4 (p + 1)^{2} - 4 p^{2} \geq 0$
$8 p + 4 \geq 0 \Rightarrow p \geq - \frac{1}{2}$
Smallest value: $p = - \frac{1}{2}$
Equation: $x^{2} - 2 (\frac{1}{2}) x + \frac{1}{4} = 0 \Rightarrow x^{2} - x + \frac{1}{4} = 0$
$(x - \frac{1}{2})^{2} = 0 \Rightarrow x = \frac{1}{2}, \frac{1}{2}$

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q325 marksLong AnswerQuadratic Equations CBSE 2024 · Basic 430/2/1

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. Find the total number of toys produced on that day.

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Answer: 25 or 30 toys

Let the number of toys be $x$ . Cost of each toy $= ₹ (55 - x)$ .
$x (55 - x) = 750$
$x^{2} - 55 x + 750 = 0$
$(x - 25) (x - 30) = 0$
$x = 25$ or $x = 30$
So 25 or 30 toys were produced that day.

Q355 marksLong AnswerQuadratic Equations CBSE 2024 · Basic 430/2/2

A cottage industry produces a certain number of pottery articles in a day. It was observed that on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Show answer & solution

Answer: 6 articles; ₹ 15 each

Let the number of articles be $x$ . Cost of each $= ₹ (2 x + 3)$ .
$x (2 x + 3) = 90$
$2 x^{2} + 3 x - 90 = 0$
$(2 x + 15) (x - 6) = 0$
$x = 6$ ( $x = - \frac{15}{2}$ is rejected)
Cost of each article $= 2 (6) + 3 = ₹ 15$

Q355 marksLong AnswerQuadratic Equations CBSE 2024 · Basic 430/2/3

The area of a rectangular plot is 528 m $^{2}$ . The length of the plot (in metres) is one more than twice the breadth. Find the length and breadth of the plot. Also, find the cost of levelling the plot at the rate of ₹ 80 per square metre.

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Answer: Length = 33 m, breadth = 16 m; cost = ₹ 42240

Let breadth = $x$ m. Length $= (2 x + 1)$ m.
$x (2 x + 1) = 528 \Rightarrow 2 x^{2} + x - 528 = 0$
$x = \frac{- 1 \pm 1 + 4224}{4} = \frac{- 1 \pm 65}{4}$
$x = 16$ (negative value rejected)
Breadth = 16 m, length $= 33$ m
Cost $= 528 \times 80 = ₹ 42240$

Q325 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/3/1

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.

Show answer & solution

Answer:

3 \frac{1}{2}

hours (3 h 30 min)

Let the original speed be $x$ km/h.
$\frac{2800}{x - 100} - \frac{2800}{x} = \frac{1}{2}$ .
$2800 \times \frac{100}{x ( x - 100 )} = \frac{1}{2} \Rightarrow x^{2} - 100 x - 560000 = 0$ .
$(x - 800) (x + 700) = 0 \Rightarrow x = 800$ (speed cannot be negative).
Original duration = $\frac{2800}{800} = 3.5$ hours = 3 hours 30 minutes.

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/3/1

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.

Show answer & solution

Answer:

\frac{3}{7}

Let the numerator be $x$ ; the denominator is $2 x + 1$ .
$\frac{x}{2 x + 1} + \frac{2 x + 1}{x} = \frac{58}{21}$ .
$21 [x^{2} + (2 x + 1)^{2}] = 58 x (2 x + 1) \Rightarrow 21 (5 x^{2} + 4 x + 1) = 116 x^{2} + 58 x$ .
$105 x^{2} + 84 x + 21 = 116 x^{2} + 58 x \Rightarrow 11 x^{2} - 26 x - 21 = 0$ .
$(x - 3) (11 x + 7) = 0 \Rightarrow x = 3$ (taking the natural-number value).
Fraction = $\frac{3}{7}$ . Check: $\frac{3}{7} + \frac{7}{3} = \frac{58}{21} = 2 \frac{16}{21}$ .

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q345 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/4/1

A train travels a distance of 90 km at a constant speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Show answer & solution

Answer: 45 km/h

Let the original speed be $x$ km/h.
$\frac{90}{x} - \frac{90}{x + 15} = \frac{1}{2}$ .
$\frac{90 \times 15}{x ( x + 15 )} = \frac{1}{2}$ , so $x^{2} + 15 x = 2700$ .
$x^{2} + 15 x - 2700 = 0$ , i.e. $(x + 60) (x - 45) = 0$ .
Speed cannot be negative, so $x = 45$ km/h.

Also asked in: 2024 Standard 30/4/3

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/4/1

Find the value of ‘c’ for which the quadratic equation
$(c + 1) x^{2} - 6 (c + 1) x + 3 (c + 9) = 0$ ; $c \neq = - 1$
has real and equal roots.

Show answer & solution

Answer:

c = 3

For real and equal roots, $b^{2} - 4 a c = 0$ .
$[6 (c + 1)]^{2} - 4 (c + 1) \cdot 3 (c + 9) = 0$ .
$12 (c + 1) [3 (c + 1) - (c + 9)] = 0$ , i.e. $12 (c + 1) (2 c - 6) = 0$ .
Since $c \neq = - 1$ , $2 c - 6 = 0$ , so $c = 3$ .

Also asked in: 2024 Standard 30/4/3

Q325 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/4/2

Find the value of ‘k’ for which the quadratic equation $(k + 1) x^{2} - 2 (3 k + 1) x + (8 k + 1) = 0$ has real and equal roots.

Show answer & solution

Answer:

k = 0

k = 3

For real and equal roots, $b^{2} - 4 a c = 0$ .
$4 (3 k + 1)^{2} - 4 (k + 1) (8 k + 1) = 0$ .
$(9 k^{2} + 6 k + 1) - (8 k^{2} + 9 k + 1) = 0$ , i.e. $k^{2} - 3 k = 0$ .
$k (k - 3) = 0$ , so $k = 0$ or $k = 3$ (both keep $k + 1 \neq = 0$ ).

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/4/2

A 2-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

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Answer: 92

Let the tens digit be $x$ and the units digit $y$ ; the number is $10 x + y$ .
$10 x + y - 63 = 10 y + x$ gives $9 x - 9 y = 63$ , so $x - y = 7$ , i.e. $y = x - 7$ .
$x y = 18$ : $x (x - 7) = 18$ , so $x^{2} - 7 x - 18 = 0$ .
$(x - 9) (x + 2) = 0$ , so $x = 9$ (a digit cannot be negative) and $y = 2$ .
The number is 92. Check: $9 \times 2 = 18$ and $92 - 63 = 29$ .

Q325 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/5/1

Find the value of 'k' for which the quadratic equation $(k + 1) x^{2} - 6 (k + 1) x + 3 (k + 9) = 0$ , $k \neq = - 1$ has real and equal roots.

Show answer & solution

Answer:

k = 3

For equal roots, $D = b^{2} - 4 a c = 0$ .
$36 (k + 1)^{2} - 12 (k + 1) (k + 9) = 0$ .
$12 (k + 1) [3 (k + 1) - (k + 9)] = 0 \Rightarrow 12 (k + 1) (2 k - 6) = 0$ .
Since $k \neq = - 1$ , $2 k - 6 = 0$ , so $k = 3$ .
Check: the equation becomes $4 x^{2} - 24 x + 36 = 0$ , i.e. $(x - 3)^{2} = 0$ , with equal roots 3, 3.

Also asked in: 2024 Standard 30/5/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/5/1

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find their present ages.

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Answer: Son: 4 years, man: 32 years

Let the son's present age be $x$ years; the man's age is $2 x^{2}$ years.
After 8 years: $2 x^{2} + 8 = 3 (x + 8) + 4$ .
$2 x^{2} - 3 x - 20 = 0 \Rightarrow (2 x + 5) (x - 4) = 0$ .
Age cannot be negative, so $x = 4$ .
Son is 4 years old and the man is $2 \times 16 = 32$ years old. (Check: in 8 years, 40 = 3 × 12 + 4.)

Also asked in: 2024 Standard 30/5/3

Q355 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/5/2

A 2-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits are reversed. Find the number.

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Answer: 27

Let the tens digit be $x$ and the units digit be $y$ ; $x y = 14$ .
$10 x + y + 45 = 10 y + x \Rightarrow 9 y - 9 x = 45 \Rightarrow y = x + 5$ .
$x (x + 5) = 14 \Rightarrow x^{2} + 5 x - 14 = 0 \Rightarrow (x + 7) (x - 2) = 0$ .
A digit cannot be negative, so $x = 2$ and $y = 7$ .
The number is 27. (Check: $27 + 45 = 72$ .)

Q35 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2024 · Standard 30/5/2

The side of a square exceeds the side of another square by 4 cm and the sum of the areas of the two squares is 400 $c m^{2}$ . Find the sides of the squares.

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Answer: 12 cm and 16 cm

Let the smaller side be $x$ cm; the other side is $(x + 4)$ cm.
$x^{2} + (x + 4)^{2} = 400 \Rightarrow 2 x^{2} + 8 x - 384 = 0 \Rightarrow x^{2} + 4 x - 192 = 0$ .
$(x + 16) (x - 12) = 0$ , and a side cannot be negative, so $x = 12$ .
The sides are 12 cm and 16 cm. (Check: $144 + 256 = 400$ .)

Q325 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/1/1

The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 80 m more than the shorter side, find the length of the sides of the field.

Show answer & solution

Answer: No such field exists: the resulting equation

x^{2} + 40 x + 2800 = 0

has no real roots.

Let the shorter side be $x$ m. Then the longer side is $(x + 80)$ m and the diagonal is $(x + 60)$ m.
By Pythagoras: $x^{2} + (x + 80)^{2} = (x + 60)^{2}$
$x^{2} + x^{2} + 160 x + 6400 = x^{2} + 120 x + 3600$
$x^{2} + 40 x + 2800 = 0$
$D = 4 0^{2} - 4 (1) (2800) = 1600 - 11200 = - 9600 < 0$
The equation has no real roots, so no rectangle has these measurements.
(With the NCERT data, longer side 30 m more than the shorter side, $x^{2} - 60 x - 2700 = 0$ gives sides 90 m and 120 m.)

Also asked in: 2023 Basic 430/1/3

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/1/1

The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present age.

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Answer: Father 36 years, son 9 years

Let the father's present age be $x$ years; the son's age is $(45 - x)$ years.
Five years ago: $(x - 5) (40 - x) = 124$
$40 x - x^{2} - 200 + 5 x = 124$
$x^{2} - 45 x + 324 = 0$
$(x - 36) (x - 9) = 0$ , so $x = 36$ or $x = 9$ .
The father must be older, so $x = 36$ .
Father = 36 years, son = 45 − 36 = 9 years.

Also asked in: 2023 Basic 430/1/3

Q345 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/1/2

The sum of reciprocals of Roohi’s age (in years) 3 years ago and 5 years hence from now is $\frac{1}{3}$ . Find her present age.

Show answer & solution

Answer: 7 years

Let Roohi's present age be $x$ years.
$\frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3}$
$\frac{2 x + 2}{( x - 3 ) ( x + 5 )} = \frac{1}{3}$ , so $6 x + 6 = x^{2} + 2 x - 15$
$x^{2} - 4 x - 21 = 0$ , so $(x - 7) (x + 3) = 0$
$x = 7$ (age cannot be negative)
Roohi's present age is 7 years.

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/1/2

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Show answer & solution

Answer: 40 km/hr

Let the speed be $x$ km/hr.
$\frac{360}{x} - \frac{360}{x + 5} = 1$
$\frac{360 \times 5}{x ( x + 5 )} = 1$ , so $x^{2} + 5 x - 1800 = 0$
$(x + 45) (x - 40) = 0$
$x = 40$ (speed cannot be negative)
Speed of the train = 40 km/hr.

Q355 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/1

The difference of two numbers is 5 and the difference of their reciprocals is $\frac{1}{10}$ . Find the numbers.

Show answer & solution

Answer: 10 and 5, or

- 5

and

- 10

Let the numbers be $x$ and $x - 5$ (larger first).
$\frac{1}{x - 5} - \frac{1}{x} = \frac{1}{10}$ , so $\frac{5}{x ( x - 5 )} = \frac{1}{10}$ .
$x^{2} - 5 x - 50 = 0$ , i.e. $(x - 10) (x + 5) = 0$ .
$x = 10$ gives numbers 10 and 5; $x = - 5$ gives numbers $- 5$ and $- 10$ .
The numbers are 10 and 5 (or $- 5$ and $- 10$ ).

Q35 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/1

Find all the values of k for which the quadratic equation $2 x^{2} + k x + 8 = 0$ has equal roots. Also, find the roots.

Show answer & solution

Answer:

k = 8

(roots

- 2, - 2

) or

k = - 8

(roots 2, 2)

Equal roots: $b^{2} - 4 a c = 0$ , so $k^{2} - 64 = 0$ and $k = \pm 8$ .
For $k = 8$ : $2 x^{2} + 8 x + 8 = 0$ , i.e. $2 (x + 2)^{2} = 0$ , roots $- 2, - 2$ .
For $k = - 8$ : $2 x^{2} - 8 x + 8 = 0$ , i.e. $2 (x - 2)^{2} = 0$ , roots 2, 2.

Q325 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/2

A 2-digit number is four times the sum of its digits and twice the product of its digits. Find the number.

Show answer & solution

Answer: 36

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$10 x + y = 4 (x + y)$ gives $6 x = 3 y$ , so $y = 2 x$ .
$10 x + y = 2 x y$ gives $10 x + 2 x = 2 x (2 x)$ , i.e. $4 x^{2} - 12 x = 0$ .
$4 x (x - 3) = 0$ ; $x \neq = 0$ , so $x = 3$ and $y = 6$ .
The number is 36. (Check: 4 × 9 = 36 and 2 × 18 = 36.)

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/2

The length of the rectangle exceeds its breadth by 8 cm and the area of the rectangle is 240 cm $^{2}$ . Find the dimensions of the rectangle.

Show answer & solution

Answer: Length = 20 cm, breadth = 12 cm

Let the breadth be $x$ cm; then the length is $(x + 8)$ cm.
$x (x + 8) = 240$ , i.e. $x^{2} + 8 x - 240 = 0$ .
$(x + 20) (x - 12) = 0$ , so $x = 12$ (breadth cannot be negative).
Breadth = 12 cm and length = 20 cm.

Q325 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/3

Divide 16 into two parts such that twice the square of the greater part, exceeds the square of the smaller part by 164.

Show answer & solution

Answer: 10 and 6

Let the greater part be $x$ ; the smaller part is $16 - x$ .
$2 x^{2} - (16 - x)^{2} = 164$ , i.e. $2 x^{2} - 256 + 32 x - x^{2} = 164$ .
$x^{2} + 32 x - 420 = 0$ , i.e. $(x + 42) (x - 10) = 0$ .
$x = 10$ (a part of 16 cannot be negative), so the smaller part is 6.
Check: $2 \times 100 - 36 = 164$ . The parts are 10 and 6.

Q32 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Basic 430/5/3

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream, than to return to the same point. Find the speed of the stream and total time of the journey.

Show answer & solution

Answer: Speed of the stream = 6 km/h; total time = 3 hours

Let the speed of the stream be $x$ km/h. Upstream speed $= 18 - x$ , downstream speed $= 18 + x$ .
$\frac{24}{18 - x} - \frac{24}{18 + x} = 1$ , so $24 \times 2 x = 324 - x^{2}$ .
$x^{2} + 48 x - 324 = 0$ , i.e. $(x + 54) (x - 6) = 0$ , so $x = 6$ .
Upstream time $= \frac{24}{12} = 2$ h, downstream time $= \frac{24}{24} = 1$ h.
Speed of the stream = 6 km/h and total time of the journey = 3 hours.

Q345 marksLong AnswerQuadratic Equations CBSE 2023 · Standard 30/5/1

A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the journey, what was its first average speed ?

Show answer & solution

Answer: 36 km/h

Let the first speed be x km/h.
$\frac{54}{x} + \frac{63}{x + 6} = 3$ .
$54 (x + 6) + 63 x = 3 x (x + 6)$ , so $117 x + 324 = 3 x^{2} + 18 x$ .
$3 x^{2} - 99 x - 324 = 0$ , i.e. $x^{2} - 33 x - 108 = 0$ .
$(x - 36) (x + 3) = 0$ ; speed is positive, so x = 36.
First average speed = 36 km/h.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q34 (OR) (OR)5 marksLong AnswerQuadratic Equations CBSE 2023 · Standard 30/5/1

Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes 2 hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.

Show answer & solution

Answer: Smaller pipe: 5 hours; larger pipe: 3 hours

Let the smaller pipe take x hours; the larger takes (x – 2) hours.
$\frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15}$ .
$15 (2 x - 2) = 8 x (x - 2)$ , so $8 x^{2} - 46 x + 30 = 0$ , i.e. $4 x^{2} - 23 x + 15 = 0$ .
$(x - 5) (4 x - 3) = 0$ , so x = 5 or $x = \frac{3}{4}$ .
$x = \frac{3}{4}$ makes x – 2 negative, so x = 5.
Smaller pipe 5 hours, larger pipe 3 hours.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

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