# Class 10 Case based problem of Chapter 4 Quadratic eq 2

# Chapter 4:Quadratic Equation

Class 10 Case based problem of Chapter 4 quadratic eq 1

Class 10 Case based problem of Chapter 4 Quadratic eq 2

# Case Based: 4

**Â Â Â The two gears shown in the figure resemble two circle which touch each other externally. The sum of their areas is 130Ï€ sq. cm and distance between their centres is 14 cm.**

**(A) Taking the radius of one circle to be x cm, the quadratic equation for the situation described above is:**

(a) xÂ² – 16x + 63 = 0

(b) xÂ² – 16x – 63 = 0

(c) xÂ² – 14x + 33 = 0

(d) xÂ² + 14x – 33 = 0

**(B) The radii of two circles are:**

(A)Â 3 cm and 11 cm

(b) 5 cm and 9 cm

(c) 4 cm and 10 cm

(d) 6 cm and 8 cm

**(C) The roots of the quadratic equation abxÂ² + (bÂ² – ac)x – bc = 0 are:**

(a) Real and equal

(b) Real and distinct

(c) No real roots exist

(d) None of the above

**(D) The roots of the quadratic equation aÂ²xÂ² – 3abx + 2bÂ² = 0 are:**

(a)

(b)

(c)

(d)

**(E) The value of p for which the quadratic equation x(x – 4) + p = 0 has real and distinct roots, is:**

(a) p < 4Â Â Â Â Â Â Â Â Â Â (b) p > 4

(c) p < -4Â Â Â Â Â Â Â Â Â (d) p > -4

## Solution:

**(A) Answer (c) xÂ² – 14x + 33 = 0**

**Explanation:Â **Radius of one circle = x cm.

and the radius of second circle = (14 – x) cm

The area of first circle = Ï€ xÂ² cmÂ²

The area of the second circle = Ï€(14 – x)Â² cmÂ²

According to the question,

Ï€ xÂ² + Ï€(14 – x)Â² = 130 Ï€

â‡’ Ï€(xÂ² + 196 + xÂ² – 28x) = 130Ï€

â‡’Â 2xÂ² – 28 x + 196 = 130

â‡’ 2xÂ² – 28 x + 196 – 130 = 0

â‡’ 2xÂ² – 28 x + 66 = 0

â‡’ xÂ² – 14 x + 33 = 0 ….(i)

**(B) Answer (a) 3 cm and 11 cm**

**Explanation:** Solving equation (i)

xÂ² – 14 x + 33 = 0

â‡’ xÂ² – 11 x – 3 x + 33 = 0

â‡’ x(x – 11) – 3(x – 11) = 0

â‡’ (x – 11)(x – 3) = 0

â‡’ x = 3, 11

Radius of one circle = 3 cm

Radius of second circle = 11 cm

**(C) Answer (b) Real and distinct**

**Explanation: **Given quadratic equation

abxÂ² + (bÂ² – ac)x – bc = 0

A = ab, B = bÂ² – ac, C = – bc

Discrimnant (D) = BÂ² – 4AC

= (bÂ² – ac)Â² – 4(ab).(-bc)

The square of every integer is always positive

Hence, roots of the equation is Real and Distinct

**(D) Answer (d)**

**Explanation: aÂ²xÂ² – 3abx + 2bÂ² = 0**

â‡’ aÂ²xÂ² – 2abx – abx + 2bÂ² = 0

â‡’ ax(ax – 2bx) -b(ax – 2b) = 0

â‡’ (ax – 2b)(ax – b) = 0

**(E) Answer (a) p < 4**

**Explanation:Â **The given quadratic equation

x(x – 4) + p = 0

â‡’ xÂ² – 4x + p =0

Roots of the equation is Real and distinct

Then, D > 0

bÂ² – 4ac > 0

â‡’ (-4)Â² – 4Ã—1Ã—p > 0

â‡’ 16 – 4p > 0

â‡’ 16 > 4p

**â‡’ p < 4**

# Case Based: 5

Due to some technical problems, an aeroplane started late by one hour from its starting point. The pilot decided to increase the speed of the aeroplane by 100 km/h from its usual speed., to cover a journey of 1200 km in time.

**(A) The usual speed of the aeroplane is:**

(a) 200 km/hÂ Â Â Â Â Â Â Â (b) 300 km/h

(c) 400 km/hÂ Â Â Â Â Â Â Â Â (d) 600 km/h

**(B) The roots of the quadratic equation formed in this situation are:**

(a) Real and Distinct

(b) Real and Equal

(c) No Real roots exist

(d) None of the above

**(C) Which of the following is correct if the roots of the equation (a – b)xÂ² + (b – c)x + (c – a) = 0 are equal ?**

(a) 2c = b + aÂ Â Â Â Â (b) 2b = c + a

(c) 2a = b – cÂ Â Â Â Â Â (d) 2a = b + c

**(D) The values of k for which the difference between the roots of the equation xÂ² + kx + 3 = 0 is 2, is:**

(a) 0Â Â Â Â Â Â Â Â Â Â Â Â Â (b) Â± 2

(c) Â± 4Â Â Â Â Â Â Â Â Â Â Â Â (d) Â± 8

**(E) The values of m and n if x = 2 and x = 3 are solutions of the equation 3xÂ² – mx + 2n = 0; are:**

(a) m = 9, n = 15Â Â Â Â Â Â (b) m = 15, n = 9

(c) m = 3, n = 6Â Â Â Â Â Â Â (d) m = 6, n = 15

## Solution:

**(A) Answer (b) 300 km/h**

**Explanation:** Let the usual speed of planeÂ = x km/h

increased speed of planeÂ = (x + 100) km/h

According to the question

Usual speed of plane = 300 km/h

**(B) Answer (a) Real and distinct**

**Explanation: **Since, the roots are 300, -400Â

They are Real and Distinct

**(C) Answer (d) b + c = 2a**

**Explanation:**Given quadratic equation

(a – b)xÂ² + (b – c)x + (c – a) = 0

A = a – b, B = b – c, C = c – a

Roots are equal

Then, D = BÂ² – 4AC = 0

**(D) Answer (d) Â± 4**

**Explanation: **The given quadratic equation

xÂ² + kx + 3 = 0

a = 1, b = k, c = 3

Roots of the equation,

The difference between the roots = 2

Squaring both side

**(E) Answer (b) m = 15, n = 9**

**Explanation: The solution of the quadratic equation 3xÂ² – mx + 2n = 0 are**

x = 2 and x = 3. they satisfy the equation

……..(i)

………..(ii)

Substracting (ii) to (i) after multiplied by 2

putting in (i)

m – n = 6

# Case Based : 6

**Â Â Â Â Â The speed of a motor boat is 20 km/h. For covering the distance of 15 km, the boat took 1 hour more for upstream than downstream.**

**(A) Let speed of the stream be x km/h. Then speed of the motor boat in upstream will be:**

(a) 20 km/hÂ Â Â Â Â Â Â Â Â (b) (20 + x) km/h

(c) (20 – x) km/hÂ Â Â Â Â (d) 2 km/h

**(B) What is the relation between speed, distance and time ?**

(a) Speed = (distance)/time

(b) distance = (speed)/time

(c) time = speed Ã— distance

(d) speed = distance Ã— time

**(C) Which is the correct quadratic equation for the speed of the current ?**

(a) xÂ² + 30 x – 200 = 0

(b) xÂ² + 20 x – 400 = 0

(c) xÂ² + 30 x – 400 = 0

(d) xÂ² – 20 x – 400 = 0

**(D) What is the speed of current ?**

(a) 20 km/hÂ Â Â Â Â (b) 10 km/h

(c) 15 km/hÂ Â Â Â Â Â (d) 25 km/h

**(E) How much time boat took in down-stream ?**

(a) 90 minutesÂ Â Â Â (b) 15 minutes

(c) 30 minutesÂ Â Â Â (d) 45 minutes

## Solution:

**(A) Answer (c) (20 – x) km/h**

**Explanation: Speed of motor boat = 20km/h**

Let the speed of streamÂ = x km/h

Hence, speed of boat in upstream = (20 – x) km/h

**(B) Answer (a) Speed = Distance/Time**

**(C) Answer (c) xÂ² + 30 x – 400 = 0**

**Explanation:** Let the speed of streamÂ = x km/h

speed of boat in upstream = (20 – x) km/h

Speed of boat in down-stream = (20 + x) km/h

According to the question

**(D) Answer (b) 10 km/h**

**Explanation:** Solving the quadratic equation Â xÂ² + 30 x – 400 = 0

â‡’ xÂ² +40x – 10x – 400 =0$

â‡’ x(x + 40) – 10(x + 40) = 0

â‡’ (x + 40)(x – 10) = 0

â‡’ x = 10, -40

Speed can not be negative

**Hence, speed of the stream = 10 km/h**

**(E) Answer (c) 30 minutes**

**Explanation: **Speed of boat in down-stream = (x + 20) km/h

= (10 + 20) = 30 km/h

Time taken in down-stream = Distance/speed

= 15/30 = 1/2 hour

= 30 minutes

## Some other Case Based Question

Class 10: Case based problem of Chapter 4 quadratic eq 1

Class 10: Case based problem of Chapter 3 Pair of Linear eq 1

Class 10 :Case based problem of Chapter 3 Pair of Linear eq 2