Quadratic Equations- Class X ICSE Mathematics MCQs

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An equation with one variable in which the highest power of the variable is two is called a quadratic equation. The general form of the equation is    a x² + b x + c = 0 where a, b, c  ∈ R and a ≠ 0.

Click here to open MCQs of Quadratic Equations

      Solutions of the above MCQs are given below

                    Solutions 

Question 1: The roots of 6 − 4 x²+5 x = 0 are 

  • (a)  − 2, −(3/4)
  • (b)  2,  3/4
  • (c) 2,  −(3/4)
  • (d) none of these

Solution:   6 − 4 x² + 5 x = 0

⇒  4 x² − 5 x − 6 = 0

⇒ 4 x² − 8 x + 3 x − 6 = 0

⇒ 4 x (x − 2) + 3 ( x − 2) = 0

⇒ (x – 2) (4 x + 3) = 0

∴ x = 2, −(3/4)

Option (c) 

Question 2:  The sum of two numbers is 9 and the sum of their

squares are 41. The numbers are 

  • (a)  7,  2
  • (b)  6,  3
  • (c)  5,  4
  • (d)  1,  8

Solution:  Let the numbers be x and 9 − x.

∴   x² + (9 − x)² = 41

⇒   x² + 81 – 18 x + x² = 41

⇒   2 x² −18 x + 40 = 0

⇒ x² − 9 x + 20 = 0

⇒ x² − 5 x − 4 x + 20 = 0

⇒ (x − 5) ( x − 4) = 0

∴ x = 5,  4  :    If x = 5, other no 9 −5 = 4;  if x = 4, other no = 5

the numbers are 5, 4

Option (c) 

Question 3: Which of the following is a quadratic equation?

  • (a) (x+4) (2x -1) = (2x+3) (x − 2) 
  • (b) (x−1)³= x³ − 2 x²
  • (c) x(x+1) = (x+3) (x −3)
  • (d)  (x −3)² +1 = x² + 2 x + 4

Solution:    (a)   (x+4) (2x −1) = (2 x + 3) (x − 2)

⇒ 2 x²+8x−x−4 = 2x² + 3x −4 x – 6

⇒  8 x + 2 = 0    which is not a quadratic equation

(b)    (x −1)³ = x³ − 2 x²

⇒  x³ − 3 x² + 3 x − 1 = x³ − 2 x²

⇒  x² − 3 x + 1 = 0  which is a quadratic equation

(c)     x (x + 1) = (x + 3) (x − 3)

⇒ x² + x = x² − 9

⇒ x + 9 = 0  which is not a quadratic equation

(d)    (x − 3)² + 1 = x² + 2 x + 4

⇒ x² − 6 x + 9 + 1 = x² + 2x + 4

⇒ 8 x − 6 = 0      which is not a quadratic equation

Option (b)

Question 4: If  − (2/3) is a root of the equation

 k x² − 13 x − 10 = 0, the value of k is

  • (a)  1
  • (b)  2
  • (c)  3
  • (d) 4

Solution:      Putting k = −(2/3) in the equation k x² −13 x − 10 = 0

k (−2/3)² − 13 (−2/3) −10 = 0

⇒ k(4/9) + (26/3) – 10 = 0

⇒ k (4/9) = 10 − (26/3)

⇒ k (4/9) = 4/3    ⇒ k = 3

Option (c) 

Question 5:    The roots of the equation x² − 5 x + 6 = 0 are 

  • (a)  6,  −1
  • (b)  −2, 3
  • (c)  1, −6 
  • (d)  2,  3 

Solution:    x² – 5 x + 6 = 0

⇒ x² −2 x −3 x + 6 = 0

⇒ (x − 2) (x − 3) = 0

⇒ x = 2, 3

Option (d) 

Question 6: The solution of the equation 2 x² − 9 x + 10 = 0,  x ∈ Z 

  • (a)  2,  (5/2)
  • (b)  – 2,  (5/2)
  • (c)  2
  • (d)  (5/2) 

Solution:      2 x² − 9 x + 10 = 0,  x ∈ Z

⇒ 2 x² − 4 x −5 x + 10 = 0

⇒ 2 x (x − 2) − 5 ( x − 2) = 0

⇒ (x − 2) (2 x − 5) = 0

⇒ x = 2, (5/2)   but x ∈ Z

so x = 2 

Option (c)

Question 7:     The solution of the quadratic equation x² – 3(x + 3) = 0

 correct to two significant figures 

  • (a)  − 4.854,  1.854
  • (b)   4.85,  −1.854
  • (c)   4.85,  −1.85
  • (d)  4.9,  − 1.9 

Solution:    x² − 3 (x + 3) = 0

⇒   x² − 3 x − 9 = 0

⇒ x = 4.9,  − 1.9 (correct to two significant figures)

Question 8:   The discriminant of the equation 4 x² − 5 x − 3 = 0 is 

  • (a)  −23
  • (b)    23
  • (c)    73
  • (d)  none of these

Solution:    Discriminant = (−5)²− 4 × 4 × (−3) = 25 + 48 = 73

Option (c) 

Question 9:  The quadratic equation   3 x² −4 √3 x + 4 = 0  has 

  • (a) two distinct real roots
  • (b)  two equal real roots
  • (c)  no real roots
  • (d) two imaginary roots 

Solution:    Discriminant = (−4√3)²  − 4 ×3×4 = 48 −48 = 0

Two roots are real and equal     

Option (b) 

Question 10:      A trader bought a number of articles for ₹ 1200.

Ten were damaged and he sold each of the rest at ₹ 2 more than

what he paid for it, thus clearing a profit of ₹ 60 on the whole

transaction. The number of articles bought by the trader

  • (a)  40
  • (b)  60
  • (c)  80
  • (d)  100

Solution:  1st case:    Let no of articles be x. ∴ original price of each article = ₹ (1200/x)

2nd case:  No of articles = x − 10 and S.P = ₹ 1260  (net profit = ₹ 60)

Price of each article = ₹ {1260/(x −10) }

so the equation is    1260/(x −10) −(1200/x )  = 2

⇒  60 x + 12000 = 2 x (x − 10)

⇒ 30 x + 6000 = x (x − 10)

⇒ x² − 10 x − 30 x − 6000 = 0

⇒ x² − 40 x − 6000 = 0

⇒ x² − 100 x + 60 x − 6000 = 0

⇒ x (x − 100) + 60 (x − 100) = 0

⇒ (x − 100) (x + 60) = 0

∴ x = 100

Option (d)

Question 11:  The distance by road between two towns A and B is 216 km

 and by rail, it is 208 km. A car travels at a speed of x km/h and the train

   travels at a speed that is 16km/h faster than the car.

(i)        the time taken by the car to reach town B from A in terms of x  is 

  •    (a)   (208/x) hr
  •    (b)    {216/(x+16)}  hr 
  •    (c)    (216/x) hr
  •    (d)    {208/(x+16)} hr       

Solution:   (216/x) hr

Option (c)

(ii)         The time taken by the train  to reach town B from A in terms of x is 

  •  (a)   (208/x) h
  •   (b)    {216/(x+16)}  h  
  •   (c)    (216/x) h
  •   (d)    {208/(x+16)} h    

Solution:    {208/(x+16)} h 

Option (d)

(iii)      If the train takes 2 hours less than the car to reach town B,

 then the quadratic equation formed is 

  • (a)   x² −12 x − 1728 = 0 
  • (b)   x² + 12 x + 1728 = 0
  • (c)  x² −12 x +  1728 = 0 
  • (c)   x² + 12 x − 1728 = 0

Solution:  (216/x) − (208/(x+16)) = 2

⇒ (216 x + 216 × 16  − 208 x) / (x (x +16))  = 2

⇒  8 x + 216 × 16 = 2 (x² + 16 x)

⇒ 4 x + 216 × 8 = x² + 16 x

x² + 12 x − 1728 = 0     

Option (c) 

(iv) the speed of the car is

  • (a)   36 km/h
  • (b)   52 km/h
  • (c)  16 km/h
  • (d)  68 km/h

Solution:   x² + 12 x – 1728 = 0

⇒  x² + 48 x − 36 x – 1728 = 0

⇒  x(x + 48) − 36(x + 48) = 0

⇒ (x + 48) ( x − 36) = 0

∴ x = − 48 (not possible)  or  x = 36

∴ the speed of the car is 36 km/h     

Option (a)

Question 12: A two-digit number is such that the product of the

 digits is 12. When 36 is added to this number the digits interchange

  their places. Find the number. 

  • (a)   43
  • (b)  62
  • (c)  26
  • (d)  34 

Solution:  Let the unit’s digit of the two-digit number be x. ∴ its ten’s digit = (12/x)

∴ the number is  10 × (12/x) + x = (120/x) +x

On interchanging the digits, the number = 10 x + (12/x)

According to question       10 x + (12/x) = (120/x) + x + 36

⇒  9 x = (120/x) − (12/x) + 36

⇒ 9 x = (108/x) + 36

⇒ 9 x² = 108 + 36 x

⇒ x² = 12 + 4 x

⇒ x² − 4 x − 12 = 0

⇒ x² − 6 x + 2 x −12 = 0

⇒ x(x − 6) + 2(x −6) = 0

⇒ (x − 6) (x + 2)

⇒ x = 6  or x = −2 ( x being a digit of a number can’t be negative)

⇒ x = 6

∴ unit digit = 6 and ten’s digit = 12/6 = 2.

Hence the number is 26

Option (c)

Question 13:  At an annual function of the school, each student gives

  a gift to every other student. If the number of gifts is 4830, find the

   number of students.

  • (a)    50
  • (b)    60
  • (c)    70
  • (d)    80

Solution:    Let the number of students be x.

∴   x (x −1) = 4830

⇒    x² − x − 4830 = 0

⇒ x² − 70 x + 69 x − 4830 = 0

⇒  x(x − 70) + 69(x − 70) = 0

⇒   (x − 70) ( x + 69) = 0

⇒  x = 70   or  x = − 69 (number of students can’t be negative)

Hence number of students = 70 

Option (c) 

Question 14:      If the perimeter of a rectangular plot is 32 m and the

  length of its diagonal is 10 m, the area of the plot is 

  • (a)   39 m²
  • (b)   78 m²
  • (c)   30 m²
  • (d)  none of these

Solution:   Let the length and breadth of the rectangle be x and y respectively.

∴   2(x + y) = 32      ⇒ x + y = 16    ——— (i)

According to Pythagoras theorem,     x² + y²  = 10²  ⇒ x² + y² = 100  ——-  (ii)

squaring (i),         x² + y² + 2 x y =  256

⇒  100 + 2 x y = 256

⇒    2 x y =  156

⇒  x y = 78

the area of the rectangle = length × breadth = x y = 78 m² 

Option (b) 
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                 MCQ               Class 10            Mathematics

Banking (Set 1)
Banking (Set 2)
GST
Linear Inequations (Set 1)
Linear Inequations (Set 2)
Quadratic Equations
Ratio and Proportion
Matrices
Arithmetic Progression (A.P.)
Similarity (Set 1)
Similarity (Set 2)
Reflection
Section Formula
Straight Lines
Probability
Circle (Set 1)
Circle (Set 2)
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