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) |
ICSE – Class 10 – Mathematics – Sem 1 – 2021 – Sample Paper
|
ICSE – Class 10 -Mathematics – Sem 2 – 2022 – Sample Paper
|
Formula Mathematics
Formula and Theorems for Class X |
MCQ Class 10 Physics
Sound |
ISC / ICSE Board Paper
ISC – Class 12 – 2020 – Mathematics Question Paper |
ISC – Class 12 – 2019 – Mathematics Question Paper |
ICSE – Class 10 – 2020- Mathematics Question Paper |
ICSE – Class 10 – 2019 – Mathematics Question Paper |
ICSE – Class 10 – 2018 – Mathematics Question Paper |
ICSE – Class 10 – 2019 – Physics Question Paper |
ICSE – Class 10 – 2018 – Physics Question Paper |
ICSE – Class 10 -2020- Mathematics Question Paper– Solutions |
Contents – Tapati’s Classes |
2 thoughts on “Quadratic Equations- Class X ICSE Mathematics MCQs”