Students preparing for CBSE Class 10 Maths Board Examination are suggested to practice the given problems for Mathematics:
Important 2 Marks Questions for Class 10 Maths Board are as follows-
Q.1: Find the value of k for which the roots of the quadratic equation
, will have equal value.Solution:
Given,
2x2 + kx + 8 = 0
Comparing with the standard form ax2 + bx + c = 0,
a = 2, b = k, c = 8
Condition for the equal roots is:
b2 – 4ac = 0
k2 – 4(2)(8) = 0
k2 – 64 = 0
k2 = 64
k = ±8
Q.2: Determine the AP whose third term is 5 and the seventh term is 9.
Solution:
Let a be the first term and d be the common difference of an AP.
Given,
Third term = 5
a + 2d = 5….(i)
Seventh term = 9
a + 6d = 9….(ii)
Subtracting (i) from (ii),
a + 6d – a – 2d = 9 – 5
4d = 4
d = 1
Substituting d = 1 in (i),
a + 2(1) = 5
a = 5 – 2 = 3
Therefore, the AP is: 3, 4, 5, 6,…
Q.3: Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear.
Solution:
Let the given points be:
A(x, y) = (x1, y1)
B(-4, 6) = (x2, y2)
C(-2, 3) = (x3, y3)
If three points are collinear then the area of the triangle formed by these points is 0.
i.e. (½) |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)| = 0
(½) |x(6 – 3) + (-4)(3 – y) + (-2)(y – 6)| = 0
x(3) – 4(3 – y) – 2(y – 6) = 0
3x – 12 + 4y – 2y + 12 = 0
3x + 2y = 0
Or
x = -2y/3
Q.4: In the figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T, prove that ∠PTQ = 2OPQ.
Solution:
Given that two tangents TP and TQ are drawn to a circle with centre O from an external point T
Let ∠PTQ = θ.
Now, by using the theorem “the lengths of tangents drawn from an external point to a circle are equal”, we can say TP = TQ. So, TPQ is an isosceles triangle.
Thus,
∠TPQ = ∠TQP = ½ (180°− θ ) = 90° – (½) θ
By using the theorem, “the tangent at any point of a circle is perpendicular to the radius through the point of contact”, we can say ∠OPT = 90°
Therefore,
∠OPQ = ∠OPT – ∠TPQ = 90° – [90° – (½) θ]
∠OPQ = (½)θ
∠OPQ = (½) ∠PTQ
⇒ ∠PTQ = 2 ∠OPQ.
Hence proved.
Q.5: A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle. [Use Ï€ = 22/7]
Solution:
Length of the arc = Length of wire
(θ/360°) 2Ï€r = 22
(60°/360°) × 2 × (22/7) × r = 22
(⅙) × (2/7) × r = 1
r = (7/2) × 6
r = 21
Therefore, the radius of the circle is 21 cm.
Q.6: If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3, what is the probability that x2 ≤ 4?
Solution:
Sample space = S = {-3, -2, -1, 0, 1, 2, 3}
n(S) = 7
Let E be the event of choosing a number x such that x2 ≤ 4.
i.e. x ≤ ±2
E = {-2, -1, 0, 1, 2}
n(E) = 5
P(E) = n(E)/n(S) = 5/7
Hence, the required probability is 5/7.
Q.7: The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Solution:
Let x and (x + 18°) be the supplementary angles.
That means,
x + (x + 18°) = 180°
2x = 180° – 18°
2x = 162°
x = 162°/2
x = 81°
Now, x + 18° = 81° + 18° = 99°
Therefore, the supplementary angles are 81° and 99°.
Q.8: Find the mean of the following distribution:
Class | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 | 11 – 13 |
Frequency | 5 | 10 | 10 | 7 | 8 |
Solution:
Class | Frequency (fi) | Midpoint (xi) | fixi |
3 – 5 | 5 | 4 | 20 |
5 – 7 | 10 | 6 | 60 |
7 – 9 | 10 | 8 | 80 |
9 – 11 | 7 | 10 | 70 |
11 – 13 | 8 | 12 | 96 |
Total | ∑fi = 40 | ∑xi = 40 | ∑fixi = 326 |
Mean = ∑fixi/ ∑fi
= 326/40
= 8.15
Therefore, the mean of the given distribution is 8.15.
Q.9: Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.
Solution:
Let (5 + 3√2 ) be a rational number.
5 + 3√2 = p/q (Where q ≠ 0 and p and q are co- prime numbers)
3√2 = (p/q) – 5
3√2 = (p – 5q )/q
√2 = (p – 5q )/ 3q
p and q are integers and q ≠ 0
Thus, (p -5q) / 3q is rational number.
Also, √2 is a rational number.
However, it is given that √2 is an irrational number.
This is a contradiction, and thus, our assumption that (5 + 3√2 ) be a rational number is wrong.
That means (5+3√2) is an irrational number.
Hence proved.
Q.10: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.
Solution:
Given quadratic equation is:
px2 – 14x + 8 = 0
Let α and 6α be the roots of the given quadratic equation.
Sum of the roots = -coefficient of x/coefficient of x2
α + 6α = -(-14)/p
7α = 14/p
α = 2/p….(i)
Product of roots = constant term/coefficient of x2
(α)(6α) = 8/p
6α2 = 8/p
Substituting α = 2/p from (i),
6 × (2/p)2 = 8/p
24/p2 = 8/p
3/p = 1
p = 3
Therefore, the value of p is 3.
Important 2 Marks Questions for Class 10 Maths for Practice
- Two different dice are tossed together. Find the probability:(i) of getting a doublet(ii) of getting a sum of 10, of the numbers on the two dice.
- Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?
- Find the area of a triangle whose vertices are given as (1, -1), (-4, 6) and (-3, -5).
- Find the mode of the following data:
CI 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 120 – 14- f 6 8 10 12 6 5 3 - The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar.
- Find the sum of the first 8 multiples of 3.
- A circle touches all four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.
- A line intersects the y-axis at the points P and Q respectively. If (2,-5) is the midpoint of PQ, then find the coordinates of P and Q.
- Which term of the A.P. 8, 14, 20, 26, …….. Will be 72 more than its 41st term?
- In the figure below, the radius of incircle ofof area 84 cm2 is 4 cm and the lengths of the segment AP and BP into which side AB is divided by the point of contact are 6 cm and 8 cm. Find the lengths of the sides AC and BC.
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