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

**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 = Î¸.

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°− Î¸ ) = 90° – (½) Î¸

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) Î¸]

∠OPQ = (½)Î¸

∠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

(Î¸/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 Î± and 6Î± be the roots of the given quadratic equation.

Sum of the roots = -coefficient of x/coefficient of x2

Î± + 6Î± = -(-14)/p

7Î± = 14/p

Î± = 2/p….(i)

Product of roots = constant term/coefficient of x2

(Î±)(6Î±) = 8/p

6Î±2 = 8/p

Substituting Î± = 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.

## No comments:

## Post a Comment