Q1 The first three terms of an AP respectively are

(A) –3 (B) 4 (C) 5 (D) 2

Solution: (C) 5

Q2 In Fig. 1, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is :

(A) 3.8 (B) 7.6 (C) 5.7 (D) 1.9

Solution: (B) 7.6

Q3 In Fig. 2, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then ∠QOR equals:

(A) 67° (B) 134° (C) 44° (D) 46°

Solution: (B) 134°

Q4. A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is:

(A) 43 (B) 43 (C) 22 (D) 4

Solution: (D) 4

Q5 If two different dice are rolled together, the probability of getting an even number on both dice, is:

(A) 136 (B) 12 (C) 16 (D) 14

Solution: (D) 14

Q6 A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is:

(A) 23 (B) 16 (C) 13 (D) 1130

Solution: (C) 13

Q7 If the points A(x, 2), B(−3, −4) and C(7, − 5) are collinear, then the value of x is:

(A) −63 (B) 63 (C) 60 (D) −60

Solution: (A) −63

Q8 The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:

(A) 3 (B) 5 (C) 4 (D) 6

Solution: (B) 5

SECTION- B

Q9 Solve the quadratic equation 2x2 + ax − a2 = 0 for x.

Solution: Comparing the given equation with the standard quadratic equation (ax2 + bx + c = 0),

We get:

a = 2, b = a and c =-a2

Using the quadratic formula,

= - b ± /2a,

we get: x = - a ± / 2×2 =- a± /4 = (-a ± 3a)/4⇒ x= (- a+3a)/4 Þ

Þ x = a/2 or x= - a -3a / 4 = - a

So, the solutions of the given quadratic equation are x=a/2 or x = -a.

Q10. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

Let a be the first term and d be the common difference.

Given: a = 5 Tn = 45 Sn = 400

We know:

Tn = a + (n − 1)d ⇒ 45 = 5 + (n − 1)d ⇒ 40 = (n − 1)d ... (i)

and Sn = n/2[a + Tn]

⇒400=n/2[5+45]⇒n/2=400/50 ⇒ n = 16

On substituting n = 16 in (i), we get:

40 = (16 − 1)d ⇒ 40 = (15)d ⇒ d = 40/15=8/3 Thus, the common difference is 8/3.

Q11 Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Solution:

Now, XB∥AO

⇒ ∠XBO + ∠AOB = 180° sum of adjacent interior angles is 180°)

Now, ∠XBO = 90° (A tangent to a circle is perpendicular to the radius through the point of contact)

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = 180° − 90° = 90°

Similarly, ∠AOC = 90°

∴ ∠AOB + ∠AOC = 90° + 90° = 180°

Hence, BOC is a straight line passing through O.

Thus, the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Q12 If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that ∠QPR = 120°, prove that 2PQ = PO.

Solution: We know that the radius is perpendicular to the tangent at the point of contact.

∴∠OQP=90∘

We also know that the tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.

∴∠QPO=60∘

Now, in ∆QPO:

cos60∘ =PQ/ PO⇒1/2=PQ/PO⇒2PQ=PO

Q13 Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.

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3y – 1, 3y + 5 and 5y + 1. Then y equals:

(A) –3 (B) 4 (C) 5 (D) 2

Solution: (C) 5

Q2 In Fig. 1, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is :

(A) 3.8 (B) 7.6 (C) 5.7 (D) 1.9

Solution: (B) 7.6

Q3 In Fig. 2, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then ∠QOR equals:

(A) 67° (B) 134° (C) 44° (D) 46°

Solution: (B) 134°

Q4. A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is:

(A) 43 (B) 43 (C) 22 (D) 4

Solution: (D) 4

Q5 If two different dice are rolled together, the probability of getting an even number on both dice, is:

(A) 136 (B) 12 (C) 16 (D) 14

Solution: (D) 14

Q6 A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is:

(A) 23 (B) 16 (C) 13 (D) 1130

Solution: (C) 13

Q7 If the points A(x, 2), B(−3, −4) and C(7, − 5) are collinear, then the value of x is:

(A) −63 (B) 63 (C) 60 (D) −60

Solution: (A) −63

Q8 The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:

(A) 3 (B) 5 (C) 4 (D) 6

Solution: (B) 5

SECTION- B

Q9 Solve the quadratic equation 2x2 + ax − a2 = 0 for x.

Solution: Comparing the given equation with the standard quadratic equation (ax2 + bx + c = 0),

We get:

a = 2, b = a and c =-a2

Using the quadratic formula,

= - b ± /2a,

we get: x = - a ± / 2×2 =- a± /4 = (-a ± 3a)/4⇒ x= (- a+3a)/4 Þ

Þ x = a/2 or x= - a -3a / 4 = - a

So, the solutions of the given quadratic equation are x=a/2 or x = -a.

Q10. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

Let a be the first term and d be the common difference.

Given: a = 5 Tn = 45 Sn = 400

We know:

Tn = a + (n − 1)d ⇒ 45 = 5 + (n − 1)d ⇒ 40 = (n − 1)d ... (i)

and Sn = n/2[a + Tn]

⇒400=n/2[5+45]⇒n/2=400/50 ⇒ n = 16

On substituting n = 16 in (i), we get:

40 = (16 − 1)d ⇒ 40 = (15)d ⇒ d = 40/15=8/3 Thus, the common difference is 8/3.

Q11 Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Solution:

Now, XB∥AO

⇒ ∠XBO + ∠AOB = 180° sum of adjacent interior angles is 180°)

Now, ∠XBO = 90° (A tangent to a circle is perpendicular to the radius through the point of contact)

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = 180° − 90° = 90°

Similarly, ∠AOC = 90°

∴ ∠AOB + ∠AOC = 90° + 90° = 180°

Hence, BOC is a straight line passing through O.

Thus, the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Q12 If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that ∠QPR = 120°, prove that 2PQ = PO.

Solution: We know that the radius is perpendicular to the tangent at the point of contact.

∴∠OQP=90∘

We also know that the tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.

∴∠QPO=60∘

Now, in ∆QPO:

cos60∘ =PQ/ PO⇒1/2=PQ/PO⇒2PQ=PO

Q13 Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.

**SA–II Subject Mathematics class 10 CBSE Board 2014[Delhi] All Sections**

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