HBSE Class 10th Maths Solved Question Paper 2020

Q1. Express 0.375 in the form p/q. 

Sol. 

0.375 = 375/1000 = 3/8 

Q2. The zeroes of 6x² – 7x – 3 are :

(A) -1/3, 3/2 

(B) -7/3, -3/6 

(C) 7/6, -3/6 

(D) None of these 

Ans. (A) -1/3, 3/2 

6x² – 7x – 3 = 0 

6x² – 9x + 2x – 3 = 0

3x(2x – 3) +1(2x – 3) = 0

(2x – 3)(3x + 1) = 0 

Either 3x + 1 = 0  or  2x – 3 = 0 

x = -1/3, 3/2 

Q3. Solve :  x + y = 14,  x – y = 4 

Sol. 

x + y= 14 ……..(i) 

x – y= 4 ………(ii) 

Add eqn.(i) and eqn.(ii), 

2x = 18

x = 18/2 = 9 

Put x = 9 in eqn.(i), 

9 + y = 14 

y = 14 – 9 = 5 

So, x = 9 and y = 5 

Q4. Which one is an A. P. series ?

(A) 2, 4, 8, 12, ………

(B) 0.2, 0.22, 0.222, …….

(C) -10, -6, -2, 2, ………

(D) 1, 3, 9, 27, ………. 

Ans. (C) -10, -6, -2, 2, ………

Common difference = d 

d = -6 – (-10) = -2 – (-6) = 2 – (-2) = 4 

Q5. Find the 10th term of A. P. 2, 7, 12, …….. 

Sol. 

Here, a = 2,  d = 7 – 2 = 5,  n = 10 

Tn or an = a + (n-1)d 

a10 = 2 + (10-1)5 = 2 + (9)5 = 2 + 45 = 47 

Q6. Fill in the blank using correct word given in bracket :

All circles are …………. (congruent, similar)

Ans. similar

Q7. Sides of two similar triangles are in the ratio 4 : 9. Areas of their triangles are in the ratio :

(A) 16 : 81

(B) 8 : 18

(C) 81 : 16

(D) 12 : 27 

Ans. (A) 16 : 81

If two triangles are similar to each other, then the ratio of the area of triangles will be equal to the square of the ratio of the corresponding sides of triangle.

So, Areas of their triangles are in the ratio = (4)² : (9)² = 16 : 81 

Q8. If the areas of two similar triangles are 36 m² and 121 m² respectively, the ratio of there corresponding sides is : 

(A) 11 : 6

(B) 6 : 11

(C) 9 : 11

(D) None of these 

Ans. (B) 6 : 11

If two triangles are similar to each other, then the ratio of the corresponding sides of triangles will be equal to the square root of the ratio of the corresponding areas of triangle.

So, Sides of triangles are in the ratio = √36 : √121 = 6 : 11 

Q9. From a point Q, the length of tangent to a circle is 24 cm and distance of Q from centre is 25 cm. Find the radius of circle. 

Sol. 

Using Pythagoras Theorem, 

H² = P² + B² 

25² = r² + 24² 

r² = 25² – 24² = 625 – 576 = 49 

r = √49 = 7 cm 

Q10. Find the distance between the points (2, 3) and (4, 1). 

Sol. 

A(2, 3) and B(4, 1) 

Here x1 = 2, x2 = 4, y1 = 3, y2 = 1 

Using Distance Formula, 

AB = √x2-x1)²+(y2-y1)²

      = √(4-2)²+(1-3)²

      = √(2)²+(-2)²

      = √4+4 = √8 = 2√2 unit

 

Q11. If (3, 4) is mid point of the line segment whose one end is (7, -2), then find the coordinates of the other end point. 

Sol.

O is Mid point of AB, 

A(x, y) and O(3,4) and B(7, -2) 

Using Mid Point Formula,

O(3, 4) = (x+7)/2 , (y-2)/2 

(x+7)/2 = 3 and (y-2)/2 = 4 

x + 7 = 6 and y – 2 = 8 

x = 6 – 7 = -1 and y = 8 + 2 = 10 

Co-ordinates of A(x, y) are (-1, 10). 

Q12. Find the value of  tan65°/cot25°.

Sol. 

tan65°/cot25° = tan(90°-25°)/cot25° = cot25°/cot25° = 1 

Q13. If sinA = 3/4 , then cosA is :

(A) 4/√7

(B) √7/4 

(C) 3/√7

(D) None of these 

Ans. (B) √7/4 

sin²A + cos²A = 1  

cos²A = 1 – sin²A 

cos²A = 1 – (3/4)² = 1 – 9/16 = 7/16 

cosA = √7/4 

Q14. Find the area of a sector of a circle with radius 4 cm, if angle of the sector is 30°. (Use π = 3.14) 

Sol. 

Area = θ/360 × πr² = 30/360 × 3.14 × 4 × 4 = 4.19 cm² 

Q15. The diameter of the base of right circular cylinder is 2r and its height is h. The curved surface area is :

(A) πr²h

(B) 2πrh 

(C) 2πr(r+h)

(D) None of these 

Ans. (B) 2πrh 

Q16. A bag contains 3 blue balls, 2 white balls and 4 red balls. If one ball is taken out at random from the bag. What is the probability that it will be White ? 

Sol. 

Total balls = 3 + 2 + 4 = 9

White balls = 2 

P(white) = 2/9 

Q17. Prove that √2 is an irrational number. 

Sol.

Let us assume that √2 is a rational number.

∴ √2 = p/q,  where p, q are co-prime integers and q ≠ 0 

p = √2q  

Squaring both side, we get   

p² = 2q² ……….. (i)  

∴ 2 is a factor of p²  

∴ 2 divides p² 

∴ 2 divides p

Put, p = 2m, m is integer

Now from (i), 

4m² = 2q² 

q² = 2m² 

∴ 2 is factor of q²   

∴ 2 divides q² 

∴ 2 divides q 

Hence, p,q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

So, √2 is irrational number.

Q18. Divide the polynomial p(x) = 3x³ + x² + 2x + 5 by the polynomial q(x) = x² + 2x + 1. Find the quotient and remainder. 

Sol. 

Q19. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. 

Sol. 

ED = height of girl = 0.9 m  

BD = 1.2 × 4 = 4.8 m 

AB = 3.6 m  

AF = 3.6 – 0.9 = 2.7 m  

EF = BD = 4.8 m  

Let CD = x m  

In similar ∆ABC and ∆AFE,  

BC/EF = AB/AF 

(x+4.8)/4.8 = 3.6/2.7

x = 1.6 m  

Q20. Prove that :

secA(1-sinA)(secA+tanA) = 1 

Sol. 

LHS = secA(1-sinA)(secA+tanA) 

= 1/cosA × (1-sinA) × (1/cosA + sinA/cosA) 

= 1/cosA × (1-sinA) × [(1+sinA)/cosA]

= (1-sin²A)/cos²A

= cos²A/cos²A = 1 = RHS 

Q21. Find the circumference of a circle whose area is 6.16 cm². 

Sol.

Area of circle = 6.16 cm² 

πr² = 6.16 

r² = 6.16 × 7/22 = 196/100 = 1.96 

r = √1.96 = 1.4 m 

Circumference of circle = 2πr = 2 × 22/7 × 1.4 = 8.8 m 

Q22. Solve : 

5/(x-1) + 1/(y-2) = 2  and  6/(x-1) – 3/(y-2) = 1 

Sol. 

5/(x-1) + 1/(y-2) = 2 ……..(i) 

6/(x-1) – 3/(y-2) = 1 ……..(ii) 

Take 1/(x-1) = p and 1/(y-2) = q 

5p + q = 2 ………(iii) 

6p – 3q = 1 ……..(iv) 

Multiply eqn.(iii) by 3 and add in eqn.(iv), 

21p = 7 

p = 1/3 

Putting p = 1/3 in eqn.(iii), 

q = 1/3 

Now 1/(x-1) = p and 1/(y-2) = q 

1/(x-1) = 1/3  and  1/(y-2) = 1/3 

x – 1 = 3 and y – 2 = 3 

So, x = 4 and y = 5 

Q23. The speed of motor-boat is 18 km/h in still water. It takes one hour more to 24 km upstream than to return downstream the same distance. Find the speed of the stream. 

Sol. 

Let speed of stream = x km/h  

Speed of Motor-boat in still water = 18 km/h  

Speed of Motor-boat upstream = 18-x km/h  

Speed of Motor-boat down-stream = 18+x km/h

According to question, 

24/(18-x) – 24/(18+x) = 1 

x² + 48x – 324 = 0 

x² + 54x – 6x – 324 = 0 

x(x + 54) + 6(x + 54) = 0 

(x + 54)(x – 6) = 0 

So, x – 6 = 0 (x≠-54, speed never negative) 

x = 6 km/h 

Q24. How many terms of A. P. 24, 21, 18, …….. should be taken so that their sum is 78 ? 

Sol. 

Here, a = 24,  d = 21 – 24 = -3,  Sn = 78 

Sn = n/2 [2a + (n-1)d] 

78 = n/2 [2×24 + (n-1)(-3)] 

78×2 = n [48 – 3n + 3]

156 = -3n² + 51n 

3n² – 51n + 156 = 0 

n² – 17n + 52 = 0 

n² – 13n – 4n + 52 = 0

n(n – 13) – 4(n – 13) = 0

(n – 13)(n – 4) = 0 

Either n – 13 = 0 or n – 4 = 0 

So, n = 13 or 4  

  

Q25. Prove that the length of tangents drawn from an external point to a circle are equal. 

Sol. 

Let AP and BP be the two tangents to the circle with centre O.

To Prove : AP = BP

Proof :

In ΔAOP and ΔBOP

OA = OB (radii of the same circle)

∠OAP = ∠OBP = 90° (since tangent at any point of a circle is perpendicular to the radius through the point of contact)

OP = OP (common)

∴ΔAOP ≅ ΔBOP (by R.H.S. congruence criterion)

AP = BP (corresponding parts of congruent triangles, CPCT)

Hence the length of the tangents drawn from an external point to a circle are equal.

Q26. In what ratio does the point (-4, 6) divides the line segment joining the points A(-6, 10) and B(3, -8). 

Sol. 

Using Section Formula, 

x = (m1x2 + m2x1) ÷ (m1 + m2) 

– 4 = [3k+1(-6)] ÷ (k+1) 

– 4k – 4 = 3k – 6 

7k = 2 

k = 2/7 

Ratio = 2 : 7 

Q27. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will (i) ace (ii) not ace. 

Sol. 

Total cards = 52 

Ace cards = 4 

Not Ace cards = 52-4 = 48 

(i) P(ace) = 4/52 = 1/13 

(ii) P(not ace) = 48/52 = 12/13 

Q28. Solve the equation 2x² – 5x + 3 = 0 by completing the square method. 

Sol.

2x² – 5x + 3 = 0 

dividing both side by 2, 

x² – 5/2 x + 3/2 = 0 

x² – 2(x)(5/4) + (5/4)² – 25/16 + 3/2 = 0 

(x-5/4)² – 1/16 = 0 

(x-5/4)² = 1/16 = (1/4)² 

x – 5/4 = ± 1/4 

take x – 5/4 = 1/4 and x – 5/4 = -1/4 

x = 3/2, 1 

Q29. Construct a triangle whose sides are 4 cm, 5 cm and 6 cm and construct a similar triangle whose sides are 2/3th of the corresponding sides of given triangle. 

Sol.

Steps of construction– 

Step I- Let us draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

Step II- On considering the point A as centre, and draw an arc of radius 5 cm.

Step III- Let us, take the point B as its centre, and draw an arc of radius 6 cm.

Step IV- The arcs drawn will intersect each other at point C.

Step V- Now, we get AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.

Step VI- Now drawa ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

Step VII- Position 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

Step VIII- Join the point BA3 and draw a line through A2 which is parallel to the line BA3 that intersect AB at point B’.

Step IX- Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.

Step: X- Therefore, ΔAB’C’ is the required triangle.

Q30. Prove that :

(sinθ+cosθ-1)/(sinθ-cosθ+1) = 1/(secθ+tanθ) 

Sol.

L.H.S. = (sinθ+cosθ-1)/(sinθ-cosθ+1) 

dividing numerator and denominator by cosθ, 

= (tanθ+1-secθ)/(tanθ-1+secθ) 

= [1-(secθ-tanθ)]/(tanθ-1+secθ) 

= [(sec²θ-tan²θ)-(secθ-tanθ)]/(tanθ-1+secθ)

From numerator and denominator (tanθ-1+secθ) cancelled,  

= secθ-tanθ

Multiplied numerator and denominator by (secθ+tanθ), 

= (sec²θ-tan²θ)/(secθ+tanθ) 

= 1/(secθ+tanθ) = R.H.S. 

                                           OR

A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. 

Sol. 

AE = BG = 88.2 – 1.2 = 87 m 

In ∆ACE, 

AE/CE = tan60° 

87/CE = √3 

CE = 29√3 ………(i) 

In ∆BCG, 

BG/CG = tan30° 

87/CG = 1/√3 

CG = 87√3 

Balloon travelled distance = EG = GC – EC = 87√3 – 29√3 = 58√3 m 

Q31. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. 

Sol. 

It is given that

Volume of 1 cube = 64 cm³

(side)³ = 64 cm³

a³ = 64 = 4³ 

side (a) = 4 cm

Length of cuboid (l) = a + a = 4 + 4 = 8 cm

Breadth (b) = a = 4 cm

Height (h) = a = 4 cm

Surface Area of cuboid = 2(lb + bh + hl)

= 2(8×4 + 4×4 + 4×8)

= 2(32 + 16 + 32) = 2(80) = 160 cm² 

Hence, surface area of cuboid = 160 cm²

Q32. The distribution below gives the weight of 30 students of a class. Find the median weight of the students. 

Sol. 

Median = L + (N/2 – C)/f × h = 55 + (15-13)/6 × 5 = 55 + 10/6 = 55 + 1.67 = 56.67 

                                             OR 

The table below shows daily expenditure on food of 25 households in a locality. Find the mean daily expenditure.               

SET-B 

Q1. Express 0.104 in the form p/q. 

Sol. 

0.104 = 104/1000 = 13/125 

Q2. The zeroes of 4x² – 4x + 1 are :

(A) -1/2, -1/2 

(B) 1/2, 1/2

(C) 1, 1/4 

(D) -1/4, 1 

Ans. (B) 1/2, 1/2

4x² – 4x + 1 = 0 

4x² – 2x – 2x + 1 = 0

2x(2x – 1) – 1(2x – 1) = 0

(2x – 1)(2x – 1) = 0

(2x – 1)² = 0 

x = 1/2, 1/2 

Q3. Solve :  x – y = 3,  x/3 + y/2 = 6 

Sol. 

x – y = 3 ……(i) 

x/3 + y/2 = 6 ……(ii) 

Multiply eqn.(ii) by 6, 

2x + 3y = 36 …..(iii) 

Multiply eqn.(i) by 3 and add in eqn.(iii), 

5x = 45 

x = 45/5 = 9 

Put x = 9 in eqn.(i), 

9 – y = 3

y = 9 – 3 = 6 

So, x = 9 and y = 6 

Q4. Which one is an A. P. series ?

(A) 1, 2, 4, 8, ………

(B) 5, 8, 12, 15, …….

(C) 3, 6, 9, 12, ………

(D) None of these 

Ans. (C) 3, 6, 9, 12, ………

Common difference (d) = 6-3 = 9-6 = 12-9 = 3 

Q5. Find the 12th term of A. P. 7, 10, 13, 16, …….. 

Sol. 

Here, a = 7,  d = 10 – 7 = 3,  n = 12 

an = a + (n-1)d = 7 + (12-1)3 = 7 + (11)3 = 7 + 33 = 40 

Q6. Fill in the blank using correct word given in bracket :

All squares are …………. (similar, congruent)

similar 

Q7. Sides of two similar triangles are in the ratio 3 : 5. Areas of their triangles are in the ratio :

(A) 5 : 3 

(B) 9 : 25 

(C) √3 : √5 

(D) None of these 

Ans. (B) 9 : 25 

If two triangles are similar to each other, then the ratio of the area of triangles will be equal to the square of the ratio of the corresponding sides of triangle.

So, Areas of their triangles are in the ratio = (3)² : (5)² = 9 : 25 

Q8. If DE || BC and AD : BD = 2 : 3, then area (ΔADE) : area (ΔABC) is : 

(A) 2 : 5 

(B) 3 : 5 

(C) 4 : 25 

(D) 2 : 3 

Ans. (C) 4 : 25 

AD = 2, BD = 3, AB = AD + BD = 2 + 3 = 5 

area (ΔADE) : area (ΔABC) = (AD)² : (AB)² = (2)² : (5)² = 4 : 25 

Q9. The length of tangent from a point which is at a distance of 5 cm from the centre of circle is        4 cm. Find the radius of circle. 

Sol. 

Using Pythagoras Theorem, 

H² = P² + B² 

5² = r² + 4² 

r² = 5² – 4² = 25 – 16 = 9 

r = √9 = 3 cm 

Q10. Find the distance between the points (-5, 7) and (-1, 3). 

Sol. 

A(-5, 7) and B(-1, 3) 

x1 = -5, x2 = -1, y1 = 7, y2 = 3 

Using Distance Formula, 

AB = √x2-x1)²+(y2-y1)²

      = √(-1+5)²+(3-7)²

      = √(4)²+(-4)²

      = √16+16 = √32 = 4√2 unit

Q11. If origin is at one end of a line segment whose mid point is (1, 0), find the coordinates of other end of segment. 

Sol. 

C is Mid point of AB, 

A(0, 0) and C(1, 0) and B(x, y) 

Using Mid Point Formula,

C(1, 0) = (0+x)/2 , (0+y)/2 

(0+x)/2 = 1 and (0+y)/2 = 0 

x = 2 and y = 0

Co-ordinates of B(x, y) is (2,0). 

Q12. Find the value of  tan26°/cot64°. 

Sol.

tan26°/cot64° = tan(90°-64°)/cot64° = cot64°/cot64° = 1 

Q13. If sinA = 3/4 , then tanA is :

(A) 4/3

(B) 3/√7

(C) 4/√7

(D) None of these 

Ans. (B) 3/√7

sin²A + cos²A = 1  

cos²A = 1 – sin²A 

cos²A = 1 – (3/4)² = 1 – 9/16 = 7/16 

cosA = √7/4 

tanA = sinA/cosA = 3/4 ÷ √7/4 = 3/√7

Q14. Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 60°. 

Sol. 

Area = θ/360 × πr² = 60/360 × 22/7 × 6 × 6 = 132/7 cm² or 18.84 cm² 

Q15. The diameter of the base of right circular cylinder is 2r and its height is h. The total surface area is :

(A) 2πrh

(B) 2πr(r+h) 

(C) πr²h

(D) None of these 

Ans. (B) 2πr(r+h) 

Q16. A bag contains 3 blue balls, 2 white balls and 4 red balls. If one ball is taken out at random from the bag. What is the probability that it will be Blue ?

Sol.

No. of Blue balls = 3 

No. of White balls = 2 

No. of Red balls = 4 

Total balls = 3 + 2 + 4 = 9 

P(Blue) = 3/9 = 1/3 

Q17. Prove that √3 is an irrational number. 

Sol.

Let us assume that √3 is a rational number.

∴ √3 = p/q,  where p, q are co-prime integers and q ≠ 0 

p = √3q  

Squaring both side, we get   

p² = 3q² ……….. (i)  

∴ 3 is a factor of p²  

∴ 3 divides p² 

∴ 3 divides p 

Put  p = 3m, m is integer

Now from (i), 

9m² = 3q² 

q² = 3m² 

∴ 3 is factor of q²   

∴ 3 divides q² 

∴ 3 divides q

Hence, p,q have a common factor 3. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

So, √3 is irrational number.

Q18. Divide the polynomial p(x) = 3x² – x³ – 3x + 5 by the polynomial q(x) = x – 1 – x². Find the quotient and remainder. 

Sol.

Q19. CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, prove that : 

                           ΔAMC ~ ΔPNR 

Sol. 

Proof : 

CM is median of ∆ABC, 

AM = MB = 1/2AB ……(i) 

RN is median of ∆PQR, 

PN = QN = 1/2PQ ……(ii) 

Given, ΔABC ~ ΔPQR 

∠A = ∠P ……(iii) 

AB/PQ = BC/QR = CA/RP   (Corresponding sides of similar triangle are proportional)

AB/PQ = CA/RP 

2AM/2PN = CA/RP  (from (i) & (ii))

AM/PN = CA/RP …….(iv) 

In ΔAMC and ΔPNR, 

∠A = ∠P  (from (iii))

AM/PN = CA/RP  (from (iv)) 

So, ΔAMC ~ ΔPNR  (by SAS similarity criteria) 

Hence Proved. 

Q20. Prove that : 

(cotA-cosA)/(cotA+cosA) = (cosecA-1)/(cosecA+1) 

Sol. 

L.H.S. = (cotA-cosA)/(cotA+cosA) 

= (cosA/sinA – cosA)/(cosA/sinA + cosA) 

= cosA(1-sinA)/cosA(1+sinA) 

= (1-sinA)/(1+sinA)

= (1 – 1/cosecA)/(1 + 1/cosecA) 

= (cosecA-1)/(cosecA+1) = R.H.S. 

Q21. What is the area of the circle, the circumference of which is equal to the perimeter of a square of side 11 cm ? 

Sol. 

Circumference of circle = Perimeter of square 

2πr = 4 × side 

2 × 22/7 × r = 4 × 11 

r = 7 cm 

Area of circle = πr² = 22/7 × 7 × 7 = 154 cm² 

Q22. Solve : 

2/x + 3/y = 13  and  5/x – 4/y = -2 

Sol. 

2/x + 3/y = 13 …….(i)  

5/x – 4/y = -2 …….(ii) 

Take 1/x = p and 1/y = q 

2p + 3q = 13 ……(iii) 

5p – 4q = -2 ……(iv) 

Multiply eqn.(iii) by 4 and eqn.(iv) by 3, and Add each other

8p + 15p = 52 – 6 

23p = 46 

p = 46/23 = 2 

Put p = 2 in eqn.(iii), 

2×2 + 3q = 13 

3q = 13 – 4 = 9 

q = 9/3 = 3 

Now 1/x = p and 1/y = q 

1/x = 2 and 1/y = 3 

So, x = 1/2 and y = 1/3 

Q23. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. 

Sol. 

Let speed of train = x km/h 

Time = Distance/speed 

A.T.Q. (according to question), 

360/x – 360/(x+5) = 1 

x² + 5x – 1800 = 0 

x² + 45x – 40x – 1800 = 0 

x(x + 45) – 40(x + 45) = 0

(x + 45)(x – 40) = 0 

Either x – 40 = 0 or x + 45 = 0 

x = 40 km/h (speed never negative, so x≠-45)  

Q24. How many terms of A. P. 9, 17, 25, …….. should be taken so that their sum is 636 ? 

Sol. 

Here a = 9,  d = 17 – 9 = 8,  Sn = 636

Sn = n/2 [2a + (n-1)d] 

636 = n/2 [2×9 + (n-1)8] 

636×2 = n [18 + 8n – 8]

1272 = 8n² + 10n 

8n² + 10n – 1272 = 0 

4n² + 5n – 636 = 0 

After solving quadratic equation, 

n = 12 (numbers of terms never negative, so n≠-53/4)

So, numbers of terms, n = 12

Q25. Prove that the length of tangents drawn from an external point to a circle are equal. 

Sol. 

Let AP and BP be the two tangents to the circle with centre O.

To Prove : AP = BP

Proof :

In ΔAOP and ΔBOP

OA = OB (radii of the same circle)

∠OAP = ∠OBP = 90°(since tangent at any point of a circle is perpendicular to the radius through the point of contact)

OP = OP (common)

∴ΔAOP ≅ ΔBOP (by R.H.S. congruence criterion)

AP = BP (corresponding parts of congruent triangles, CPCT)

Hence the length of the tangents drawn from an external point to a circle are equal.

Q26. Find the coordinates of the point which divide the line segment joining the points (4, -3) and (8, 5) in a ratio 3 : 1 internally. 

Sol. 

Here x1 = 4, x2 = 8, y1 = -3, y2 = 5, m1 = 3, m2 = 1 

Using Section Formula, 

x = (m1x2 + m2x1) ÷ (m1 + m2) 

   = [3(8) + 1(4)] ÷ (3 + 1) 

   = (24+4) ÷ 4 = 28 ÷ 4 = 7  

y = (m1y2 + m2y1) ÷ (m1 + m2) 

   = [3(5) + 1(-3)] ÷ (3 + 1) 

   = (15-3) ÷ 4 = 12 ÷ 4 = 3

So, C(x, y) = (7, 3) 

Q27. A box contains 4 blue, 3 white and 2 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) White (ii) Blue (iii) Red. 

Sol. 

No. of blue marbles = 4 

No. of white marbles = 3 

No. of red marbles = 2 

Total marbles = 4 + 3 + 2 = 9 

(i) P(White) = 3/9 = 1/3 

(ii) P(Blue) = 4/9 

(iii) P(Red) = 2/9 

Q28. Solve the equation 5x² – 6x – 2 = 0 by completing the square method. 

Sol. 

5x² – 6x – 2 = 0

dividing both side by 5, 

x² – 6/5 x – 2/5 = 0

x² – 2(x)(3/5) + (3/5)² – (3/5)² – 2/5 = 0 

(x – 3/5)² = 9/25 + 2/5 

(x – 3/5)² = 19/25 

x – 3/5 = ± √19/5 

x = (3+√19)/5  and  x = (3-√19)/5 

Q29. Construct a triangle whose sides are 5 cm, 6 cm and 7 cm and construct a similar triangle whose sides are 7/5th of the corresponding sides of given triangle. 

Sol. 

Steps of construction–  

Step I- Firstly draw a line segment AB of length 5 cm.

Step II- Now cut an arc of radius 6 cm from point A and an arc of 7 cm from point B.

Step III- Name the point of intersection of arcs to be point C. 

Step IV- Now join point AC and BC. Thus ABC is the required triangle.

Step V- Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

Step VI- Cut seven equal parts of line AD namely  AA1, AA2, AA3, AA4, AA5, AA6, AA7. 

Step VII- Now join A5 to B. Draw a line A7B’ parallel to A5B.

Step VIII- And then draw a line B’C’ parallel to BC.

Hence AB’C’ is the required triangle.

Q30. Prove that :

(cosA-sinA+1)/(cosA+sinA-1) = cosecA+cotA 

Sol. 

L.H.S. = (cosA-sinA+1)/(cosA+sinA-1) 

dividing numerator and denominator by sinA, 

= (cotA-1+cosecA)/(cotA+1-cosecA) 

= (cosecA+cotA-1)/(cotA+1-cosecA) 

= [(cosecA+cotA-(cosec²A-cot²A)]/(cotA+1-cosecA)

= [(cosecA+cotA)(1-cosecA+cotA)]/(cotA+1-cosecA) 

= (cosecA+cotA)(1-cosecA+cotA)]/(1-cosecA+cotA) 

= (cosecA+cotA) = R.H.S. 

                                         OR

From a point on a bridge across a river, the angle of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, find the width of the river. 

Sol. 

Let AB be the width of the river  

Height of the bridge = CD = 3m   

In ∆ACD, 

tan30° = CD/AC 

1/√3 = 3/AC  

AC = 3√3 m 

In ∆CDB,  

tan45° = CD/BC 

1 = 3/BC 

BC = 3 m 

Width of the river = AB = AC + BC = 3√3 + 3 = 3(√3+1) m 

Q31. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and total height of the vessel is 13 cm. Find the inner surface area of the vessel. 

Sol. 

Radius of cylinder = Radius of hemisphere = 7 cm  

height of cylinder = 13 – 7  = 6 cm 

inner surface area of vessel = surface area of cylinder + surface area of hemisphere 

= 2πrh + 2πr² = 2πr(h+r) = 2 × 22/7 × 7(6+7) = 44 × 13 = 572 cm² 

Q32. The table below gives the percentage distribution of female teachers in the schools of rural areas. Find the mean percentage of female teachers. 

Sol. 

Mean =A + Σfd/N = 50 + (-360)/35 = 50 – 10.3 = 39.7 

                                          OR 

The following distribution gives the monthly consumption of consumers of a locality. Find the median of the distribution.     

Sol. 

Median = L + (N/2 – C)/f × h = 125 + (30-22)/20 × 20 = 125 + 8 = 133 

SET-C 

Q1. Express 0.0875 in the form p/q. 

Sol. 

0.0875 = 875/10000 = 7/80 

Q2. The zeroes of 3x² – x – 4 are :

(A) -1, -4/3

(B) 1, -4/3

(C) -1, 4/3 

(D) 1, 4/3 

Ans. (C) -1, 4/3 

3x² – x – 4 = 0

3x² + 3x – 4x – 4 = 0 

3x(x + 1) – 4(x + 1) = 0 

(x + 1)(3x + 4) = 0

Either x + 1 = 0  or  3x – 4 = 0 

So, x = -1, 4/3 

Q3. Solve :  x + y = 5,  2x – 3y = 4 

Sol. 

x + y = 5 …….(i)

2x – 3y = 4 ………(ii) 

Multiply eqn.(i) by 3 and add in eqn.(ii), 

3x + 2x = 15 + 4 

5x = 19 

x = 19/5 

Put x = 19/5 in eqn.(i), 

19/5 + y = 5

y = 5 – 19/5 = (25-19)/5

y = 6/5 

So, x = 19/5 and y = 6/5 

Q4. Which one is an A. P. series ?

(A) 1, 3, 9, 27, ………

(B) a, 2a, 3a, 4a, ………

(C) a, a², a³, a⁴, ………

(D) 1², 2², 3², 4², ……… 

Ans. (B) a, 2a, 3a, 4a, ………

Common difference (d) = 2a-a = 3a-2a = 4a-3a = a 

Q5. Find the 30th term of A. P. 10, 7, 4, …….. 

Sol. 

Here, a = 10,  d = 7 – 10 = -3, n = 30 

an = a + (n-1)d 

a30 = 10 + (30-1)(-3) = 10 + (29)(-3)= 10 – 87 = – 77 

Q6. Fill in the blank using correct word given in bracket :

All ………. triangles are similar. (isosceles, equilateral)

equilateral

Q7. Areas of two similar triangles are in the ratio of 4 : 5, then the ratio of their corresponding sides is :

(A) 16 : 25

(B) 5 : 4 

(C) 4 : 5

(D) 2 : √5 

Ans. (D) 2 : √5 

If two triangles are similar to each other, then the ratio of the corresponding sides of triangles will be equal to the square root of the ratio of the corresponding areas of triangle.

So, Sides of triangles are in the ratio = √4 : √5 = 2 : √5  

Q8. In figure, AB || QR, AB = 3 cm, QR = 9 cm and PR = 6 cm, the length of PB is : 

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) None of these 

Ans. (C) 2 cm

AB/QR = PB/PR 

3/9 = PB/6 

PB = 2 cm 

Q9. From a point P, the length of tangent to a circle is 12 cm and distance of P from centre is 13 cm. Find the radius of circle. 

Sol. 

Using Pythagoras Theorem, 

H² = P² + B² 

13² = r² + 12² 

r² = 13² – 12² = 169 – 144 = 25

r = √25 = 5 cm 

Q10. Find the distance between the points (-1, 5) and (7, 3). 

Sol. 

A(-1, 5) and B(7, 3) 

x1 = -1, x2 = 7, y1 = 5, y2 = 3 

Using Distance Formula, 

AB = √x2-x1)²+(y2-y1)²

      = √(7+1)²+(3-5)²

      = √(8)²+(-2)²

      = √64+4 = √68 = 2√17 unit 

Q11. Find the coordinates of A, where AB is diameter of circle whose centre is (2, -3) and B is (1, 4). 

Sol. 

M is Mid point of AB, 

A(x, y) and M(2, -3) and B(1, 4) 

Using Mid Point Formula,

M(2, -3) = (x+1)/2 , (y+4)/2 

(x+1)/2 = 2 and (y+4)/2 = -3

x+1 = 4 and y+4 = -6 

x = 3 and y = -10 

Co-ordinates of A(x, y) are (3, -10). 

Q12. Find the value of cos37°/sin53°. 

Sol. 

cos37°/sin53° = cos(90°-53°)/sin53° = sin53°/sin53° = 1 

Q13. If cotA = 8/15 , then sinA is :

(A) 15/17

(B) 8/17 

(C) 15/8 

(D) None of these 

Ans. (A) 15/17

cotA = 8/15 = Base/Perpendicular = QR/PR 

PQ = √15²+8² = √225+64 = √289 = 17 

sinA = Perpendicular/Hypotenuse= PR/PQ = 15/17 

So, sinA = 15/17 

Q14. Circumference of a circle is 22 cm, find the radius of the circle. 

Sol. 

Circumference of circle = 2πr 

2πr = 22 

2 × 22/7 × r = 22 

r = 22 × 7/22 × 1/2 = 7/2 cm = 3.5 cm

r = 3.5 cm 

Q15. The diameter of the base of right circular cylinder is 2r and its height is h. The volume of cylinder is :

(A) 2πr(r+h)

(B) πr²h

(C) 2πrh 

(D) None of these 

Ans. (B) πr²h 

Q16. A bag contains 3 blue balls, 2 white balls and 4 red balls. If one ball is taken out at random from the bag. What is the probability that it will be Red ?

Sol. 

No. of Blue balls = 3 

No. of White balls = 2 

No. of Red balls = 4 

Total balls = 3 + 2 + 4 = 9 

P(Red) = 4/9 

Q17. Prove that √5 is an irrational number. 

Sol. 

Let us assume that √5 is a rational number.

∴ √5 = p/q,  where p, q are co-prime integers and q ≠ 0 

p = √5q  

Squaring both side, we get   

p² = 5q² ……….. (i)  

∴ 5 is a factor of p²  

∴ 5 divides p² 

∴ 5 divides p 

Put  p = 5m, m is integer

Now from (i), 

25m² = 5q² 

q² = 5m² 

∴ 5 is factor of q²   

∴ 5 divides q² 

∴ 5 divides q

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

So, √5 is irrational number.

Q18. Divide the polynomial p(x) = x³ – 3x² + 5x – 3 by the polynomial q(x) = x² – 2. Find the quotient and remainder. 

Sol. 

Quotient = x – 3

Remainder = 7x – 9

Q19. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. 

Sol. 

Let height of tower = x 

In similar ∆ PQR and ∆ ABC, 

PQ/AB = QR/BC

x/6 = 28/4

x = 28/4 × 6 = 42 

So, height of  tower = 42 m

Q20. Prove that : 

(sinθ-cosθ+1)/(sinθ+cosθ-1) = 1/(secθ-tanθ) 

Sol. 

L.H.S. = (sinθ-cosθ+1)/(sinθ+cosθ-1) 

dividing numerator and denominator by cosθ, 

= (tanθ-1+secθ)/(tanθ+1-secθ) 

= (secθ+tanθ-1)/(tanθ+1-secθ) 

= [(secθ+tanθ)-(sec²θ-tan²θ)]/(tanθ+1-secθ)

= (secθ+tanθ)[1-secθ+tanθ]/(1-secθ+tanθ)

From numerator and denominator (1-secθ+tanθ) cancelled,  

= (secθ+tanθ) 

Multiply numerator and denominator by (secθ-tanθ), 

= (secθ+tanθ)(secθ-tanθ)/(secθ-tanθ) 

= (sec²θ-tan²θ)/(secθ-tanθ) 

= 1/(secθ-tanθ) = R.H.S.

Q21. The circumference of semi-circular piece of design is 72 cm. Find its area. 

Sol. 

Let r be the radius of semi-circle  

Circumference of semi-circle = πr + 2r 

πr + 2r = 72 

r(22/7 + 2)= 72 

r(36/7) = 72 

r = 72 × 7/36 = 14 cm

Area of semi-circle = 1/2 πr² = 1/2 × 22/7 × 14 × 14 = 308 cm² 

Q22. Solve : 

1/2x + 1/3y = 2  and  1/3x + 1/2y = 13/6 

Sol. 

1/2x + 1/3y = 2 …….(i) 

1/3x + 1/2y = 13/6 ……..(ii) 

Multiply eqn.(i) by 6 and multiply eqn.(ii) by 6, we get 

3/x + 2/y = 12 …….(iii) 

2/x + 3/y = 13 …….(iv) 

Take 1/x = p and 1/y = q

3p + 2q = 12 ……(v)

2p + 3q = 13 …….(vi) 

Multiply eqn.(v) by 3 and multiply eqn.(vi) by 2, and subtract each other, we get

9p – 4p = 36 – 26 

5p = 10 

p = 10/5 = 2 

Put p = 2 in eqn.(v), 

3(2) + 2q = 12 

6 + 2q = 12 

2q = 12-6 = 6  

q = 6/2 = 3 

But 1/x = p and 1/y = q 

1/x = 2 and 1/y = 3 

So, x = 1/2 and y = 1/3

Q23. The sum of the reciprocal of Rehman’s age (in years) 3 years ago and 5 years after from now is 1/3 . Find his present age. 

Sol. 

Let x be the present age of Rehman, 

3 years ago, age of Rehman = x – 3 

5 years after, age of Rehman = x + 5 

According to question (A.T.Q.), 

1/(x-3) + 1/(x+5) = 1/3 

x² – 4x – 21 = 0  (after solving the equation) 

x² – 7x + 3x – 21 = 0

x(x-7) + 3(x-7) = 0 

(x-7)(x+3) = 0

Either x-7 = 0  or  x+3 = 0 

x = 7  (age never negative, so x≠-3) 

Hence, present age of Rehman = 7 years 

Q24. Find the sum of first 40 positive integers divisible by 6. 

Sol. 

A.P 6, 12, 18, ……… to 40 terms 

Here, a = 6,  d = 12 – 6 = 6,  n = 40 

Sn = n/2 [2a + (n-1)d] 

= 40/2 [2×6 + (40-1)6] 

= 20 [12 + (39)6] 

= 20 [12+234]

= 20 × 246 = 4920 

Q25. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord. 

Sol. 

From the figure,

O is the centre of the circle and AB is a chord of the circle.

PA and PB are the tangents drawn at the ends of a chord of the circle. The tangents intersect at P.

Join OA and OB.

OA = OB = radius of the circle.

Considering triangle OAB,

OA = OB = radius of the circle.

Since the two sides of the triangle are equal, OAB is an isosceles triangle.

We know that the angles opposite to the equal sides of a triangle are equal.

So, ∠OAB = ∠OBA

i.e., ∠1 = ∠2 …………(i)

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

So, OA ⟂ PA and OB ⟂ PB

∠OAP = ∠OBP = 90°

i.e., ∠2 + ∠3 = ∠1 + ∠4 ……….(ii)

Substituting (i) in (ii),

∠1 + ∠3 = ∠1 + ∠4

∠1 + ∠3 – ∠1 = ∠4

∠3 = ∠4

Therefore, ∠PAB = ∠PBA.

Q26. In what ratio does the point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8). 

Sol. 

Using Section Formula, 

x = (m1x2 + m2x1) ÷ (m1 + m2) 

-1 = [6k+1(-3)]/(k+1) 

– k – 1 = 6k – 3

7k = 2 

k = 2/7 

Ratio = 2 : 7 

Q27. A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number. 

Sol. 

Total numbers = 6  (1,2,3,4,5,6) 

(i) prime numbers = 3  (2,3,5) 

P(prime number) = 3/6 = 1/2 

(ii) number lying between 2 and 6 = 3  (3,4,5) 

P(number lying between 2 and 6) = 3/6 = 1/2 

(iii) odd numbers = 3  (1,3,5) 

P(odd number) = 3/6 = 1/2 

Q28. Solve the equation 2x² – 7x + 3 = 0 by completing the square method. 

Sol. 

2x² – 7x + 3 = 0 

dividing both side by 2, 

x² – 7/2 x + 3/2 = 0 

Add and subtract (7/4)² in equation, 

x² – 2(x)(7/4) + (7/4)² – (7/4)² + 3/2 = 0 

(x-7/4)² = 49/16 – 3/2 

(x-7/4)² = 25/16 

x-7/4 = √25/16 

x-7/4 = ±5/4

x = ±5/4 + 7/4 

So, x = 3, 1/2 

Q29. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle. 

Sol. 

Q30. Prove that :

cosθ/(1+sinθ) + cosθ/(1-sinθ) = 2secθ 

Sol. 

L.H.S. = cosθ/(1+sinθ) + cosθ/(1-sinθ) 

= cosθ[1+sinθ+1-sinθ]/(1-sin²θ)

= 2cosθ/cos²θ 

= 2/cosθ 

= 2secθ = R.H.S. 

                                            OR

A tree breaks due to storm and the broken part bands so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. 

Sol. 

Q31. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area of the toy. 

Sol. 

Q32. 100 surnames were picked from a telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as below. Determine the median number of letters in the surnames. 

Sol. 

Median = L + (N/2 – C)/f × h = 7 + (50-36)/40 × 3 = 7 + 21/20 = 7 + 1.05 = 8.05 

                                             OR 

The following data gives the information on the life-time (in hours) of 75 electrical instruments. Find the mean life time of the instruments. 

Sol. 

Mean = A + Σfd/N = 50 + 580/75 = 50 + 7.73 = 57.73 

SET-D