# HBSE Class 10th Maths Solved Question Paper 2020

HBSE Class 10th Maths Solved Question Paper 2020

HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2020. HBSE 10th Question Paper Download 2020. HBSE Class 10 Maths Paper Solution 2020. Haryana Board Class 10th Maths Question Paper 2020 Pdf Download with Answer.

SET-A

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)

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

= √(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, A(as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A= 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.

Sol.

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

= √(-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

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.

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

= √(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.

Steps of construction–
Step I- Draw BC = 8cm.
Step II- Through D, the mid-point of BC, draw the perpendicular to BC and cut an arc from D on it such that DA = 4cm.
Step III- Join BA and CA. ΔABC is obtained.
Step IV- Draw the ray BX so that ∠CBX is acute.
Step V- Mark three points B1, B2, B3 on BX such that BB1 = B1B2 = B2B3
Step VI- Join B2 to C and draw B3C’ parallel to B2C, intersecting BC extended at C’.
Step VII- Through C’ draw C’A’ parallel to CA to intersect BA extended to A’. Now, ΔA’BC’ is the required triangle similar to ΔABC where BA’/BA = C’A’/CA = BC’/BC = 3/2
Thus, ∆A’BC’ is the required triangle, each of whose sides is 3/2 times of the corresponding sides of ∆ABC.

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.

Let height of tree (h) = AB
According to question (A.T.Q.),
AB = x + y
In ∆OBC,
x/8 = tan30°
x/8 = 1/√3
x = 8/√3
and
y/8 = sec30°
y/8 = 2/√3
y = 16/√3
AB = h = x + y = 8/√3 + 16/√3 = 24/√3
Rationalise the denominator (multiply and divide by √3),
= 24/√3 × √3/√3 = 8√3
Height of tree (h) = 8√3 m

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.

height of toy = 15.5 cm
height of cone = 12 cm
Let l be the slant height of cone,
l² = r² + h² = (3.5)² + (12)² = 12.25 + 144 = 156.25
l = √156.25 = 12.5 cm
Surface area of cone = πrl = 22/7 × 3.5 × 12.5 = 137.5 cm²
Surface area of hemisphere = 2πr² = 2 × 22/7 × 3.5 × 3.5 = 77 cm²
Total surface area of toy = 137.5 + 77 = 214.5 cm²

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

Q1. Express 0.125 in the form p/q.

Sol.

0.125 = 125/1000 = 1/8

Q2. The zeroes of 3x² + 4x + 1 are :

(A) -1, -1/3

(B) -1, 1/3

(C) 1, -1/3

(D) 1, 1/3

Ans. (A) -1, -1/3

3x² + 4x + 1 = 0

3x² + 3x + x + 1 = 0

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

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

Either x + 1 = 0 or 3x + 1 = 0

x = -1, -1/3

Q3. Solve :  3x + 4y = 10 and x – y = 1

Sol.

3x + 4y = 10 ……..(i)

x – y = 1 ……..(ii)

Multiply eqn.(ii) by 4 and add in eqn.(ii),

3x + 4x = 10 + 4

7x = 14

x = 14/7 = 2

Put x = 2 in eqn.(ii),

2 – y = 1

y = 2-1 = 1

So, x = 2 and y = 1

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

(A) -1.2, -3.2, -5.2, ………

(B) 2, 5, 7, 9, ………

(C) 1, 2², 3², 4², ……….

(D) None of these

Ans. (A) -1.2, -3.2, -5.2, ………

Common difference (d) = -3.2-(-1.2) = -5.2-(-3.2) = -2

Q5. Find the 11th term of A. P. -3, -1/2, 2, ……..

Sol.

Here, a = -3, d = -1/2 -(-3) = -1/2 + 3 = 5/2, n = 11

an = a + (n-1)d = – 3 + (11-1)5/2 = – 3 + (10)5/2 = – 3 + 25 = 22

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

Two polygons of the same number of sides are similar, if their corresponding angles are …………. (equal, proportional)

Ans. equal

Q7. Areas of two similar triangles are in the ratio of 7 : 3, then the ratio of their corresponding sides is :

(A) √7 : √3

(B) 7 : √3

(C) √7 : 3

(D) None of these

Ans. (A) √7 : √3

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 = √7 : √3

Q8. If sides of two similar triangles are in the ratio 5 : 7, then areas of their triangles are in the  ratio :

(A) 15 : 14

(B) 25 : 49

(C) 49 : 25

(D) None of these

Ans. (B) 25 : 49

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 = 5² : 7² = 25 : 49

Q9. Find the length of tangent drawn from a point whose distance from the centre of a circle is 25 cm. Given the radius of circle is 7 cm.

Sol.

Let length of tangent = x

Using Pythagoras Theorem,

H² = P² + B²

QO² = PQ² + PO²

25² = x² + 7²

x² = 25² – 7² = 625 – 49 = 576

x = √576 = 24 cm

So, Length of tangent = 24 cm

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

Sol.

A(1, -3) and B(4, 1)

Here x1 = 1, x2 = 4, y1 = -3, y2 = 1

Using Distance Formula,

AB = √x2-x1)²+(y2-y1

= √(4-1)²+(1+3)²

= √(3)²+(4)²

= √9+16 = √25 = 5 unit

Q11. Find the mid point of the line segment whose end points are (4, 5) and (2, –1).

Sol.

Take C is Mid point of AB,

A(4, 5) and B(2, -1) and C(x, y)

Here x= 4, x2 = 2, y1 = 5, y2 = -1

Using Mid Point Theorem,

C(x, y) = (x1+x2)/2, (y1+y2)/2

= (4+2)/2, (5-1)/2

= 6/2, 4/2

= 3, 2

Co-ordinates of C(x, y) is (3, 2).

Q12. Find the value of tan25°/cot65°.

Sol.

tan25°/cot65° = tan(90°-65°)/cot65° = cot65°/cot65° = 1

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

(A) 17/8

(B) 8/17

(C) 15/17

(D) None of these

Ans. (A) 17/8

Using Pythagoras Theorem,

H² = P² + B² = 15² + 8² = 225 + 64 = 289

H = √289 = 17

secA = Hypotenuse/Base = 17/8

Q14. Find the area of sector of circle with radius        21 cm, if angle of the sector is 60°.

Sol.

Area = θ/360 × πr² = 60/360 × 22/7 × 21 × 21 = 231 cm²

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

(A) πrl

(B) πr²h

(C) 1/3 πr²h

(D) None of these

Ans. (C) 1/3 πr²h

Q16. A bag contains 3 red and 5 black balls. If one ball is taken out at random from the bag. What is the probability that it will be Red ?

Sol.

No. of Red balls = 3

No. of Black balls = 5

Total balls = 3 + 5 = 8

P(Red) = 3/8

Q17. Prove that 3√2 is an irrational number.

Sol.

Let us suppose 3√2 is rational

3√2 = p/q , where p and q are co-prime integers and q ≠ 0

:. √2 = p/3q

since 3, p, q are integers, therefore p/3q is rational

√2 is also rational

But it is contradiction to the fact that √2 is irrational.

Hence 3√2 is irrational.

Q18. Divide the polynomial p(x) = x⁴ – 3x² + 4x + 5 by the polynomial q(x) = x² – x + 1. Find the quotient and remainder.

Sol.

Quotient = x² + x – 3

Remainder = 8

Q19. A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Sol.

Let AB be the length of ladder

In ∆ABC,

AB² = AC² + BC² = 6² + (2.5)² = 36 + 6.25 = 42.25

AB = √42.25

AB = 6.5 m

Q20. Prove that :

tanθ/(1-cotθ) + cotθ/(1-tanθ) = 1 + secθcosecθ)

Sol.

L.H.S = tanθ÷(1-cotθ) + cotθ÷(1-tanθ)

= (sinθ/cosθ)÷(1-cosθ/sinθ) + (cosθ/sinθ)÷(1-sinθ/cosθ)

= sin²θ÷cosθ(sinθ-cosθ) + cos²θ÷sinθ(cosθ-sinθ)

= (sin³θ-cos²θ)÷sinθcosθ(sinθ-cosθ)

= (sinθ-cosθ)(sin²θ+cos²θ+sinθcosθ) ÷ sinθcosθ(sinθ-cosθ)

= (1+sinθcosθ)/(sinθcosθ)

= 1 + secθcosecθ = R.H.S.

Q21. The circumference of a circle exceeds the diameter by 33.6 cm. Find the area of the circle.

Sol.

According to question (A.T.Q.),

Circumference of circle = 2r + 33.6

2πr = 2r + 33.6

2r(π-1) = 33.6

r(22/7 – 1) = 33.6/2 = 16.8

r(15/7) = 16.8

r = 16.8 × 7/15 = 7.84 cm

Hence area of circle = πr² = 22/7 × 7.84 × 7.84 = 193.18 cm²

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),

15p + 6p = 6 + 1

21p = 7

p = 7/21 = 1/3

Put p = 1/3 in eqn.(iii) ,

5/3 + q = 2

q = 2 – 5/3 = 1/3

But 1/(x-1) = p and 1/(y-2) = q

So, 1/(x-1) = 1/3 and 1/(y-2) = 1/3

x-1 = 3 and y-2 = 3

x = 3+1 = 4 and y = 3+2 = 5

So, x = 4 and y = 5

Q23. A motor boat whose speed is 15 km/h in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.

Sol.

Let speed of water = x  km/h

Speed of Motor-boat in still water = 15 km/h

Speed of Motor-boat upstream = 15-x km/h

Speed of Motor-boat down-stream = 15+x km/h

Distance= 30 km, Time = 4h + 30min = 4h + 1/2h = 9/2 hours

According to question (A.T.Q),

30/(15+x) + 30/(15-x) = 9/2

30(15-x+15+x) ÷ [15²-x²] = 9/2

30(30) ÷ (225-x²) = 9/2

30×30×2 = 9(225-x²)

1800/9 = 225 – x²

200 = 225 – x²

x² = 225 – 200 = 25

x = √25 = ± 5

x = 5  (speed never negative, so x≠-5)

Speed of water = 5 km

Q24. How many terms of sequence 1, 4, 7, 10, ….… should be taken so that their sum is 176 ?

Sol.

Here, a = 1,  d = 4 – 1 = 3,  Sn = 176

Sn = n/2 [2a + (n-1)d]

176 = n/2 [2×1 + (n-1)3]

176 × 2 = n [2 + 3n -3]

352 = 3n² – n

3n² – n – 352 = 0

3n² – 33n + 32n – 352 = 0

3n(n-11) + 32(n-11) = 0

(n-11)(3n+32) = 0

Either n-11 = 0 or 3n+32 = 0

n = 11  (number of terms never negative, so n≠-32/3)

No. of terms (n) = 11

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

Sol.

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 ratio in which the line segment joining the points A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Sol.

Let C divided AB on x-axis in ratio k : 1

A(1, -5),  B(-4, 5),  C(x, 0)

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

Using Section Formula,

x = (m1x2+m2x1)/(m1+m2

x = [k(-4)+1(1)] ÷ (k+1)

x = (-4k+1) ÷ (k+1) …………(i)

and

y = (m1y2+m2y1)/(m1+m2)

0 = [k(5)+1(-5)] ÷ (k+1)

0 = (5k-5) ÷ (k+1)

5k -5 = 0

5k = 5

k = 5/5 = 1

Put k = 1 in eqn.(i),

x = [-4(1)+1] ÷ (1+1)

x = (-4+1) ÷ (2)

x = -3/2

Co-ordinates of C(x, 0) is (-3/2, 0).

Q27. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) king of red colour (ii) face card.

Sol.

Total cards = 52

(i) king of red colour = 2

P(king of red colour) = 2/52 = 1/26

(ii) face card = 12

P(face card) = 12/52 = 3/13

Q28. Solve the equation 2x² + x – 4 = 0 by completing the square method.

Sol.

2x² + x – 4 = 0

Dividing both side by 2,

x² + x/2 – 2 = 0

Add and subtract (1/4)² in equation,

x² + 2(x)(1/4) + (1/4)² – (1/4)² -2 = 0

(x + 1/4)² = (1/4)² + 2

(x + 1/4)² = 1/16 + 2

(x + 1/4)² = 33/16

x + 1/4 = √33/√16 = ±√33/4

x = (-1±√33)/4

x = (-1-√33)/4  and  (-1+√33)/4

Q29. Construct a triangle ABC whose sides are BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a similar triangle whose sides are 3/4th of corresponding sides of given triangle.

Sol.

Steps of constructions–

Step I- Draw a line BC = 6 cm.

Step II- At B, make ∠C = 60° and cut an arc at A on the same line so that BA = 5 cm.

Step III- Join AC, ΔABC is obtained.

Step IV- Draw the ray BX such that ∠CBX is acute.

Step V- Mark 4 points B1, B2, B3, B4 on BX such that BB1 = B1B2 = B2B3 = B3B4

Step VI- Join B4 to C and draw B3C’ parallel to B4C to intersect BC at C’.

Step VII- Draw C’A’ parallel to CA to intersect BA at A’.

Now, ΔA’BC’ is the required triangle similar to ΔABC where BA’/BA = BC’/BC = C’A’/CA = 3/4

Q30. Prove that :

√(1+sinA/1-sinA) = secA + tanA

Sol.

L.HS. = √(1+sinA/1-sinA)

Rationalise the denominator means divide and multiply by (1+sinA),

= √(1+sinA)²/(1-sin²A)

= √(1+sinA)²/cos²A

= (1+sinA)/cosA

= 1/cosA + sinA/cosA

= secA + tanA = R.H.S.

OR

A man 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from his eyes is 45°. What is the height of chimney ?

Sol.

BC = DE = 28.5 m

Let AB = h

In ∆AED,

AE/ED = tan45°

AE/28.5 = 1

AE = 28.5 m

Now AB = h = AE + ED = 28.5 + 1.5 = 30 m

Height of chimney (h) = 30 m

Q31. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the hemisphere can have ? Find the total surface area of the solid.

Sol.

Side of cubical box (a) = 7 cm

Diameter of hemisphere = 7 cm

Total surface area of solid = 6a² + 2πr² – πr² = 6a² + πr²

= 6×7×7 + 22/7 × 7/2 × 7/2 = 294 + 38.5 = 332.5 cm²

Q32. The following distribution shows the daily pocket money of children of a school. Find the mean of daily pocket money of children.

Sol.

error: MsEducationTv.com