# HBSE Class 10th Maths Solved Question Paper 2019

HBSE Class 10th Maths Solved Question Paper 2019

HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2019. HBSE 10th Question Paper Download 2019. HBSE Class 10 Maths Paper Solution 2019. Haryana Board Class 10th Maths Question Paper 2019 Pdf Download with Answer.

SET-A

Q1. Express 0.375 in the form p/q.

Sol.

0.375 = 375/1000 = 3/8

Q2. Find the sum of zeroes of quadratic polynomial x² + 7x + 10.

Sol.

Sum of zeroes = -(coefficient of x) ÷ (coefficient of x²)

α + β = = -b/a = -7/1 = -7

Q3. The values of x and y from the equations x + y = 14 and x – y = 4 are :

(A) x = 9, y = 4

(B) x = 9, y = 5

(C) x = 5, y = 9

(D) None of these

Ans. (B) x = 9, y = 5

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. 30th term of the A.P. 10, 7, 4, ……….. is :

(A) 77

(B) 87

(C) –77

(D) None of these

Ans. (C) –77

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

Q5. Find the common difference of the A.P. :  3, 1, –1, –3, …………….

Sol. Common difference (d) = a2 – a= 1 – 3 = -2

Q6. Fill in the blank using the 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. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is :

(A) 15 cm

(B) 12 cm

(C) 24.5 cm

(D) 7 cm

Ans. (D) 7 cm

Using Pythagoras Theorem,

H² = P² + B²

25² = r² + 24²

r² = 25² – 24² = 625 – 576 = 49

r = √49 = 7 cm

Q9. Fill in the blank :

A tangent to a circle intersects it in ……… point(s).

Ans. One

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

Sol.

A(2, 3) and B(4, 1)

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. Find the mid point of the line joining the points (7, 6) and (–3, –4).

Sol.

Take C is Mid point of AB,

A(7, 6) and B(-3, -4) and C(x, y)

Here x= 7, x2 = -3, y1 = 6, y2 = -4

Using Mid Point Theorem,

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

= (7-3)/2, (6-4)/2

= 4/2, 2/2

= 2, 1

Co-ordinates of C(x, y) is (2, 1).

Q12. Evaluate : sin18°/cos72°

Sol. sin18°/cos72° = sin(90°-72°)/cos72° = cos72°/cos72° = 1

Q13. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of sinA is :

(A) 7/25

(B) 7/24

(C) 24/25

(D) None of these

Ans. (A) 7/25

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625

AC = √625 = 25

sinA = BC/AC = 7/25

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 = 18.84 cm²

Q15. The volume of the cuboid, whose length, breadth and height are 12 m, 10 m and 8 m respectively is :

(A) 592 m³

(B) 960 m³

(C) 480 m³

(D) None of these

Ans. (B) 960 m³

Volume of cuboid = length × breadth × height = 12 × 10 × 8 = 960 m³

Q16. Two dice are thrown at the same time. Find the probability of getting the sum on the dice is 8.

Sol.

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

P(sum 8) = 5/36

Q17. Prove that √2 is irrational.

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) = x³ – 3x² + 5x – 3 by the polynomial g(x) = x² – 2. Find the quotient and remainder.

Sol.

Quotient = x – 3

Remainder = 7x – 9

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. In ∆OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sinQ and cosQ.

Sol.

OQ – PQ = 1

In ∆OPQ,

OQ² = OP² + PQ²

(1+PQ)² = OP² + PQ²

1 + PQ² + 2PQ = OP² + PQ²

1 + 2PQ= 49

2PQ = 49 – 1 = 48

PQ = 48/2 = 24 cm

OQ = 1 + PQ = 1 + 24 = 25 cm

sinQ = 7/25

cosQ = 24/25

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 :

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. Find the roots of the equation 5x² – 6x – 2 = 0 by the method of completing the square.

Sol.

5x² – 6x – 2 = 0

dividing both side by 5,

x² – 6/5 x – 2/5 = 0

Add and subtract (3/5)² in equation,

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

Q24. Find the sum of the 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. Construct a triangle ABC whose sides are BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a similar triangle whose sides are 3/4 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, ∆ABC 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, ∆A’BC’ is the required triangle similar to ∆ABC where BA’/BA = BC’/BC = C’A’/CA = 3/4

Q26. Find the co-ordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4).

Sol.

Using Section Formula,

x = (m1x2 + m2x1) ÷ (m1 + m2

y = (m1y2 + m2y1) ÷ (m1 + m2

x= (1×(-7) + 2×2) ÷ (1+2) = (-7+4)÷3 = -3÷3 = -1

y1 = {1×4+2(-2)} ÷ (1+2) = (4-4) ÷ 3 = 0÷3 = 0

:. C (-1, 0)

x2 = {2×(-7)+1×2} ÷ (2+1) = (-14+2)÷3 = -12÷3 = -4

y2 = {2×4+1×(-2)} ÷ (2+1) = (8-2)÷3 = 6÷3 = 2

:. D (-4, 2)

Q27. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will :

(i) be an ace

(ii) not be an 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. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Sol.

Let shorter side = x

BD = x+60 metre and AB = x+30 metre

In Rt. angle triangle,

(x+60)² = (x+30)² + x²

x² + 3600 + 120x = x² + 900 + 60x + x²

x² – 60x – 2700 = 0

(x-90)(x+30) = 0

x = 90  (height never negative, so x≠-30)

Shorter side = 90 metre

Longer side = 90+30 = 120 metre

Q29. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Sol.

Steps of Construction–

Step I- A circle with radius 6 cm is drawn taking O as centre.

Step II- Point P is marked at 10 cm away from centre of circle.

Step III- With the half of compass mark M which is the midpoint of OP.

Step IV- Draw a circle with centre M, taking radius MO or MP which intersects the given circle at Q and R.

Step V- Now join PQ and PR. These are the tangents of the circle.

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

:. In ∆OPQ, OQ⊥QP

and in ∆OPR, OR⊥PR

Hence, both ∆OPQ and ∆OPR are right angle triangles.

Applying Pythagoras theorem to both ∆s, we get:

OP² = OQ² + PQ²

and OP² = OR² + PR²

OQ = OR = radius = 6 cm and OP = 10 cm

10² = 6² + PQ²

and 10² = 6² + PR²

PQ² = 100 – 36 = 64

and PR² = 100 – 36 = 64

:. PQ = PR = 8 cm

Hence, the length of the tangents to a circle of radius 6 cm, from a point 10 cm away from the centre of the circle, is 8 cm.

Q30. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Sol.

Let BC = 15 m

∠ACB = 60°

In ∆ACB,

tan60° = AB/BC

√3 = AB/15

AB = 15√3 m

Hence height of the tower = 15√3 m

OR

Prove that :

(cosecθ-cotθ)² = (1-cosθ)/(1+cosθ)

Sol.

L.H.S. (cosecθ-cotθ)²

= (1/sinθ – cosθ/sinθ)²

= [(1-cosθ)/sinθ]²

= (1-cosθ)²/sin²θ

= (1-cosθ)²/(1-cos²θ)

= (1-cosθ)/(1+cosθ) = R.H.S.

Q31. A vessel open at the top is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the 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. Find the mean of the following frequency distribution :

Sol.

Mean = A + Σf.u/Σf × h= 150 + (-12)/50 × 20 = 150 – 24/5 = 150 – 4.8 = 145.2

Note : May be solved by another method.

OR

The median of the following data is 525. Find the value of x and y, if the total frequency is 100 :

Sol.

Median = L + (N/2 – Cf)/f × h

525 = 500 + (50-36-x)/20 × 100
525 – 500 = 5(14-x)
25 = 70-5x
5x = 70-25 = 45
x = 45/5 = 9
76 + x + y = 100
76 + 9 + y = 100
y = 15

SET-B

Q1. Express 0.104 in the form p/q.

Sol.

0.104 = 104/1000 = 13/125

Q2. Find the product of zeroes of quadratic polynomial x² + 7x + 10.

Sol.

Product of zeroes = constant/coefficient of x²

αβ = c/a = 10/1 = 10

Q3. The values of x and y from the equations x – y = 3 and x/3 + y/2 = 6 are :

(A) x = 6, y = 9

(B) x = 9, y = 6

(C) x = 8, y = 5

(D) None of these

Ans. (B) x = 9, y = 6

x – y = 3 …….(i)

x/3 + y/2 = 6

Multiply both side by 6, we get

2x + 3y = 36 ……(ii)

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

3x + 2x = 9 + 36

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. 11th term of the A.P. :  -3, -1/2, 2, ……….. is :

(A) -38

(B) 28

(C) 22

(D) None of these

Ans. (C) 22

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

an = a + (n-1)d

a11 = -3 + (11-1)(5/2) = -3 + 25 = 22

Q5. Find the common difference of the A.P. :  -5, -1, 3, 7, ……………

Sol. Common difference (d) = a2 – a= -1-(-5) = -1+5 = 4

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

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

Ans. Similar

Q7. Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, then BC is :

(A) 11.2 cm

(B) 11.4 cm

(C) 12.4 cm

(D) None of these

Ans. (A) 11.2 cm

Area of ∆ABC/Area of ∆DEF = (BC/EF)²

64/121 = (BC/15.4)²

8/11 = BC/15.4

BC = 8/11 × 15.4 = 11.2 cm

Q8. If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is :

(A) 80°

(B) 90°

(C) 70°

(D) 60°

Ans. (C) 70°

∠PTQ + ∠P +  ∠Q + ∠POQ = 360°

∠PTQ + 90° + 90° + 110° = 360°

∠PTQ + 290° = 360°

∠PTQ = 360° – 290° = 70°

Q9. Fill in the blank :

A line intersecting a circle in two points is called a ……………..

Ans. Secant

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. Find the mid point of the line joining the points (3, 4) and (5, 2).

Sol.

Take C is Mid point of AB,

A(3, 4) and B(5, 2) and C(x, y)

Here x= 3, x2 = 5, y1 = 4, y2 = 2

Using Mid Point Theorem,

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

= (3+5)/2, (4+2)/2

= 8/2, 6/2

= 4, 3

Co-ordinates of C(x, y) is (4, 3).

Q12. Evaluate : tan26°/cot64°

Sol.

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

Q13. In ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. The value of cosA is :

(A) 7/25

(B) 24/25

(C) 7/24

(D) None of these

Ans. (B) 24/25

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625

AC = √625 = 25 cm

cosA = AB/AC = 24/25

Q14. Find the area of a sector of a circle with radius 4 cm and of angle 30°. (use π = 3.14)

Sol.

Area = θ/360 × πr² = 30/360 × 3.14 × 4 × 4 = 4.19 cm² (appro.)

Q15. The volume of the cuboid, whose length, breadth and height are 10 m, 8 m and 6 m respectively is :

(A) 460 m³

(B) 480 m³

(C) 520 m³

(D) None of these

Ans. (B) 480 m³

Volume of cuboid = length × breadth × height = 10 × 8 × 6 = 480

Q16. Two dice are thrown at the same time. Find the probability of getting the sum on the dice is 13.

Ans. P(sum 13) = 0/36 = 0   (:.Maximum sum=6+6=12)

Q17. Prove that √3 is irrational.

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

Sol.

Quotient = x² + x – 3

Remainder = 8

Q19. A vertical pole of length 6 m casts a shadow 4 m long on the ground and 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. In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP.

Sol.

PQ = 5 cm

PR + QR = 25 cm

In ∆PQR, using Pythagoras theorem

PR² = QR² + PQ²

(25-QR)² = QR² + PQ²

625 + QR² – 50QR = QR² + 5²

600 = 50QR

QR = 600/50 = 12 cm

PR = 25 – 12 = 13 cm

sinP = QR/PR = 12/13

Q21. What is area of circle, the circumference of which is equal to the perimeter of a square of side 11 cm?  (π = 22/7)

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 :

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. Find the real roots of 4x² + 3x + 5 = 0 by the method of completing the square.

Sol.

4x² + 3x + 5 = 0

dividing by 4, we get

x² + 3/4 x + 5/4 = 0

x² + 2(x)(3/8) + (3/8)² – (3/8)² + 5/4 = 0

(x+3/8)² – 9/64 + 5/4 = 0

(x+3/8)² = -71/64 < 0

(x+3/8)² can not b negative.

:. Equation has no real roots.

Q24. Find the sum of the first 15 positive multiples of 8.

Sol.

A.P 8, 16, 24, 32, ……… to 15 terms

Here, a = 8, d = 16-8 = 8, n = 15

Sn = n/2 [2a + (n-1)d]

= 15/2 [2×8 + (15-1)8]

= 15/2 [16 + (14)8]

= 15/2 [16+112]

= 15/2 × 128 = 960

Q25. Construct a triangle similar to a given triangle ABC with its sides equal to 5/3 of the corresponding sides of the triangle ABC.

Sol.

Steps of Construction :

Step I- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

Step II- From B cut off 5 arcs B1, B2, B3, B4 and B5 on BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5.

Step III- Join B3 to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C’.

Step IV- Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then, A’BC’ is the required triangle.

Q26. Find the co-ordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4).

Sol.

Using Section Formula,

x = (m1x2 + m2x1) ÷ (m1 + m2

y = (m1y2 + m2y1) ÷ (m1 + m2

x= (1×(-7) + 2×2) ÷ (1+2) = (-7+4)÷3 = -3÷3 = -1

y1 = {1×4+2(-2)} ÷ (1+2) = (4-4) ÷ 3 = 0÷3 = 0

:. C (-1, 0)

x2 = {2×(-7)+1×2} ÷ (2+1) = (-14+2)÷3 = -12÷3 = -4

y2 = {2×4+1×(-2)} ÷ (2+1) = (8-2)÷3 = 6÷3 = 2

:. D (-4, 2)

Q27. A box contains 3 blue, 2 white and 4 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.

Total Balls = 3 + 2 + 4 = 9

(i) P(White) = 2/9

(ii) P(Blue) = 3/9 = 1/3

(iii) P(Red) = 4/9

Q28. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Sol.

Let large number = x

and smaller number = y

x² – y² = 180 …….(i)

y² = 8x ……….. (ii)

From (i) and (ii), we get

x² – 8x = 180

x² – 8x – 180 = 0

x² – 18x + 10x – 180 = 0

x(x-18) + 10(x-18) = 0

(x-18)(x+10) = 0

Either x-18 = 0 or x+10 = 0

x = 18  or  -10

y² = 8x = 8×18 = 144

y = √144 = 12

and y² = 8(-10) = -80 (not possible)

So, x = 18 and y = 12

Q29. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Sol.

Steps of construction–

Step I- Draw a circle of radius 5 cm

Step II- Draw horizontal radius OQ

Step III- Draw angle 120° from point O

Let the ray of angle intersect the circle at point R

Step IV- Now draw 90° from point Q

Step V- Draw 90° from point R

Step VI- Where the two arcs intersect, mark it as point P

So, PQ and PR are the tangents at an angle of 60°.

Q30. An electrician has to repair an electric fault on a pole of height 5 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use which, when inclined at an angle of 60° to the horizontal, would enable him to reach the required position?  (Take √3 = 1.73)

Sol.

AB = 1.3 m

BD = AD – AB = 5 – 1.3 = 3.7 m

In ∆BDC

BD/BC = sin60°

BC = BD/sin60° = (3.7×2)/√3 = 7.4/1.73 = 4.28 m (approx.)

Length of ladder = BC = 4.28 m (approx.)

OR

Prove that :

cosA/(1+sinA) + (1+sinA)/cosA = 2secA

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.

Q31. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of 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. Find the mean of the following frequency distribution :

Sol.

Mean = A + Σf.u/Σf × h= 75.5 + 4/30 × 3 = 75.5 + 0.4 = 75.9

Note : May be solved by another method.

OR

Find the median of the following frequency distribution :

Sol.

Median = L + (N/2 – Cf)/f × h = 125 + (34-22)/20 × 20 = 125 + 12 = 137

SET-C

Q1. Express 0.15 in the form p/q.

Sol.

0.15 = 15/100 = 3/20

Q2. Find the sum of zeroes of quadratic polynomial x² – 2x – 8.

Sol.

Sum of zeroes = -(coefficient of x) ÷ (coefficient of x²) =

α + β = –b/a = -(-2)/1 = 2/1 = 2

Q3. The values of x and y from the equations 3x – y = 3 and 9x – 3y = 9 are :

(A) one solution

(B) no solution

(C) infinite number of solutions

(D) None of these

Ans. (C) infinite number of solutions

Here, a1 = 3, a2 = 9, b= -1, b2 = -3, c1 = -3, c2 = -9

a1/a= 3/9 = 1/3

b1/b2 = -1/-3 = 1/3

c1/c2 = -3/-9 = 1/3

so, a1/a2 = b1/b2 = c1/c2 = 1/3

:. Equations has infinite number of solutions.

Q4. 10th term of the A.P. :  2, 7, 12, ……….. is :

(A) 45

(B) 43

(C) 47

(D) None of these

Ans. (C) 47

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

Q5. Find the common difference of the A.P. :  1/3, 5/3, 9/3, 13/3, is …………….

Ans. Common difference (d) = a2 – a= 5/3 – 1/3 = 4/3

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

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

Ans. Equilateral

Q7. Triangles ABC and DEF are similar. If area of ∆ABC = 16 cm², area of ∆DEF = 25 cm² and BC = 2.3 cm, then EF is :

(A) 2.875 cm

(B) 2.758 cm

(C) 2.578 cm

(D) None of these

Ans. (A) 2.875 cm

Area of ∆ABC/Area of ∆DEF = (BC/EF)²

16/25 = (2.3/EF)²

4/5 = 2.3/EF

EF = (2.3×5)/4 = 11.5/4 = 2.875 cm

Q8. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is :

(A) 80°

(B) 50°

(C) 40°

(D) 70°

Ans. (B) 50°

∠APO = ∠APB/2 = 80/2 = 40°

∠POA + ∠PAO + ∠APO = 180°  (in ∆APO)

∠POA + 90° + 40° = 180°

∠POA + 130° = 180°

∠POA = 180° – 130° = 50°

Q9. Fill in the blank :

A circle can have …………. parallel tangents at the most.

Ans. Two

Q10. Find the distance between the points (4, 7) and (10, -1).

Sol.

A(4, 7) and B(10, -1)

Here, x1 = 4, x2 = 10, y1 = 7, y2 = -1

Using Distance Formula,

AB = √x2-x1)²+(y2-y1

= √(10-4)²+(-1-7)²

= √(6)²+(-8)²

= √36+64 = √100 = 10 unit

Q11  Find the mid point of the line joining the points (4, 7) and (2, 3).

Sol.

Take C is Mid point of AB,

A(4, 7) and B(2, 3) and C(x, y)

Here, x= 4, x2 = 2, y1 = 7, y2 = 3

Using Mid Point Theorem,

C(x, y) = (x1+x2)/2,  (y1+y2)/2

= (4+2)/2,  (7+3)/2

= 6/2, 10/2

= 3, 5

Co-ordinates of C(x, y) is (3, 5).

Q12. Evaluate : cos48° – sin42°.

Sol.

cos48°-sin42° = cos(90°-42°) – sin42° = sin42° – sin42° = 0

Q13. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of sinC is :

(A) 24/25

(B) 7/25

(C) 7/24

(D) None of these

Ans. (A) 24/25

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625

AC = √625 = 25 cm

sinC = AB/AC = 24/25

Q14. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of arc.

Sol.

Length of arc = θ/360 × 2πr = 60/360 × 2 × 22/7 × 21 = 22 cm

Q15. The volume of the cuboid, whose length, breadth and height are 10 m, 8 m and 5 m respectively is :

(A) 400 m³

(B) 200 m³

(C) 300 m³

(D) None of these

Ans. (A) 400 m³

Volume of cuboid = length × breadth × height = 10 × 8 × 5 = 400

Q16. Two dice are thrown at the same time. Find the probability of getting the sum on the dice is less than or equal to 12.

Sol.

P(sum less than or equal 12) = 36/36 = 1   (:.Maximum sum=6+6=12)

Q17. Prove that √5 is irrational.

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) = x⁴ – 5x + 6 by the polynomial g(x) = x² – 2. Find the quotient and remainder.

Sol.

Quotient = x² + 2

Remainder = -5x + 10

Q19. A person goes 10 m due east and then 30 m due north. Find the distance from the starting point.

Sol.

Let OA = 10 m

AB = 30 m

In ∆ OAB,

OB² = OA² + AB²

OB² = 10² + 30² = 100 + 900 = 1000

OB = √1000 = 10√10 m

Q20. If sin(A-B) = 1/2 and cos(A+B) = 1/2, 0° < A+B ≤ 90°,  A > B, find A and B.

Sol.

sin(A-B) = 1/2  →  A-B = 30° ……..(i)

cos(A+B) = 1/2  →  A+B = 60° ……..(ii)

Add eqn.(i) and (ii), we get

2A = 90°

A = 90°/2 = 45°

Put A=45° in eqn.(ii), we get

45° + B = 60°

B = 60° – 45° = 15°

So, A = 45° and B = 15°

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 :

1/2 x – 1/y = -1  and  1/x + 1/2 y = 8

Sol.

Let 1/x = u  and  1/y = v,  we get

1/2 u – v = -1  →  u – 2v = -2 …….(i)

and  u + 1/2 v = 8  →  2u + v = 16 …….(ii)

Multiply eqn.(ii) by 2 and add in eqn.(i), we get

5u = 30

u = 30/5 = 6

Putt u=6 in eqn.(ii), we get

2(6) + v = 16

12 + v = 16

v = 16-12 = 4

But 1/x = u = 6  and  1/y = v = 4

So, x = 1/6 and y = 1/4

Q23. Find the roots of quadratic equation 2x² – 7x + 3 = 0 by the method of completing the square.

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 = ±5/4

x = ±5/4 + 7/4 = (7±5)/4

So, x = 3, 1/2

Q24. Find the sum of the odd numbers between 0 and 50.

Sol.

1, 3, 5, 7, ……….., 49

a = 1, d = 3-1 = 2, an = 49

Tn or an = a + (n-1)d

49 = 1 + (n-1)2

49-1 = (n-1)2

(n-1)2 = 48

n-1 = 48/2 = 24

n = 24+1 = 25

Sn = n/2 [2a + (n-1)d]

S25 = 25/2 [2×1 + (25-1)2]

= 25/2 [2+48] = 25/2 × 50 = 25 × 25 = 625

Q25. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Sol.

To construct a triangle from given conditions-

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.

Q26. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

Sol.

Let Point C intersect line AB in the Ratio k : 1 and at Y-axis

A(5, -6), B(-1, -4), C(0, y)

Here,  x1= 5, x2 = -1, y1 = -6, y2 = -4, m1 = k, m= 1

Using Section Formula,

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

0 = [k(-1)+1(5)] ÷ (k+1) =(-k+5) ÷ (k+1)

-k+5 = 0

k = 5

y = (m1y2+m2y1) ÷ (m1+m2) = [k(-4)+1(-6)] ÷ (k+1)

y = [5(-4)-6] ÷ (5+1) = (-20-6) ÷ 6 = -26 ÷ 6 = -13/3

Ratio k : 1 = 5 : 1

Co-ordinates of C(0, y) is (0, -13/3).

Q27. A die is thrown once. Find the probability of getting :

(i) prime number

(ii) number lying between 2 and 6

(iii) an odd number

Sol.

Total number on dice is 1, 2, 3, 4, 5, 6

(i) prime number = 2, 3, 5

P(prime number) = 3/6 = 1/2

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

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

(iii) odd number = 1, 3, 5

P(odd number) = 3/6 = 1/2

Q28. 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 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)

Q29. Construct two tangents to a circle of radius 3.5 cm from a point P at a distance of 6 cm from its centre. Measure the lengths of these tangents.

Sol.

Steps of Construction :

Step I- A circle with radius 3.5 cm is drawn taking O as centre.

Step II- Point P is marked at 6 cm away from centre of circle.

Step III- With the half of compass mark M which is the midpoint of OP.

Step IV- Draw a circle with centre M, taking radius MO or MP which intersects the given circle at Q and R.

Step V- Now join PQ and PR. These are the tangents of the circle.

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

:. In ∆OPQ, OQ⊥QP

and in ∆OPR, OR⊥PR

Hence, both ∆OPQ and ∆OPR are right angle triangles.

Applying Pythagoras theorem to both ∆s, we get:

OP² =OQ² + PQ²

and OP² = OR² + PR²

OQ = OR = radius = 3.5 cm and OP = 6 cm

6² = 3.5² + PQ²

and 6² = 3.5² + PR²

PQ² = 36-12.25 = 23.75

and PR² = 36-12.25 = 23.75

:. PQ = PR = 4.8 cm (approx.)

Hence, the length of the tangents to a circle of radius 3.5 cm, from a point 6 cm away from the centre of the circle, is 4.8 cm.

Q30. A tree breaks due to storm and the broken part bends 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 = 8√3 m

OR

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.

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

Sol.

Side of cubical box (a) = 7 cm

diameter of hemisphere (d) = 7 cm

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

=6 x 7 x 7 + 22/7 × 7/2 × 7/2

=294 + 38.5 = 332.5 cm²

Q32. Find the mean of the following frequency distribution :

Sol.

Mean = A + Σf.u/Σf × h= 57 + 25/400 × 3 = 57 + 0.19 = 57.19

Note : May be solved by another method.

OR

If the median of the distribution given below is 28.5, find the value of x and y of total number is 60 :

Sol.

Median= 28.5

x + y + 45 = 60

x + y = 60 – 45 = 15

Median = L + (N/2 – Cf)/f × h

28.5 = 20 + (30-5-x)/20 × 10

28.5 – 20 = (30-5-x)/20 × 10

8.5 = (25-x)/2

8.5×2 = 25 – x

25 – x = 17

x = 25 – 17 = 8

x + y = 15

8 + y = 15

y = 15 – 8 = 7

SET-D

Q1. Express 15.75 in the form p/q.

Sol.

15.75 = 1575/100 = 63/4

Q2. Find the product of zeroes of the quadratic polynomial x² – 2x – 8.

Sol.

Product of zeroes = constant/coefficient of x²

αβ = c/a = -8/1 = -8

Q3. The values of x and y from the equations 2x – y = 3 and 4x + y = 3 are :

(A) x = 1, y = -1

(B) x = 2, y = 1

(C) x = -1, y = 1

(D) None of these

Ans. (A) x = 1, y = -1

2x – y = 3 …….(i)

4x + y = 3 …….(ii)

Add eqn.(i) and (ii), we get

6x = 6

x = 6/6 = 1

Put x = 1 in eqn.(ii), we get

4(1) + y = 3

y = 3 – 4 = -1

So, x = 1 and y = -1

Q4. 10th term of the A.P. :  -0.1, -0.2, -0.3, ……….. is :

(A) -0.9

(B) -0.8

(C) -1.0

(D) -1.1

Ans. (C) -1.0

a = -0.1,  d = -0.2 – (-0.1) = -02 + 0.1 = -0.1,  n = 10

Tn or an = a + (n-1)d

a10 = –0.1 + (10-1)(-0.1) = -0.1 – 0.9 = -1.0

Q5. Find the common difference of the A.P. :  0.6, 1.7, 2.8, 3.9, …………….

Ans.

Common difference (d) = a2 – a= 1.7-0.6 = 1.1

Q6. Fill in the blank using the 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. Triangles ABC and DEF are similar. If AC = 19 cm and DF = 8 cm, the ratio of the area of two triangles are :

(A) 19/8

(B) 361/64

(C) 38/65

(D) None of these

Ans. (B) 361/64

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.

Area of ∆ABC : Area of ∆DEF = AC² : DF² = 19² : 8² = 361 : 64  or  361/64

Q8. PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then ∠QOR is :

(A) 160°

(B) 150°

(C) 135°

(D) 134°

Ans. (D) 134°

∠QOR + ∠P +  ∠Q + ∠QPR = 360°

∠QOR + 90° + 90° + 46° = 360°

∠QOR + 226° = 360°

∠QOR = 360° – 226° = 134°

Q9. Fill in the blank :

The common point of a tangent to a circle and the circle is called ………….

Ans. Point of contact

Q10. Find the distance between the points (-1, -4) and (3, 5).

Sol.

A(-1, -4) and B(3, 5)

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

Using Distance Formula,

AB = √x2-x1)²+(y2-y1

= √(3+1)²+(5+4)²

= √(4)²+(9)²

= √16+81 = √97 unit

Q11. Find the mid point of the line joining the points (3, -4) and (7, 10).

Sol.

Take C is Mid point of AB,

A(3, -4) and B(7, 10) and C(x, y)

Here, x= 3, x2 = 7, y1 = -4, y2 = 10

Using Mid Point Theorem,

C(x, y) = (x1+x2)/2,  (y1+y2)/2

= (3+7)/2,  (-4+10)/2

= 10/2, 6/2

= 5, 3

Co-ordinates of C(x, y) is (5, 3).

Q12. Evaluate : cosec 31° – sec 59°.

Sol.

cosec31° – sec 59° = cosec(90°-59°) – sec59° = sec59° – sec59° = 0

Q13. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of cosC is :

(A) 7/25

(B) 7/24

(C) 24/25

(D) None of these

Ans. (A) 7/25

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625

AC = √625 = 25 cm

cosC = BC/AC = 7/25

Q14. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the sector formed by the arc.

Sol.

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

Q15. The volume of the cuboid, whose length, breadth and height are 13 m, 10 m and 8 m respectively is :

(A) 1040 m³

(B) 1060 m³

(C) 1020 m³

(D) None of these

Ans. (A) 1040 m³

Volume of cuboid = length × breadth × height = 13 × 10 × 8 = 1040

Q16. Two dice are thrown at the same time. Find the probability of getting the sum on the dice is 10.

Sol.

(4,6), (5,5), (6,4)

P(sum 10) = 3/36

Q17. Prove that 3√2 is irrational.

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 3x³ + x² + 2x + 5 by x² + 2x + 1. Find the quotient and remainder.

Sol.

Quotient = 3x – 5

Remainder = 9x + 10

Q19. A ladder 17 m long reaches a window of a building 15 m above the ground, find the distance of the foot of the ladder from the building.

Sol.

height of window, AB = 15 m

In ∆ABC, using Pythagoras theorem

AC² = AB² + BC²

17² = 15² + BC²

BC² = 17² – 15² = 289 – 225 = 64

BC = √64 = 8 m

Distance of foot of ladder from the building = 8 m

Q20. If tan(A+B) = √3 and tan(A-B) = 1/√3, 0° < A+B ≤ 90°,  A > B, find A and B.

Sol.

tan(A+B) = √3  →  A+B = 60° ……..(i)

tan(A-B) = 1/√3  →  A-B = 30° ……..(ii)

Add eqn.(i) and (ii), we get

2A = 90°

A = 90/2 = 45°

Put A = 45° in eqn.(i), we get

45° + B = 60°

B = 60° – 45° = 15°

So, A = 45° and B = 15°

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

Sol.

Circumference= Diameter+ 33.6

2πr = 2r + 33.6

2 × 22/7 × r – 2r = 33.6

44r – 14r = 33.6 × 7

30r = 235.2

r = 235.2/30 = 7.84 cm

Area of circle = πr² = 22/7 × 7.84 × 7.84 = 193.18 cm²

Q22. Solve :

5/x – 4/y = -2  and  2/x + 3/y = 13

Sol.

Let 1/x = u and 1/y = v

5u – 4v = -2 …….(i)

2u + 3v = 13 …….(ii)

Multiply eqn.(i) by 3 and eqn.(ii) by 4, then Add

15u + 8u = – 6 + 52

23u = 46

u = 46/23 = 2

Put u = 2 in eqn.(ii), we get

2(2) + 3v = 13

3v = 13 – 4

v = 9/3 = 3

But 1/x = u and 1/y = v

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

x = 1/2 and y = 1/3

Q23. Find the roots of the quadratic equation 2x² + x – 4, by the method of completing the square.

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

Q24. Find the sum of the first 1000 positive integers.

L. SoS

1, 2, 3, 4, 5, ……….., 1000

a = 1,  d = 2 – 1 = 1,  an = 1000, n = 1000

Sn = n/2 [2a + (n-1)d]

S1000 = 1000/2 [2×1 + (1000-1)1]

= 500 [2+999] = 500 × 1001 = 500500

Q25. Construct a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the ∆ABC.

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, ∆ABC 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, ∆A’BC’ is the required triangle similar to ∆ABC where BA’/BA = BC’/BC = C’A’/CA = 3/4

Q26. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Sol.

Let Point C intersect line AB in the Ratio k : 1 and at Y-axis

A(-3, 10), B(6, -8), C(-1, 6)

Here,  x= -3, x2 = 6, y1 = 10, y2 = -8, x = -1, y = 6, m1 = k, m= 1

Using Section Formula,

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

-1 = (6k-3) ÷ (k+1)

-k – 1 = 6k – 3

7k = 2

k = 2/7

Ratio is 2 : 7

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.

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

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

Q28. Find two consecutive odd positive integers, sum of whose square is 290.

Sol.

Let x, x+2 be two +ve odd integers

According to question :

x² + (x+2)² = 290

x² + x² + 2x + 4 = 290

2x² + 2x + 4 – 290 = 0

2x² + 2x – 286 = 0

Dividing by 2, we get

x² + 2x – 143 = 0

x² + 13x – 11x – 143 = 0

x(x+13) – 11(x+13) = 0

(x+13)(x-11) = 0

Either x + 13 = 0  or  x – 11 = 0

So, x = -13 and  x = 11

Q29. Draw a circle of radius 4 cm and construct tangents to it from an external point using the centre.

Sol.

Steps of constructions–

Step I- Take given circle and a point P outside the circle. O is centre of the circle.

Step II- Joint OP.

Step III- Bisect OP and get its mid-point M.

Step IV- Draw circle with centre M and radius = PM = MO.

Step V- Circle drawn meets the given circle at Q above PO and at Q’ below PO.

Step VI- Join PQ and PQ’.

Step VII- PQ and PQ’ are the required tangents drawn to the circle from the point P.

We observe that PQ = PQ’.

Q30. The shadow of a tower standing on a level ground is found to be 40 m longer when the sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Sol.

Given tower be AB

When Sun’s altitude is 60°, ∠ACB = 60°

When Sun’s altitude is 30°, ∠ADB = 30°

Shadow is 40 m when angle changes from 60° to 30°,  CD = 40 m

Since tower is vertical to ground, ∠ABC = 90°

In right angle triangle ∆ABC,

tanC = AB/CB

tan60° = AB/CB

√3 = AB/CB

CB = AB/√3 …….(i)

In right angle triangle ∆ABD,

tanD = AB/DB

tan30° = AB/DB

1/√3 = AB/DB

DB = √3AB

DC + CB = √3AB

40 + CB = √3AB

CB = √3AB – 40 ……..(ii)

From equation (i) & (ii),

AB/√3 = √3AB – 40

AB = √3(√3AB) – 40√3

AB = 3AB – 40√3

3AB – AB = 40√3

2AB = 40√3

AB = 20√3

Hence, Height of the tower = 20√3 m

OR

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)

Multiply and divide by (tanθ-secθ),

= (tanθ-1+secθ)/(tanθ+1-secθ) × (tanθ-secθ)/(tanθ-secθ)

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

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

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

= -1 ÷ (tanθ-secθ) = 1/(secθ-tanθ) = R.H.S.

Q31. The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m surmounted by a cone of equal base and slant height 12.5 m. Find the capacity of the building.

Sol.

Base radius of cylinder = 12m

Height of cylinder = 3.5 m

Volume of cylinder = πr²h = 22/7 × 12 × 12 × 3.5 = 1584 m³

In a cone, l² = r² + h²

h² = l² – r² = (12.5)² – (12)² = 156.25 – 144 = 12.25

h = √12.25 = 3.5 m

Volume of cone = 1/3 πr²h = 1/3 × 22/7 × 12 × 12 × 3.5 = 528 m³

Total capacity of building = Volume of cylinder + Volume of cone = 1584 + 528 = 2112 m³

Q32. Find the mean of the following frequency distribution :

Sol.

Mean = A + Σf.u/Σf × h= 225 – 7/25 × 50 = 225 – 14 = 211

Note : May be solved by another method.

OR

Find the median of the following frequency distribution :

Sol.

Median = L + (N/2 – Cf)/f × h = = 55 + (15-13)/6 × 5 = 55 + 2/6 × 5 = 55 + 1.7 = 56.7 (approx.)

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