# HBSE 10th Maths Solved Sample Paper 2023

HBSE 10th Maths Solved Sample Paper 2023

Q1. Which of the following is an irrational number?

(A) 2√4

(B) √9

(C) 2 + √3

(D) 2 + √4

Q2. Zeroes of the polynomial 6x² – 7x – 3 are :

(A) -1/3, 3/2

(B) -2/3, 3/2

(C) 5/2, 3/2

(D) -1/2, 5/2

6x² – 7x – 3

= 6x² + 2x – 9x – 3

= 2x(3x+1) – 3(3x+1)

= (3x+1)(2x-3)

Either 3x+1 = 0  or  2x-3 = 0

x = -1/3  or  x = 3/2

Roots of quadratic equation are  -1/3 and 3/2

Q3. Which of the following is a quadratic equation?

(A) x² + 2x-¹ = 3

(B) x(x² – 1) = 4

(C) x + 1/x = 5

(D) (x² – 1)² = 2

x + 1/x = 5

x² + 1 = 5x

x² – 5x +1 = 0

Q4. 15th term of A.P.  5, 6½, 8, 9½, ………… is

(A) 15½

(B) 14½

(C) 26

(D) 27½

Here  a = 5, d = 6½ – 5 = 1½ = 3/2

nth term of AP,  an = a + (n-1)d

a15 = 5 + (15-1) × 3/2 = 5 + 21 = 26

Q5. If the third term of an A.P. is 12 and tenth term is 26, then its 20th term is :

(A) 46

(B) 52

(C) 50

(D) 44

nth term of AP,  an = a + (n-1)d

3rd term of the A.P. = 12

a + 2d = 12 ……(i)

10th term of the A.P. = 26

a + 9d = 26 …….(ii)

Subtract eqn.(i) from eqn.(ii),

7d = 14

d = 2

Puting d=2 in eqn.(i),

a + 2(2) = 12

a + 4 = 12

a = 12-4 = 8

So, the 20th term of A.P. = a + 19d

= 8 + 19(2) = 8 + 38 = 46

Q6. If areas of two similar triangles are 1 : 2 then ratio of their sides are :

(A) 1 : 4

(B) 1 : √2

(C) 1 : 2

(D) 2 : 1

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 = √1 : √2 = 1 : √2

Q7. If the vertices of a ∆ABC are A(4, -6), B(3, -2) and C(5, 2), then the coordinates of the mid point of BC are :

(A) (4,0)

(B) (-1,-2)

(C) (3/2, 5/2)

(D) (1,2)

Take D is Mid point of BC,

B(3, -2) and C(5, 2) and D(x, y)

Here x= 3, x2 = 5, y1 = -2, y2 = 2

Using Mid Point Theorem,

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

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

= 8/2, 0/2

= 4, 0

Co-ordinates of D(x,y) is (4, 0).

Q8. Distance between the points (-1, 7) and (4, -3) is :

(A) 7√7

(B) 5√5

(C) 25

(D) 3√3

A(-1, 7) and B(4, -3)

Here,  x1 = -1, x2 = 4, y1 = 7, y2 = -3

Using Distance Formula,

AB = √(x2-x1)²+(y2-y1

= √(4+1)²+(-3-7)²

= √(5)²+(-10)²

= √25+100 = √125 = 5√5

AB = 5√5 unit

Q9. The value of (2tan²30°)/(1+tan²30°) is :

(A) √3

(B) 3/2

(C) √3/2

(D) 1/2

(2tan²30°)/(1+tan²30°)

= 2(1/√3)² ÷ [1+(1/√3)²]

= 2(1/3) ÷ (1+1/3)

= 2/3 ÷ 4/3

= 2/3 × 3/4

= 1/2

Q10. If cosecA = 5/4, then value of tanA is :

(A) 3/4

(B) 3/5

(C) 4/3

(D) 5/3

cosecA = 5/4

1/sinA = 5/4

sinA = 4/5

sin²A + cos²A = 1

(4/5)² + cos²A = 1

cos²A = 1 – (4/5)² = 1 – 16/25 = 9/25

cosA = √9/25 = 3/5

tanA = sinA/cosA = (4/5) ÷ (3/5) = 4/5 × 5/3 = 4/3

Q11.  Number of tangents drawn from a point inside the circle is :

(A) 1

(B) 2

(C)  3

(D) 0

Q12. All circles are ……………… (Similar/Congruent)

Similar

Q13. ∆ABC and ∆BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is :

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Ratio of area of similar triangles is equal to ratio of square of their sides.

Area of ∆ABC/Area of ∆BDE = BC²/BD²

BC = 2BD (due to the mid point theorem)

Area of ∆ABC/Area of ∆BDE = (2BD)²/BD²

= 4BD²/BD² = 4 : 1

So, the ratio of the areas of ∆ABC and ∆BDE is 4 : 1

Q14. The diameter of the base of a cylinder is 4.2 cm and height 5 cm, then volume is :

(A) 22.05 π

(B) 7.35 π

(C) 21 π

(D) 19 π

Diameter = 4.2 cm,  Radius = 4.2/2 = 2.1 cm

Volume of the cylinder = πr²h = π x 2.1 x 2.1 x 5 = 22.05 π

Q15. H.C.F. of 12, 21 and 84 is :

(A) 2

(B) 3

(C) 7

(D) 4

12 = 2 × 2 × 3

21 = 3 × 7

84 = 2 × 2 × 3 × 7

HCF = 3

Q16. For solution of equations x + 2y + 3 = 0 and 2x + 4y + 6 = 0, which of the following is true?

(A) Unique Solution

(B) infinitely many solutions

(C) No Solution

(D) None of these

x + 2y + 3 = 0,  2x + 4y + 6 = 0

Here, a= 1,  a2 = 2,  b= 2,  b2 = 4,  c= 3,  c= 6

a1/a2 = 1/2

b1/b2 = 2/4 = 1/2

c1/c2 = 3/6 = 1/2

a1/a2 = b1/b2 = c1/c2

so, equations have infinitely many solutions.

Q17. Show that √2 is an irrational number.

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. Find a quadratic polynomial whose zeroes are 3 and -2.

α + β = 3 + (-2) = 3 – 2 = 1,  αβ = 3(-2) = – 6

= x² – (α+β)x + αβ

= x² – (1)x + (-6)

= x² – x – 6

Q19. Solve the following pair of linear equations :

2x + 3y = 7  and  6x – 5y = 11

2x + 3y = 7 ……..(i)

6x – 5y = 11 ………(ii)

Multiply eqn.(i) by 3 subtract from eqn.(ii),

(6x – 5y) – (6x + 9y) = 11 – 21

6x – 5y – 6x – 9y = -10

-14y = -10

y = (-10) ÷ (-14) = 5/7

Put y = 5/7 in eqn.(i),

2x + 3(5/7) = 7

2x + 15/7 = 7

2x = 7 – 15/7 = (49-15)/7 = 34/7

2x = 34/7

x = 17/7

So, x = 17/7  and  y = 5/7

Q20. Find two numbers whose sum is 37 and product is 300.

Let two numbers are x and y.

According to question,

x + y = 37 ……(i)

xy = 300 ……(ii)

From eqn.(i),  y = 37-x

Put  y = 37-x  in eqn.(ii),

x(37-x) = 300

37x – x² = 300

x² – 37x + 300 = 0

x² – 25x – 12x + 300 = 0

x(x-25) – 12(x-25) = 0

(x-25)(x-12) = 0

Either  x-25=0  or  x-12=0

x = 25  or  x = 12

If x = 25 then y = 12

If x= 12 then y = 25

So, numbers are 25 and 12.

Q21. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of TP.

Since OT is perpendicular bisector of PQ.  Therefore, PR = RQ = 4 cm
Now, OR = √OP²-PR² = √5²-4² = √25-16 = √9 = 3 cm
Now, ∠TPR + ∠RPO = 90°  (∠TPO=90°)
∠TPR + ∠PTR = 90° (∠TRP=90°)
∠RPO = ∠PTR
So, Right ∆TRP is similar to the right ∆PRO.
∆TRP~∆PRO  [By A-A Rule of similar triangles]
TP/PO = RP/RO
TP/5 = 4/3
TP = 20/3 cm

Q22. A die is thrown once, find the probability of getting a prime number.

Total numbers = 6 (1,2,3,4,5,6)

Prime numbers = 3 (2,3,5)

P(prime no.) = 3/6 = 1/2

Q23. For which value of a and b does the following pair of linear equations have infinitely number of solutions.

2x + 3y = 7  and  (a-b)x + (a+b)y = 3a + b – 2

2x + 3y – 7 = 0 and (a-b)x + (a+b)y – (3a+b-2) = 0

Here a1=2,  a2=(a-b),  b1=3,  b2=(a+b),  c1=-7,  c2=-(3a+b-2)

a1/a2 = 2/(a-b)

b1/b2 = 3/(a+b)

c1/c2 = -7/[-(3a+b-2)]

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

2/(a-b) = 3/(a+b) =  -7/[-(3a+b-2)]

Taking  2/(a-b) = 3/(a+b)

3(a-b) = 2(a+b)

3a – 3b = 2a + 2b

a = 5b …….(i)

Taking  2/(a-b) = -7/[-(3a+b-2)]

2/(a-b) = 7/(3a+b-2)

7(a-b) = 2(3a+b-2)

7a – 7b = 6a + 2b – 4

a – 9b = – 4 …….(ii)

Put a=5b in eqn.(ii),

5b – 9b = -4

-4b = -4

b = 1

Put b=1 in eqn.(i),

a = 5b = 5(1) = 5

So, a = 5 and b = 1

Q24. One side of a rectangle exceeds its other side by 2 cm. If its area is  195 cm², then determine the sides of the rectangle.

Let one of the side be x

Other side = x + 2

Area of rectangle = length × breadth

x(x+2) = 195

x² + 2x – 195 = 0

x² + 15x – 13x – 195 = 0

x(x+15) – 13(x+15) = 0

(x+15)(x-13) = 0

Either  x+15=0  or  x-13=0

x = 13   [x=-15, not possible]

So,one of the side = x = 13 cm

Other side = x + 2 = 13 + 2 = 15 cm

Q25. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.

First term, a = 5

Last term, l = 45

Sum of n terms, Sn = 400

Sum of n terms,  Sn = n/2 [a+l]

400 = n/2 (5+45)

400 = n/2 × 50

400 = 25n

n = 400/25 = 16

Last term, l = an = 45

nth term of A.P.,  l = an = a + (n-1) d

45 = 5 + (16-1)d

45-5 = 15d

15d = 40

d = 40/15

d = 8/3

So, the number of terms, n = 16

common difference, d = 8/3

Q26. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

We know that, in a rhombus, diagonals bisect each other at 90°.

In rhombus ABCD, AC and BD are the diagonals.

AC ⊥ BD, OA = OC and OB = OD

In ΔAOB, ∠AOB = 90°

Using Pythagoras theorem,

AB² = OA² + OB² ……………(i)

Similarly,

BC² = OB² + OC² ……………..(ii)

CD² = OC² + OD² ………………(iii)

AD² = OD² + OA² ……………….(iv)

Adding (i), (ii), (iii) and (iv)

AB² + BC² + CD² + AD² = OA² + OB² + OB² + OC² + OC² + OD² + OD² + OA²

AB² + BC² + CD² + AD²  = 2OA² + 2OB² + 2OC² + 2OD²

AB² + BC² + CD² + AD² = 2[OA² + OB² + OC² + OD²]

AB² + BC² + CD² + AD² = 2 [(AC/2)² + (BD/2)² + (AC/2)² + (BD/2)²]

[Since, OA = OC = AC/2 and OB = OD = BD/2]

AB² + BC² + CD² + AD² = 2 [(AC² + BD² + AC² + BD²)/4]

AB² + BC² + CD² + AD² = 2[(2AC² + 2BD²)/4]

AB² + BC² + CD² + AD² = 4[(AC² + BD²)/4]

AB² + BC² + CD² + AD² = AC² + BD²

Hence Proved.

Q27. If tan2A = cot(A-18°), where 2A is an acute angle, then find the value of A.

tan2A = cot (A-18°)

cot(90°-2A) = cot(A-18°)   [cotθ = tan(90°-θ)]

90° – 2A = A – 18°

3A = 108°

A = 108°/3

A = 36°

Q28. Find the ratio in which the line joining (3,4) and (-4,7) is divided by Y-axis. Also find the co-ordinates of the point of intersection.

x= 3, x2 = -4, y1 = 4, y2 = 7

Using Section Formula,

x = (m1x2 + m2x1) ÷ (m1 + m2

0 = [k(-4) + 1(3)] ÷ (k + 1)

0 × (k+1) = -4k + 3

0 = -4k + 3

k = 3/4

So,  m1 = 3,  m2 = 4

y = (m1y2 + m2y1) ÷ (m1 + m2

= [3(7) + 4(4)] ÷ (3+4)

= (21+16) ÷ 7 = 37/7

So, (x,y) = (0, 37/7)

Q29. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm, sweeping through angle of 120°. Find the total area cleaned at each sweep of the blades.

Area of sector = θ/360° × πr²

Total Area cleaned by two wipers blade = 2 ×  θ/360° × πr²

= 2 × 120°/360° × 22/7 × 25 × 25 = 1309.5 cm²

Q30. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a perfect square number

(ii) a number divisible by 5

(i) Perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64, 81)

Total numbers = 90

P(perfect square number) = 9/90 = 1/10

(ii) Numbers divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90)

P(number divisible by 5) = 18/90 = 1/5

Q31. An observer 1.6m tall is 20m away from a tower. The angle of elevation of the top of the tower from his eyes is 60°. Find the height of the tower.

BC = DE = 20 m

Let AB = h

In ∆AED,

AE/ED = tan60°

AE/20 = √3

AE = 20√3 m

Now AB = h = AE + ED = 20√3 + 1.6 = 20(1.7) + 1.6 = 34 + 1.6 = 35.6 m

Height of tower = 35.6 m

OR

Prove that :

(cosecA-sinA)(secA-cosA)(tanA+cotA)=1

L.H.S. (1/sinA – sinA) × (1/cosA – cosA) × (sinA/cosA + cosA/sinA)

= [(1-sin²A)/sinA] × [(1-cos²A)/cosA] × [(sin²A+cos²A)/cosAsinA]

= cos²A/sinA × sin²A/cosA × 1/cosAsinA

= 1 = R.H.S.

Q32. Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6cm.

Steps of construction :

(i) Draw a circle of radius 4 cm from the centre O. With same centre and radius equal to 6 cm, construct another circle.

(ii) Take any point P on the circumference of the outer circle and join OP.

(iii) Construct perpendicular bisector for the line segment PO, which intersect OP at point M.

(iv) Now, with M as centre, construct a circle of radius equal to PM or MO.

(v) This circle now intersects the inner circle at points Q and R. Join PQ and PR.

(vi) Thus, tangents have been constructed from outer circle to the inner circle.

Q33. A vessel is in the form of hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere and cylinder is 21cm. If total height of the vessel is 28cm, find the volume of the vessel.

Radius of cylinder = Radius of hemisphere = 21/2 = 10.5 cm

height of cylinder = 28 – 10.5 = 17.5 cm

Volume of vessel = volume of cylinder + volume of hemisphere

= πr²h + 2/3 πr³ = (22/7 × 10.5 × 10.5 × 17.5) + (2/3 × 22/7 × 10.5 × 10.5 × 10.5)

= 6063.75 + 2425.5 = 8,489.25 cm³

OR

A well of diameter 2 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of circular ring of width 5m to form an embankment. Find the height of the embankment.

height of the embankment = h

The shape of the embankment will be like the shape of a cylinder of internal radius 1 m and external radius (5+1)m = 6m

Now, the volume of the earth dug = volume of the cylindrical well = π × 1² ×14 = 14π m³

Also, the volume of the embankment = π(R²-r²)h

= π(6²-1²)h = π(36-1)h = 35πh m³

Hence, we have

35πh = 14π

h = 14π/35π = 2/5 = 0.4 m

Hence, the height of the embankment = 0.4 m

Q34. The median of the following data is 28.5. Find the values of x and y, if the total frequency is 60.

Median = 28.5
n = ∑fi = 60
n/2 = 60/2 = 30
Since, median is 28.5, median class is 20−30.
Hence,  l = 20,  h = 10,  f = 20,  cf = 5+x
Median = l + (n/2 – cf)/f × h
28.5 = 20 + (30-5-x)/20 × 10
28.5 = 20 + (25-x)/2
28.5 – 20 = (25-x)/2
8.5 × 2 = 25 – x
17 = 25 – x
x = 25 – 17 = 8
Also,  45 + x + y = 60
y = 60−45−x = 15-8 = 7

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