HBSE Class 10 Maths Question Paper 2021 Answer Key

HBSE Class 10 Maths Question Paper 2021 Answer Key

HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2021. HBSE 10th Question Paper Download 2021. HBSE Class 10 Maths Paper Solution 2021. Haryana Board Class 10th Maths Question Paper 2021 Pdf Download with Answer.

Subjective Questions

Q1. Find two consecutive positive integers, the sum of whose squares is 365.

Sol.

Let First integer = x

So, Second integer = x+1

Also given that

Sum of squares = 365

(First number)² + (Second number)² = 365

x² + (x+1)² = 365

x² + x² + 1² + 2x = 365

2x² + 2x +1 – 365 = 0

2x² + 2x – 364 = 0

divide both side by 2,

x² + x – 182 = 0

x² + 14x – 13x – 182 = 0

x(x+14) – 13(x+14) = 0

(x+14)(x-13) = 0

so, x = -14, 13

x is positive, so use x=13

First integer = x = 13

Second integer = x+1 = 1+13 = 14

Q2. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Sol.

We know that diagonals of parallelogram bisect each other

So, O is the mid-pint of AC & BD

We find x co-ordinate of O from both AC & BD

Finding mid-point of AC,

x-coordinate of O = (x1+x2)/2 = (6+9)/2 = 15/2

Finding mid-point of BD,

x-coordinate of O = (x1+x2)/2 = (8+p)/2

Comparing x-coordinate of both,

(8+p)/2 =15/2

8+p = 15

p = 15-8 = 7

Q3. Evaluate :  [7sin²30° + 6cosec²60° – cot²45°] ÷ [sin²60° + cos²60°]

Sol.

= [7(1/2)² + 2(2/√3)² – (1)²] ÷ [(√3/2)² + (1/2)²]

= [7/4 + 8 – 1] ÷ [3/4 + 1/4]

= [(7+32-4)/4] ÷ [(3+1)/4]

= [35/4] ÷ [4/4]

= 35/4

Q4. Two cubes each of volume 64cm³ 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 cm + 4 cm = 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.

Q5. A bag contains 7 red balls and 3 green balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) green.

Sol.

No. of Red balls = 7

No. of green balls = 6

so, Total balls = 7 + 6 = 13

(i) P(red) = 7/13

(ii) P(green) = 6/13

Q6. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes fraction 1/2 if we only add 1 to the denominator. What is the fraction?

Sol.

Let Numerator = x

and Denominator = y

so, Fraction = x/y

Given that,

(Numerator+1) ÷ (Denominator-1) =1

(x+1) ÷ (y-1) = 1

x+1 = y-1

x – y = -2 ………(i)

Given that,

(Numerator) ÷ (Denominator+1) = 1/2

(x) ÷ (y+1) = 1/2

2x = y+1

2x – y = 1 ………(ii)

Subtract equation (i) from equation (ii),

x = 3

use x = 3 in equation (i)

3 – y = -2

y = 3+2 = 5

Hence, Original Fraction = x/y = 3/5

Q7. Find the 20th term from the last of the A.P. 3, 8, 13, …….., 253.

Sol.

Given AP 3, 8, 13, …….., 253

Now, 20th term from last of AP 3, 8, 13, ….., 253 = 20th term of AP 253, 248, 243, …….., 8, 3

Thus, our AP is 253, 248, 243, ….., 8, 3

Here a = 253, d = 248 – 253 = – 5, n = 20

an = a + (n-1)d

an = 253 + (20-1)(- 5)

an = 253 + 19(- 5)

an = 253 – 95 = 158

So the 20th term from the Last is 158.

Q8. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Sol.

Height of window = AC = 8m

Length of ladder = AB = 10m

Since the wall will be perpendicular to ground ACB = 90°

∆ACB is a right angle triangle

So, Using Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

AB² = AC²+ BC²

10² = 8² + BC²

100 = 64 + BC²

100 – 64 = BC²

36 = BC²

BC = √36 = 6 m

Hence, distance of the foot of the ladder from base of the wall (BC) = 6m

Q9. Find the area of the shaded region as shown in figure, where ABCD is a square of side 16 cm.

Sol.

Area of shaded region = Area of square ABCD – Area of 4 circles

Area of square ABCD = (side)² = 16² = 256 cm²

Diameter of each circle = AB/2 = 16/2 = 8 cm

Radius of each circle = Diameter/2 = 8/2 = 4 cm

Area of one circle = πr² = 3.14 x 4 x 4 = 50.24 cm²

So, Area of 4 circle = 4 x Area of one circle

= 4 x 50.24 = 200.96 cm²

Area of shaded region = Area of square ABCD – Area of 4 circles = 256 – 200.96 = 55.04 cm²

Area of shaded region = 55.04 cm²

Q10. Find the median of the following data :

Sol.

Here  cf = 13,  n = 60,  l = 20,  f = 20,  h = 10

Median = l + (n/2 – cf)/f × h

= 20 + (30 – 13)/20 × 10

= 20 + 17/2 = 20 + 8.5 = 28.5

Q11. Solve the following pair of equations by reducing them to a pair of linear equations :

6x + 3y = 6xy and 2x + 4y = 5xy

Sol.

Divide both equations by xy, then

6/y + 3/x = 6 ……..(i)

2/y + 4/x = 5 ……..(ii)

Let  1/x = u,  1/y = v

So, our equations become

6v + 3u = 6 ……..(iii)

2v + 4u = 5 ……..(iv)

Multiple equation (iv) by 3, then it subtract from equation (iv),

(6v-6v) + (3u-12u) = 6-15

-9u = -9

u = 1

Put u = 1 in equation (iv)

2v + 4(1) = 5

2v = 5 – 4 = 1

v = 1/2

Hence,  u = 1, v = 1/2

Now u = 1/x , x = 1/u = 1/1 = 1

and  v = 1/y, y = 1/v = 2

Hence x = 1, y = 2 is the solution of the given equation.

OR

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages ?

Sol.

Let Present age of Jacob = x  years

& Present age of Jacob’s son = y  years

Five years hence (later),

Jacob’s Age = x + 5

Jacob son’s Age = y + 5

Age of Jacob will be three times of his son,

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x – 3y = 15 – 5

x – 3y = 10 ……..(i)

Also,

Five years ago,

Jacob’s Age = x – 5

Jacob son’s Age = y – 5

Age of Jacob was seven times of his son,

x – 5 = 7(y – 5)

x – 5 = 7y – 35

x – 7y = -35 + 5

x – 7y = -30 ……..(ii)

Subtract equation (ii) from (i),

4y = 40

y = 10

Putting y = 10 in equation (i),

x – 3(10) = 10

x = 10 + 30 = 40

So, Present age of Jacob = x = 40 years

& Present age of Jacob’s son = y = 10 years

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