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