Class 10

To keep the lawn green and cool, Sadhna uses . . .

Question : To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area. The diagram below shows the circular areas covered by two sprinklers:

Two circles touch externally. The sum of their areas is 130π sq. m and the distance between their centres is 14 m.

Based on the above information, answer the following questions :

(i) Obtain a quadratic equation involving R and r from above. [1 Mark]

(ii) Write a quadratic equation involving only r. [1 Mark]

(iii) (a) Find the radius r and the corresponding area irrigated. [2 Marks]

(b) Find the radius R and the corresponding area irrigated. [2 Marks]

Solution :

(i) Circular Area covered by First Sprinkler + Circular Area covered by Second Sprinkler = 130π m² (Given)
πR²+πr²=130π
π(R²+r²)=130π
R²+r²=130 — (1)

which is the required quadratic equation involving R and r.

(ii) R+r=14 m (Given)
R=14-r — (2)
Substituting in equation (1)

R²+r²=130
(14-r)²+r²=130
(14)²+(r)²-2(14)(r)+r²=130
196+r²-28r+r²=130
2r²-28r+196=130
2r²-28r+196-130=0
2r²-28r+66=0
2[r²-14r+33]=0
r²-14r+33=0/2
r²-14r+33=0

which is the required quadratic equation involving only r.

(iii)

(a) r²-14r+33=0

Solving by Factorisation Method (Splitting the Middle Term)

r²-(11+3)r+33=0
r²-11r-3r+33=0
r(r-11)-3(r-11)=0
(r-11)(r-3)=0
r-11=0
r=11

but we know R+r=14 m and r is smaller radius so r≠11

r-3=0
r=3 m

Hence, the required value of r is 3 m
Corresponding Area irrigated
= πr²
= π(3)²
=9π m²

(b) We know,
r = 3 m
Also
R+r=14
R+3=14
R=14-3
R=11 m
Corresponding Area Irrigated
=πR²
=π(11)²
=π(121)
=121π m²

A backyard is in the shape of a triangle ABC with right angle . . .

Question : A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = x m.

Based on the above information, answer the following questions :

(i) Find the length of AR in terms of x. [1 Marks]

(ii) Write the type of quadrilateral BQOR. [1 Marks]

(iii)

(a) Find the length PC in terms of x and hence find the value of x. [2 Marks]

(b) Find x and hence find the radius r of the circle. [2 Marks]

CBSE Delhi 2024 (30/1/1)

Solution :

This is a case study based question from circles.

In Δ ABC
∠ABC=90°
AB=7 m
BC=15 m

AP=x m

(i) Find the length of AR in terms of x.
Ans :

AP=x m (Given)

AP=AR (Length of tangent drawn from an external point to a circle are equal in length)
∴ AR = x m

(ii) Write the type of quadrilateral BQOR.

Ans :

In BQOR
BR=BQ (Length of tangent drawn from an external point to a circle are equal in length)

OR=OQ (Radii of the same circle)

∠ABC=90° (Given)
∠BRO=∠BQO=90° (Tangent at any point of the circle is perpendicular to the radius through the point of contact)

∠ROQ=90° (By ASP of a quadrilateral)

∵ All sides are equal and all angles are of 90°.

∴ BQOR must be a square.

(iii)

(a) Find the length PC in terms of x and hence find the value of x.

Ans :
We know, length of tangent drawn from an external point to a circle are equal in length

AP=AR = x
BR=BQ = (7-x) cm
PC=QC
= 15-(7-x)
= (15-7+x) cm
= (8+x) cm

∴ PC=(8+x) cm

In Rt. ΔABC

AC²=AB²+BC² (By Pythagoras Theorem)
AC²=(7)²+(15)²
AC²=49+225
AC²=274
AC=√274
AC=√274 cm
AP+PC=√274
x+8+x=√274
2x+8=√274
2x=(√274)-8
x=[(√274)-8]/2 cm

(b) Find x and hence find the radius r of the circle.

Ans :
x=[(√274)-8]/2 cm

Radius (r)
= BQ=BR

= (7-x) cm
= 7-{(√274)-8]/2} cm

A rectangular floor area can be completely tiled with . . .

Question : A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.

(i) Assuming the original length of each side be x units, make a quadratic equation from the above information. [1 Mark]

(ii) Write the corresponding quadratic equation in standard form. [1 Mark]

(iii) (a) Find the value of x, the length of side of a tile by factorisation. [2 Marks]

(b) Solve the quadratic equation for x, using quadratic formula. [2 Marks]

CBSE Outside Delhi 2024 (30/3/1)

Solution :

This is a case study question from Quadratic Equations.

Here, it should be understood that the area covered by 200 square shaped tiles of each side 'x' units must be equal to the area covered by 128 different square shaped tiles of each area (x+1) units.

(i) Assuming the original length of each side be x units, make a quadratic equation from the above information.

Ans :

200×Area of each tiles of side x units = 128×Area of each tiles of side (x+1) units.
200×(x)²=128(x+1)²

(ii) Write the corresponding quadratic equation in standard form.

Ans :

200×(x)²=128(x+1)²
Dividing both sides by 8
25x²=16(x²+2x+1)
25x²=16x²+32x+16
25x²-16x²-32x-16=0
9x²-32x-16=0
which is the required quadratic equation.

(iii) (a) Find the value of x, the length of side of a tile by factorisation.
Ans :

9x²-32x-16=0
9x²-(36-4)x-16=0
[By Splitting the middle term]
9x²-36x+4x-16=0
9x(x-4)+4(x-4)=0
(x-4)=(9x+4)=0
(x-4)=0 or (9x+4)=0
x=4 or 9x=-4
x=4 or x=-4/9
But Side of a tile can't be negative. So, x≠0.
Hence, x=4

(b) Solve the quadratic equation for x, using quadratic formula.

Ans :

9x²-32x-16=0

Here

a=9
b=-32
c=-16
D=b²-4ac
D=(-32)²-4(9)(-16)
D=1024+576
D=1600

D>0
∴ Roots are real and equal.

Using Quadratic Formula

x=[-b±√D]/2a
x=[-(-32)±√1600]/2(9)
x=[32±40]/18
x=[32±40]/18
x=[32+40]/18 OR x=[32-40]/18
x=[72]/18 OR x=[-8]/18
x=4 OR x=-4/9

But Side of a tile can't be negative. So, x≠0.
Hence, x=4

A stable owner has four horses. He usually tie these horses . . .

Question : A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But trying with rope sometimes result in injuries to his horses, so he decided to built fence around the area so that each horse can graze.

Based on the above, answer the following questions :

(i) Find the area of the square shaped grass field. [1 Mark]

(ii)
(a) Find the area of the total field in which these horses can graze. [2 Marks]
OR
(b) If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. [Use π=3.14] [2 Marks]

(iii) What is the area of the field that is left ungrazed, if the length of the rope of each horse is 7 cm. [1 Marks]

CBSE Outside Delhi [2024] (30/3/1)

Solution :

This is a case study based question from Area Related to circles (Mensuration).

Here side of the square (a) = 20 m

Length of the rope (r) = 7 m

(i) Find the area of the square shaped grass field.

Ans :

Area of square shaped grass field
= (side)²
= a²
= (20)²
= 400 m²

(ii)

(a) Find the area of the total field in which these horses can graze.

Ans :

Required Area
=4×Area of sector
=4×[θ/360]×[π](r)²
=4×[90/360]×[22/7](7)²
=4×[1/4]×22×7
=22×7
=154 m²

OR
(b) If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. [Use π=3.14]

Ans :

Here,
New Radius (r') =10 m

Area grazed by one horse when the rope is increased from 7 m to 10 m

= [θ/360]×[π](r)²

= [90/360]×[3.14](10)²
= [1/4]×314
= 78.5 m²

(iii) What is the area of the field that is left ungrazed, if the length of the rope of each horse is 7 cm.
Ans :

Area of the field left ungrazed
= Area of field - area of field grazed by 4 horses
= 400 m² - 154 m²
= 246 m²

Teaching Mathematics through activities is a powerful . . .

Question : Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms Mukta planned a prime number game for class 5 students. She announces the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to second student. Second student also multiply it by a prime number and passed it to third student. In this way by multiplying to a prime number, the last student got 173250.

Now, Mukta asked some questions as given below to the students :

(i) What is the least prime number used by the students? [1 Mark]

(ii)

(a) How many students are in the class? [2 Marks]
OR

(b) What is the highest prime number used by the students? [2 Marks]

(iii) Which prime number has been used maximum times? [1 Mark]

CBSE Outside Delhi [2024] (30/3/1)

Solution :

Here the final number obtained by the last students is 173250.

Since all the students have multiplied prime numbers only so we can get all the multiplied prime numbers back by prime factorisation of final number obtained which is 173250.

173250=2×3×3×5×5×5×7×11

173250=2×3²×5³×7×11

(i) What is the least prime number used by the students?

Ans : Least Prime number is 2 but it is given by the teacher not the student. So the least prime number used by the student is 3 and not 2.

(ii)

(a) How many students are in the class?
Ans : Number of students in the class will be equal to prime factors of 173250 except the first number 2.

173250=2×3×3×5×5×5×7×11

So, there are 7 students in the class.

(b) What is the highest prime number used by the students?

Ans :

173250=2×3×3×5×5×5×7×11

Clearly, highest prime number used by the student is 11.

(iii) Which prime number has been used maximum times?

Ans :

173250=2×3×3×5×5×5×7×11

From the above prime factorisation it is clear that the prime factor 5 has been used maximum times (5 times).

BINGO is game of chance. The host has 75 . . .

Question : BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participants cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'

Numbers Announced	Number of times
0-15	8
15-30	9
30-45	10
45-60	12
60-75	9

Based on the above information, answer the following:

(i) Write the median class. [1 Mark]

(ii) When first ball was picked up, what was the probability of calling out an even number? [1 Mark]

(iii) (a) Find median of the given data. [2 Marks]

(iv) Find mode of the given data. [2 Marks]

CBSE Delhi 2024 (30/1/1)

Solution :

This is a case study question based on Statistics and Probability.

(i)

Numbers Announced	Number of times	Cumulative Frequency (CF)
0-15	8	8
15-30	9	17
30-45	10	27
45-60	12	39
60-75	9	48
	∑f_i=48

Here N=∑f_i=48
N/2=48/2=24

Median Class = 30-45

(ii) Total Number of Balls {1,2,3,. . . , 75}
= 75-1+1 = 75
Number of balls having even number {2,4,6. . .74}
=37

P(E) = [Number of favorable outcomes] / [Total number of possible outcomes]

P(Calling out an even number)
= [No. of balls having even number]/[Total number of balls]
= 37/75

(iii)

(a) Median

Numbers Announced	Number of times	Cumulative Frequency (CF)
0-15	8	8
15-30	9	17
30-45	10	27
45-60	12	39
60-75	9	48
	∑f_i=48

Here N=∑f_i=48
N/2=48/2=24

Median Class = 30-45

Lower Limit of the Median Class (L) = 30
Class Size (h)
=45-30
=15
CF=17
f=10

Median
= L+[(N/2-CF)/f]×h
= 30+[(24-17)/10]×15
= 30+[7/10]×15
= 30+[0.7×15]
= 30+10.5
= 40.5

Hence, the median of the given data is 40.5

(b)

Mode

Numbers Announced	Number of times
0-15	8
15-30	9
30-45	10=f0
45-60	12=f1
60-75	9=f2

Here, the maximum frequency is 12
Modal Class = 45-60

Lower limit of the modal class (L) = 45

Class size (h) = 60-45 = 15

f₀=10

f₁=12

f₂=9

Mode
= L+[(f₁-f₀)/(2f₁-f₀-f₂)]×h
= 45+[(12-10)/(2{12}-10-9)]×15
= 45+[(2)/(24-19)]×15
= 45+[2/5]×15
= 45+[2]×3
= 45+6
= 51
Hence, the mode of the given data is 51.

Vocational training complements traditional education . . .

Question : Vocational training complements traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing employability thus making the student self reliant. Keeping this in view, a teacher made the following table giving the frequency distribution of students/adults undergoing vocational training from the training institute.

Age (in years)	15-19	20-24	25-29	30-34	35-39	40-44	45-49	50-54
No. of participants	62	132	96	37	13	11	10	4

From the above answer the following questions :

(i) What is the lower limit of the modal class of the above data? [1 Mark]

(ii) (a) Find the median class of the above data. [2 Marks]

(b) Find the number of participants of age less than 50 years who undergo vocational training.

(iii) Give the empirical relationship between the mean, median and mode. [1 Mark]

CBSE 2024 (30/3/1)

Solution :

This is a case study based question form Class 10th Statistics.

Here the class intervals are not continuous so first of all we have to make them continuous.

As we can see that here the gap is 1 (15-19, 20-24)

So we will subtract 0.5 (1/2=0.5) from the lower limit and add 0.5 to the upper limit.

Modified continuous frequency distribution table is given below

Age (in years)	14.5-19.5	19.5-24.5	24.5-29.5	29.5-34.5	34.5-40.5	39.5-44.5	44.5-49.5	49.5-54.5
No. of participants	62	132	96	37	13	11	10	4

i) Maximum frequency = 132
Modal Class = 19.5-24.5
Lower Limit of the modal class = 19.5

ii)

Age (in years)	14.5-19.5	19.5-24.5	24.5-29.5	29.5-34.5	34.5-40.5	39.5-44.5	44.5-49.5	49.5-54.5
No. of participants	62	132	96	37	13	11	10	4
Cumulative Frequency (CF)	62	194	290	327	340	351	361	365

(a) Total number of observations (N) = 365
N/2 = 365/2 = 182.5

Hence, the required Median Class is 19.5-24.5

OR

(b) Number of participants of age less than 50 years = 365-4 = 361

(iii) 3Media = 2Mean+Mode
which is the required empirical relation between mean, median and mode.

If one zero of the quadratic polynomial f(x)=4x2-8kx-9 . . .

Question : If one zero of the quadratic polynomial f(x)=4x²-8kx-9 is negative of the other, then find the value of k.

Doubt by Ankita

Solution :

f(x)=4x²-8kx-9

Here

a=4

b=-8k

c=-9

Let

α=α

β=-α

Sum of zeroes

α+β=-b/a

α+(-α)=-(-8k)/4

0=2k

0/2=k
0=k

k=0

Hence, the required value of k is 0.

If one zero of f(x)=4x2-8kx+8x-9 is negative of the . . .

Question : If one zero of f(x)=4x²-8kx+8x-9 is negative of the other, then find zeroes of kx²+3kx+2.

Doubt by Angel

Solution :

f(x)=4x²-8kx+8x-9
f(x)=4x²-8(k-1)x-9
Here

a=4
b=-8(k-1)
c=-9

Let

α=α
β=-α

Sum of zeroes
α+β=-b/a
α+(-α)=-[-8(k-1)]/4
0=8(k-1)/4
0=2(k-1)
0/2=k-1

0=k-1
1=k
k=1 — (1)

=kx²+3kx+2
=(1)x²+3(1)x+2 [From equation (1)]
=x²+3x+2
=x²+(2+1)x+2
=x²+2x+x+2
=x(x+2)+1(x+2)
=(x+1)(x+2)

x+1=0
x=-1

x+2=0
x=-2

Hence, required zeroes of kx²+3kx+2 are -1 and -2.

A spherical glass vessel has a cylindrical neck 8 cm long . . .

Question : A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

Doubt by Swarna

Solution :