If x=ab3 and y=a3b, where a and b are prime numbers, then . . .

Question : If x=ab³ and y=a³b, where a and b are prime numbers, then [HCF(x,y)-LCM(x,y)] is equal to :

(a) 1-a³b³

(b) ab(1-ab)

(c) ab-a⁴b⁴

(d) ab(1-ab)(1+ab)

CBSE 2024

Solution : 

x=ab³
y=a³b

HCF = ab
LCM = a³b³

HCF(x,y)-LCM(x,y)
=ab-a³b³
=ab[1-a²b²]
=ab[1²-(ab)²]
=ab[(1-ab)(1+ab)]
=ab(1-ab)(1+ab) 

Hence, (d) ab(1-ab)(1+ab), would be the correct option. 

Case Study Based Questions on Some Applications of Trigonometry | CBSE |

A drone was used to facilitate movement of an ambulance on a straight highway to a point P on the ground where there was a accident. The ambulance was travelling at a speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be 30° at a particular instant. 

Based on the above information, answer the following questions:

(i) Represent the above situation with the help of a diagram. [1 Mark]

(ii) Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use √3=1.73) [1 Mark]

(iii) (a) Find the time (in seconds) in which the angle of depression changes from 30° to 45°. [2 Marks]

OR

(iii) (b) How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is 60° from the drone? [2 Marks]

Case Study Based Question on Aritemetic Progression (AP) | CBSE

Cable cars at hill stations are one of the major tourist attractions. On a hill station, the length of cable car ride from base point to top most point on the hill is 5000 m. Poles are installed at equal intervals on the way to provide support to the cables on which car moves. 

The distance of first pole from base point is 200 m and subsequent poles are installed at equal interval of 150 m. Further, the distance of last pole from the top is 300 m. 

Based on above information, answer the following questions using Arithemtic Progression : 

(i) Find the distance of 10th pole from the base. [1 Mark]

(ii) Find the distance between 15th pole and 25th pole. [1 Mark]

(iii) (a) Find the time taken by cable car to reach 15th pole from the top if it is moving at the speed of 5 m/sec and coming from top. [2 Marks]

OR

(iii) (b) Find the total number of poles installed along the entire journey. [2 Marks] 

Case Study Based Question on Area Related to Circles | CBSE 10th Board

Case Study Based Question on Area Related to Circles | Class 10 Board

The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of class X took part in various activities organised by the school. One such group of students made 5 circular rings in the school lawn with the help of ropes. Each circular ring required 44 m of rope. 

Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that ΔOAB is an equilateral triangle and all unshaded regions are congruent.

Based on above informatiomn, answer the following questions :

(i) Find the radius of each circular ring. [1 Mark]

(ii) What is the measure of each circular ring. [1 Mark]

(iii) (a) Find the area of shaded region R1. [1 Mark]

OR

(iii) (b) Find the length of rope around the unshaded regions. [2 Marks]

The denominator of a fraction is one more than twice the numerator. If the sum of the . . .

Question : The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2 16/21 , find the fraction. [CBSE D 2024] [5 Marks]

Solution : 

Let the numerator be x
and denominator be 1+2x
Required fraction = x/(1+2x)
Reciprocal of fraction = (1+2x)/x

ATQ
x/(1+2x) + (1+2x)/x=2 (16/21)
x/(1+2x) + (1+2x)/x=58/21

Let x/(1+2x) = y
and (1+2x)/x = 1/y

y+1/y=58/21

(y²+1)/y=58/21
21y²+21=58y
21y²-58y+21=0
21y²-(49+9)y+21=0
21y²-49y-9y+21=0
7y(3y-7)-3(3y-7)=0
(3y-7)(7y-3)=0

3y-7=0
3y=7
y=7/3

Or 

7y-3=0
7y=3
y=3/7

Now, 
x/(1+2x) = 7/3
3x=7(1+2x)
3x=7+14x
3x-14x=7
-11x=7
x=7/-11
x=-7/11
Rejected because x can't be negative.

x/(1+2x) = 3/7
7x=3+6x
7x-6x=3
x=3

Required Fraction = x/(1+2x)
=3/[1+2(3)]
=3/[1+6]
=3/7

Hence, the required fraction is 3/7.

Video Solution :

An arc of a circle is of length 5p cm and the sector bounds an . . .

Question : An arc of a circle is of length 5p cm and the sector bounds an area of 20p cm². Then, the radius of the circle is

(a) 4 cm 

(b) 8 cm 

(c) 12 cm 

(d) 16 cm

Doubt by Shravya

Solution : 

Length of arc = 5p cm
Area of sector = 20p cm²
Radius  = ?

Formula Used
Area of sector = ½×length of arc × Radius
2 [Area of sector / length of arc] = Radius
Radius = 2 [Area of sector / length of arc] 
Radius = 2[20p/5p]
Radius = 40p/5p
Radius = 8 cm 

Hence, (b) 8 cm, would be the correct option. 

Similar Question : 

1.) An arc of a circle is of length 5π cm and the sector bounds an area of 20π cm². Then, the radius of the circle is

(a) 4 cm 

(b) 8 cm 

(c) 12 cm 

(d) 16 cm

2.) Area of sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector. [3π cm]

3.) The length of an arc of a circle is 4 cm and its radius is 6 cm. Find the area of this sector of the circle. [12 cm²]

If 1/b+c, 1/a+c, 1/a+b are in AP . . .

Question : If 1/b+c, 1/a+c, 1/a+b are in AP, prove that a², b², c² are also in AP.

Doubt by Divya

Solution : 

1/b+c, 1/a+c, 1/a+b
d₁=d₂ (Given)
a₂-a₁=a₃-a₂
1/a+c-1/b+c = 1/a+b-1/a+c
(b+c)-(a+c)/(a+c)(b+c)=(a+c)-(a+b)/(a+c)(a+b)
(b+c-a-c)/(b+c)= (a+c-a-b)/(a+b)
(b-a)/(b+c) = (c-b)/(a+b)
(b-a)(a+b)=(c-b)(b+c)
ab+b²-a²-ab=bc+c²-b²-bc
b²-a²=c²-b²
d₁'=d₂'
Hence, a², b², c² are also in AP

If the arithmetic mean of first n natural numbers is 15. . .

Question : If the arithmetic mean of first n natural numbers is 15, then n is equal to 

(a) 29

(b) 14

(c) 15

(d) 20

Doubt by Divya

Solution : 

Formula Used 
Arithmetic Mean (x̄) 
= Sum of observations / Total Number of Observations
=(x₁+x₂+. . . +x_n)/n

Sum of first n natural numbers
= [n(n+1)]/2

ATQ

x̄ = 15 (Given) 

x̄=(x₁+x₂+. . . +x_n)/n
15×n=(x₁+x₂+. . . +x_n)
15n=n(n+1)/2
15=(n+1)/2
30=n+1
30-1=n
n=29

Hence, (a) 29, would be the correct option.

In the given figure, ∠AEF=∠AFE and E is the mid point of . . .

Question: In the given figure, ∠AEF=∠AFE and E is the mid point of CA. Prove that BD/CD = BF/CE.

Doubt by Suraj 

Solution : 

Given : ∠AEF=∠AFE i.e. ∠1=∠2

E is the mid point of CA i.e AE=CE

To Prove : BD/CD = BF/CE

Construction : Take point G on AB such that CG||DF

Proof : 

∠1=∠2 (Given) 
AE=AF (Sides opposite to equal angles of a triangle are equal) — (1) 

Also, 
AE=CE (Given) — (2) 

In ΔAGC
E is the mid point of AC (Given)
CG||EF [∵CG||DF]
⇒F is the mid point of GA
(By Mid Point Theorem)
 AF=GF — (3)

In ΔBDF
CG||DF (By Const.) 
BC/CD = BG/GF (By BPT)
Adding 1 both sides
[BC/CD] + 1 = [BG/GF] +1
[BC+CD]/CD = [BG+GF]/GF
BD/CD = BF/GF
BD/CD = BF/CE

[Using equation (1), (2) and (3)
GF=AF=AE=CE]

Hence Proved.

A solid iron pole consists of a solid cylinder of height 200 cm . . .

Question : A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm, which is surmounted by another cylinder of height 50 cm and radius 7cm. Find the mass of the pole, given that 1cm³ of iron has approximately 8g mass.

CBSE 2024

Solution : 

Dimensions of Bigger Cylinder 

H = 200 cm 
D = 28 cm
R = D/2 = 28/2 = 14 cm

Dimensions of smaller Cylinder
h = 50 cm 
r = 7 cm 

Volume of the Iron pole  = πR²H+r²h
= π[R²H+r²h]
= 22/7 [(14)²×200+(7)²×50]
= (22/7)×(7)²[2²×200+1²×50]
= 22×7[4×200+50]
= 154×[800+50]
= 154×850
= 130900 cm³

Mass of 1cm³ iron = 8g
Mass of 130900 cm³ iron = 8×130900
= 1047200 g
= [1047200/1000] kg
= 1047.2 kg