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

Tamper-proof tetra-packed milk guarantees . . .

Case Study Based Question on Surface Area and Volume

Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals.

500 mL milk is packed in a cuboidal container of dimensions 15cm×8cm×5cm. These milk packets are then packed in cuboidal cartons of dimensions 30cm×32cm×15cm.

Based on the above given information, answer the following questions : 

(i) Find the volume of the cuboidal carton. [1 Mark]

(ii) (a) Find the total surface area of a milk packet. [2 Marks]

OR

(b) How many milk packets can be filled in a carton? [2 Marks]

(iii) How much milk can the cup (as shown in figure) hold? [1 Mark]

CBSE 2024

Solution : 

Dimensions of cuboidal container 
l = 15 cm 
b = 8 cm 
h = 5 cm

Dimensions of cuboidal cartons 
L = 30 cm 
B = 32 cm 
H = 15 cm

Dimensions of cylindrical cup 
r = 5 cm
h'= 7cm

(i) Volume of the cuboidal carton
= LBH
= 30×32×15
= 14400 cm³

(ii)
(a) Total Surface Area of a milk packet 
=2(lb+bh+hl)
=2[(15×8)+(8×5)+(5×15)]
=2[120+40+75]
=2[235]
=470 cm²

OR

(b) No. of milk packets filled in a carton 
= Volume of carton / Volume of milk packet 
= LBH/lbh
= (30×32×15) / (15×8×5)
= 2×4×3
= 24

(iii) Capacity of the cup = Volume of the cylinder
= πr²h
=(22/7)×5×5×7
= 22×25
= 550 cm³

A solid toy is in the form of a hemisphere surmounted by a . . .

Question : A solid toy is in the form of a hemisphere surmounted by a right circular cone. Ratio of the radius of the cone to its slant height is 3:5. If the volume of the toy is 240π cm³, then find the total height of the toy.

CBSE 2024 (July Session)

Solution : 
For Right Circular Cone
r:l=3:5
Let
r=3x
l=5x

Height of the cone
h=√[l²-r²]
h=√[(5x)²-(3x)²]
h=√[25x²-9x²]
h=√[16x²]
h=4x

Also the radius of hemisphere would be equal to that of cone. 

Volume of the toy= Volume of cone + Volume
f hemisphere

240π = (1/3)πr²h + (2/3)πr³
240π = (1/3)πr²[h+2r]
240=(1/3)(3x)²[4x+2(3x)]
240=3x²[4x+6x]
240=3x²[10x]
240=30x³
240=30x³
240/30=x³
8=x³
x³=2³
x=2

Total height of the toy
= h+r
= 4x+3x
=7x
=7(2) [∵x=2]
=14 cm

Amita buys some books for Rs 1920 . . .

Question : Amita buys some books for ₹1920. If she had bought 4 more books for the same

amount each book would cost her ₹ 24 less. How many books did she buy? What was the initial price of one book?

Doubt by Yana

Solution : 

Let the original number of books purchased be x. 

Cost of each book = 1920/x

Number of books when she bought 4 more books = x+4

New cost of each book = 1920/(x+4)

ATQ 
1920/x-1920/(x+4)=24
1920[1/x-1/(x+4)]=24
[(x+4)-x]/[x(x+4)]=24/1920
4/[x²+4x]=1/80
x²+4x=320
x²-4x-320=0
x²+(20-16)x-320=0
x²+20x-16x-320=0
x(x+20)-16(x+20)=0
(x-16)(x+20)=0
x-16=0 OR x+20=0
x=+16 OR x=-20
Number of books can't be negative. 
So, x=-20 (Rejected)
x=16
Hence, she bought 16 Books.

Initial Price of each Book = 1920/x = 1920/16 

=₹120

If the sum of the zeroes of 5x²+(p+q+r)x+pqr is zero . . .

Question : If the sum of the zeroes of 5x²+(p+q+r)x+pqr is zero, then find p³+q³+r³.

Doubt by Suraj

Solution : 

5x²+(p+q+r)x+pqr
Here
a=5
b=p+q+r
c=pqr

α+β=-b/a
α+β=-(p+q+r)/5
But α+β=0 (Given)
0=-(p+q+r)/5
0×5=-(p+q+r)
p+q+r=0

Using Identity
If a+b+c=0 then a³+b³+c³=3abc

Here 
p+q+r=0
So, 
p³+q³+r³ = 3pqr