Given:

Three solid metallic spheres of radii 1 cm, 6 cm, and 8 cm, respectively, are melted and recast into a single solid sphere.

Concept used:

Volume of sphere is = 4/3πr³

Calculation :

Let the radius be R

As per the question,

4/3 π (1³ + 6³ + 8³ ) = 4/3πR³ 

R³ = 729

R³ = 9³ 

R = 9 cm

∴ The correct option is 1

As we can see in the figure, RT and QR are the radius of the bigger circle and equal to 7 cm,

⇒ QT² = 49 + 49

⇒ QT = 7√2

Now suppose radius of smaller circle is r cm,

⇒ QO = r√2

Since QT = TU + UO + OQ

⇒ QT = 7 + r + r√2

⇒ 7√2 = 7 + r(√2 + 1)

⇒ r = [7(√2 - 1)/(√2 + 1)]

⇒ r = 7(3 - 2√2)

⇒ r = 21 - 14√2

NOTE:

The rectangular figure when divided in makes two squares and thus the diagonal of the square will be r√2.

To calculate the radius of the small circle we will equate the lengths of the diagonal.

Let the radius of circle having centre at ‘O’ be ‘r’

⇒ OA = (8 + r) cm and OE = (8 – r) cm

By Pythagoras theorem,

⇒ (8 + r)² = (8 – r)² + 4²

⇒ 64 + r² + 16r = 64 + r² – 16r + 16

⇒ 32r = 16

⇒ r = 1/2 cm

⇒ Area of the circle = πr²

⇒ Area of the circle = (22/7) × (1/2)²

⇒ Area of the circle = 22/ (7 × 2 × 2)

∴ Area of the circle = 11/14 cm

Formula used:

Area of circle = π(radius)²

Calculation:

As we can see, a shaded region AOB which is added in the another half part of the circle is

the same as region OCD

Hence, required area = area of semicircle only

⇒ A = 7 × 11

⇒ A =77 cm²

∴ The area of the shaded region in the given figure is 77 cm2.

First take the LCM of 12, 10 and 11.

The LCM is 660

Now, we can see that in each number there is a difference of 1.

⇒ 12 - 11 = 1

⇒ 10 - 9 = 1

⇒ 11 - 10 = 1

⇒ So, subtract the LCM by 1

⇒ 660 - 1 = 659

∴ The smallest number is 659.

As we know, sum of an infinite geometric progression when r is less than 1 for infinite terms = a/(1 – r)

Here, a = 1 and r = 1/2

∴ Sum of geometric progression = 1/(1 – 1/2) = 2.

Given:

1 + 3 + 5 + 7 + __________(2n - 1)

Formula used:

S_n = (n/2) × [2a + (n – 1)d] = (n/2)[a + l]

Calculation:

First term(a) = 1,

Common difference(d) = 3 – 1 = 5 – 3 = 7 – 5 = 2

last term(l) = 2n – 1

Number of terms = n

Sum of terms = (n/2)[2a + (n – 1)d] = (n/2)[a + l]

⇒ (n/2) × (1 + 2n – 1)

⇒ (n/2) × (2n)

⇒ n × n

∴ The value of 1 + 3 + 5 + 7 + __________(2n - 1) is n × n

Shortcut Trick

Put n = 2

Sum of 2 term = 1 + 3 = 4

Check the option

1. (2n – 1)(2n – 1)

⇒ [(2 × 2) – 1][(2 × 2) – 1]

⇒ 3 × 3

⇒ 9

2. n/2 = 2/2 = 1

3. n(n + 1)/2

⇒ 2 × 3/2

⇒ 3

4. n × n

⇒ 2 × 2

⇒ 4

∴ Option (4) is right.

Hint

Sum of n odd number = n²

Let the positive no. be x².

x² - x = 30

⇒ x² - x - 30 = 0

⇒ x² - 6x + 5x - 30 = 0

⇒ (x - 6)(x + 5) = 0

⇒ x = 6       (x is positive)

∴ The no. is 6² = 36

Given:

(1³ + 2³ + ... + 9³) = 2025

Calculation :

We have,

(0.11³ + 0.22³ + ... + 0.99³)

⇒ [(0.11 × 1)³ + (0.11 × 2)³ + ... +(0.11 × 9)³]

⇒ 0.11³ [ 1³ + 2³ + ... + 9³]

⇒ 0.001331 × 2025 = 2.695275

∴ The correct answer is option 1.