Surface area sphere = 4π x (10)²
volume sphere=4π/3 x (10)³

Correct Option: C

Surface area sphere = 4π x (10)²
volume sphere=4π/3 x (10)³
= (4000π/3) cm³

∴ Required percentage = {(400π x 3) / (4000π)} x 100 %
= 30%

Let original rate = Rs. P per kg
New rate = 79% of Rs. P per kg
= Rs. (79 x P)/100) per kg
∵ {(100/79P)/100} - {100/P} = 10.5
⇒ 10000/79P - 100/P = 10.5
⇒ 10000 - 7900 = 10.5 x 79P
∴ P = 2100/(10.5 x 79)
∴ P = 2100/(10.5 x 79)
Reduced rate = Rs. (79/100) x 2100/(10.5 x 79) per kg = Rs. 2 per kg.

Correct Option: A

Let original rate = Rs. P per kg
New rate = 79% of Rs. P per kg
= Rs. (79 x P)/100) per kg
∵ {(100/79P)/100} - {100/P} = 10.5
⇒ 10000/79P - 100/P = 10.5
⇒ 10000 - 7900 = 10.5 x 79P
∴ P = 2100/(10.5 x 79)
∴ P = 2100/(10.5 x 79)
Reduced rate = Rs. (79/100) x 2100/(10.5 x 79) per kg = Rs. 2 per kg.

Required number = 800 x (1 + 15/100)²

Correct Option: C

Required number = 800 x (1 + 15/100)²
= 800 x (23/20) x (23/20)
= 1058

Consumption Decreases = (common %)²/100

Correct Option: D

Consumption Decreases = (common %)²/100
= 225/100
= 2.25%.

Let length of a rectangle = 100 m,
Breadth = 100 m
New length = 160 m,
New breadth = B meters
Then, 160 x B = 100 x 100

Correct Option: A

Let length of a rectangle = 100 m,
Breadth = 100 m
New length = 160 m,
New breadth = B meters
Then, 160 x B = 100 x 100
⇒ B = (100 x 100) / 160 = 125/2
Decrease in breadth = (100 - 125/2)% = 37¹/₂%