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New answer posted

7 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Let L be the vertical distance between the centers of two adjacent balls.

L = [ ( 2 r ) 2 r 2 ] = r 3

Also, if the box can hold n sphere then (n – 1) L + 2r

≤ 9r [ the last ball height = 2r]

( n 1 ) 3 r 7 r

n 7 3 + 1

n ≤ 5.04

Since, n is a natural number.

n = 5

So, the box can hold maximum of 5 spheres.

New answer posted

7 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(b):Total time required to empty it =

1 (1313.5)=21hours.

New answer posted

7 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

 

17 * 5 = 85

New answer posted

7 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

(a):Let, s be the speed of stream.

So, 8 + s = 2 (8 – s)

s = 8/3 mph

New answer posted

7 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(a):Let G be (X, Y) then X = { 3 + 5 + ( 3 ) } 3 = 5 3

Y = ( 7 + 5 + 2 ) 3 = 1 4 3

Gis  (53, 143)

New answer posted

7 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

 

17 – 12 = 5

New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

(c) :As per the problem:

Average score of 20 candidate = 25 marks

Total score of 20 candidate = 500 marks

Let the score of the topper be x. Then,

( 5 0 0 x ) 1 9 = 2 3

500 – x = 437

x = 500 – 437 = 63 marks

New answer posted

7 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

 

12 * 9 – 17 * 5 = 23

New answer posted

7 months ago

0 Follower 11 Views

P
Payal Gupta

Contributor-Level 10

f (3n) – f (3n– 3) = n

n = 1

f (3) – f (0) = 1

f (0) = 0

n = 2

f (6)f (3)=2? ? ? ?

f (3n) – f (3) = 2 + 3 + 4+…………….+ n

f (3n) = 1 + 2 + 3 +…………….+ n

n (n+1)2

?  f (1312) – f (3 * 311)

3112 (311+1)=322+3112

Hence, remainder  [322+311210] = 8

New answer posted

7 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

1log10105=13 [log1010x*51/32]

3 [log101010 – log10105] = log1010x*51/32

log101023 = log1010x*51/32

23 = x*51/32

x = 2451/3

Hence, m + 3n = 4+3*13 = 5

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