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**Ques No.2**

Let the radii of the 3 spheres be r1, r2 and r3

r1 =6, r2 = 8 and r3= 10

let the radius of the new sphere formed after melting the above 3 spheres be r. the volume of the new sphere will be the total volume of the 3 spheres with radius r1, r2 and r3

ie., 4/3 pi r cube = 4/3 pi ( r1 cube+r2 cube+r3 cube)

dividing by 4/3 pi

r cube = r1 cube+r2 cube+ r3 cube = 6*6*6 + 8*8*8 + 10*10*10 = 216+512+1000 =1728

Since 12* 12*12= 1728

r = 12

r1 =6, r2 = 8 and r3= 10

New sphere Area = πr

^{3}= π(6×6×6 + 8×8×8 +10×10×10)

r

^{3}= 216+512+1000

r

^{3}= 1728

r = 12

diameter = 2×r = 2×12= 24 cm

^{3}+4/3*22/7*b

^{3}+4/3*22/7*c

^{3}=4/3*22/7*R

^{3}

4/3*22/7*(a

^{3}+b

^{3}+c

^{3})=4/3*22/7*R

^{3}

6

^{3}+8

^{3}+10

^{3}=R

^{3}

R

^{3}=1738

R=12

total marks of all students = 80 *100 = 8000 no. of students

10% students scored= 95 marks

total marks of 10% students = 95 * (10/100*100) = 950

similarly,

total marks of 20% students = 90*20 = 1800

marks of remaining students = 8000-1800-950 = 5250 which are the total marks of remaining students

remaining students = 70

Av. of remaining= 5250/70

= 75 ANS.

area of an equilateral triangle, A = √3a^2/4 where a is the length of a side

Area as a function of x, A(x) = √3a^2/4

A'(x) = √3a/2

when a=12, A'(x) = √3 * 12/2 = 6√3

Required rate = 6√3 cm

^{2}/min

=(1/2)*base* height

=(1/2)*x*(√3/2)x=(√3/4)x^2.........(let x be the side of the equilateral triangle).

now as it is given that the edge is increasing at a rate of √3 cm/min

so (dx/dt) i.e., rate of increase in length with respect to rate of change of time

so ,as area or A =(√3/4)x^2

now differentiate it with respect to time i.e.,

(dA/dt) = rate of change in are with respect to rate of change of time

upon doing that we get (dA/dt)= 2*(√3/4)*x*(dx/dt).

so as we have got all the values i.e., x=12 and (dx/dt)=√3 cm/min

upon putting those we get the value of dA/dt=18 cm^2/minute

the person above has forgotten to multiply the value of dx/dt

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