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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 = πr3 = π(6×6×6 + 8×8×8 +10×10×10)
r3 = 216+512+1000
r3 = 1728
r = 12
diameter = 2×r = 2×12= 24 cm
4/3*22/7*(a3+b3+c3)=4/3*22/7*R3
63+83+103=R3
R3=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 cm2/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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