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Quantitative Aptitude Questions and Answers with Explanation
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In this section, you can find numerous aptitude questions with answers and explanation. The quantitative aptitude questions with answers mentioned above covers various categories and extremely helpful for competitive exams. All the answers are explained in detail with very detailed answer descriptions.
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Comments(188)
sagar2015-07-04 03:46:50
Yes
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rugmani2015-06-04 17:10:08
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
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
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Reshma2015-07-13 06:47:13
Let the radii of the 3 spheres be r1, r2 and r3
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
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
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2>4/3*22/7*a3+4/3*22/7*b3+4/3*22/7*c3=4/3*22/7*R3
4/3*22/7*(a3+b3+c3)=4/3*22/7*R3
63+83+103=R3
R3=1738
R=12
4/3*22/7*(a3+b3+c3)=4/3*22/7*R3
63+83+103=R3
R3=1738
R=12
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Pinky Rambo2015-04-24 15:58:04
6) let no. of students =100
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.
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.
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User2015-07-05 21:03:19
Shorter method(just on numbers):
10 students
1 got 95: difference is 95-80=15
2 got 90: diff is 10*2 = 20
Avg=80
7 got diff of -20-15=-35
Each got -5
I.e. 80-5=75
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richard2015-04-07 18:47:09
equilateral triangle is increasing at a rate of √3 cm/min. find the rate
at which its area increasing when its edge is 12 cm long.
Dev2015-04-07 18:59:34
Assuming that the question says each side of equilateral triangle is increasing at a rate of √3 cm/min
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
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
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ananya2015-05-18 06:31:56
i didnt got how u solved it....plz explain a bit more
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SOMARAY2015-06-07 20:06:21
area of an equilateral triangle
=(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
=(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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