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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.

The quantitative aptitude questions mentioned above also contain aptitude questions asked for various placement exams and competitive exams. These will help students who are preparing for any type of competitive examinations.

Quantities aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT),MAT, GMAT, IBPS Exam, CSAT, CLAT , Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams , Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

sagar

2015-07-04 03:46:50

Yes

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rugmani

2015-06-04 17:10:08

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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Reshma

2015-07-13 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 = 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

r1 =6, r2 = 8 and r3= 10

New sphere Area = πr

r

r

r = 12

diameter = 2×r = 2×12= 24 cm

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Rex Roy

2015-06-01 11:18:59

2>4/3*22/7*a^{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

4/3*22/7*(a

6

R

R=12

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Pinky Rambo

2015-04-24 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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User

2015-07-05 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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richard

2015-04-07 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.

Dev

2015-04-07 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 cm^{2}/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 cm

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ananya

2015-05-18 06:31:56

i didnt got how u solved it....plz explain a bit more

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SOMARAY

2015-06-07 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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