Latest TCS Aptitude Question SOLUTION: The average of three numbers is 45.The first number is much more than the average as the second number is less than the average.Find the third number

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Algebra

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40033183) TCS Numerical Ability question TCS aptitude question Keep an EYE

The average of three numbers is 45.The first number is much more than the average as the second number is less than the average.Find the third number

Asked In TCS Ginjala Achuta Lahari (10 years ago)

Solved RAKESH Read Solution (10)

Is this Puzzle helpful?   (21)   (13) Submit Your Solution Algebra

Read Solution (10) :

RAKESH  (10 years ago)

if x,y,z are 1st,2nd & 3rd no.s respectively then,

x+y+z=3*45 .......(1)
x-45=45-y => x+y=90 .......(2)

(1)-(2) => z=45

Is this solution Helpfull? Yes (59) | No   

Vivek Jain  (10 years ago)

Let First number(F),Second(S) and Third(T).
=>Given, (F+S+T)/3 = 45
=> Also, F=45+x ...(1)
=> and S=45-x ....(2)
where x is the amount by which first number is more than the average as the second number is less than the average.
=>(1)+(2)
=>F+s=90
=>Substituting in avg equation we get Third number T as,
=>(90+T)/3=45
So, answer T=45

Is this solution Helpfull? Yes (14) | No   

sindhu  (10 years ago)

let f be the first number,s be the second number and t be the third number
given,
(f+s+t)/3=45
f+s+t=135.....(1)
as it is also given that f is more than average and let it be x
so f=45+x
as it is also given that s is less than average and let it be x
so s=45-x
from the equation
45+x+45-x+t=135
90+t=135
t=135-90
third number t=45

Is this solution Helpfull? Yes (2) | No   

shanthi  (10 years ago)

take three numbers as x,y,z respectively
(x+y+z)/3=45
given that x is greater than 45 by some a value
y is lesser than 45 by same a value
therefore (45+a+45-a+z)/3=45
=> 90+z=135
=>hence z=45

Is this solution Helpfull? Yes (2) | No   

nikita  (10 years ago)

45
let the numbers be x, y ,z
given (x+y+z)/3=45
and
x-45=45-y
=> x+y=90......(1)
90+z=135.....(2)
solving these, we get
z=45

Is this solution Helpfull? Yes (2) | No   

anushri  (10 years ago)

let the first no. b x more than the average.. i.e first no is 45+x
then secnd no. will be 45-x
let the third no. be y
therefore average will be.. {(45+x)+(45-x)+y}/3= 45
x will be cancelled..on solving we'l get y =45

Is this solution Helpfull? Yes (1) | No   

phanindra yellapragada  (10 years ago)

numbers are 45+x,45-x,k
k=150-90
=60

Is this solution Helpfull? Yes | No (2)  

B Ziyaulla  (10 years ago)

a+b+c=45
a>45 b

Is this solution Helpfull? Yes | No   

sasidhar  (10 years ago)

if x,y,z are 1st,2nd&3rd no.s respectively then x=45,y=45,z=45,then x+y+z/3=45.
given first number must be more than average and second number less than average.
take x=46>average.
take y=44

Is this solution Helpfull? Yes | No (2)  

Nishant Varshney  (7 years ago)

(a+b+c)/3=45...........(1)
(a+b+c)=135
a=45+x (Given)
c=45-x (Given)

45+x+b+45-x=135
90+b=135
b=45.................ans

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.