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Numerical Ability
Geometry
Mean of three numbers is 10 times more than the least of the numbers and 15 times less than greatest of the three numbers is five. Find the sum of the three numbers.
Read Solution (Total 10)
-
- let the 3 nos. be x,y,z
median=5
i.e. y=5
mean=(x+5+z)/3=x+10----------------------- (1)
=>(x+5+z)/3=z-15-----------------------------------(2)
solve both equations:
x=0,z=25
sum=X+Y+Z=>0+5+25= 30------------->ANSWER - 10 years agoHelpfull: Yes(12) No(3)
- re chutiya, a66a se question to dal.....
- 9 years agoHelpfull: Yes(9) No(0)
- since mean is given thus
mean=10=(X+Y+Z)/3or X+Y+Z=30 which is the required sum
- 10 years agoHelpfull: Yes(3) No(3)
- let the 3 nos. be x,y,z
mean=(x+y+z)/3=x+10----------------------- (1)
=>(x+y+z)/3=z-15-----------------------------------(2)
solve both equations:
x=0,z=25
from (1)
(x+y+z)/3=10
x+y+z=30 - 9 years agoHelpfull: Yes(1) No(0)
- 15 as (a+b+c)/3=5.......therefore a+b+c=5*3=15
- 9 years agoHelpfull: Yes(1) No(0)
- median must be provided for this ques. data insufficient.
- 10 years agoHelpfull: Yes(0) No(0)
- question is not complete median is given and median is 5 so ans will be 30.
- 10 years agoHelpfull: Yes(0) No(1)
- no..question is cmplt....bt hv no idea to slv it
- 10 years agoHelpfull: Yes(0) No(2)
- Let m be the mean of the three numbers. Then the least of the numbers is m-10 and the greatest is m + 15. The middle of the three numbers is the median, 5. So 1/3[(m-10) + 5 + (m + 15)] = m, which implies that m=10. Hence, the sum of the three numbers is 3(10) = 3
- 9 years agoHelpfull: Yes(0) No(0)
- let no be a,b,c that means mean=10a,mean =c/15 and given c=5 on equating i got sum as 1
- 9 years agoHelpfull: Yes(0) No(0)
TCS Other Question
A number when successively divided by 1,5,8 leaves remainders 1,4,7resp. Find the respective remainders if the order of divisors be reversed.
If F(n) = {x/x is a prime factor of n};
1. F(110) ∩ F(165)
(1) {5, 11} (2) {1, 5, 11} (3) {1, 2, 3, 5, 11} (4) {2, 3, 5, 11}
2. The least value of x when
F(x) ∪ F(15) = F(30)
(1) 2 (2) 0 (3) 1 (4) 30