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Abhishek kaushik 7 years ago

There are 9 three digit numbers . Each number is reversed and the reversed number is subtracted from the original number. The results were found to have an average of 0. If for each number, hundreds digit is not less than its unit digit , find the average of the hundreds digit of the greatest and least numbers ?

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Varun 7 years ago

101 is the least 3 digit number to have "hundreds digit not less than unit digit"

Result = 101-101 = 0

999 is the greatest 3 digit number.
Result = 999-999 = 0

Rest 7 numbers take any palindrome numbers like 252, 373, 444, 505, 666, 767, 868
Or any set giving average result zero like:
234, 432, 566, 775, 476, 988, 777

So required answer = (9-1)/2 = 5

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Abhishek kaushik 7 years ago

There are eleven people in a group. The average age of the oldest and youngest is 11 years. If any one person leaves the group the maximum and minimum average of the remaining address 12 years and 11 years respectively. Find the average age of the entire group ??

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Abhishek kaushik 7 years ago

A basketball player played nine matches. The average number of points he scored per match was 16. His points in the ith match were two less than that in (i-1)th match. Find the average number of points scored in the second and eighth matches ??

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Varun 7 years ago

Let points scored in first match be x

Then, his points are:
x, x-2, x-4,..., x-14, x-16
Average = [9x- 2(1+2+3...+8)]/9
= [9x-2(8*9)/2]/9
= [9x-72]/9
= x-8 = 16...., Given
=> x = 24
So second match score = 24-2 = 22
Eight match score = x-14 = 24-14 =10
Average points = (22+10)/2
= 16

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Abhishek kaushik 7 years ago

There are N students in a class.their class teacher gave them a task of finding the average of the first N natural numbers and each of them left out a different number and found the average of remaining numbers.the average of the averages obtained by all the students was 21.find N .

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Manav 7 years ago

sum of sum of numbers given by all N students = N*( 21*(N-1) )

Each student leaves 1 number,,,
let the numbers left by students be k1,k2,k3, ....,kn,,,,so for student1 it will be

21(N-1) = sum of N numbers - k1

=N(N+1)/2 - k1

for every student, it will be N(N+1)/2 - ki
Adding sum obtained by all N students , we get

N*( 21*(N-1) ) = N* ( N(N+1)/2 ) - (k1 -k2 ...... -kn)

adding k1 to kn, we get sum of N numbers

so it will be,, 21*N*(N - 1 ) = (N - 1)*N*(N+1)/2

solving it, we get N=41

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Abhishek kaushik 7 years ago

There are 2 containers having mixture of HCL acid and water.in container 1, the ratio of HCL and water is 1:2 and in container 2 the ratio of HCL acid and water is 4:1.find the amount of the mixture that should be taken from container 1 in order to make 28 litres of a mixture containing equal amount of water and HCL ? Can any1 solve this ..??

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Mihir 7 years ago

18 Ltrs mixture from container 1 and 10.0 ltrs of mixture 2 should be taken to make 28 litres of mixture containing equal amount of water and HCL.

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