Latest Elitmus Aptitude Question SOLUTION: 33 coins are there which are to be distributed among 3 sons A, B, C. No son gets more than 15, since C is youngest, he gets more than A or B & they are in arithematic progression.

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(#M40030076) ELITMUS Verbal Ability question Keep an EYE

33 coins are there which are to be distributed among 3 sons A, B, C. No son gets more than 15, since C is youngest, he gets more than A or B & they are in arithematic progression. How many distributions are possible

Asked In Elitmus Sagar Dasari (10 years ago)

Unsolved Read Solution (18)

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kumar gaurav  (10 years ago)

A B C
7 11 15
8 11 14
9 11 13
10 11 12
hence 4 distribution.

Is this solution Helpfull? Yes (36) | No (9)  

usha  (10 years ago)

C A B
15 11 7
15 7 11
14 11 8
14 8 11
13 9 11
13 11 9
12 11 10
12 10 11
So possibilities are 8
bcz we dont know whether a is younger or b is younger

Is this solution Helpfull? Yes (26) | No (12)  

Sagar Dasari  (10 years ago)

The answer is 8

for C 15, then there are 2 chances like C=15, B=11, A=7 if B is younger than A, or C=15, A=11, B=7, if A is younger than B, similarly for all others.
This is because we dont know who is younger among A & B

Is this solution Helpfull? Yes (20) | No (4)  

koka easwarendra  (10 years ago)

let the share of youngest be n
and the difference of AP be r
the shares of A B and C are
n-2r, n-r, n
sum= (n-2r) + (n-r) + (n) =3*(n-r)=33
r=n-11 must atleast 1.
n can be more than 15.
n=12,13,14,15

Is this solution Helpfull? Yes (10) | No   

MANPREET SINGH  (10 years ago)

let a,b,c, be in A.P so b=a+c/2 also c>a,b and a,b,ca so possibilities are
A B C
7 11 15
8 11 14
9 11 13
10 11 12

Is this solution Helpfull? Yes (5) | No   

Sasanka Banerjee  (10 years ago)

hence it is in a.p so the minimum diff ll be 1.if so then
A B C
10 11 12
now c may varies between 12-15 i.e. 12,13,14,15 so that the distribution becomes
A B C
9 11 13 diff is 2
8 11 14 diff is 3
7 11 15 diff is 4
hence 4 distribution.

Is this solution Helpfull? Yes (3) | No (1)  

Sagar Dasari  (10 years ago)

I think the answer is 8 as follows

15 for C & 7, 11 for A & B in 2 ways
14 for C & 8, 11 for A & B in 2 ways
similarly for 13 for C ... & 12 for C ....

Is this solution Helpfull? Yes (1) | No (3)  

Rajesh T N  (10 years ago)

By AP,

a,a+d,a+2d=33
wkt,a=15
so,15,15+d,15+2d=33
45+3d=33
3d=12
d=4
so 4 is the answer...

Is this solution Helpfull? Yes (1) | No (1)  

Jeevana  (10 years ago)

8 ways.. 2!*4 ways as either of A and B can get more coins..

Is this solution Helpfull? Yes (1) | No (1)  

anujnandwana  (10 years ago)

11 distributions are possible

Is this solution Helpfull? Yes (1) | No   

Anjali Sadhukhan  (8 years ago)

to those who thinking this question is a hard one---
among A,B and C ....C will always get more than that of other two,and it is given that the max value of C can be 15,therefore,keeping in mind that c is always greater ,,thus when the value of C will reach to 11(decrementing by one unit value),,the solution will no longer be possible as the value of A and B becomes greater..hence 8 is the possible solutions that we can derive.:)

Is this solution Helpfull? Yes (1) | No (3)  

Anurag patel  (7 years ago)

only 4 combination is possible bcoz they are in AP.
A B C
7 11 15
8 11 14
9 11 13
10 11 12

Is this solution Helpfull? Yes (1) | No   

LOKESH SAMANT  (10 years ago)

C:13
B:11
A:9

Is this solution Helpfull? Yes | No (3)  

pratendra kumar  (10 years ago)

26 distributions

Is this solution Helpfull? Yes | No (3)  

PRIYA SATSANGI  (10 years ago)

9,11,13
coz 9,9+2,9+4

Is this solution Helpfull? Yes | No (1)  

ADITYA RANJAN  (9 years ago)

x+x+d+x+2d=33
x+d=11
so,x=10, x+d=11, x+2d=12

Is this solution Helpfull? Yes | No   

chandan kumar  (8 years ago)

suppose a-d ,a and a+d are coins

according to question (a-d)+a+(a+d)=33 => a=11
now d may be 1,2,3,4 in order to a+d

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chanda gupta  (6 years ago)

8 distributions..according to given condition ..because it is not mention the younger in A &B so we take both the possibilities...

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