Elitmus
Exam
Verbal Ability
Synonyms
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
Read Solution (Total 18)
-
- A B C
7 11 15
8 11 14
9 11 13
10 11 12
hence 4 distribution. - 11 years agoHelpfull: Yes(36) No(9)
- 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
- 11 years agoHelpfull: Yes(26) No(12)
- 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 - 11 years agoHelpfull: Yes(20) No(4)
- 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 - 11 years agoHelpfull: Yes(10) No(0)
- 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 - 11 years agoHelpfull: Yes(5) No(0)
- 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. - 11 years agoHelpfull: Yes(3) No(1)
- 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 .... - 11 years agoHelpfull: Yes(1) No(3)
- 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... - 11 years agoHelpfull: Yes(1) No(1)
- 8 ways.. 2!*4 ways as either of A and B can get more coins..
- 11 years agoHelpfull: Yes(1) No(1)
- 11 distributions are possible
- 11 years agoHelpfull: Yes(1) No(0)
- 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.:) - 9 years agoHelpfull: Yes(1) No(3)
- 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 - 8 years agoHelpfull: Yes(1) No(0)
- C:13
B:11
A:9 - 11 years agoHelpfull: Yes(0) No(3)
- 26 distributions
- 11 years agoHelpfull: Yes(0) No(3)
- 9,11,13
coz 9,9+2,9+4 - 11 years agoHelpfull: Yes(0) No(1)
- x+x+d+x+2d=33
x+d=11
so,x=10, x+d=11, x+2d=12 - 9 years agoHelpfull: Yes(0) No(0)
- 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 - 9 years agoHelpfull: Yes(0) No(0)
- 8 distributions..according to given condition ..because it is not mention the younger in A &B so we take both the possibilities...
- 7 years agoHelpfull: Yes(0) No(0)
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