Money trouble!
A,B,C and D share a certain amount amongst themselves. B sees that the other three gets 3 times what he himself gets.C sees that the other three gets 4 times what he.gets,while D sees that other three get 5 times what he gets.If the sum of the largest and smallest shares is 99,what is the sum of other two shares?
The answer is 81.
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1 solution
Let A,B,C&D be the shares A+C+D=3B,
A+B+C+D=4B--------------------------a or,B=(A+B+C+D)/4-----------------------------------equation i similarly we will get from A+B+D=4C---------------b and A+B+C=5D-----------------------c C=(A+B+C+D)/5-------------------equation ii D=(A+B+C+D)/6----------------------equation iii Adding a,b and c we get 3A+2B+2C+2D=3B+4C+5D or,3A=B+2C+3D or,3A=B+C+D+2D Adding A on both sides or,4A=A+B+C+D+C+2D or,A= (A+B+C+D)/4 + (C+2D)/4----------------------iv
So from i, ii, iii and iv we know that A is the greatest share and D is the smallest share
Therefore A+D=99
Putting the value of B from equation a in b A+ (A+C+D)/3+ D= 4C or, (3A+A+C+D+3D)/3 = 4C or,4A+4D+C= 12C or, 4(A+D)=11C or,C= 4*99/11=36 or,C=36
Putting the value of C in equation a 3B=A+D+C or,3B=A+D+36 or,3B=99+36=135 or,B=45
THEREFORE C+B=81---------------------------------------------------------ANSWER