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A,B,C are having some marbles with each of them.
A has given B and C the same number of marbles each of them already have.
Then, B gave C and A the same number of marbles they already have.
Then C gave A and B the same number of marbles they already have.
At the end A, B, and C have equal number of marbles.
11. If x,y,z are the marbles initially with A,B,C respectively,
then the number of marbles B has at the end are ____
1] 2(x-y-z) 2] 4(x-y-z) 3] 2(3y-x-z)
4] x + y-z
12. If the total number of marbles are 72, then the number of marbles with A at the starting were
1] 20 2] 30
3] 32 4] 39
Read Solution (Total 3)
-
- Answers 2(3y-x-z),, 39
11)question
given that three numbers are x,y,z
first A given to B,C
so B=2y
C=2z
A=x-y-z
second B gives C,A
C=4z
A=2(x-y-z)
B=3y-z-x
third C gives A,B
A=4(x-y-z)
B=2(3y-z-x)
C=5z-x-3y
so ans is 2(3y-z-x)
question 12
at final A,B,C Same numers
so 4x-4y-4z=72/3=24-------(1)
x+y+z=72----------(2)
solve (1) and (2)
A at starting has 39
- 10 years agoHelpfull: Yes(3) No(0)
- Answers 2(3y-x-z),, 39
11)question
given that three numbers are x,y,z
first A given to B,C
so B=2y
C=2z
A=x-y-z
second B gives C,A
C=4z
A=2(x-y-z)
B=3y-z-x
third C gives A,B
A=4(x-y-z)
B=2(3y-z-x)
C=5z-x-3y
so ans is 2(3y-z-x)
question 12
at final A,B,C Same numers
so 4x-4y-4z=72/3=24-------(1)
x+y+z=72----------(2)
solve (1) and (2)
A at starting has 39
- 10 years agoHelpfull: Yes(1) No(1)
- Answers 2(3y-x-z),, 39
11)question
given that three numbers are x,y,z
first A given to B,C
so B=2y
C=2z
A=x-y-z
second B gives C,A
C=4z
A=2(x-y-z)
B=3y-z-x
third C gives A,B
A=4(x-y-z)
B=2(3y-z-x)
C=5z-x-3y
so ans is 2(3y-z-x)
question 12
at final A,B,C Same numers
so 4x-4y-4z=72/3=24-------(1)
x+y+z=72----------(2)
solve (1) and (2)
A at starting has 39
- 10 years agoHelpfull: Yes(1) No(0)
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