3 Solved Sibling Problems To Prepare For Infosys Aptitude Tests
1) Mr.X has one older brother and three younger sisters.
2) Mr.Y has two older brothers and two younger sisters
3) Mr.Z has three older brothers and one younger sister.
X has one older brother while Y has two. Therefore, X is older than Y. Similarly Y is older than Z. Therefore, among X,Y and Z
X is the oldest, Y is next and Z is the youngest.
X has an older brother and Z has a younger sister.
Then, it is clear that X is male and Y is female.
M M F M F F 1st 2nd 3rd 4th 5th 6th X Y Z
1) A & C say " We have 2 more brothers than the number of sisters"
2) B says " I have 2 more brothers than twice the sisters "
Let b be the number of boys in the family.
Therefore, b-2 = g+2
Or, b-g = 4 -----(1)
Therefore, Twice the number of sisters = 2(g-1) which is equal to 2 + number of brothers.
Then we have, 2(g-1) + 2 = b
2g - 2 + 2 = b
b-2g = 0 -------(2)
Solving (1) and (2), we have g = 4 and b = 8.
Hence the number of siblings = 4+8 = 12.
1)Aravind's sister has thrice as many sisters of Bavya as brothers.
2)Aravind has 1 more brother than than that of Bavya
3)Twice Bavya's brother's sisters equals the number of boys in Aravind's family.
Let the number of girls in Aravind's family be b
Let the number of boys in Bavya's family be c
Let the number of girls in Bavya's family be d.
Thrice the Bavya's sisters = 3(d-1)
Aravind's sister has 'a' number of brothers.
Therefore, 3(d-1) = a
d-1 = a/3....(1)
Number of Bavya's brother = c
Since, Aravind has 1 more brother than Bavya then a-1 = c+1
a-c = 2...(2)
Number of boys in Aravind's family = a
Therefore 2d = a
Or d = a/2
Then, d-1 = a/2 - 1 ...(3)
a/3 = a/2 - 1
a = 6
Put a value in (2), we have c = 4
Put a value in (3), we have d = 3
And Bavya has c = 4 brothers.
Hence, Aravind has 1 more brother than Bavya.