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Numerical Ability
Age Problem
If A, B and C are three positive integers such that A is greater than B and B is greater than C, then which of the following is definitely true? i. A% of B is greater than B% of C. ii. B% of A is greater than C% of B iii. C% of A is greater than B% of C
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- Let us assume A=100, B= 50 and C=10.
Now, A% of B = 100% of 50 = 50 and B% of C = 5. Hence, A%(B)>B%(C)
Again B% of A = 50 and C% of B = 5. FALSE
Similarly, C% of A = 10 and B% of C = 5. FALSE
Hence (i) is Definitely True - 5 years agoHelpfull: Yes(12) No(25)
- ALL Conditions are true
Let assume A=20,B=10,C=5;
now,A% OF B=2 > B% of C=0.5. Hence A is TRUE. ie)(20/100)*10=2 && (10/100)*5=0.5.
now B % of C=2 > C% of B=0.5. Hence B is TRUE.
now C% of A=1 > B% of C=0.5. Hence C is TRUE.
Hence, all the 3 conditions are true; - 5 years agoHelpfull: Yes(10) No(3)
- Let us take A=10, B=5 and C=2
1) Let us check for 1st condition , A%B ie: 10%5=0 and B%C ie: 5%2=1 Hence first condition is false.
2) Let us check for 1st condition , B%A ie: 5%10=5 and C%B ie: 2%5=2 Hence second condition is true.
3) Let us check for 1st condition , C%A ie: 2%10=2 and B%C ie: 5%2=1 Hence third condition is true.
Therefore 2nd and 3rd options are definitely True. - 5 years agoHelpfull: Yes(4) No(11)
- Lets proceed mathematically instead of numerical examples which are drops in the ocean.
So, A>B>C>0,
i. A% of B = A*B/100, B% of C = B*C/100. Clearly it is a case of A*B>B*C => A>C (since B>0)
Which is given, thereby, true.
Proceeding similarly you can prove the other two to be true.
ty - 5 years agoHelpfull: Yes(2) No(0)
- all option are true....check it by above method
- 5 years agoHelpfull: Yes(1) No(1)
- (I) and (ii) same
A%of B =B%of A
AB>BC true
(III) AC>BC = A>B definitely true - 5 years agoHelpfull: Yes(1) No(0)
- taking three numbers we can solve the following problem easily. Take three numbers 4,5,6 and then check the conditions third one is true for this one.
- 5 years agoHelpfull: Yes(0) No(1)
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