Logical Reasoning - Logical Problems - Discussion
Discussion Forum : Logical Problems - Type 1 (Q.No. 1)
Directions to Solve
Each problem consists of three statements. Based on the first two statements, the third statement may be true, false, or uncertain.
1.
Tanya is older than Eric.
Cliff is older than Tanya.
Eric is older than Cliff.
If the first two statements are true, the third statement is
Cliff is older than Tanya.
Eric is older than Cliff.
If the first two statements are true, the third statement is
Answer: Option
Explanation:
Because the first two statements are true, Eric is the youngest of the three, so the third statement must be false.
Discussion:
30 comments Page 2 of 3.
C P THIRU VIKRAM said:
8 years ago
Good explanation. Thanks to all.
Shweta said:
8 years ago
@Nemo.
Why you have taken base 2?
One can take any base?
Why you have taken base 2?
One can take any base?
Shashwath said:
8 years ago
Hi, everyone ,nice to see you all.
Good explanation. Thanks to all and Special Thanks, @Sundar.
But even will give you an example.
My name is Shashwath.
My younger name is Abhi.
My elder name is Ramunjun.
So,
Abhi > Shashwath > Ramunjun.
So we can say it in that way also.
If you understand you can do all sums like this!
Good explanation. Thanks to all and Special Thanks, @Sundar.
But even will give you an example.
My name is Shashwath.
My younger name is Abhi.
My elder name is Ramunjun.
So,
Abhi > Shashwath > Ramunjun.
So we can say it in that way also.
If you understand you can do all sums like this!
Smita said:
9 years ago
Good explanation. Thanks to all.
Shekharrao said:
9 years ago
Well explained @Suresh.
Suresh JP said:
9 years ago
T > E.
C > T.
E > C .
E < T & C,
So answer is FALSE.
C > T.
E > C .
E < T & C,
So answer is FALSE.
Sasi said:
9 years ago
Yup it's false because if I am older than brother and my sister is older than me then then of course.
My brother is older than sister is false.
My brother is older than sister is false.
Angelena said:
9 years ago
Very helpful for the coat so much logical work helps you get in to gifted program.
Nemo said:
9 years ago
@Somnath.
Base 2 (Binary should be able to help you with your problem).
Bag 1 = 1 chocolate = 000 000 001.
Bag 2 = 2 chocolates = 000 000 010.
Bag 3 = 4 chocolates = 000 000 100.
Bag 4 = 8 chocolates = 000 001 000.
And so on, all the way to:
Bag 10 = 512 chocolates = 100 000 000.
Assuming that a 1000 chocolates are requested by the first customer, you will use bags 10, 9, 8, 7, 6, and 4; i.e 1.
512 + 256 + 128 + 64 + 32 + 8 = 1000.
So the answer is 10 bags.
This problem is easy as only addition (Sum of chocolates) is involved.
Here is a harder question, how many pieces should a grocer break his 10 Kg weight piece into, if he wishes to be able to measure weights from 100 gm to 10 Kg at intervals of 100 gm?
i.e 100 gm, 200 gm, 300 gm, . All the way too 9.9 kg, 10 kg.
Base 2 (Binary should be able to help you with your problem).
Bag 1 = 1 chocolate = 000 000 001.
Bag 2 = 2 chocolates = 000 000 010.
Bag 3 = 4 chocolates = 000 000 100.
Bag 4 = 8 chocolates = 000 001 000.
And so on, all the way to:
Bag 10 = 512 chocolates = 100 000 000.
Assuming that a 1000 chocolates are requested by the first customer, you will use bags 10, 9, 8, 7, 6, and 4; i.e 1.
512 + 256 + 128 + 64 + 32 + 8 = 1000.
So the answer is 10 bags.
This problem is easy as only addition (Sum of chocolates) is involved.
Here is a harder question, how many pieces should a grocer break his 10 Kg weight piece into, if he wishes to be able to measure weights from 100 gm to 10 Kg at intervals of 100 gm?
i.e 100 gm, 200 gm, 300 gm, . All the way too 9.9 kg, 10 kg.
Slazzy said:
10 years ago
@Somnath.
Your question has no base. Kindly please go through it again yourself and see if you have made a mistake or not.
Your question has no base. Kindly please go through it again yourself and see if you have made a mistake or not.
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