I solved it algebraically ... Using the given equation (either only "yes" or only "no") finding the definite yes or definite no for the given two choices ...I used the definite "yes" of the given equation that is a^3 > a^2 . for (1) we shall always get a definite NO. but for (2) we shall get both No and Yes.
Hey Mrugen, Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
Hi, Let's take examples when a < 0, which satisfies statement 2. a = -1/5 (-1/5)^5 = -0.00032 (-1/5)^3 = -0.008 So, here a^5 > a^3 as -0.00032 > -0.008 If 'a' is a negative integer, then statement 2 itself will not be satisfied (which is why we have also not considered it)
@@Wizako Hello sir, I believe what Aaryan is pointing out is in the video explanation, you forgot to mention when using a negative fraction, you will satisfy that a^5>a^3, however the statement 2 will still not be sufficient, as you have worked out in the response above. Leaving this info may confuse some viewers like me. :)
Hello Sir, for SII: What I did was divide on both sides by a^3 which leave with a^2>1 implies a can be either +ve or -ve so cannot determine if a^3>a^2. Is this also a possibility? Can I approach in this way?
You cannot divide by a3 because you don't know if a3 is positive or negative . If you divide by negative then signs of inequalities change , if you divide by positive , signs remain same
Statement 2 can also give a definite answer because a5>a3 for all positive numbers.,so a is not a negative number If a is positive then we will get a definite answer. Please correct me if I am wrong.
Hey Pulkit, Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
so in statement 1 when you put a=2 so it becomes 1/2>2 which is not correct and sometimes it gives the opposite when you take fraction and negative numbers so how is it a definite NO?
Please I don't think you're correct. The answer is d... For the second statement, a^5>a^3 can only work If a is positive. The instruction is that we should assume all statements are true. If you doubt it please read the official guide. The same way you dismissed the fractions for not meeting this condition is the same way you should dismiss the negatives too because they don't meet the condition of statement B.. Please correct yourself
Hey Eminue, Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
this is so confusing... first in statement 1 you did not consider when a>1 and gave a definite NO and in statement 2 you considered all the possibilities and said the value changes, logic doesn't make any sense or you did not provide good explanation, its very confusing.
You are correct mate. Moreover when a statement is given, you don't need to validate it, you need to consider that statement is true and validate whether that validates the question. You have started validating the statement the statement has to be taken as true and valid that is basic logic.
I solved it algebraically ... Using the given equation (either only "yes" or only "no") finding the definite yes or definite no for the given two choices ...I used the definite "yes" of the given equation that is a^3 > a^2 . for (1) we shall always get a definite NO. but for (2) we shall get both No and Yes.
Brilliant stuff Gaurav!! Cheers!!
If statement 2 says a5>a3, means a is positive integer greater than 1. And in that case a3>a2. So can't we say statement 2 also gives definite answer
a5>a3 in the interval -1
Hey Mrugen,
Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
Hey Satvik,
If a= -1/2 then a^5 and a^3 would have values -1/32 and -1/8.
Here, -1/32>-1/8, since -1/32 is closer to zero on the number-line.
@@Wizako got it. Thanks
@@Wizako SO YOU CANNOT SAY, I AM NOT EVEN GOING TO LOOK WHEN a^5 is greater than^3. You have to find some values where this condition holds true.
Hi sir, if we take aa^3 first?
Hi,
Let's take examples when a < 0, which satisfies statement 2.
a = -1/5
(-1/5)^5 = -0.00032
(-1/5)^3 = -0.008
So, here a^5 > a^3 as -0.00032 > -0.008
If 'a' is a negative integer, then statement 2 itself will not be satisfied (which is why we have also not considered it)
@@Wizako Hello sir, I believe what Aaryan is pointing out is in the video explanation, you forgot to mention when using a negative fraction, you will satisfy that a^5>a^3, however the statement 2 will still not be sufficient, as you have worked out in the response above. Leaving this info may confuse some viewers like me. :)
Hello Sir, for SII: What I did was divide on both sides by a^3 which leave with a^2>1 implies a can be either +ve or -ve so cannot determine if a^3>a^2. Is this also a possibility? Can I approach in this way?
You cannot divide by a3 because you don't know if a3 is positive or negative . If you divide by negative then signs of inequalities change , if you divide by positive , signs remain same
Statement 2 can also give a definite answer because a5>a3 for all positive numbers.,so a is not a negative number If a is positive then we will get a definite answer. Please correct me if I am wrong.
You're right
Hey Pulkit,
Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
@@eminueaaron6791 nope, you're wrong.
If you take a being -1/2 , then a5 > a3 ?
-1/32 > -1/8 the condition holds true . Lets check for a3>a2 . -1/8 > 1/4 . ( Not true )
Yes for positive numbers .
No for numbers lying between -1 to 0
Statement is not sufficient
so in statement 1 when you put a=2 so it becomes 1/2>2 which is not correct and sometimes it gives the opposite when you take fraction and negative numbers so how is it a definite NO?
First statement says that 1/a > a , and if we put a >1 which is a= 2 that means 1/2> 2 ( which is not true so the condition does not match
I din understand how a can lie between 0 and 1 only? since its not given!
Please I don't think you're correct. The answer is d... For the second statement, a^5>a^3 can only work If a is positive. The instruction is that we should assume all statements are true. If you doubt it please read the official guide. The same way you dismissed the fractions for not meeting this condition is the same way you should dismiss the negatives too because they don't meet the condition of statement B.. Please correct yourself
Hey Eminue,
Please do check the validity of statement a^5>a^3 for the value of a in the interval -1 < a < 0. Hence, a can be both positive and negative. For positive values of 'a', the answer would be a YES, but for negative values of 'a' the answer would be NO. Hence statement 2 is not sufficient.
this is so confusing... first in statement 1 you did not consider when a>1 and gave a definite NO and in statement 2 you considered all the possibilities and said the value changes, logic doesn't make any sense or you did not provide good explanation, its very confusing.
You are correct mate. Moreover when a statement is given, you don't need to validate it, you need to consider that statement is true and validate whether that validates the question. You have started validating the statement the statement has to be taken as true and valid that is basic logic.