Hey guys after a long wait finally NMTC was conducted on 22nd August in all parts of the country.SO how well did you fare? What is expected score of yours and the cut-off you expect. .Kindly share your views about the Question paper or the interesting questions.Don't forget to share your expected marks.
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2 \times 3
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Top NewestHi! Mohit , the most interesting question and easy question according to me was -
Find the number of integeral solutions of the following inequality,
\((n^{2}-2)(n^{2}-20)<0 \)
Which one was yours?
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For me the easiest question was q7. A nom divided by 899 gives a remainder 63. The remainder when this no. Is divided by 29 is
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The answer was 6 na?
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No the ans. Was 6 .I am pretty sure about it.
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Ha Ha ..... it was the \(easiest\) question.
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I got selected for stage 2.
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is the result declared
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Yeah I gave it from FIITJEE South Delhi Center and they declared the result on their site for only FIITJEE students. From the result it seems that the cut off was 15. It may be lower elsewhere because fiitjee's cut off is always higher even last time in NMTC it was higher than what it should have been.
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Hey kushagra great. ...........CONGRATS..... wonderful job done. ...... i'm still waiting for my results. ......again CONGRATS....
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Thnx mate. I wish you best of luck for the results.
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can anyone tell what was the answer of the log question...... is it 890 or 841+\(\sqrt{7}\) on AMTI site it is 890 Go here and on Allen site it is 841+\(\sqrt{7}\) Go here TO all csquarians,Ad sir first gave the answer as 890 then we have consulted to sir then he gave answer as 841+\(\sqrt{7}\) still Its so confusing can anyone tell me the answer with well surity...
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Answer is \(\boxed{841+\sqrt{7}}\)
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Yeah the answer is surely 841+ root 7
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Answer is 841+root7 you can confirm it.I AM 120 PERCENT SURE.....
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Did you get selected?
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I am expecting 27 or 28. What about you?
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Hello Mohit!
The question 14 was very interesting.
I solved like this:
I set \({ 2 }^{ x }=a\) and \({ 3 }^{ x }=b\) and then simplified to get
\(\huge\ \frac { { a }^{ 3 }+{ b }^{ 3 } }{ { a }^{ 2 }b+a{ b }^{ 2 } } =\frac { 7 }{ 6 }\)
i.e, \(\large\ \frac { { a }^{ 2 }-ab+{ b }^{ 2 } }{ a{ b } } =\frac { 7 }{ 6 }\)
i.e, \(6{ a }^{ 2 }-13ab+6{ b }^{ 2 }=0\)
i.e, \((2a-3b)(3a-2b)=0\)
which on solving yields \(x=1\) and \(x=-1\).
How you solved that? Also how much you expected?
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Hey did anyone notice a controversial question. The question which has a cube root and a sixth root.This question has two answers 1 & -1
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No answer was 1. Because -1 would be an extraneous root.
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Hey come on man....cube root of -1 has three values...two lie in complex plane and one lie in real plane i.e. -1
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Which question are you saying?
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