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MATHMAA

KVPY2013solution6

Probability Introduction :

KVPY2013solution6 contains the detail explanation for the questions 13 , 14 and 15.

13) How many natural numbers n are there such that n! + 10 is perfect square?
A) 1      B)  2      C)  4      D)  infinitely many

Solution:

For  n= 1, 2 , 4  n!+10 is not perfect square.

n=3 , 3!+10=16 is a perfect square.

If n $$\geq$$5 then, we write  n!+10= 10*(1*3*4*6*......+ 1) .

1*3*4*.... + 1 is even+odd = odd .

odd *10 is not perfect square

Except n=3 there is no natural number exist to satisfy n! + 10 is perfect square.

Therefore option (A) is the answer.

14) Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions each containing four of the remaining points is
A)  1         B)  5
C) 10         D) dependent on the configuration of points

Solution:

Therefore we get 5 lines which satisfies the given condition.

Hence the option B is correct answer.

15)In a city, the total income of all people with salary below Rs. 10000 per annum is less than the total income of all people with salary above Rs. 10000 per annum. If the salaries of people in the first group increases by 5% and salaries of the people in the second group decreases by 5% then the average income of all people
A) increases           B) decreases
C)remains the same    D) cannot be determined from the data

Solution: Let total number of people with salary below 10000 per annum be x and above 10000 be y .

Let the salary of x  people be A and y people be  B then, xA < yB.

Before changes average income of people = xA + yB .

If xA increased by 5% and yB decreased by 5% then average income of total people

= $$\frac{\frac{100+5}{100}*xA +\frac{100-5}{100}*yB}{xA+yB}$$

=$$1 +\frac{5}{100}*\frac{xA-yB}{xA+yB}$$ < 1 as (  xA -yB <0   ).

Means after changing the average income decreases.

Hence option (B) is correct answer

This ends part 1 Mathematics of KVPY 2013 paper. Part II questions and solutions with detail explanation will be continued in  next page. Please click Back to go to previous solutions and click Next to go to further solutions of part II. Visit the page and leave your review on these solutions. I am always with students and ready to update if necessary.

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