online user counteronline user counter

MATHMAA

Only Search Your Site

KYPY2013solution

This page contains KYPY2013solution 2 and 3 in detail. We have three different cases which you can see in detail below.

KYPY2013solution 2

2) If a,b are natural numbers such that 2013+a² = b², then the minimum possible value of ab is
    a) 671        b) 668        c) 658      d) 645

Solution:-
    First we have to factor 2013+a² = b². 
               2013 + a² = b²
    Subtracting a² on both sides, we get,
               b² - a² = 2013
    Applying the formula x² - y² = (x+y)(x-y), we get,
               (b+a)(b-a) = 2013
    Now we have to split 2013 as prime factors. We get,
               2013 = 11 * 3 * 61.
    We have to write these factors as the product of two numbers so we can write them as
         61 times 33  or  183 times 11  or  671 times 3 
     Case I:
               (b+a)(b-a) = 61*33
               b+a = 61   
               b-a  = 33
          Solving these, we get,
             b = 47 and a = 14 so ab = 47*14 = 658
     Case II:
               (b+a)(b-a) = 183*11
               b+a = 183
               b-a  = 33
          Solving these, we get,
             b=108 and a=75 so ab = 108*75 = 8100
     Case III:
               (b+a)(b-a) = 671 times 3
               b+a = 671
               b-a  = 3 
           Solving these, we get,
             b=337 and a=334 so ab = 337*334 = 112558
     Now the minimum value of ab is 658 therefore the correct option is (C).

KVPY2013solution 3

3) The number of values of b for which there is an isosceles triangle with side lengths b+5, 3b-2 and 6-b is
     a) 0            b) 1          c) 2           d)3

Solution:
    The triangle in which we have lengths of any two sides equal is called an isosceles triangle. So here we are given the lengths of an isosceles triangle b+5, 3b-2, 6-b.
     Here the possible equal sides are either b+5 and 3b-2 or 3b-2 and 6-b or b+5 and 6-b.
  Case I:
      The possible isosceles triangle with equal side lengths b+5 and 3b-2.
                   b+5 = 3b-2
                   2b = 7
                   b = \(\frac{7}{2}\)
                   b = 3.5
      
Now the length of sides of the triangle are 8.5, 8.5, 2.5. So triangle is possible with these measurements.  

    Case II:
       The possible isosceles triangle with equal side lengths of 3b-2 and 6-b.
                   3b-2 = 6-b
                   4b = 8
                   b = 2
       
Now the length of sides of the triangle are 7, 4, 4. So triangle is possible with these measurements. 
   Case III:
        The possible isosceles triangle with equal side length of b+5 and 6-b.
                   b+5 = 6-b
                   2b = 1
                   b = \(\frac{1}{2}\)
                   b = 0.5
           Now the length of sides of the triangle are 5.5, -0.5, 5.5 which is not possible as the length cannot be negative.

      Hence possible values of b are 3.5, 2. There are 2 possible values. So the correct option is (C).

To go the previous solution click Back and further solutions click Next. Share your views to update if necessary.

Back<<                                                                                                   Next>>


SHARE YOU ENORMOUS EFFORT AND SMART EXAMPLES HERE

!! NEED MORE HELP !!

SBI! Case Studies