ANSWERS For Questions directions-for-questions-from-1-to-2

Answer for question  1 & 2 :- 

Let p be the number of problems on the test. 

Let n be the number of problems solved by the tenth student. 

Then n <= p. 

The number of problems solved, counted with multiplicity, is 7p 

since each student solved seven questions. 

However, we also know that the first nine students solved four questions each and the tenth solved n questions, 

so 

      7p = (4×9) +n. 

Re-arranging gives us 

p = (36 + n)/7 

Recalling that n <= p, we have

n <= p ⇒ 7n <= 7p ⇒ 7n <= 36 + n ⇒ 6n <= 36 ⇒ n <= 6.

Therefore, we are looking for non-negative n such that p = (36 + n)/7 and n <= 6

Therefore n = 6 and so p = 6. That is, the number of problems solved by the tenth 

student was six and there were six questions total on the test. Hence, option (2) is the correct choice

3.)    Given

                 (xy)^z = 64

            

            

since  positive integral 

           

 Whenz = 1, xy = 64. The total number of factors of 64 is 7.
z = 2, xy = 8. The total number of factors of 8 is 4.
z = 3, xy = 4. The total number of factors of 4 is 3.
z = 6, xy = 2. the total number of factors of 2 is 2
Hence, total number of positive integral solutions is 16.

4.)  Diameter of the circle is equal to the side of the smallest

   square is πr² = 77
  ⇒r = 3.5 (sqrt(2))
  ⇒ 2r = 7 sqrt(2) , side of the smallest square.
  Diagonal of the innermost square = 14 units.
  Diagonal of the largest square will be 14+ [(1.5) × 2] × 4
  = 14 + 12 = 26 units
  ⇒ Area of the largest square
  =  0.5 × (diagonal)² = 338 sq units and area of the smallest square =  (7 sqrt(2) )² sq. units.
 Required difference = 338 – 98 = 240 square units.

 5 .)  Total number of ways of dividing 24 boys in two distinct sub-sections is ²⁴C₁₂ = 24!/(12! 12!)
There can be two ways in which two shortest boys lie in different sub-section and correspondingly the remaining 22 boys can be divided into two distinct sub-sections in ²²C₁₁ ways.
∴ Probability = (2 x ²²C₁₁ )/²⁴C₁₂ = 12/23

6.)


7 ) here are 18 possible values of x and 9 possible values of y.
Since, BC is parallel to the X – axis, therefore AB will be parallel to the Y – axis.
The X – coordinate of points B and A will be same and the Y – coordinate of points B and C will be same.
For, every chosen co – ordinate of point B, the Y – coordinate of point A can be chosen in
9 – 1 = 8 ways and the X – coordinate of point A can be chosen in 18 – 1 = 17 ways.
Number of ways in which co – ordinate of point B can be chosen = 18 x 9
Total number of distinct triangles = 18 x 9 x 17 x 8 = 22032

8 .) Question

 The maximum possible area (in square units) of such a triangle ABC is ?

For maximum possible area of such a triangle ABC, the co – ordinates of point B, A and C should be


(- 8, 9), (- 8, 17) and (17, 9) respectively.


Maximum Area = (1/2) x 25 x 8 = 100 square units.


9 .)



10 )  Let, the average age of the family when the first child was born be 'x' years.
Therefore, the total age of the family when the first child was born is 3x.

Let, after a years, the second child was born:
Therefore, (3x + 3a) = 18 x 4 = 72

Let, after another b years, the third child was born
Therefore, 72 + 4b = 16 x 5
b = 2,
Let, after another c years, the fourth child was born
Therefore, 80 + 5c = 20 x 6 = 120
c = 8

Let, after another d years, the fifth child was born
Therefore, 120 + 6d = 24 x 7 = 168
d = 8.

Let, after another 'e' years the average age of the family is 28 years.
Also, 168 + 7e = 28 x 8 = 224
7e = 56
e = 8

Also, a + b + c + d + e = Present age of the eldest child = 27
a = 27 – 26 = 1 year.

3x + 3 = 72
x = 23. 

 I AM  EXTREMELY SORRY FOR DELAY IN ANSWERS  

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