practice problems pending

Q1. The age of a person in 2025 is a perfect square. His age was also a perfect square in 2012. His age will be a perfect cube (m) years after 2025. Determine the smallest value of (m).

S1.
Age in 2012 = p^2
p^2 + 13 = q^2
=> 13 = q^2 - p^2 = (q-p)(q+p)
=> q - p = 1, q + p = 13
=> q = 7, p = 6
=> Age in 2012 = 36
=> 49 + m = r^3
Smallest m  = 15 so that r^3 = 4^3 = 64
Answer = 15

Q2. The sum of two real numbers is a positive integer (n) and the sum of their squares is (n + 1012). Find the maximum possible value of (n).
S2.
a + b = n
a^2 + b^2 = n + 1012

Root mean square inequality tells us that
sqrt[(x1^2 + x2^2 ...xn^2)/n] >= (x1 + x2 ... xn)/n
=> sqrt[(a^2 + b^2)/2] >= (a+b)/2
=> (n+1012)/2 >= n^2/4
=> n^2 - 2n - 2024 <= 0
=> n^2 - 2n - 1 <= 2025
=> (n-1)^2 <= 2025
=> |n-1| <= 45
=> -45 <= n-1 <= 45
=> -44 <= n <= 46
Max(n) = 46
Does it satisfy original equations?
a + b = 46
(a + b)^2 - 2ab = 46 + 1012 = 46^2 - 2ab
=> 2ab = 46^2 - 1058 = (1600 + 36 + 480) - 1058 = 2116 - 1058 = 1058
=> ab = 529 = 23^2 => a = b = 23 which is expected since RMS inequality attains equality at a = b.

Answer n = 46

Q3. Four sides and a diagonal of a quadrilateral are of lengths 10, 20, 28, 50, 75, not necessarily in that order. Which amongst them is the only possible length of the diagonal?

S3. Let's try few cases:
Let 10 be the diagonal. Then there are 2 triangles which have 2 sides and a diagonal.
Let's check:
20,28,10 and 10,50,75
The second triangle doesn't work out since 50 + 10 < 75.
In fact 75 can only be overcome by 28 + 50.
So one triangle has to be 28,50,75.
So diagonal has to be one of 28,50,75.

Let's try d = 28
10,20,28 28,50,75 both are valid.
d = 50
10,20,50 doesn't work
d = 75
10,20,75 doesn't work.

So answer = 28.

Q4. A quadrilateral has four vertices (A, B, C, D). We want to colour each vertex in one of the four colours — red, blue, green or yellow, so that every side of the quadrilateral and the diagonal (AC) have end points of different colours. How many ways can we do this?

S4.
Answer  = 48
Why?
We can color A in 4 ways. And C in 3 ways.
Now B,D can have 2 options because both remaining colors will work.
4*3*2*2 = 48.

Common mistake:
A => 4
B => 3
C => 2
D => 1

To avoid this, always try the vertices with maximum edges. In this case A,C. And then you are likely to avoid this mistake.



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