practice problems

Q1). Andy and Bethy are at same point. Andy leaves at 1:30 toward north at a steady 8 miles/hr speed. Bethy leaves at 2:30, toward east at a steady 12 miles/hr speed. At what time they will be exactly the same distance away from their starting point?


Q2). A box contains 10 pounds of a nut mix i.e., 50% peanuts, 20% cashews, 30% almonds. A 2nd nut mix containing 20% peanuts, 40% cashews, 40% almonds is added to the box resulting in a new nut mix i.e., 40% peanuts. How many pounds of cashews are now in the box?

Q3). How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 2025?

Q4) 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, ... What is the 2025th term?

Q5) Suppose a and b are real numbers. When the polynomial (x^3 + x^2 + ax + b) is divided by (x - 1), the remainder is 4, and when divided by (x - 2), the remainder is 6.
b - a = ?

Q6) The sequence (1, x, y, z) is arithmetic. The sequence (1, p, q, z) is geometric. Both sequences are strictly increasing, and contain only integers, and z is as small as possible.
(x + y + z + p + q = ?)

Solution 1:
4:30.
Since
t*8 = (t-1)*12
=> 4t = 12
=> t = 3
1:30 + 3 hours = 4:30

Solution 2:
Initially we had 5,2,3 pounds of peanuts,cashews,almonds.
Now we add x pounds of new nut mix.
5 + x*.2 = (10+x)*.4
=> 5 + .2x = 4 + .4x
=> 1 = .2x
=> x = 1/.2 = 5

Cashews = 10*.2 + 5*.4 = 4 pounds = Answer


Solution 3:
Case 1: 2025,2025,x where x < 2025
Case 2: x,x,2025 again with x < 2025

Case 1:
2025*2 > x and x < 2025 => x < 2025
2025 + x > 2025 => always true
So 1 <= x <= 2024 is the range => 2024 such triangles.

Case 2:
2x > 2025 => x > 2025/2 => 1013 <= x <= 2024 => 1012 such triangles 
x + 2025 > x => always true

Total = 1012 + 2024 + 1(equilateral) = 3037


Solution 4:
The pattern is like this:
1,2,1
1,2,3,2,1
1,2,3,4,2,1

But the last 1 of the previous sequence and first 1 of the current sequence are the same.
We can write it like this:
1
2,1
2,3,2,1
2,3,4,3,2,1
....
Ignore the first sequence for a while which has 1 term.
All the next sequences contribute 2,4,6,8... terms.

2 + 4 + 6 ... 2n >= 2025- 1 = 2024
First sequence peaks at 2, second at 3 and so on..
=> 2*n*(n+1)/2 >= 2024
=> n^2 + n >= 2024
We know that 45^2 = 2025
So let's try n = 44
LHS = 44*45 = 180*11 = 1980
So we are short by 44 terms.
44th sequence would have peaked at 45.
Start the next sequence which will peak at 46.
First 44 terms of it would be:
2,3,4....45
Answer 45


Solution 5:
Let the given polynomial be f(x).
Then,
f(x) = (some polynomial)(x-1) + 4
=> f(x) - 4 = some polynomial)(x-1) 
=> f(x) - 4 has one of its roots as 1
=> 1 + 1 + a + b - 4 = 0
=> a + b = 2

Similarly
2a + b = -6

=> a = -8, b = 10 => b - a = 18 = Answer.
Answer 18


Solution 6:
1.r^3 = z
1 + 3d = z

So z is a cube as well as = 1 mod 3
Try all the cubes since that's quicker
8,27,64
64 works
r = 4
d = 21
1,x,y,z = 1,22,43,64
1,p,q,z = 1,4,16,64
So 22 + 43 + 4 + 16 + 64 = 149 = Answer



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