Geometry
Algebra
Sequences & Series
Number Theory
Miscellaneous
100
A big 
We find the area of the big rectangle to be 
100
If 64 is divided into three parts proportional to 2, 4, and 6, the smallest part is:
If the three numbers are in proportion to 
, then they should also be in proportion to 
. This implies that the three numbers can be expressed as 
, 
, and 
. Add these values together to get:
Divide each side by 6 and get that
which is 
.
100
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57, 60, and 91. What is the fourth term of this sequence?
(A) 190
(B) 194
(C) 198
(D) 202
(E) 206
Let the arithmetic sequence be a, a + d, a + 2d, a + 3d and the geometric sequence be b, br, br , bro.
We are given that
a + b= 57,
a + d + br = 60,
a + 2d + b72 = 91,
and we wish to find a + 3d + b3
Subtracting the first equation from the second and the second equation from the third, we get
d+b（r-1）=3，
d + brr - 1) = 31.
Subtract these results, we get
6(r- 1)? = 28.
Note that either (r - 1)? = 1 or (r - 1)? = 4. We proceed with casework:
- If (r - 1) = 1, then r = 2, b = 28, a = 29, and d = -25. The arithmetic sequence is 29, 4, -21, -46, arriving at a contradiction.
- If (r - 1)? = 1, then r = 3,b = 7, a = 50, and d = - 11. The arithmetic sequence is 50, 39, 28, 17, and the geometric sequence is 7,21, 63, 189. This case is valid.
Therefore, The answer is a + 3d + br3 = 17 + 189 = (E) 206.
100
Find all primes that can be represented as sums and differences between two primes.
Let x be a prime that can be represented both as a sum and as a difference of 2 primes.
For the given statement, we must have x > 2.
∴ x is an odd prime number.
Also, one of those prime numbers must be 2.
That means we must have x = y + 2 = z – 2, where y and z are prime numbers.
So, z = x + 2 and y = z + 2
Thus, we can say that x, y and z will be three consecutive odd primes.
As we know, there is only one such set of prime numbers, and they are 3, 5, and 7.
x = 5 = 3 + 2 = 7 – 2
Hence, there is only one prime number .i.e 5 can be represented as sums and differences between two primes.
100
What is the value of 1- (-2) -3 - (-4) - 5 - (-6)
A) -20 B) -3 C) 3 D) 5 E) 21
We know that when we subtract negative numbers, 
The equation becomes 
200
The sides of triangle 
By the Angle Bisector Theorem, we have 
200
Cities A and B are 45 miles apart. Alicia lives in A and Beth lives in B. Alicia bikes towards B at 18 miles per hour. Leaving at the same time, Beth bikes toward A at 12 miles per hour. How many miles from City A will they be when they meet?
This is a 
problem, so let be the time it takes to meet. We can write the following equation:
Solving gives us 
. The 
is Alicia so 
200
Problem
Define a function on the positive integers recursively by f (1) = 2, f(n) = f(n - 1) + 1 if n is even, and /(n) = f(n - 2) + 2 if n is odd
and greater than 1. What is f (2017)?
(A) 2017
(B) 2018
(C) 4034
(D) 4035
(E) 4036
This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the
base case, f (1) = 2. We also know that when n is odd, f(n) = f(n - 2) + 2. Thus we know that f (2017) = f(2015) + 2. Thus we know
that n will always be odd in the recursion of f (2017), and we add 2 each recursive cycle, which there are 1008 of. Thus the answer is
1008 * 2 + 2 = 2018, which is answer | (B) .
200
What is the value of (612 − 392) ÷ (512 − 492)?
Solution:
(612 − 392) ÷ (512 − 492)
Consider (612 – 392)
Using the identity a2 – b2 = (a + b)(a – b)
(612 – 392) = (61 + 39) (61 – 39)
= 100 × 22
(512 – 492) = (51 + 49) (51 – 49)
= 100 × 2
Therefore, (612 − 392) ÷ (512 − 492) = (100 × 22)/ (100 × 2) = 11
200
Carl has 5 cubes each having side length 1, and Kate has 5 cubes each having side length 2. What is the total volume of these 10 cubes?
(A) 24 (B) 25 (C) 28 (D) 40 (E) 45
A cube with side length 
A cube with side length 
300
The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
Let the length of the edges of this box have lengths 
300
Reduced to lowest terms, 
is equal to:
We start off by factoring the second fraction.
Now create a common denominator and simplify.
300
The first three terms of an arithmetic sequence are 2it - 3,5x - 11, and 3x + 1 respectively. The nth term of the sequence is 2009. What is n
?
(A) 255
(B) 502
(C) 1004
(D) 1506
(E) 8037-
As this is an arithmetic sequence, the difference must be constant: (5x - 11) - (2x - 3) = (3x + 1) - (5x - 11). This solves to x = 4.
The first three terms then are 5, 9, and 13. In general, the nth term is 1 + 4n. Solving 1 + 4n = 2009, we get n = 502).
300
Find the number of trailing zeros in the 100!.
Solution:
Let us divide 100 by 5.
100/5 = 20
20/5 = 4
Now, 4 is less than 5. So, we stop the division here.
Also, the number of trailing zeros = 20 + 4 = 24
300
The ratio of w to x is 4:3, the ratio of y to z is 3:2, and the ratio of z to x is 1:6. What is the ratio of w to y?
(A) 4:3 (B) 3:2 (C) 8:3 (D) 4:1 (E) 16:3
We have
400
As the number of sides of a polygon increases from 3 to n, the sum of the exterior angles formed by extending each side in succession:
(A) Increases (B) Decreases (C) Remains constant (D) Cannot be predicted (E) Become (n-3) straight angles
By the Exterior Angles Theorem, the exterior angles of all convex polygons add up to 
so the sum 
400
If r and s are the roots of the equation 
, the value of 
is:
Solution 1
Note that 
.
By Vieta's, this is 
400
In the eight term sequence A, B, C, D, E, F, G, II, the value of C is 5 and the sum of any three consecutive terms is 30. What is A + H?
(A) 17
(B) 18
(C) 25
(D) 26
(E) 43
Let A = x. Then from A + B + C = 30, we find that B = 25 — 7. From B + C + D = 30, we then get that D = t. Continuing this pattern, we find E = 25 - 1, F = 5, G = t, and finally H = 25 - 1. So A + H = x + 25 - x = 25 → C
C
400
What is the value of (11! - 10!)/9!
(A) 99 (B) 100 (C) 110 (D) 121 (E) 132
(11! - 10!)/9! = [9!(11 x 10 - 10)]/9! = 110 - 10 = (B) 100
400
How many positive even multiples of 3 less than 2020 are perfect squares?
(A) 7 (B) 8 (C) 9 (D) 10 (E) 12
Any even multiple of 
500
Rhombus 
Triangle DAB is equilateral so triangles 
Note: A quick way to visualize this method is to draw the line 
500
The coefficient of 
in the expansion of 
is:
By the Binomial Theorem, each term of the expansion is 
.
We want the exponent of to be 
, so
If 
, then the corresponding term is
The answer is 
.
500
500
The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller?
(A) 5/4 (B) 3/2 (C) 9/5 (D) 2 (E) 5/2
x + y = 5(x - y), thus 4x = 6y.
(B) 3/2
500
Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?
(A) 1/36 (B) 7/72 (C) 1/9 (D) 5/36 (E) 1/6
We split this problem into 
First, we calculate the probability that all four are the same. After the first dice, all the numbers must be equal to that roll, giving a probability of 
Second, we calculate the probability that three are the same and one is different. After the first dice, the next two must be equal and the third different. There are 
Adding these up, we get 
