1. SAT MATH PREPERATION

2. Problems Related to SAT
Geometry
Algebra
Precalculus
Additional problems in the blog

3. Problems 1.
Question: In an isosceles triangle ABC, AM & CM are angle bisectors of angles BAC and BCA respectively. What is the measure of angle AMC?
(A)110’ (B)115’ (C)120’ (D)125’ (E)130’

4. Solution:
Isosceles triangle means that any 2 sides are equal
Here we consider AB and CB to be equal
Therefore angles BAC and BCA are equal and the value is (180-40)/2 = 70’
We are given that AM and CM are angle bisectors. Angles MAC and MCA = 35’
Finally angle AMC is 180-(2*35) = 110’

5. 2.
Question: In a circle a square is inscribed. What is the degree measure of arc ST?
(A)45’ (B)60’ (C)90’ (D)120’ (E)180’

6. Solution:
OS and OT are angle bisectors
Angles of sides of a square are 90’
Therefore the angle bisectors cut the square at 45’ each.
The angle of arc ST is 180-(2*45)=90’

7. 5/x = (5 + a)/(x + a) ; If a not equal to 0, find the value of x?
(A)-5 (B)-1 (C)1 (D)2 (E)5
Solution:
Multiply x on both sides to get
5 = (5 + a)*x/(x + a)
Multiply (x + a) on both sides to get
5(x + a) = x*(5 + a)
Simplify to get 5x + 5a = 5x + x*a
5a – x*a = 0

8. a(5 - x) = 0
Given that a is not zero. Therefore (5 - x)=0
This implies x = 5

9. If y = 2x + 3 and x < 2 which of the following represents all possible values of y?
(A)y<7 (B)y>7 (C)y<5 (D)y>5 (E)5<y<7
Solution:
Consider that x = 2
Substitute the value of x in main equation to get the value of y
y = 2*2 + 3 = 7
Since x is less than 2 therefore y must be less that 7.
y < 7

10. Let the function f(x) = 5x – 2a, where a is a constant
Let the function f(x) = 5x – 2a, where a is a constant. If f(10) + f(5) = 55, what is the value of a?
(A)-5 (B)0 (C)5 (D)10 (E)20
Solution:
f(10) = 5*10 – 2a = 50 – 2a
f(5) = 5*5 – 2a = 25 – 2a
Substitute in the above results in the Given function f(10) + f(5) = 55
50 – 2a + 25 – 2a = 55
-4a = -20
a = 5

11. If f(x) = 2x and g(x) = x + 2. Find the answer for fog(x)?
(A)2x + 4 (B)x + 2 (C)2x + 2 (D)3x + 2 (E)x + 2
Solution:
The above problem is a function within function
fog(x) means f(g(x)). This means we substitute the value of g(x) in place of the x term in the f(x) equation
We get f(x + 2) = 2(x + 2) = 2x + 4

12. Extra Problems in the blog
The following problems have the solutions also being provided

13. Given that x + k = 6 and p(x + k) = 36. Find the value of p?
(A)8 (B)6 (C)-8 (D)-6 (E)-2
Solution:
Given that (x + k) = 6
Substitute this value in the second equation
p * 6 = 36
Therefore p = 6

14. 2. y = h/x ; h is a constant. Initially the value of y is 3 and x is 4
2. y = h/x ; h is a constant. Initially the value of y is 3 and x is 4. Find the value of y when x is 6?
(A)-2 (B)5 (C)-1 (D)1 (E) 2
Solution:
We find the value of h by substituting values of y = 3 and x = 4 in the main equation
We get h = 4 * 3 = 12
Now we find y = h/x = 12/6 = 2

15. 3. A rectangle is 4 times as long as it is wide
3. A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle?
Solution:
Draw a rectangle with the required dimensions

16. Let x = original width of rectangle
Area of rectangle = Length * Width
Plug in the values from the question and from the sketch = (4x + 4)(x –1)
Solving we get 4x2 – 4 – 60 = 0
(2x – 8)(2x + 8) = 0
x= 4 or x = -4
We consider only the positive value. So the width of the original rectangle is 4 and the length is 16

17. 4. Find the domain and range for function 
f(x) = x2 + 2.
Solution:
The function f(x) = x2 + 2 is defined for all real values of x.
Therefore the domain is all real values of x
Since x2 is never negative, x2 + 2 is never less than 2. Hence, the range of f(x) is "all real numbers f(x) ≥ 2".

19. 5. Find the domain and range for the function 
Solution:
g(s) is not defined for real numbers greater than 3 which would result in imaginary no.
Hence, the domain for g(s) is "all real numbers, s ≤ 3".
Also >= 0. The range of g(s) is "all real numbers g(s) ≥ 0"

21. 6. Graph the function y = x − x2
Solution:
Determine the y-values for a typical set of x-values and write them in a table.
Plot the values on a x-y plane.
For a better graph have more number of points

23. 7. Graph the function 
Solution:
Note: y is not defined for values of x < -1
We determine x and its corresponding y-values and write them in a table
Draw the graph

25. 8. Consider the function
Solution:
Factoring the denominator gives
We observe that the function is not defined for x = 0 and x = 1.
Here is the graph of the function.

26. We say the function is discontinuous when x = 0 and x = 1.