SAT MATH PREPERATION. P roblems Related to SAT Geometry Algebra Precalculus Additional problems in the blog.

SAT MATH PREPERATION

Problems Related to SAT

• Geometry• Algebra• Precalculus• Additional problems in the blog

Problems1.

• 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’

• 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’

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’

• 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’

3. 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

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

4. 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

5. 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

6. 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

Extra Problems in the blog

• The following problems have the solutions also being provided

1. 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

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

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

• Let x = original width of rectangle• Area of rectangle = Length * Width• Plug in the values from the question and from

the sketch 60 = (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

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".

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"

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

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

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.

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