SAT MATH PREPERATION Problems Related to SAT Geometry

SAT MATH PREPERATION

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

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’

• 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 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 5 x + 5 a = 5 x + x*a • 5 a – x*a = 0

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

4. If y = 2 x + 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) = 5 x – 2 a, 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 – 2 a = 50 – 2 a • f(5) = 5*5 – 2 a = 25 – 2 a • Substitute in the above results in the Given function f(10) + f(5) = 55 • 50 – 2 a + 25 – 2 a = 55 • -4 a = -20 • a=5

6. If f(x) = 2 x and g(x) = x + 2. Find the answer fog(x)? • (A)2 x + 4 (B)x + 2 (C)2 x + 2 (D)3 x + 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 + 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 = (4 x + 4)(x – 1) • Solving we get 4 x 2 – 4 – 60 = 0 (2 x – 8)(2 x + 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) = x 2 + 2. • Solution: • The function f(x) = x 2 + 2 is defined for all real values of x. • Therefore the domain is all real values of x • Since x 2 is never negative, x 2 + 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 − x 2 • Solution: • Determine the y-values for a typical set of xvalues 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 yvalues 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.