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Question distributions

Pre-algebra - 14 questions

1. Which of the following value of nwill make the following fraction undefined?

2n24n2−6

A.−14

B. 0

C.116

D.14

E.12

Question 1. Correct answer is E.A fraction is not defined when its denominator is zero, therefore24n2−6=0. Collecting like

terms on opposite sides of the equal sign and dividing by 24 gives n2= 624

=14 . Taking the

square roots on both sides gives an equivalent equation n=√ 14=± 12

.

If you chose A, you would have applied the equality in subtraction incorrectly and also forgotten to determine the square root. If you chose B, you may have thought that a fraction is not

defined when the numerator is zero. If you chose C, you may have determined the square of 14

instead of determining its square root. If you chose D, you would have forgotten to determine its square root.

2. Tiffany realizes that after 3 months of intensive exercise, her weight reduced by 48 lbs from from168 lbs. What is the ratio of her initial weight to the current one?

A. 2 :7B. 5 :7C. 7 :9D. 7 :5E. 7 :2

Question 2. Correct answer is D.Begin by determining the new weight after the reduction then determine the ratio. The new

weight is 168−48=120lbs . The ratio, Initial weightNewweight

equals 168120

=75=7 :5.

If you chose A, you may have may have forgotten to find the actual value of the new weight but proceed to determine the ratio. If you chose B, you may have ordered the ratio components incorrectly. If you chose C, you may have may added the decrease in value to the initial value then determined the ratio. If you chose E, you may have may have forgotten to find the actual value of the new weight but proceed to determine the ratio then interchanged the ratio components.

3. Given that 4−2 (1−x )=3−3 ( x−2 )+4 x+9what is the value of2−x?

A. -12B. 12C. 14D. 16E. 24

Question 3. Correct answer is A.The best approach is to determine the value of x then use it to evaluate 2−x . Opening the bracket gives us an equivalent equation 4+2 x−2=3−3 x+6+4 x+9. Collecting like terms and simplifying the equation equals x=18−4=14. The required value is 2−x=2−14=−12.If you chose B, you may have solved the problem x−2 instead of 2−x . If you chose C, you would have not evaluated 2−x but simply gave the value ofx . If you chose D, you may have added 2 to the value of x instead of subtracting it from 2. If you chose D, you may have multiplied 2 to the value ofx .

4. What is the difference between the second largest and the third smallest factor of 48?

A. 20B. 21C. 24D. 27E. 47

Question 4. Correct answer is E.Factors are numbers that divide another. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

The second largest factor is 24 while the third smallest one is 3. The difference is 24−3=21.

If you chose A, you would have ignored 1 as a factor then picked 4 as the third smallest factor. If you chose C, you would have simply provided the value of the second largest factor without considering the thirds smallest one. If you chose D, you would have added the two factors. If you chose E, you would have picked the smallest and the largest factors.

5. 40% of a three quarters of a number equals 155 less 2. What is 25% of the number?A. 127.5B. 501C. 510

D.15503

E.15703

Question 5. Correct answer is A.

Let the number be x. Three quarter of the number is 34

x and 40% of that is ( 40100 )( 34 ) x or

310

x. This must equal (155−2 ), giving 310

x=155−2=153. Multiply 103 on both sides of the

equation to find x=510. 25% of 510 is ( 25100 ) (510 )=127.5.

If you chose B, you would have provided the value of the number but again interchanges the position of the digits in the once and tens positions. If you chose C, you would have provided the value of the number. If you chose D, you may have forgot to include -2 on the right hand side of the equation formed. If you chose E, you may have forgot to include added 2 instead of subtracting 2 on the right hand side of the equation formed.

6. Two years ago Steve was 6 more than seven times the son’s age. Three years to come, Steve would be 3 years more than four times the age of his son. What is the current age of the son?

A. 4B. 6C. 9D. 34E. 36

Question 6. Correct answer is B.Let the current age of Steve’s son be x, hence, his age two years ago is x−2. Two years ago, Steve’s age was 6+7 (x−2) and two years to come, his age would be 6+7 ( x−2 )+2+3=11+7 (x−2). On the other hand, the son’s age three years to come would be x+3 implying that at the same time Steve would be 4 ( x+3 )+3 years old. The two expressions for the age of Steve must be equal, therefore, 11+7 ( x−2 )=4 (x+3 )+3. Opening the brackets and solve for x equals 11+7 x−14=4 x+12+3 which is equivalent to 3 x=18. Thus x=6.

If you chose A, you would have provided the age of the son two years ago. If you chose C, you would have incorrectly solved the equation 3 x=18. If you chose D, you would have provided the age of Steve two years ago. If you chose E, you would have provided the age of Steve.

7. The probability that George wins two rounds of a game is 0.32. If the probability that he wins the first one is 0.4, what is the probability that he wins the second one if the two events are independent?

A. 0.08B. 0.128C. 0.6D. 0.72E. 0.8

Question 7. Correct answer is E.Let the probability of winning the first game be A, and that of the second game be B. Then P (A∧B )=0.32 P ( A )=0.4 . Since the two events are independent P (A∧B )=P (B ) ×P ( A ) ,

therefore, the probability of wining in the second round, P (B )=P ( A∧B )P(A)

=0.320.4

=0.8.

If you chose A, you would have solved the division problem incorrectly by providing more decimal places. If you chose B, you would have multiplied the two probabilities rather than diving one by another. If you provided C through that 4 divided 32 6 times. If you chose D, you would have added the probabilities rather than dividing one by another.

8. What is|2−7|−|−3−4 (9−2 )|=¿?A. -42B. -36C. -26D. 26E. 42

Question 8. Correct answer is C.First, simplify the numerical expressions between the absolute value functions;|2−7|−|−3−4 (9−2)|=|−5|−|−31|. The absolute value of a number is equal to the positive value of that number hence|−5|−|−31|=5−31=−26.

If you chose A, you would have multiplied -4 by -2, when opening the brackets, to get 8 instead of 8. If chose B, may have added the values in the two from the absolute value functions. If you chose D, you would have subtracted 5 from 31 instead of subtracting 31 from 5. If you chose E, you would have multiplied -4 by -2, when opening the brackets, to get 8 instead of 8 then proceeded to subtract the first value of the numerical expression from the second one.

9. The median seven numbers is 45. If two values, 44 and 47, are introduced in the data, what would be the new median? A. 44B. 44.5C. 45D. 46E. 47

Question 9. Correct answer is C.The median is the number at the central position with respect to other when the data is arranged in ascending or descending order. When two values are added to the data, one on one side of the median (lower side) and the other value on the other side (higher value), the median remains the same value because the either sides of the median will have the same number of value.

If you chose A, you may have though that the median value shifts to the lower value. If you chose B, you may have though that the median value would be the mean of the median of the initial data and the smaller value added to the data. If you chose D, you may have though that the median value would be the mean of the median of the initial data and the larger value added to the data. If you chose E, you may have though that the median value shifts to the lower value.

10. What is √32−√18=¿?

A. −24B. −12√2C. −√2D. √2E. 7√2

Question 10. Correct answer is B.Express the values under the root as a product of a square of an integer where possible; √32=√16×2 and √18=√9×2. Determining the square root where possible equals √32=√16×2=4√2 and √18=√9×2=3√2. Upon substituting the final results into the initial expression, we get √32−√18=4√2−3√2=√2. If you chose A, you would have multiplied the two final values. If you chose B, you would have multiplied the coefficients of the roots. If you chose C, you would have subtracted the first term from the second one. If you chose E, you would have added the two final values.

11. A class is composed of 10 girls and 9 boys. A class teacher would like to choose 3 girls and 2 boys to represent the class in the school choir. How many ways can she choose the five from the class given that all students have equal chances of being chosen?

A. 159B. 4320C. 11628D. 51840E. 1395360

Question 11. Correct answer is B.In the selection process, order does not matter hence, this is a combination problem. The of ways of selecting 3 girls from the total, 10 is 10C 3 and the number of ways of selecting 2 boys from the total, 9, is 9C2. For each girl selected, 2 boys are selected, hence for all the combinations 10C3, there are 9C2 ways of selecting boys. Therefore, all possible ways of selecting the 5students is equal to

10C3×9C2= 10!3 ! (10−3 ) !

× 9 !2 ! (9−2 )!

= 10 !3 !7 !

× 9 !2!7 !

=10×9×86

× 9×82

=4320.

If you chose A, you would have added the two combination expressions. If you chose C, you would have ignored the existence of groups and applied the combination formula on the whole set of 19 students. If you chose D, you would have applied a permutation formula to the two situations. If you chose E, you would have ignored the existence of groups and applied the permutation formula on the whole set of 19 students.

12. If

2 x+ x+23

=23−2(1−x)

what is the value of2−x?

A. - 4B. - 6C. 4D. 6E. 8

Question 12. Correct answer is E.

The first step is to multiply throughout the equation by 3 to do away with the fractions; 6 x+x+2=2−6 (1−x ) . Opening the bracket and collecting like terms leads to equation 6 x+x+2=2−6+6 x which is equivalent to x=−6. The value of 2−x is 2− (−6 )=2+6=8.

If you chose A, you would have treated the problem2−−6 as2−6. If you chose B, you would have not evaluated 2−x but simply gave the value of x as the required answer. If you chose A, you would have treated the problem2−−6 as2−6 then evaluated the opposite one, 6−2. If you chose D, you would have gotten 6 instead of -6 while multiplying -6 by 1 when opening the brackets and proceed to provide the value of x as the required answer.

13. Two business partners Halid and Boen share their annual profit in the ratio of 5 :7 respectively. Boen got $3640 more than Halid. The profit shared represents 60% of the total profit that they made from their business. What was the total profit? A. $6,240B. $10,400C. $21,840D. $34,944E. $36,400


If we take k to be the common factor between their shares, then Halid and Boen got 5k and 7k respectively. The difference is 2k=3640, hence k=$ 1820. The total profit shared is 5k+7 k=12k=12 (1820 )=$21,840. Since the figure represents 60% of the total profit, the total profit would be equivalent to 100% which is equal to

10060

×21840=$ 36,400.

If you chose A, you would have treated the difference, $3640, as Boen’s share then proceeded to provide the shared profit as the total profit. If you chose B, you would have treated the difference, $3640, as Boen’s share then proceeded to get the total profit. If you chose C, you would have treated the shared profit as the total profit. If you chose D, you would have treated the shared profit as 100% then proceeded to get the total profit which is equivalent to 160%.

14. Two values aand y are such that 5a=6 y where y is an integer. Which of the following is a possible value of a?

A. 10B. 21C. 24D. 40E. 52

Question 14. Correct answer is C.

The number, 5a is expressed as a product of 6 and y , implying that both 6 and y divides 5a . Since 5 and 6, do not have a common factor other than ±1, 6divides a. In other words, 6 is a factor of a. Therefore, a is a multiple of 6. The answer is number that is a multiple of 6 , that is, 24.

If you chose A, you would have through that a is a factor of 5 and 2 only. If you chose B, you would have through that a is a factor of 3 only. If you chose D, you would have through that a is a factor of 5 and 2 only and since it is a multiple of 8, it would be the required number. If you chose E, you would have through that a is a factor of 2 only.

Elementary algebra - 10 questions

15. A firm wishes to fence its 1.96 million square yards compound with a perimeter wall for security purposes. What will be the total length of the wall?A. 1400 ydB. 4800 ydC. 5600 ydD. 6400 ydE. 560000 yd


The area of the compound of side, s , is s2=1.96millon squre yards=1,960,000. Taking the square roots, gives us the length of one side, s, as s=√1,960,000=1400 yd . The length of the wall is equal to the perimeter of the compound = 4 s=4×1400=5600 yd.

If you chose A, you would have provided the length of one side as the perimeter. If you chose B, you would have incorrectly got the value from the product of 4∧1400. If you chose D, you would have incorrectly got the value from the product of 4∧1400. If you chose E, you would have got the square root of the area incorrectly as 140000 then proceeded to get the perimeter.

16. Ifx2+1< 3

2−x, which of the following value is less than

1x ?

A.52

B.273

C.183

D.103

E.72


The first step is to multiply throughout the inequality by 2 to do away with the fractions;

x+2<3−2 x. Collecting like terms and simplifying gives us 3 x<1 which is equivalent to x<13.

Taking the reciprocal changes the direction of the inequality, therefore, 1x>3. Based on the

options given, 52=2.5 is one of the required values.

If you chose B, you would have thought that 1x is compared to the reciprocal of

273

=9 . If you

chose C, you would have thought that 1x is compared to the reciprocal of

183

=6. If you chose

D, you would thought that 103 is less than 3. If you chose E, you would thought that

72 is less

than 3 because its denominator is less than 3.

17. Two polynomials P1(x) and P2(x) are given as P1 (x )=a x6+8x3−6 x2+bx+1 and

P2 (x )=d x3−4 x2+e x2−3. If f ( x )=t xn+s xm+g(x ) where f ( x )=P1 (x )P2 (x )

, what is the value of

n if n>m and the degree of g(x ) is less than m?

A.12

B. 2C. 3D. 6E. 9


When dividing polynomials, the degree of the quotient is equal to the degree of the numerator less that of the denominator. The degree of the numerator is 6 while that of the denominator is

3. Therefore, the degree of the quotient f ( x )=P1 (x )P2 (x )

is 6−3=3. The fact that n>m and the

degree of g(x ) is less than m, n is power of the leading terms consequently, the degree of f ( x ) . Thus, n=3.

If you chose A, you would have thought the degree of the quotient is the ratio of that of the denominator to that of the numerator. If you chose B, you would have thought the degree of the quotient is the ratio of the degree of the numerator to that of the denominator. If you chose D, you would thought that the degree of the dividend does not change when dividing by any polynomial. If you chose E, you would have thought that the degree of the quotient is the sum of the degree of the numerator and the denominator.

18. The product of two numbers is 168. If their sum is 2, what is the value of the smaller number?A. −21B. -14C. -12D. 12E. 14

Question 18. Correct answer is B.

Assigning the numbers the variables x and y; the statement above implies that xy=168 and x+ y=2. The later is equivalent to x= y−2. Substituting into the first equation equal y ( y−2 )= y2−2 y=168 or y2−2 y−168=0. Two numbers whose product is −168 and sum is −2 are −14 and 12. The fact that coefficient of y2 is 1, the equation equal ( y+14 ) ( y−12 )=0 which implies that y+14=0 or y−12=0. Therefore, y=−14 or y=12. When y=−14 , x=12 and when y=12 , x=−14, therefore, the smallest number is -14.

If you chose A, you would have picked the smallest number among the multiple choice without making any calculation. If you chose C, you would have erroneously exchanged the signs on 12 and −14. If you chose D, you would have eliminated the negative value as part of your solution of the quadratic equation then proceeded to provide it as the answer. If you chose E, you would have erroneously forgot the negative sign on the number after correctly identifying it as the correct answer.

19. 4−2 x×32=1. What is the value of 1−x−1?

A. −14

B. 14

C. 15

D. 45

E. 54


The first step is to express the numbers under the same base, 2; (22)−2x×25=20. Applying the laws of indices gives the equation 2− 4x+5=20 which implies that −4 x+5=0. Simplifying the

equation gives the resultx=54 . The value of the expression 1−x−1=1−( 54 )−1

=1−45=15

.

If you chose A, you would have forgotten to determine the reciprocal of 54. If you chose B, you

would have erroneously interpreted 1−x−1 and x−1−1 then forgot to take the reciprocal of 54 .

If you chose D, you may have solved the equation −4 x+5=0 , erroneously and forgot to determine the value of the expression1−x−1 . If you chose E, you would have provided the value of x as that of the expression 1−x−1 .

20. 16x2+44 x−128 x−2

=?

A. −2 x−6

B.x+32

C. x+3D. 2 x+3E. 2 x+6

Question 20. Correct answer is E.The first step is to factorize the numerator as well as the denominator; 16x2+44 x−12

8 x−2=4(4 x2+11 x−3)2(4 x−1)

.

Expanding then factorizing the numerator gives the expression 4 (4 x2+11 x−3 )2 (4 x−1 )

=4 (4 x2+12x−x−3 )

2 (4 x−1 )=4 (4 x (x+3 )−1 ( x+3 ) )

2 (4 x−1 )

¿4 (4 x−1)(x+3)2 (4 x−1)

=2 (x+3 )=2x+6

If you chose A, you would have erroneously factorized the denominator. If you chose B, you

would have erroneously carried out division of 4 by 2 to get 12 instead of 2. If you chose C,

would have erroneously factorized the denominator to get 4 (4 x−1). If you chose D, you would have incorrectly expanded the bracket in the expression 2(x+3) to get 2 x+3.

21. Hibert is a businessman who understand the market patterns of his products throughout the years. During the end of October, he usually pumps a lot of funds into the business so that he can buy more stock. During the last month of the years, demand for his products, usually, increases, as a result, he records the highest profit that other months. How dependent variables does this, Hibert’s, case have?A. 0B. 1C. 2D. 3E. 4


When funds are pumped into the business, stock increases. When demand increases during the last month of October, higher profits are made because more stock is bought. Therefore, Stock depends on funds and profit depends on stock and demand. In the statement, there is no clear description of time as a factor affecting demand, hence demand cannot be considered a dependent variable. Therefore, the dependent variables are two.

If you chose A, you would not have identified any dependent variable due to lack of understanding of what they are. If you chose B, you may have ignored the intermediate variables and considered the final one only, profit. If you chose D, you may have included demand among the dependent variable. If you chose E, you have considered all variables mentioned are dependent.

22. A newspaper vendor takes 3 minutes more to deliver her products to a town, 10 miles away, when he drives at a 10 mi/hr speed faster than the usual speed. What is her new speed?A. 30 mi/hB. 40 mi/hC. 50 mi/hD. 60 mi/hE. 90 mi/h


We use the formula for time, e=distancespeed , to solve the problem. Taking the usual speed to be

s , the new speed is s+10. Time taken when travelling at the usual speed is 10s while that taken

when traveling at the new speed is 10

s+10 . When moving faster, it takes her fewer

hours/minutes to get to the destination, hence 10s

= 10s+10

+ 360 , equivalently,

10s

= 10s+10

+ 120

.

Multiplying throughout the equation by 2 s (s+10) results into the equation20 (s+10 )=20 s+0.1(s (s+10)). Expanding and collecting like terms results into

20 s+200=20 s+0.1 s2+s which is equivalent to0.1 s2+s−200=0. Multiplying through by 10 gives uss2+10 s−2000=0. By the quadratic formula,

s=−10±√102−4 (−2000)2

=−10±902

=−5± 45.

The speed is a positive quantity hences=−5+45=40mi /s. The news speed is 40+10=50mi /h.

If you chose A, you would have subtracted instead of adding 10 mi/hr to the usual speed. If you chose B, you would have provided the value of the usual speed not knowing that what is required is the new speed. If you chose D, you would have ignored the negative values in the expression s=−5±45 the proceeded to determine the new speed takings=5+45mi /hr. If you chose E, you would have forgotten to divide the numerator by 2 then proceeded to find the new speed.

23. Which of the following is the smallest interval having the number√72?A. 8 - 8.5B. 8.5 - 9C. 8 - 9D. 8.5 - 10E. 7 - 9

Question 23. Correct answer is A.We begin by selecting two squares of integers closer to 72 where one is less while another is more than 72. The numbers are 64 and 81. Therefore, 64<72<81. Taking square roots, we have √64<√72<√81which is equivalent to 8<√72<9. To determine the smallest interval, we find out in there are numbers between 8 and 9 that can allow the above representation. Using the multiple choices given, we try to find out the direction in which 8.5 will group the value. The square of 8.5 is 72.25 hence √72.25>√72implying that 8.5>√72 . The smallest interval having √75 is 8 to 8.5.

If you chose B, you may have thought that 8.5<√72. If you chose C, you may have thought that the interval must have integers as its boundaries. If you chose D, you may have thought that 8.5<√72 and that any interval will do so long as the value falls within it. If you chose E, you may have misread the statement; something that may have led you to determine the largest interval rather than the smallest one.

24. What is the largest value of p that satisfies the following equation?25 p2+p=0

A.−125

B. −25C. 0

D.125

E. 25

Question 24. Correct answer is A.The first step is to factorize the equation; 25 p2+p=p(25 p+1)=0 implies p=0 or

25 p+1=0. If 25 p+1=0, then p=−125

. Between 0 and −125 , 0 is the largest.

If you chose A, you may have overlooked the negative sign on the number. If you chose B, you have factorized the equation incorrectly and overlooked the negative sign. If you chose D, you may have erroneously simplified the equation25 p+1=0. If you chose E, you have factorized the equation incorrectly.

Intermediate algebra - 9 questions

25. Given that A=[1 −43 −1 ]B=[3 −1

5 12 2 ]C=[ 0 5 2

−3 4 −6 ] and D=[2 −13 5 ], which of the

following operation is not a defined matrix multiplication?

A. ADB. BCC. CBD. CDE. BA

Question 25. Correct answer is D

Matrix multiplication PQ is defined when the number of column of P is equal to the number of rows of Q. Using this fact, the only multiplication where the number of columns of the first matrix is not equal to the number of rows of the second matrix is CD.

If you chose A, you may have through that the questions requires one to provide a defined matrix multiplication. Since this is an obvious yes, you easily picked it. If you chose B, you may have thought that the matrices must be unequal number of columns. If you chose C, you may have thought that the matrices must be unequal number of rows. If you chose E, you may have thought that the matrices must be unequal dimension.

26. Which of the following equations represents the horizontal asymptote of the following rational function?

f ( x )= 6 x2−63 x2−27

A. y=0

B. y=12

C. y=1D. y=2E. y=3

Question 26. Correct answer is D.

The horizontal asymptotes for a rational function whose degree of polynomials in the numerator

is equal to that of the numerator is y=ab where a and b are the coefficients of the leading

terms in the numerator and the denominator respectively. In the function above, a=6 and b=3

hence y=63=2.

If you chose A, you have thought that when the denominator and the numerator have the same degree, then the horizontal asymptote is automatically y=0. If you chose B, you may have thought that when the denominator and the numerator have the same degree, then the

horizontal asymptote is y= pq where p and q are the coefficients of the denominator and

numerator respectively. If you chose C, you may have through that the horizontal asymptote is y=c where c is the root of the numerator. If you chose E, you may have through that horizontal asymptote is y=c where c is the root of the denominator.

27. What is 2i+3i−1

=¿?

A. −5 i+12

B. −5 i+52

C. 5i+12D. 5i+12E. 5i+12

Question 27. Correct answer is A .The first procedure is to rationalize the fraction.

2i+3i−1 =

2i+3i−1 ×−i−1

−i−1=−i (2 i+3 )−1(2 i+3)

1+1 =2−3 i−2 i−3

2 =−5 i+12

If you chose B, you would have erroneously got 2 from the multiplication of 2 i by i in the numerator. If you chose C, you would have i−1 as your conjugate and erroneously got 2 from the multiplication of 2 i by i in the numerator. If you chose D, you would have used i−1 as your conjugate. If you chose E, you may have got −2 from the simplification of the denominator.

28. Which of the following numbers is not a solution of the inequality,2 x2−8 x≤0?A. -1B. 0C. 1D. 3E. 4


The first procedure is to factorize the equality; 2 x2−8 x=2x ( x−4 )=x ( x−4 )≤0. When we have the inequality, then x=0 and x=4. Taking these two points as the boundary points of three sections of the real number set, we determine the section(s) that will satisfy the inequality. In (−∞,0), pick x=−1. When x=−1 , then −1 (−1−4 )=5≤0 , false . Thus, no point in (−∞,0) can satisfy the inequality. In (0,4 ) , take x=1, then 1 (1−4 )=−3≤0 , true . Thus, all points in (0,4 )satisfy the inequality. In (4 ,∞), take x=5, then 5 (5−4 )=5≤0 , false . Thus, no point in (4 ,∞) satisfy the inequality. When x=0 or x=4 the inequality is satisfied.

If you chose B, you would have thought that the lower boundary of the solution set is not part of the solution. If you chose C, you would have thought that (0,4) is not a solution set. If you chose D, you may have thought that 3−4=1 rather that being −1.If you chose E, you would have thought that the lower boundary of the solution set is not part of the solution.

29. What is the possible value of a that will make the equation a x2+3=6 x have two distinct real solutions?

A. 2B. 3C. 4D. 12E. ∞

Question 29. Correct answer is A.A quadratic equation has two distinct roots if the discriminant b2−4 ac>0. In the equation above (whose equivalent is a x2−6 x+3=0), a=a ,b=−6 and c=3 . Thus, b2−4 ac=36−4 ( a ) (3 )=36−12a>0. Diving through by 12 gives us 3−a>0 which implies −a>−3 or a<3. If you chose B, you would have not paid attention to the key word “distinct”. If you chose C, you would have thought that solutions are distinct when the discriminant is less than zero. If you chose D, you would have thought that the factor between the two terms of the discriminant gives the required results. If you chose E, you would have thought that no such real number exists.

30. The cost of three sets of a 3 piece cookie sheet and two tea pots is $57. Three similar tea pots and four sets of 3piece cookie sheet costs $82. Hannar would like to buy two such tea pots and a set of 3 piece cookie sheet. How much will she spend?

A. $7

B. $18C. $25D. $32E. $43

Question 30. Correct answer is E.Taking x and y to be the price of a three piece cookie sheet and one tea pot respectively, we have 3 x+2 y=57 and 4 x+3 y=82. Multiplying the first equation by 3 and the second by 2 yields 9 x+6 y=171 and 8 x+6 y=164 whose difference yields x=7. From the first equation, 2 y=57−3 x=57−3 (7 )=36. Thus, 2 y=36 implies y=18. A 3 piece cookie sheet and two tea points would cost7+2 (18 )=$ 43.

If you chose A, you would have provided the price of a set of cookie sheet without finding out what the examiner actually wants. If you chose B, you would have provided the price of a tea pot without finding out what the examiner actually wants. If you chose C, you would have provided sum of the price of a tea pot and one set of cookie sheet without finding out what the examiner actually wants. If you chose D, you would have provided sum of the price of a tea pot and two sets of cookie sheet without finding out what the examiner actually wants.

31. The fourth and fifth terms of an arithmetic sequence is 174 and 3 respectively. What would be

the first term of the sequence?

A. -8

B.−54

C. −2

D.54

E. 8


Denoting an by the nth term of an arithmetic sequence, a4=174 and a5=3. The common

difference d=a5−a4=3−174

=−54

. The fifth term a5=a+(5−1 ) d=3 where a is the first

term. Substituting for d equals a+4d=a−5=3 ,hence a=8.

If you chose A, you would have erred in determining the value of a in a−5=3. If you chose B, you would have provided the common difference as the first term. If you chose C, you would

have erred in determining the common difference by getting 54 and proceeded to get the first

term. If you chose D, you would have provided an incorrect common difference, due to mathematical error, as the first term.

32. Given that x=−2 what would be the value of |4−3x− x−12 |−2x?

A.72

B.152

C. S

D.252

E.252

F.312


The first step is to carry out substation; |4−3(−2)−(−2)−12 |−2 (−2 ).

|4−3(−2)−(−2)−12 |−2 (2 )=|4+6+ 32|+4=|232 |+4. The absolute value implies the positive

of a number hence|232 |+4=232 +4=312 .

If you chose A, you would have treated the operation −3 x=−3(−2) as a subtraction operation rather than a multiplication one. If you chose B, you would have substituted x outside the absolute sign by 2 instead of −2. If you chose C, you would have ignored the negative sign before the fraction. If you chose D, you would have simplified (−2)−1 in the numerator of the fraction to get −1 instead of -3. If

33. The function f is defined by f ( x )=12x2−4 x+2. What is the smallest value of x for which f (x)

is equal to 12.

A. -10B. -2C. 2D. 10E. 26


Solving the problem f ( x )=12 leads to 12x2−4 x+2=12 which is equivalent to x2−8 x−20=0

. Expanding and factorizing the equation gives x2−10 x+2x−20=( x−10 ) ( x+2 )=0. Therefore, x−10=0 or x+2=0 implying that x=10 or x=−2. The smallest value of x is −2.If you chose A, you would have erroneously simplified x−10=0 to get −10. If you chose 2, you would have erroneously factorized the problem to get x=2 and x=10 as the solutions. If you chose D, you would have simply chose the highest value thinking that a positive value is required. If you chose E, you would have substituted 12 into the expression.

Coordinate Geometry - 9 questions

34. A line l1 is parallel to a line whose equation is y−2=x . If l2 intersects the line y=2, at what x−¿cordinate does this happen if the line passes through (−1 ,−4)?A. -4B. -1C. 0D. 5E. 7

Question 34. Correct answer is DThe slope-intercept form of equation of y−2=x is y=x+2, hence its slope is the coefficient of x, that is, 1. The slope of l1 is 1 because parallel lines have the same slope. If l1 passes through(−1 ,−4), then y+4x+1

=1Implying that y+4=x+1. Wheny=2, x+1=2+4 hencex=5.

If you chose A, you would have used −1, the slope of the perpendicular line, to find the equation of l2 then proceeded to get the value of x . If you chose B, you would have determined

the equation of l2 erroneously as y−4x−1

=1 then proceeded to get the value of x . If you chose C,

you would have substituted 2 for y in the equation of l1. If you chose E, you would have erred in simplifying the equation, x+1=2+4.

35. What is the shortest distance from point A(−4−1) to the circle whose equation is x2+ y2−4 x+2 y−4=0?

A. NoneB. 0C. √2D. 3E. 9

Question 35. Correct answer is D

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

The first step is to get the standard form of the equation of the circle. By completing square x2+ y2−4 x+2 y−4=0 is equivalent to x2−4 x+22+ y2+2 y+1=4+22+1=9. Factorizing the equation equals ( x−2 )2+( y+1 )2=32. The circle is of radius 3 units and center (2 ,−1). Sketch the circle and the point.The shortest distance between the two is 3 units.

If you chose A, you would have located the center of your circle at (−2,1) that than (2 ,−1). If you chose C, you would have located point A at (−1 ,−4). If you chose B, you would have drawn a circle of radius 6 rather than 3 units. If you chose E, you would have determine the distance between A and the furthest point of the circle.

36. How many times does the graph of x=5 intersects that of y=12x2+2x−3?

4

3

2

1

y

A

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

A. 0B. 1C. 2D. 3E. 4


We substitute for x into the function y=12x2+2 x−3 to find out the number of the

corresponding y values the equation has. When x=5, y=12(5)2+2 (5 )−3=39

2. Therefore, the

line and the curve intersects one only.

If you chose A, you would have erred in your drawing by drawing a horizontal line below the parabola. If you chose C, you would have substituted for y by 5 instead of substituting for x . If you chose D, you would have identified the number of times the curve intersects the main axes. If you chose E, you would have identified the total number of times the curve intersects the main axes and the line x=5.

37. How far is point (6 ,−1) from (−2,3)?

A. 4 √3 unitsB. 4 √5 unitsC. 12 unitsD. 16√3 units

E. 16√5 units

Question 37. Correct answer is BThe distance between (x1 , x2) and ( y1 , y2) is d=√(x2−x1 )2+( y2− y1 )

2. Taking

(x1 , x2 )=(6 ,−1) and ( y1 , y2)=(−2,3),

d=√(−2−6 )2+(3−−1 )2=√64+16=√80=√16×5=4√5.

If you chose A, you would have applied the incorrect formula d=√(x2−x1 )2−( y2− y1 )2. If you chose C, you would have used the incorrect interpretation of square root of the sum of two numbers √64+16=√64+√16. If you chose D, you would have applied the incorrect formula

d=√(x2−x1 )2−( y2− y1 )2 then erred in determining the roots of 48. If you chose E, you would have erred in simplifying the root of 80.

38. A triangle ABC is has its vertices at A (1 ,−3 ) , B (1,5) and C (−3,2). In which ratio does a line, whose equation is y+x+1=0, divide the triangle?

A. 1 : 7

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

B. 1 : 8C. 1 : 9D. 7 : 8E. 8 : 9

Question 38. Correct answer is APlot the triangle and the line on the same axes. Label the intersections where necessary. The line y+x+1=0 which is equivalent to y=−x−1 passes through (−3,2) and (0 ,−1 ) . Let the intersection of the line AB be D. The bases of ADC and ADB are 1 unit and 7 units respectively

while its heights are 4 units each. The area of ADC is 12×1×4=2 sq .units while that of ADC is

12×7×4=14 sq . units . The ratio of the area ADC : ADB=2:14=1 :7.

D

B

A

C

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

If you chose B, you would have find the ratio of the smaller portion to the whole. If you chose C, you would have incorrectly plotted the points such that AB became a 9 unit line segment. If you chose D, you would have expressed the larger portion to the whole. If you chose E, you would have incorrectly plotted the points such that AB became a 9 unit line segment and expressed the larger portion to the whole.

39. Which kind of the transformation does the function y=2x2+ x+1undergoes to get

y=12x2+ x

2+1?

A. Horizontal compression of factor 12

B. Horizontal stretch of factor 12

C. Vertical compression of factor 12

D. Vertical stretch of factor 12

E. Horizontal left shift of 12 a unit

Question 39. Correct answer is B

Relate y=12x2+ x

2+1 and y=2x2+ x+1. Comparing the first terms, 2is multiplied by

14 to get

12 while in the second term, 1is multiplied by

12 to get

12. Since the unknown in the second term

is x and that of the first term is x2, it implies that 12 IS a factor that is tied to x such that when

the term is squared, the factor is squared as well. Thus

y=12x2+ x

2+1=2( 1

2x)2

+1( x2 )+1=f (12 x). This is a horizontal stretch of factor 12.

If you chose A, you would have thought that a compression, in all cases, must have a factor between 0 and 1. If you chose C, you would have thought that thought that vertical compressions is associated with variable x and its factor is between 0 and 1. If you chose D, you would have thought that vertical stretch is associated with variable x and not f (x). If you chose

E, you would have thought that the transformation f ( 1a x ), a>1 is a shift.

40. What would be the y intercept for the function f given by 4 y2−20=−5 x2?A. −20B. −¿5C. −4D. −√5E. 4

Question 40. Correct answer is DAt the y−¿intercept, x=0, hence, the equation becomes 4 y2−20=0. 4 y2−20=0 is equivalent to y2=5 which implies that y=±√5 .

If you chose A, you would have thought that the intercept is the constant term in the equation. If you chose B, you would have thought that the intercept is the constant term in the equation when the coefficient of y2is 1. If you chose C, you thought that required value is the negative intercept ofx. If you chose E, you would have misread the question then determined the x intercept rather than the y intercept.

41. Point S(2,1) is the midpoint of the line segment OC . If C is the midpoint of OF where O is the origin, what would be the coordinates of F?A. (1,0.5)B. (4,2)C. (4,8)D. (8,2)E. (8,4)

Question 41. Correct answer is E

The midpoint formula is ( x2+x12

,y2+ y12 ) where (x1 , x2) and ( y1 , y2) are the endpoints. Given

the midpoint(m1 ,m2) and (x1 , x2 )=(0,0) the endpoint

( y1 , y2)=(2m1−x1 ,2m2−x2 )=(2m1 .2m2 ) . S(2,1) is the midpoint of (0,0) and C hence the

coordinates of C are (2 (2 ) ,2 (1 ) )=(4,2). C (4,2) is the midpoint (0,0) and F hence, the coordinates of F are F (2 (4 ) ,2 (2 ) )=(8,4 ).

If you chose A, you would have treated S as an endpoint then determine the midpoint of OS . If you chose B, you would have applied the midpoint formula once instead of doing it twice. If you

chose C, you would have interchanged the roles of x and y coordinates unknowingly. If chose D, you would have erred in applying the midpoint rule to the second coordinate.

42. P is a polynomials given by P ( x )=(2x−1 ) ( x−1 )2 ( x+1 )2 . How many distinct roots does the polynomial have?A. 1B. 2C. 3D. 4E. 5

Question 42. Correct answer is CThe roots of the polynomials are x values such that P ( x )=(2x−1 ) ( x−1 )2 ( x+1 )2=0. The equation implies 2 x−1=0 , ( x−1 )2=0 and ( x+1 )2=0. Solving for x equals

x=12, x=1 , x=1, x=−1 and x=−1. Two roots are repeated, hence, the distinct ones are

three; 12,1 and −1.

If you chose A, you would have thought that the required number is that of non-repeated roots. If you chose B, you would have thought that the required number is that of unique repeated roots. If you chose D, you would have thought that the required number is that of the total repeated roots. If you chose E, you would have thought that the required number is that of all roots.

Plane Geometry - 14 questions

43. What is the size of the supplement of angle MQN ?

Q

PN

M

L

143 °136 °

A. 9 °B. 71 °C. 109 °D. 81 °E. 99 °

Question 43. Correct answer is DThe supplement is angle MQN and its the exterior angle. Let the angle be x. x+143+136=360 because the sum of exterior angles of a polygon add up to360 ° . Therefore, x=360−143−136=81°.

If you chose A, you would have provided the complement of the supplement of angle MQN . If you chose B, you would have erred in determining the value of angle MQN which would have led to an error in its supplement. If you chose C, you would have erred in determining the value of angle MQN then proceeded to provide the angle as the required one. If you chose E, you would have provided the value of angle MQN and not its supplement.

44. In the figure below, PQRS is a parallelogram. What is the size of angle VTQ?

A. 33°B. 35 °C. 68 °D. 112°E. 145°

Question 44. Correct answer is B

V

U

T

S R

QP

68 °

33 °

∠TQS=∠QSR=33 ° because they are alternate angles. ∠TUQ+∠QUV =∠TUQ+68=180 because they are angles on a straight line. Therefore, ∠TUQ=180−68=112°. ∠VTQ=180−33−112=35° because interior angles of a triangle add up to 180 °.

If you chose A, you would have thought that triangle TUQ is an Isosceles triangle since TU and UQ appears to be equal. If you chose C, you would have thought that the angle together with ∠VUQare corresponding angles. If you chose D, you would have provided the size of angle TUQ rather than that of UTQ. If you chose E, you would have provided the size of the complement of angle VTQ and not the angle itself.

45. The exterior angle of a regular polygon is18 °. How many sides does it have?A. noneB. 5C. 10D. 20E. 40

Question 45. Correct answer is DGiven that n is the number of sides of aregular polygon whose exterior angle is A, then

nA=360. Substitution A by 18 ° gives n=36018

=20. The polygon has 20 sides.

If you chose A, you considered the angle given as an interior angle. If you chose B, considered the complement of the angle given as the exterior angle. If you chose C, you used the incorrect formula, nA=180. If you chose E, you used the incorrect formula, nA=720.

46. What is the perimeter of the following parallelogram?

A. 43 in B. 48 inC. 51 inD. 140 inE. 161 in

Question 46. Correct answer is CThe first step is to determine the size of the slant height. By the properties of parallelograms, the smallest angle is equal to 60 °.

10 in

14∈¿

120 °

10 in

60 °

10 in

120 °

Taking l to be the slant height, then sin 60=0.866=10x and x= 100.866

=11.5 in. Therefore,

the perimeter is 14+11.5+14+11.5=51∈.

If you chose A, you would have used 10∈¿ as a base instead of 14∈¿. If you chose B, you would have 10∈¿ as the slant height instead of 11.5∈¿. In you chose D, you would have determined the area instead of determining the perimeter. If you chose E, you would have determined the area using one wrong value, 11.5∈¿ instead of determining the perimeter.

47. Eight cubes of side 6 in are arranged in two different ways. In the first arrangement, they form a cuboid when arranged in a linear form. In the second arrangement, they form a cube with equal number of square sides along the length, width and height. What is the ratio of the surface area of the cuboid to the new cube formed? A. 12 :13B. 12 :17C. 1 :1D. 13 :12E. 17 :12

Question 47. Correct answer is E.In the first form, the figure is a cuboid whose length is (6×8 )=48∈¿, width is 6∈¿ and height is 6∈¿. Therefore, its surface area is 2 (lw )+2 ( lh )+2 (wh )=2 (48×6 )+2 (48×6 )+2 (6×6 )=576+576+72=1224 i n2. In the first form, the figure is a cube whose length is (6×2 )=12∈¿. Therefore, its surface area is

6 (l2 )=6 (122 )=864 i n2. The ratio of surface areas is 1224864

=1712

=17 :12.

If you chose A, you would have used a cuboid of 6 smaller cubes instead of 8 then proceeded to find the ratio of the surface area of the larger cube to that of the cuboid. If you chose B, you would have determined the ratio of the surface area of the larger cube to that of the cuboid. If you chose C, you would have determined the ratio of the volumes. If you chose D, you would have used a cuboid of 6 smaller cubes instead of 8.

48. What is the supplement of the complement of 58 °?

A. −32°B. 32 °C. 122 °D. 148 °E. 328 °

Question 48. Correct answer is D.The complement of 58 is 90−58=32 °. The supplement of this complement is 180−32=148 °.

If you chose A, you would have determined the supplement of the angle then the complement of the result. If you chose B, you would have provided the complement of the angle as the required value. If you chose C, you would have provided the supplement of the angle as the required value. If you use E, you would have determined the complement of the angle then subtracted the result from 360 ° .

49. What is the image of A (−3,1 ) when rotated through negative 90 ° about(1 ,−1)?

A. (−5 ,−1)B. (−1 ,−5)C. (−1,3)D. (2,3)

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

E. (3,2)

Question 49. Correct answer is D.Plotting the pre-image and the center of rotation then rotating the figure gives us(2,3).

If you chose A, you would have rotated the point through positive 90 ° then interchanges the coordinates of the image. If you chose B, you would have rotated the point through positive90 °. If you chose C, you would have rotated the point about (−1,1) not(1 ,−1). If you chose E, you would have interchanges the coordinates of the image.

50. A cuboid measures 6 in by 8 in by 10 in. What would be the distance between two vertices that are the furthest from one another as compared to any other two?A. 10 inB. 2√41∈¿C. 10√2∈¿D. 24∈¿E. 100√2∈¿

Question 50. Correct answer is C.The distance between the vertices that are furthest from each other is the length of the diagonal joining them. The length of this diagonal is √62+82+102=√200=10√2∈¿.

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

If you chose A, you would have determined the length of the smallest diagonal on a surface. If you chose B, you would have determined the length of the largest diagonal on a surface. If you chose D, would have simply added the length of the three mutually perpendicular axes joining the two points. If you chose E, you would have erroneously determined the roots of 200.

51. AC and BD are two diagonals of a kite intersecting at O. What would be the least measure of angle COD?A. 45 °B. 60 °C. 90 °D. 180 °E. 270 °

Question 51. Correct answer is C.Draw the kite. By properties of a kite, its diagonal are always perpendicular to each other.

The least measure of angle COD is 90 °.

If you chose A, you would have thought that the diagonals intersect at45 °. If you chose B, you would have thought that the diagonals intersect at 60°. If you chose D, you would have measured an angle of one of the straight angles at O. If you chose E, you would have measured the reflex angle COD.

52. The vertices of a parallelogram ABCD are A (1 ,−2 ) ,B (3 ,−2 ) ,C (r , s) and D(3,1). The figure is reflected about line y=x−2. What would be the new value of r after reflection?

O

D

C

B

A

4

3

2

1

y

A. (1,1)B. (3,-1)C. (3,3)D. (4,2)E. (5,1)

Question 52. Correct answer is C.Plot the points and locate C. Draw line y=x−2 which passes through (0 ,−2) and (2,0).

From the three points, it is clear that the figure is a parallelogram and not a rectangle since C must come before D. With this in mind, we locate point C at (5,1). The image of C is connected to C by a line perpendicular to the mirror line. Further, since the image C ' and the pre-image must be at equal distance from the mirror line, C ' must be at (3,3 ) .

C’

C

-4

-3

-2

-1

4

3

2

1

y

-4 -3 -2 -1 0 1 2 3 4 5 6 x

If you chose A, you would have considered the figure to be a rectangle and provided the coordinates of C are the required coordinates. If you chose B, you would have considered the figure to be a rectangle. If you chose D, you would have located the image on the mirror line rather than placing it on the opposite side. If you chose E, you would have provided the coordinates of C as the required coordinates.

53. In the figure below, MR and OQ are parallel lines. What is the measure of ∠NPQ?

A. 42 °B. 43 °C. 95 °D. 127 °E. 137 °

Question 53. Correct answer is E.∠OPN=180−42−95=43° because interior angles of a triangle add up to 180 ° . ∠NPQ=180−43=137 ° because angles on a straight line add up to 180° .

If you chose A, you would have thought that the angle together with ONP are alternate angles. If you chose B, you would have provided the measure of angle OPN, its supplement. If you chose C, you would have thought that the angle together with NOP are corresponding angles. If you chose D, you would have erred in subtracting 43 from 180 °.

54. A spherical solid of radius 8 in is melted and cast into a conical solid of the same radius. What is the surface area of the cone?A. (64+32√17 ) π in2

R Q

P

O

N

M

95 °

42 °

B. 320π i n2

C. (64+64√15 )π i n2 D. (64+64√17 )π i n2

E.20483

π

Question 54. Correct answer is D.The volume of the cone is equal to that of the sphere. Taking r and h to be radius and height of

the cone, the previous statement implies that 43

πr 3=13

π r2×h which is equivalent to 4 r=h.

Since r=8 , h=4×8=32∈. The surface area of the cone is π r2+πrl where l is the slant height. The slant height, the height and the radius makes a right triangle with l as the hypotenuse, hence l=√322+82=√1088=√64×17=8√17∈. The surface area is

π r2+πrl=(r2+rl ) π=(82+(8×8√17 ) ) π= (64+64 √17 )π in2 .

If you chose A, you would have used an incorrect slant height, 4 √17∈. If you chose B, you would have used an incorrect formula for slant height, l=√322−82. If you chose C, you would have determined the surface area using the perpendicular height. If you chose E, you would have determined the volume.

55. The ends of a 60 ° sector of radius 12 in are joined together to form a cone. What s the diameter of the cone?A. 1 inB. 2 inC. 4 inD. 8 inE. 24 in


The area of the sector = curved surface area of the cone, that is 60360

×122π=πr×12since the

slant height is the radius of the sector. The equation reduces to 16×12=2=r . Thus, the radius

of the cone is 2∈¿and its diameter is 4∈.

If you chose A, you would have divided instead of multiplying the radius by 2. If you chose B, you would have provided the measure of the radius as the diameter. If you chose D, you would have erroneously got 4∈¿ are your radius. If you chose E, you would have assumed that the cone will have the same radius as the sector.

56. The figure below shows an isosceles trapezoid. What is its perimeter?

A. 42+6√2∈¿B. 36+12√2∈¿C. 42+12√2∈¿D. 126∈¿E. 126√2∈¿

Question 56. Correct answer is C.The first step is to find the base of the smaller triangles at both ends. By the Pythagorean

theorem, the base = √ (6 (√2 ) )2−62=√72−36=√36=¿ . The longest side has a length of

15+6+6=27∈¿. The perimeter is 27+6 √2+15+6 √2=42+12√2∈¿

If you chose C, you would have added 6 to 15 once to get the length of the longest side. If you chose B, you would have added 6√2 once in the perimeter. If you chose D, you would have determined the area of the figure. If you chose E, you would have determined the area of the figure using the slant side.

Trigonometry - 4 questions

57. Which of the following is not a solution of the function4 cos2 x+4cos x+1=0?A. 30 °B. 60 °C. 210 °D. 240 °E. 300 °

Question 57. Correct answer is D.The equation is quadratic in cos x, in fact, it is a perfect square, a2+2ab+b2=( a+b )2 with a=2cos x and b=1, hence 4 cos2 x+4cos x+1= (2cos x+1 )2=0 implying that 2cos x+1=0.

The equation 2cos x+1=0 implies that cos x=−12 hence, x=120 ° or x=240 °.

6∈¿ in6√2∈¿

15 in

If you chose A, you would have got cos x=12 and thought that the cosine of 30 degree is 12. If

you chose B, you would have forgot to consider the situation when the cosine is negative. If you

chose C, you would have thought that the cosine of 30 degree is 12. If you chose E, you would

have thought that cosine is negative in the fourth quadrant.

58. What is the measure of TU?

A. √3∈¿B. 2√3∈¿C. 3∈¿D. 4 √3∈¿E. 6√3∈¿

Question 58. Correct answer is B.The fact that UTS is a right triangle, ∠UTSmust be equal to 60 ° since , ∠TSV=30 ° . This implies that triangle UTS is a 30-60-90 whose ratio of sides short leg : longer leg : hypotenuse =

x :2 x :√3 x. The fact that 6=√3x , x= 6√3

= 6√33

=2√3∈.

If you chose A, you would have rationalized the answer incorrectly. If you chose C, you would have treated TS as the longer leg. If you chose D, you would have thought that the question asks one to provide the length of the longer leg. If you chose E, you would have treated TS as the shorter leg and thought that the questions requires one to provide the measure of the hypotenuse.

59. A scientist observed the behaviors of ocean tides during high tight and modelled a function, f , to represent their motion.

6 in

U

V

ST

30 °

f=9cos (8 t+ π2 )

If distance was measured in meters, what would be the distance between the highest level and the lowest level of the tide?A. 4.5 mB. 8 mC. 9 mD. 16 mE. 18 m

Question 59. Correct answer is E.The distance between the highest level and the lowest level of the tide is twice the amplitude, 2 λ. Since the amplitude, the coefficient of the cosine function, is 9, 2 λ=18m.

If you chose A, you would have thought that the distance to be measured is that of half the amplitude. If you chose B, you would have thought that it is the amplitude that is required then gave the value of the angular velocity as the amplitude. If chose C, you would have thought that it is the amplitude that is required. If you chose D, you would have thought that the value of the angular velocity as the amplitude.

60. If cos a=0.8, what is tan(90−a)?

A. 0.6

B. 34C. 0.8

D. 43

E. 53

Question 60. Correct answer is D .

If cos a=0.8= 810 , then the hypotenuse = 10 units, the adjacent side = 8 units and the opposite

side is √102−82=√100−64=√36=6 units. The opposite of 90−a is 8 units and its adjacent

is 6 units hence tan(90−a)=86=43.

If you chose A, you would have determined the cosine and not the tangent of the angle. If you chose B, you would have determined the tangent of a and not of 90−a. If you chose C, you would have determined the sine and not the tangent of the angle. If you chose E, you would have determined the secant and not the tangent of the angle.

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