Post on 03-Jan-2022
Kotebe Metropolitan University
Department of Mathematics
Solution Manual for Ethiopian University
Entrance Examination (EUEE) Mathematics
for Social Science GINBOT 2010 & 2011/May
2018 & 2019
Prepared By:
Dr. Tsegaye Simon
Dr. Margie Balcha
Mr. Temesgen Debas
ETHIOPIAN UNIVERSITY ENTRANCE EXAMINATION (EUEE)
MATHEMATICS FOR SOCIAL SCIENCE
GINBOT 2010/May 2018
SOLUTION MANUAL
1. Which one of the following is true?
A) If the mean, mode and median of a distribution are 7, 5.5 and 6, respectively, then the
distribution is negatively skewed.
B) If the mean, median and standard deviation of a distribution are 4, 6 and 2, respectively,
then the distribution is positively skewed.
C) If a distribution is negatively skewed, then the mean of the distribution is greater than its
second quartile.
D) If the first, second and third quartiles of a distribution are 7, 10 and 13, respectively, then
the distribution is symmetrical.
Solution: Recall that the mean is pulled in the direction of skewness, that is, in the direction
of the extreme observations. For a right-skewed distribution, the mean is greater than the
median; for a symmetric distribution, the mean and the median are equal; and for a left-
skewed distribution, the mean is less than the median. The first quartile, Q1, is the number
that divides the bottom 25% of the data from the top 75%; the second quartile, Q2, is the
median, which is the number that divides the bottom 50% of the data from the top 50%; and
the third quartile, Q3, is the number that divides the bottom 75% of the data from the top
25%. Note that the first and third quartiles are the 25th and 75th percentiles, respectively. In
a symmetric distribution, Q3-Q2=Q2-Q1.
Answer: D
2. A hospital wants to buy three π-ray machines from a supplier. Suppose the price of each
machine including 15% VAT is Birr 161,000. If the hospital wants to subtract a 2%
withholding tax before VAT, what is the amount the hospital should pay (in Birr) to the
supplier after withholding 2% is subtracted?
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A) 474,600 C) 473,340
B) 475,340 D) 483,340
Solution: Let π₯ be the price of each π β-ray machine. Then, π₯ + 15%π₯ = 1.15π₯ =
161,000 βΉ π₯ = 140,000, which is the price of each π βray machine before vat. The
withholding tax before vat is 140,000 β 0.02π₯140,000 = (1 β 0.02)140,000 = 137,200.
After 15% VAT, the hospital should pay 137,200 + 15%(137,200) = 1.15(137,200) =
157, 780 for each π βray machine. The hospital should pay 3 Γ 157, 780 = 473,340 birr
for three π βray machines.
Answer: C
3. A firm deals with two kinds of fruit juices-pineapple and orange juice. These are mixed and
two types of mixtures are obtained, which are sold as soft drinks π΄ and π΅. One tin of π΄
needs 4 liters pineapple juice and 1 liter of orange juice. One tin of π΅ needs 2 liters of
pineapple and 3 liters of orange juice. The firm has only 46 liters of pineapple juice and 24
liters of orange juice. Each tin of π΄ and π΅ is sold at a profit of 4 birr and 3 birr, respectively.
How many tins of π΄ and π΅ should the firm produce to maximize its profit?
A) 6 tins of π΄ and 9 tins of π΅ C) 9 tins of π΄ and 5 tins of π΅
B) 9 tins of π΄ and 6 tins of π΅ D) 5 tins of π΄ and 9 tins of π΅
Solution: . Let π₯ and π¦ be the number of tins of type A and B respectively.
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Maximize π = 4π₯ + 3π¦
Subject to
4π₯ + 2π¦ β€ 46
π₯ + 3π¦ β€ 24
π₯, π¦ β₯ 0
By graphical method, the region of the constrained condition is as shown in the fig below.
Evaluating π§ = 4π₯ + 3π¦ at the vertices of the region, we see that the maximum value is
4(9) + 3(5) = 51.
Answer: C
4. The weights (in kg) of children in certain nursery are grouped into four class intervals of
equal width and represented by the following frequency polygon.
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Which one of the following is true about the data?
A) The mode of the weights is 20 kg.
B) The distribution of the weight is positively skewed.
C) The median weight is 25 kg.
D) The mean weight is 25 kg.
Solution: From the frequency polygon, we see that the weights of the children is grouped as
Weights of children 5-15 15-25 25-35 35-45 Total
Class mark (π₯π) 10 20 30 40
Number of children (ππ) 10 20 15 5 50
Cumulative frequency 10 30 45 50
π₯πππ 100 400 450 200 1150
The mean of the weights is οΏ½Μ οΏ½ =β π₯πππ
4π=1
β ππ4π=1
=1150
50= 23ππ.
The median of the weights is = πΏ +π
2βπΆππ
ππΓ π€ where πΏ is the lower class boundary of the
median class, π = β ππ4π=1 , πΆππ is the cumulative frequency of groups before the median
class, ππ is the frequency of the median class and π€ is class interval.
That is, the median group is 15-25. The median is then 15 +25β10
20Γ 10 = 22.5ππ. The
formula for calculating the mode of a grouped frequency distribution is given by ππππ =
πΏ +π
π+πΓ β
Where πΏ is the lower class boundary of the modal class, π difference between modal
frequency and the frequency above it, π difference between modal frequency and the
frequency below it, β class interval. The modal class is 15-25.
The mode of the weights is 15 +20β10
(20β10)+(20β15)Γ 10 = 21.67ππ.
Since the mean is greater than the median and the median is greater than the mode, the
distribution is positively skewed.
Answer: B
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5. A man bought a one bed room house for birr 250,000 and paid a down payment of 20%. The
remaining is a mortgage to be paid monthly for 15 years with 6% annual interest. What is
the monthly payment in Birr?
[Given: (1.005)-15
= 0.93, (1.005)-180
= 0.41, (1.006)-15
= 0.91, (1.006)-180
= 0.34]
A) 1694
B) 1428
C) 1333
D) 1818
Solution: The formula for calculating the payment amount is given by π¨ = π· (π
πβ(π+π)βπ)
where π΄ is payment amount per period, π initial loan, π interest rate per period and π total
number of payments.
As 20% of the price of the house is already paid, π = 0.8 Γ 250,000 = 200,000. π =
0.06
12= 0.005 and π = 12 Γ 15 = 180. Thus, π΄ = 200,000 Γ
0.005
1β(1+0.005)β180 =
200,000 Γ0.005
1β0.41= 200,000 Γ
0.005
0.59β 1694.
Answer: A
6. In the previous month, the regular prices of sugar and salt were Birr 12 and 8 per kg,
respectively. This month, the price of sugar increased by 20% and the price of salt decreased
by 30%. If a family must buy 5 kg of sugar and 3 kg of salt every month, how much more or
less does the family need to pay for the two items this month compared to their previous
month bill?
A) Birr 4.80 less
B) The same as in the previous month
C) Birr 8.40 more
D) Birr 4.80 more
Solution: Let π₯ be the price of sugar per kg and π¦ the price of salt per kg. In the previous
month, π₯ = 12 Birr and π¦ = 8 Birr. In this month, π₯ = 12 + 20%(12) = (1 + 0.2)12 =
14.4 Birr and π¦ = 8 β 30%(8) = (1 β 0.3)(8) = (0.7)(8) = 5.6 Birr.
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The family needed to pay 5(12)+3(8)=84 Birr in the previous month and need to pay
5π₯14.4 + 3π₯5.6 = 88.80 Birr this month. Thus, The family need 88.80 β 84 = 4.80
more in this month.
Answer: D
7. The grouped frequency distribution of a data is given in the following table.
Class interval 8 β 12 13-17 18-22 23-27 28-32
Frequency 4 8 10 5 3
What is its median(π), and the mean deviation (ππ·) about the median?
A) π = 20, ππ· = 4.7
B) π = 19, ππ· = 4.2
C) π = 19, ππ· = 4.7
D) π = 20, ππ· = 4.2
Solution: The class intervals given are not continuous. Convert it to continuous frequency
distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each
class interval.
Class boundaries π cumulative
frequency
Mid.value π₯π |π₯π β π| πi|π₯π β π|
7.5-12.5 4 4 10 8.67 34.68
12.5-17.5 8 12 15 3.67 29.36
17.5-22.5 10 22 20 1.33 13.3
22.5-27.5 5 27 25 6.33 31.65
27.5-32.5 3 30 30 11.33 33.99
Total 30 143
Since π/2 = 15, the class 17.5 β 22.5 is the median class.
π= πΏ + (π
2βπΆππ
ππ) π€ = 17.5 +
15β8
10π₯5 = 18.67 β 19
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The mean deviation about median = βπi|π₯πβπ|
π
ππ=1 =
143
30 = 4.7. That implies, π = 19 and ππ·
= 4.7.
Answer: B
8. What is the maximum value of π = 3π₯ β 2π¦,
Subject to βπ₯ + π¦ β€ 0
βπ₯ + 2π¦ β€ 1 ?
2π₯ β π¦ β€ 4
π₯ β₯ 0, π¦ β₯ 0
A) 5
B) 10
C) 8
D) 6
Solution: Using graphical method, the maximum value of π = 3π₯ β 2π¦ subject to the given
constraints attains at the vertices of the shaded polygon, see the fig below.
Since π(0,0) = 0, π(1,1) = 1, π(3,2) = 5 and π(2,0) = 6, the maximum value is 6 which
occurs at (2,0).
Answer: D
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9. The following histogram presents the number of HIV positive persons among five social
groups coded as ππΊ1, SG2, ππΊ3, ππΊ4 and ππΊ5, in certain town.
Moreover, the numbers of persons in each of these social groups ππΊ1, ππΊ2, ππΊ3, ππΊ4 and
ππΊ5 are 500, 2500, 4000, 5000 and 1200 respectively. If so, which of the following can be
deduced from the data?
A) ππΊ1 consists of the highest percentage of HIV positive persons followed by ππΊ5.
B) 3% of those in ππΊ4 are HIV positive.
C) ππΊ3 consists of the highest percentage of HIV positive persons.
D) 7.4% of the entire population of the five social groups are HIV positive.
Solution: The percentage of HIV positive in SG1 is 80
500Γ 100 =16%, in SG2 = 4%, in
SG3 =5%, in SG4 = 2% and in SG5 = 10%. Of the total population, 600
12700Γ 100 = 4.72%
is HIV positive.
Answer: A
10. A person is planning to make a regular monthly saving over the next five years in an
account that pays no interest. If his monthly income is Birr 4000, then what percentage of
his income should he save monthly so that his total saving will be Birr 26400 by the end of
5 years?
A) 9% B) 10% C) 12% D) 11%
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Solution: πΌ = πππ, where π-principal amount, π-number of periods and π-percentage of P.
Hence, 26400 = 4000 Γ 12 Γ 5 Γ π β π = ππ%.
Answer: D
11. Which one of the regions shaded in the figure below is the solution region of the following
system of inequalities? {
π₯ + π¦ β€ 3βπ₯ + 3π¦ β₯ 1π₯ β₯ 0, π¦ β₯ 0
A) S1 B) S3 C) S2 D) S4
Solution: Sketching the all the inequalities and taking the intersection, the region of the
solution is π2 as shown in the fig.
Answer: C
12. Birr 1000 is deposited in a saving account that pays 6% annual interest compounded
monthly. Which one of the following is the amount (in Birr) that will be in the account by
the end of 3 years?
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A) 1000 + 1000 Γ (1.005)36
B) 1000 Γ (1.005)36
C) 1000 Γ (1.05)36
D) 1000 + 1000 Γ (1.05)3
Solution: Because π΄ = π (1 + π
π)
ππ‘
= 1000(1 + 0.06
12)
36
= 1000(1 + 0.005)36.
Answer: B
13. A jacket discounted by 20% for holiday has a price tag of Birr 576. What is the amount of
discount?
(A) 144 Birr
(B) 154 Birr
(C) 135.8 Birr
(D) 115.2 Birr
Solution: Let the previous price of the jacket be π₯. Then π₯ β 20%(π₯) = π₯ β 0.2π₯ =
0.8π₯ = 576. That is π₯ =576
0.8= 720. Therefore, the discount = 720 β 576 = 144 Birr.
Answer: A
14. If π(π₯) =|π₯|
π₯ and π(π₯) =
π₯+2
π₯3β4π₯ , then what is the value of limπ₯ββ2(π(π₯) + π(π₯))?
A) β9
8 B) β C)
9
8 D)
β7
8
Solution: limπ₯ββ2(π(π₯) + π(π₯)) = limπ₯ββ2 (|π₯|
π₯+
π₯+2
π₯3β4π₯)
= limπ₯ββ2 (|π₯|
π₯) + limπ₯ββ2 (
π₯+2
π₯3β4π₯)
= limπ₯ββ2 (βπ₯
π₯) + limπ₯ββ2 (
π₯+2
π₯(π₯+2)(π₯β2)) as |π₯| = {
π₯, π₯ β₯ 0βπ₯, π₯ < 0
= limπ₯ββ2(β1) + limπ₯ββ2 (1
π₯(π₯β2)) = β1 +
1
β2(β2β2)= β1 +
1
8=
β7
8
Answer: D
15. Let = (1βπ
1+π)
18
. Then what is the value of z?
A) -1 B) π C) β π D) 1 β π
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Solution: π§ = (1βπ
1+π)
18
= ((1βπ)(1βπ)
(1+π)(1βπ))
18
= (1β2π+π2
1βπ2 )18
= (1β2πβ1
1+1)
18
= (β2π
2)
18
=
(βπ)18 = (β1)18(π)18 = (1)(π2)9 = (β1)9 = β1.
Answer: A
16. Let A be a 3x3 invertible matrix and B be any 3x3 matrix. If|π΄| = π , and |π΅| = π, then
which one of the following is NOT true?
A) |π΄ππ΄| = π2 B) |ππ΄| = π3|π΄|, for any π β π
C) |π΄β1π΅| = ππ D) If π = 0, then B is not invertible.
Solution: (A) |π΄ππ΄| = |π΄π||π΄| = π β π = π2( Property of determinant)
(B) |ππ΄| = π3|π΄| (Property of determinant)
(C) |π΄β1π΅| = |π΄β1||π΅| and |π΄β1π΄| = |π΄β1||π΄| = |πΌ| = 1 and |π΄β1| =1
π, hence
|π΄β1||π΅| =π
π. (D) If π = 0, then B is not invertible.
Answer: C
17. The time needed to type a sample of 8 business letters in an office is 7, 8, 6, 8, 9, 7, 5, 6
minutes. What are the mean(οΏ½Μ οΏ½) and the standard deviation(s) of the data in minute?
A) οΏ½Μ οΏ½ = 7, π = β1.5 C) οΏ½Μ οΏ½ = 8, π = β2
B) οΏ½Μ οΏ½ = 7, π = β2 D) οΏ½Μ οΏ½ = 8, π = β1.5
Solution: Number of sample =8 and Data:7,8,6,8,9,7,5,6
Mean = οΏ½Μ οΏ½ =7+8+6+8+9+7+5+6
8=
56
8= 7, and Standard Deviation(s)= β
β (π₯πβοΏ½Μ οΏ½)28π=1
π=
β(7β7)2+(8β7)2+(6β7)2+(8β7)2+(9β7)2+(7β7)2+(5β7)2+(6β7)2
8
= β0+1+1+1+4+0+4+1
8= β
12
8= β1.5
Answer: A
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18. A private college has 1000 students. 60% of these students are males. 45% of these
students pay their payment by credit card including 175 females. What is the probability
that the student is a male or a credit card user?
A) 0.675 B) 0.225 C) 0.325 D) 0.775
Solution: As 60% of 1000 are males, the number of males, π =60
100Γ 1000 = 600.
1000-600 =400 students are females. Let π₯ be the number of students who pay by credit
card. Then 45
100Γ 1000 = 450. The number of male students who pay their payment by
credit card is 450-175=275.
The probability that the student is a male or a credit card user is π(π ππ π₯) = π(π) +
π(π₯) β π(π β© π₯) =600
1000+
450
1000β
275
1000=
775
1000= 0.775.
Answer: D
19. Which one of the following is true about the graph of π(π₯) =2π₯3+2π₯2+3π₯
π₯2+π₯ ?
A) The vertical asymptote of the graph is only π₯ = β1 and its oblique asymptote is
π¦ = 2π₯.
B) The graph has y-intercept at(0,3).
C) The graph has at least one x- intercept.
D) The vertical asymptotes of the graph are at π₯ = 0 and π₯ = β1, but it has no horizontal
asymptote.
Solution: π(π₯) =2π₯3+2π₯2+3π₯
π₯2+π₯=
π₯(2π₯2+2π₯+3)
π₯(π₯+1)=
2π₯2+2π₯+3
π₯+1 , for π₯ β 0
The vertical asymptote is π₯ = β1, and
By long division 2π₯2 + 2π₯ + 3 Γ· π₯ + 1, y=2x is its oblique asymptote.
The graph has no y- intercept, as 0 is not in the domain of f and 2π₯2 + 2π₯ + 3 β 0 for
any x as π· = π2 β 4ππ = 22 β 4(2)(3) = β20 < 0(No solution).
Answer: A
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20. Let π(π₯) =3π₯+1
π₯β2 . Then what is the range of π(π₯)?
A) β\{2} B) β C) β\{3} D) β\ {β1
3}
Solution: range of π = β\{3} , as 3 =3π₯+1
π₯β2 , 3π₯ β 6 = 3π₯ + 1 βΊ β6 = 1(absurd!)
Answer: C
21. What is the value of β«1
π₯(πππ₯ + π₯2πβπ₯)ππ₯ ?
A) 1
2π₯2πππ₯ β (π₯ + 1)πβπ₯ + πΆ C)
1
2π₯2πππ₯ + (2 β π₯)πβπ₯ + πΆ
B) 1
2πππ₯2 β (π₯ + 1)πβπ₯ + πΆ D)
1
2πππ₯2 + (2 β π₯)πβπ₯ + πΆ
Solution:β«1
π₯(πππ₯ + π₯2πβπ₯)ππ₯ = β«
1
π₯πππ₯ππ₯ + β« π₯πβπ₯ππ₯ .
Using substitution: let π’ = πππ₯, ππ’ =1
π₯ππ₯ β β«
1
π₯πππ₯ππ₯ = β« π’ππ’ =
1
2π’2 + π1 =
1
2(πππ₯)2 + π1, and using integration by parts for β« π₯πβπ₯ππ₯, let π’ = π₯ and ππ£ = πβπ₯ππ₯ β
π£ = βπβπ₯, then
β« π₯πβπ₯ππ₯ = π’π£ β β« π£ππ’ = βπ₯ πβπ₯ β β« β πβπ₯ππ₯ = βπ₯πβπ₯ β πβπ₯ + π2.
Thus, β«1
π₯(πππ₯ + π₯2πβπ₯)ππ₯ =
1
2(πππ₯)2 + βπ₯πβπ₯ β πβπ₯ + πΆ =
1
2(πππ₯)2 β (π₯ + 1)πβπ₯ +
πΆ.
Answer: B
22. If π(π₯) =1
3π₯3 + ππ₯2 + ππ₯ + 5 has a local minimum value at π₯ = 1, then which one of
the following is true about the possible values of π and π?
A) π = 3, π = β2 C) π = β2π β 1, π < β1
B) π = β2π β 1, c any real number D) π = β2π β 1, π > β1
Solution: The function π(π₯) =1
3π₯3 + ππ₯2 + ππ₯ + 5 has local minimum at π₯ = 1.
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β πβ²(π₯) = π₯2 + 2ππ₯ + π β πβ²(1) = (1)2 + 2π(1) + π = 0
β π + 2π = β1 β π = β2π β 1 , πβ²β²(π₯) = 2π₯ + 2π β πβ²β²(1) = 2(1) + 2π >
0, ( by second derivative test)
β π > β1. Thus, π = β2π β 1 and π > β1.
Answer: D
23. Let {ππ} be a sequence
with π1 = π, π2 = π(π1) = π(π), π3 = π(π2) = π(π(π)) , β¦ . ππ+1
= π(ππ) where f is a continuous function. If limπββ ππ = 5 , what is the value of π(5)?
A) 5 B) 5π C) 5π+1 D) 1
Solution: limπββ ππ = 5 and ππ+1 = π(ππ) β limπββ ππ = limπββ ππ+1 =
limπββ π(ππ) = 5 = π(limπβΆβ ππ) since π is continuous.
= π(5) = 5.
Answer: A
24. If the sum of the first three consecutive terms of an arithmetic progression is {π΄π} with
π΄π > 0 for all n, is 9 and the sum of their squares is 35, then what is the sum ππ of the
first n terms?
A) π2 + 1 B) π2 β 1 C) π2 D) 2π2 + 1
Solution: Let π΄1, π΄2, π΄3 be three consecutive terms of an arithmetic progression {π΄π}
with π΄π > 0, for all n, let d be the common difference of the sequence:
π΄1 + π΄2 + π΄3 = 9 β π΄1 + π΄1 + π + π΄1 + 2π = 9 and π΄12 + π΄2
2 + π΄32 = 35
β 3π΄1 + 3π = 9 β π΄1 = 3 β π and (3 β π)2 + (3 β π + π)2 + (3 β π + 2π)2 = 35
β 9 β 6π + π2 + 9 + 9 + 6π + π2 = 35
β 27 + 2π2 = 35 β π2 = 4 β π = Β±2 β π = 2 as π΄π > 0 for all n.
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Thus π΄1 = 1, π΄2 = 3, π΄3 = 5, π΄π = π΄1 + (π β 1)π = 1 + (π β 1)2 = 2π β 1
ππ =π
2(π΄1 + π΄π) =
π
2(1 + 2π β 1) =
π
2(2π) = π2
Answer: C
25. Let A be a 3x3 matrix and |π΄| = β2. Then what is the value of |πππ(π΄)|?
A) -8 B) C) -1/2 D) 4
Solution: since A is square matrix and |π΄| = β2, then it has an inverse.
β π΄β1π΄ = πΌ β |π΄β1π΄| = |π΄β1||π΄| = 1 β |π΄β1| =β1
2 ,
but π΄β1 =1
|π΄|πππ(π΄) β πππ(π΄) = |π΄|π΄β1
β |ππππ΄| = |β2π΄β1| = (β2)3|π΄β1| = (β8) (β1
2) = 4
Answer: D
26. Let π΄ = {1,2,3,4,5,6,7}, π΅ = {7,8,9} and πΆ = {8,9,10}. If one of the numbers is deleted
randomly from each of these sets, what is the probability that all the three deleted
numbers are even or all are multiply of 3?
A) 4
5 B)
2
21 C)
1
9 D)
8
63
Solution: π΄ = {1,2,3,4,5,6,7}, π΅ = {7,8,9} and πΆ = {8,9,10}
ππ΄(πΈπ£ππ) =3
7 , ππ΅(πΈπ£ππ) =
1
3 , and ππΆ(πΈπ£ππ) =
2
3 ,
ππ΄(ππ’ππ‘ππππ ππ 3) =2
7 , ππ΅(ππ’ππ‘ππππ ππ 3) =
1
3 , and ππΆ(ππ’ππ‘ππππ ππ 3) =
1
3
Hence, π(πΈπ£ππ ππ ππ’ππ‘ππππ ππ 3) = π(πΈπ£ππ) + π(ππ’ππ‘ππππ ππ 3)
=3
7β
1
3β
2
3+
2
7β
1
3β
1
3=
2
21+
2
63=
8
63
Answer: D
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27. If a function f is differentiable at a, then what is the value of limπ₯βπ π(π₯)?
(A) π(π) (B) πβ(π) (C) πβ(π) β π(π) (D) 0
Solution: π(π₯) β π(π) =π(π₯)βπ(π)
π₯βπ(π₯ β π), taking the limit on both sides we have,
limπ₯βπ
(π(π₯) β π(π)) = limπ₯βπ
(π(π₯) β π(π)
π₯ β π) (π₯ β π) = πβ²(π) β 0 = 0 β lim
π₯βππ(π₯) = π(π)
Answer: A
28. Which one of the following is convergent sequence?
π΄) {1+2π
3π } B) {1
π+ sin (π)} C) {
1β3π
2π } D) {(β1)π
2}
Solution: limπββ {1+2π
3π } = limπββ
[1
3π + (2
3)
π
] = 0 + 0 = 0 and the rest choice have no
limit.
Answer: A
29. What is the value of β (1
πβ1β
1
π)20
π=2 ?
π΄) 17
20 B)
23
20 C)
21
20 D)
19
20
Solution: β (1
πβ1β
1
π)20
π=2 = (1
1β
1
2) + (
1
2β
1
3) + (
1
3β
1
4) + (
1
4β
1
5) + β― + (
1
19β
1
20) =
1 β1
20=
19
20 , as the remaining terms cancelled.
Answer: D
30. Let A and B be two events. Suppose that the probability that neither event occurs is 3
8 .
What is the probability that at-least one of the event occur?
π΄) 1
8 B)
5
8 C)
1
4 D)
3
4
Solution: π(π΄ππ΅)β² =3
8= 1 β π(π΄ππ΅) β π(π΄ππ΅) = 1 β
3
8=
5
8
Answer: B
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31. If the truth value of (πβΒ¬π) β [(π β¨ Β¬π β π)] is True, then which one of the following
must be True?
π΄) π B) π C) Β¬π D) Β¬π
Solution: (πβΒ¬π) β [(π β¨ Β¬π β π)] has truth value True means both have the same
truth values. And πβΒ¬π =False, π β¨ Β¬π=True, which implies π β¨ Β¬π β π = False, then r
= False, Β¬π=True
Answer: D
32. What is the maximum value of the function π(π₯) = π₯4 β 2π₯2 on [β2,1]?
A) 8 B) 12 C) 24 D) 40
Solution: π(π₯) = π₯4 β 2π₯2, πβ²(π₯) = 4π₯3 β 4π₯ = 0 β 4π₯(π₯2 β 1) = 0,
π₯ = β1, π₯ = 0, π₯ = 1 are critical numbers. Then evaluate the function at the critical and
end points of the interval, π(β1) = β1, π(0) = 0, π(1) = β1, π(β2) = 8 and
π(1) = β1. Hence the maximum is π(β2) = 8.
Answer: A
33. What are the greatest lower bound and least upper bound of the sequence {(β1)π (1 +
1
π)} , respectively?
π΄) β 2 and 2 B) β3
2 and 2 C) β2 and
3
2 D) β2 and β
3
2
Solution: {(β1)π (1 +1
π)} = {β2,
3
2, β
4
3,
5
4, β
6
5, β¦ . }, the lower bounds are βπ₯ β€ β2
and upper bounds are βπ₯ β₯3
2 . Hence, πππ = β2 and ππ’π = 3/2
Answer: C
34. Which one of the following is an equation of the circle whose endpoints of a diameter are
(0, β2) and (2,2)?
A) π₯2 + π¦2 = 4 C) (π₯ β 1)2 + π¦2 = 4
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B) π₯2 + π¦2 β 2π¦ β 4 = 0 D) π₯2 + π¦2 β 2π₯ β 4 = 0
Solution: Center of the circle = (0+2
2,
β2+2
2) = (1,0) and radius
= β(2 β 1)2 + (2 β 0)2 = β5 , the equation of the circle: (π₯ β 1)2 + π¦2 = 5 β π₯2 +
π¦2 β 2π₯ + 1 β 5 = 0 and π₯2 + π¦2 β 2π₯ β 4 = 0
Answer: D
35. What is the equation of the line that passes through (1,1) and is parallel to the line
3π¦ β π₯ = 1?
A) π₯ β 3π¦ + 2 = 0 C) 3π¦ β π₯ + 2 = 0
B) π₯ + 3π¦ = 4 D) 3π₯ β π¦ = 2
Solution: The line through (1,1) and parallel to 3π¦ β π₯ = 1, the slope of the required
line is π¦ =1
3 , as the two lines are parallel, the equation of the line is
π¦ β 1 =1
3(π₯ β 1) βΉ π₯ β 3π¦ + 2 = 0
Answer: A
36. Which one of the following is true about π ππππ’π, absolute value and greatest integer
functions?
(A) π ππ(π₯)=Β±|π₯|, for all π₯ β π (C) |π₯| = π₯π ππ(π₯), for all π₯ β π
(B) π ππ(π₯) β€ |π₯|, for all π₯ β€ 0 (D) π ππ(π₯) β€ |π₯|, for all π₯ β₯ 0
Solution: π ππ(π₯) = {1 , π₯ > 00 , π₯ = 0
β1 , π₯ < 0 β |π₯| = π₯π ππ(π₯), for all π₯ β π
Answer: C
37. What is the area of the triangle (in Sq. units) formed by the lines joining the vertex of the
parabola π₯2 = β36π¦ to the end points of the latus rectum?
A) 126 B) 261 C) 216 D) 162
Solution: π₯2 = β36π¦ β β4π = β36, π = 9 Focus(β, π Β± π) = (0, β9) and equation
of Latus rectum is π¦ = β9 and end point of latus rectum become: -
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π₯2 = β36(β9) = 324, π₯ = β18, π₯ = 18, (β18, β9) and (18, β9), then the length of
latus rectum is 36. Hence area of the triangle π΄ =1
2(9)(36) = 162
Answer: D
38. What is the partial fraction decomposition of π₯2+π₯+1
(π₯+2)(π₯2+1) ?
A) 3
5(π₯+2)+
2π₯+1
5(π₯2+1) (C)
2
5(π₯+2)+
3π₯+1
5(π₯2+1)
B) 5
3(π₯+2)+
2π₯+1
3(π₯2+1) (D)
2
3(π₯+2)+
2π₯+1
3(π₯2+1)
Solution: π₯2+π₯+1
(π₯+2)(π₯2+1)=
π΄
π₯+2+
π΅π₯+πΆ
(π₯2+1) β
π₯2+π₯+1
(π₯+2)(π₯2+1)=
π΄(π₯2+1)+(π₯+2)(π΅π₯+πΆ)
(π₯+2)(π₯2+1)
π₯2 + π₯ + 1
(π₯ + 2)(π₯2 + 1)=
π΄π₯2 + π΄ + π΅π₯2 + 2π΅π₯ + πΆπ₯ + 2πΆ
(π₯ + 2)(π₯2 + 1) `
=(π΄ + π΅)π₯2 + (2π΅ + πΆ) + π΄ + 2πΆ
(π₯ + 2)(π₯2 + 1)
π₯2 + π₯ + 1 = (π΄ + π΅)π₯2 + (2π΅ + πΆ) + π΄ + 2πΆ
π΄ + π΅ = 1 β π΄ = 1 β π΅, 2π΅ + πΆ = 1 and π΄ + 2πΆ = 1, then
{2π΅ + πΆ = 1
βπ΅ + 2πΆ = 0 β π΅ = 2πΆ , then 2(2πΆ) + πΆ = 1 β πΆ =
1
5 , π΅ =
2
5, and π΄ = 1 β
2
5=
3
5
Hence, π΄
π₯+2+
π΅π₯+πΆ
(π₯2+1)=
3
5(π₯+2)+
2π₯+1
5(π₯2+1) .
Answer: A
39. What is the area of the region enclosed by the graph of π¦2 = π₯ + 1 and π¦2 = βπ₯ + 1 ?
π΄) 3
8π π. π’πππ‘π B)
4
3π π. π’πππ‘π C)
8
3π π. π’πππ‘π D)
3
4π π. π’πππ‘π
Solution: The intersection of π¦2 = π₯ + 1 and π¦2 = βπ₯ + 1
π¦2 β 1 = π₯ = βπ¦2 + 1 β 2π¦2 = 2 βΊ π¦ = Β±1 , then the area
A=β« [(1 β π¦2) β (π¦2 β 1)]1
β1ππ¦ = β« (β2π¦2 + 2
1
β1)ππ¦ =
β2
3π¦3 + 2π¦
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=β2
3(1) + 2(1)-((
2
3(β1) + 2(β1)) =
β2
3+
β2
3+ 2 + 2 =
β4
3+ 4 =
8
3
Hence, π΄ =8
3 π π. π’πππ‘π
Answer: C
40. The age distribution of students in a certain class is given by
Age 10-14 15-19 20-24 25-29
No of
students
2 10 6 7
What is the modal value of the distribution?
A) 17.83 B) 17.38 C) 18.37 D) 18.73
Solution: The modal value of continuous data is given by
π₯πππ = π + π€ (π0βπ1
2π0βπ1βπ2), where π = lower class boundary of modal class,
π0 =frequency of modal class, π1 =frequency before frequency of modal class and
π€ =width of the interval.
The class mark of the data
Class Class mark
(ππ)
Frequency
10-14 12 2
15-19 17 10 (Modal class)
20-24 22 6
25-29 27 7
Hence, π€ = 5, π0 = 10, π1 = 2, and π2 = 6, πΏ = 14.5
π₯πππ = πΏ + π€ (π0βπ1
2π0βπ1βπ2)
= 14.5 + 5 (10β2
20β8) = 14.5 +
40
12
= 17.83
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Answer: A
41. If π(π₯) = π πππ₯ + ππ πππ₯ and πβ²β²(π) = β1, then what is the value of π?
A) ππ π B) π2 C) π D) 2π
Solution: πβ²(π₯) = π
π₯+ πππ π₯ ππ πππ₯
f''(x) = β π
π₯2 + πππ 2 π₯ππ πππ₯β π πππ₯πsin π₯
πβ²β²(π) = βπ
π2 + 1 = β1
βΉ βπ
π2 = β2
βΉ π = ππ π
Answer: A
42. Let π(π₯) = π₯ β π₯2 and π(π₯) =1
π₯. Then what is π (π (
1
π₯)) equal to
π΄) π₯ β π₯2 π΅) π₯2
π₯ β 1 πΆ)
1
π₯2 β π₯ π·)
π₯ β 1
π₯2
Solution: π(π₯) = π₯ β π₯2 and π(π₯) =1
π₯. Then what is π (π (
1
π₯)) ?
g (f (1
x)) = g (
1
xβ
1
x2) =
1
1π₯ β
1π₯2
=x2
x β 1, x β 1
Answer: B
43. A cylindrical tank whose inner diameter is 2 π contains 4ππ3 oil. If the oil is discharged
from the tank at the rate of 2π
3 π3/πππ, then how long (ππ πππ) does it take for the
tank to be empty?
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A) 4
3 π΅) 4 C) 12 π·) 6
Solution: The volume of cylinder is π£ = ππ2β and given π£ = 4ππ3.
Solving for β we have β = 4π for π = 1π
ππ£
ππ‘=
π
ππ‘(ππ2β),
ππ£
ππ‘=
2π
3 π3/πππ
ππ£
ππ‘= 2ππ β
ππ
ππ‘
= 8ππ3 ππ
ππ‘
2π
3
π3
πππ= 8ππ3 ππ
ππ‘
ππ
ππ‘=
2π
3
π3
πππ.
1
8ππ3
=1
12 πππ
ππ‘
ππ= 12 πππ
ππ‘ = 12 ππ πππ
β« ππ‘ = 12 min β« ππ1
0
t = 12 min
Answer: C
44. What are the values of π and π so that the function
π(π₯) = {π₯ + 1, 1 < π₯
ππ₯ + π, 1 β€ π₯ < 23π₯, π₯ β₯ 2
is continuous everywhere?
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A) a = 4, b = β2 B) π = β4, π = β2 C) π = 4, π = 2 D) π = β4, π = 2
Solution: π(π₯) is continuous if limπ₯β2+ π(π₯) = limπ₯β2β π(π₯) πππ limπ₯β1+ π(π₯) =
limπ₯β1β π(π₯)
6 = 2π + π Β·Β·Β·Β·Β·Β·Β·Β·Β· (1) πππ 2 = π + π Β·Β·Β·Β·Β·Β·Β·Β·Β· (2)
Solving equation (1) and equation (2), we obtain
{2π + π = 6 π + π = 2
Β·Β·Β·Β·Β·Β·Β·Β·Β· π = 4 πππ π = β2
Answer: A
45. Let π = π (the set of natural numbers) be the universe. Which one of the following
proposition is true?
A) (βπ₯)(π₯ + π₯ = π₯) B) (βπ₯)(βπ¦)(π¦ < π₯)
C) (βπ)(βπ)(π Γ· π = π Γ· π) D) (βπ₯)(βπ¦)(π₯ β π¦ = π₯)
Solution: A) Since there is no a natural number π₯ such that π₯ + π₯ = 2π₯ = π₯, it is false.
B) Unless (βπ₯)(βπ¦)(π¦ β€ π₯), it is false. C) (βπ)(βπ)(π Γ· π = π Γ· π) is true as it is
possible to take x=y. D) This is false.
Answer: C
46. What is the maximum possible area of a rectangle in square units with diagonal of length
16 units.
A) 48 B) 128 C) 64 D) 256
Solution: Using Pythagoras theorem we have π₯2 + π¦2 = π2
π₯2 + π¦2 = 162 , π¦ = (162 β π₯2) 1
2
Area (π΄) = π₯π¦
π΄ = π₯π¦ = π₯ (162 β π₯2)1
2
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π΄(π₯) = π₯β162 β π₯2
To find the dimension of the rectangle, we have find the critical point for π΄(π₯)
π΄β²(π₯) = β162 β π₯2 β2π₯
2β162βπ₯2
=162β2π₯2
β162βπ₯2
π΄β²(π₯) = 0
162 β 2π₯2 = 0
x = Β±β162
2= Β±
16
β2= 8β2, y = (162 β (8β2)
2)
1
2 = 8β2
Area(A) = xy = (8β2)(8β2) = 128
Answer: B
47. Let π(π₯) = ln(π₯βπ₯). Then what is πβ²(π₯) equal xto?
A) 2π₯
3 B)
βπ₯
2 C)
2
π₯βπ₯ D)
3
2π₯
Solution:
πβ²(π₯) =1
π₯βπ₯. (π₯βπ₯)
β²
=1
π₯βπ₯. (1. βπ₯ + π₯. (
1
2βπ₯) =
3
2π₯
Answer: D
48. What it is the sum of the series β 3π4βπβπ=1 ?
A) β B) 3
16 C) 3 D) 4
Solution: β 3π4βπβπ=1 = β (
3
4)
π
βπ=1 is a geometric series with π =
3
4. Since |
3
4| =
3
4< 1,
the series is convergent and it converges to πΊ1
1βπ where πΊ1 =
3
4 and π =
3
4.
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Thus, β 3π4βπβπ=1 =
3
4
1β3
4
=3
41
4
= 3.
Answer: C
49. If π§ = (1 + β3π)(1 + π), then which one of the following is the polar representation of π§?
A) π§ = 4(cos(105Β°) + ππ ππ (105Β°)) B) π§ = 2β2(cos(105Β°) + ππ ππ (105Β°))
C) π§ = 2β2(cos(15Β°) + ππ ππ (15Β°)) D) π§ = 4(cos(75Β°) + ππ ππ (75Β°))
Solution: The polar representation of π§ = π₯ + π¦π is π§π = ππ(cos(ππ) + ππ ππ (ππ))
where π = βπ₯2 + π¦2 and π = tanβ1 (π¦
π₯).
The given number π§ = (1 + β3π)(1 + π) = 1 + π + β3π + β3π2 = (1 β β3) + π(1 +
β3) which is of the form π§ = π₯ + ππ¦ where π₯ = 1 β β3 and π¦ = 1 + β3.
Then π = β(1 β β3)2
+ (1 + β3)2
= 2β2 and π = tanβ1 (1+β3
1ββ3). Note that tan(π) =
1+β3
1ββ3=
tan(45Β°)+tan(60Β°)
1β(tan(45Β°))(tan(60Β°))= tan(45Β° + 60Β°). Thus, π = 105Β°. The polar representation
of π§ = 2β2(cos(105Β°) + ππ ππ (105Β°)).
Answer: B
50. The variance of 20 observations is 5. If each observation is multiplied by 2, then what is
the variance of the resulting observations?
A) 5 B) 10 C) 20 D) 40
Solution: Let π₯1, π₯2, β¦ , π₯20 be records of the observations. Then οΏ½Μ οΏ½ = βπ₯π
20
20π=1 and
ππ₯2 = β
(π₯πβοΏ½Μ οΏ½)2
20
20π=1 = 5.
Let π¦1(= 2π₯1), π¦2(= 2π₯2), β¦ , π¦20(= 2π₯20) be the new observations. Then οΏ½Μ οΏ½ = βπ¦π
20
20π=1 =
β2π₯π
20
20π=1 = 2 β
π₯π
20
20π=1 = 2οΏ½Μ οΏ½ and ππ¦
2 = β(π¦πβοΏ½Μ οΏ½)2
20
20π=1 = β
(2π₯πβ2οΏ½Μ οΏ½)2
20
20π=1 = 4 β
(π₯πβοΏ½Μ οΏ½)2
20
20π=1 =
4(5) = 20.
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Answer: C
51. A firm has two types of overtime pay rates (payment per hour) for its employees. Normal
overtime pay rate and holiday overtime pay rate ( for Sundays and holidays ). The
holiday
overtime pay rate is 1.5 times the normal overtime pay rate; and the normal overtime pay
rate is 1.5 times the regular pay rate. The regular pay rate for an employee is given by
his weekly basic wage divided by 40. If an employee whose weekly basic wage is Birr
2400
has additionally, 6 hours of normal overtime and 3 hours of holiday overtime in a week,
how much (in Birr) is his payment for the week?
A) 3210 B) 3345 C) 3300 D) 3525
Solution: Let π₯, π¦, π§ be normal, holiday, and regular overtime pay rates respectively.
π¦ = 1.5π₯ = 1.5(1.5π§) = 2.25π§
Since π§ =2400
40= 60, π₯ = 1.5(60) = 90 and π¦ = 2.25(60) = 135.
The employee has a weekly payment of 2400 + 6(90) + 3(135) = 3345.
Answer: B
52. What is the value of β«π₯+4
π₯(π₯+2)ππ₯
2
1?
A) 2 B) ln 3 C) ln 2 D) ln 4 β ln 3
Solution: The partial fraction of π₯+4
π₯(π₯+2) takes the form
π
π₯+
π
π₯+2 . Then from the equation
π₯+4
π₯(π₯+2)=
π
π₯+
π
π₯+2, we have π(π₯ + 2) + ππ₯ = (π + π)π₯ + 2π = π₯ + 4 . Thus, π + π =
1 and 2π = 4 . then π = 2 and the value of π = β1
Then β«π₯+4
π₯(π₯+2)ππ₯
2
1= β«
2
π₯ππ₯
2
1+ β«
β1
(π₯+2)ππ₯
2
1= β«
π₯+4
π₯(π₯+2)ππ₯
2
1= (2 ln π₯ β ln(π₯ + 2))|π₯ = 2
π₯ = 1
= (2 ln 2 β ln 4) β (2 ln 1 β ln 3) = 2 ln 2 β ln 22 + ln 3 = 2 ln 2 β 2 ln 2 + ln 3 =
ln 3
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Answer: B
53. What is the slope of the tangent line to the graph of π(π₯) = 3ππ₯ + π ππ π₯ + 2 at (0,5)?
A) 2 B) 3 C) 5 D) 4
Solution: The slope of the line tangent to the graph of π at any point (π₯, π¦) is π =ππ(π₯)
ππ₯
= 3ππ₯ + cos π₯. At (0,5), π = 3π0 + cos 0 = 4.
Answer: D
54. The population of a certain city is increasing at a rate of 3% per year. If the population
was 100000 in 2010 E.C., then what will be the population in 2020E.C?
A) 134000
B) 130000
C) 1060000
D) 1378000
Solution: Let π(π‘) and π(π‘0) be the populations of the city in π‘ and π‘0 years. Then the
population dynamics is modeled by 1
π(
ππ
ππ‘) = 0.03. The solution of this is π(π‘) =
π(π‘0)π0.03π‘.
Let π‘0 = 2010 = 0. Then, π‘ = 2020 = 10. Then, π(π‘0 = 0) = 100,000 and the
population in 2020 becomes π(10) = 100,000π0.03β10 = 100,000(π0.03)10 =
100,000(1.03)10 = 100,000(1.34) = 134,000
Answer: A
55. If π§ = β3 + 4π ,and π€ = 1 + 2π ,then what is the value of 2π§
π€+ οΏ½Μ οΏ½ ?
A) 2 + 3π
B) 3 + 5π
C) 3 + 2π
D) 3 β 2π
Solution: 2π§
π€+ οΏ½Μ οΏ½ =
2(β3+4π )
1+2π+ 1 β 2π =
β6+8π
1+2π+
(1β2π )(1+2π)
1+2π=
β6+8π
1+2π+
1+2iβ2i+4
1+2π
=β6+8π
1+2π+
5
1+2π=
β1+8π
1+2π= (
β1+8π
1+2π) (
1β2π
1β2π) =
β1+2π+8π+16
5=
15+10π
5= 3 + 2π = 3 + 2π
Answer: C
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56. A salesman sold item π₯1 , π₯2 ,and π₯3 with different rates of commissions as shown in the
table below
Months Sales of unit Total commission
(in birr) π₯1 π₯2 π₯3
February 90 100 20 800
March 130 50 40 900
April 60 100 30 850
What are the rate of commission on items π₯1 , π₯2 ,and π₯3 respectively
a. 4 ,2 and 11
b. 2 ,4 and 11
c. 4 ,11 and 2
d. 11 ,2 and 4
Solution: From the table, we have the following system
90π₯1 + 100π₯2 + 20π₯3 = 800 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (1)
130π₯1 + 50π₯2 + 40π₯3 = 900 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦β¦. (2)
60π₯1 + 100π₯2 + 30π₯3 = 850 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (3)
From this system of linear equations, we do have the coefficient of matrix
π΄ = [90 100 20
130 50 4060 100 30
] , and its determinant is
βπ΄β = β90 100 20
130 50 4060 100 30
β = 90 β50 40
100 30β β 100 β
130 4060 30
β + 20 β130 5060 100
β
= β225000 β 150000 + 200000 = β175000 β 0
Let π΄π be the sub matrix obtained by replacing the ππ‘β column of the coefficient matrix
with the matrix of constants.
βπ΄1β = β800 100 20900 50 40850 100 30
β = 800 β50 40
100 30β β 100 β
900 40850 30
β + 20 β900 50850 100
β
= β2000000 + 700000 + 950000 = β350000,
βπ΄2β = β90 800 20
130 900 4060 850 30
β = 90 β900 40850 30
β β 800 β130 4060 30
β + 20 β130 90060 850
β
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= β630000 β 1200000 + 1130000 = β700000, and
βπ΄3β = β90 100 800
130 50 90060 100 850
β =
90 β50 900
100 850β β 100 β
130 90060 850
β + 800 β130 5060 100
β
= β4275000 β 5650000 + 8000000 = β1925000
Using crammers rule, we have π₯1 =βπ΄1β
βπ΄β=
β350000
β175000= 2 ,π₯2 =
βπ΄2β
βπ΄β=
β700000
β175000= 4 and
π₯1 =βπ΄3β
βπ΄β=
β1925000
β175000= 11
Answer: B
57. If A is the square matrix of order 3 and πππ‘(π΄) = 5 , then what is the value
of πππ‘(π΄. πππ(π΄))?
A) 3
B) 5
C) 125
D) 25
Solution: Recall that π΄β1 =πππ(π΄)
det (π΄) and also πππ‘(π΄β1) =
1
πππ‘(π΄).
Then πππ‘(π΄. πππ(π΄)) = det(π΄. (det(π΄)) π΄β1) = det (det(π΄) . ((π΄)(π΄β1)))
= det(det(π΄) . πΌ) = (det(π΄))3 det(πΌ) = (5)3 Γ 1 = 125.
Answer: C
58. If πβ²(π₯) = ππ₯β1 + 3π₯2 β1
π₯ ,and π(1) = 5, what is f(x) ?
A) π(π₯) = ππ₯β1 + 3π₯2 +1
π₯2 + 2
B) π(π₯) = ππ₯β1 β π₯3 + ln π₯ + 5
C) π(π₯) = ππ₯β1 + 3π₯2 β1
π₯+ 3
D) π(π₯) = ππ₯β1 + π₯3 β ln π₯ + 3
Solution: Sinceπβ²(π₯) = ππ₯β1 + 3π₯2 β1
π₯, π(π₯) = β« πβ²(π₯)ππ₯
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= β« (ππ₯β1 + 3π₯2 β1
π₯) ππ₯
= ππ₯β1 + π₯3 β ln π₯ + π
As π(1) = π1β1 + 13 β ln 1 + π = 5, 2 + π = 5. That is, π = 3
Thus, π(π₯) = ππ₯β1 + π₯3 β ln π₯ + 3
Answer: D
59. Suppose the following are the premises of an argument.
He is healthy and he is not angry
He is angry or his plan fails
His plan does not fail if he does not travel abroad
Given that the premises are true, which one of the following can be a conclusion that
makes the argument valid?
A) His plane fails and he is angry
B) His plane does not fail
C) He travels abroad
D) His plan fails and he does not travel abroad
Solution: Suppose that P: he is healthy, Q: he is not angry, R: .his plan fails, S: he travels
abroad. According to the given premises, we have
π β© π is true. That is, both P and Q are true.
Β¬π βͺ π is true. SinceΒ¬π is false, R is true.
Β¬π βΉ Β¬π is true. Since Β¬π is false, Β¬π is false or S is true.
The conclusion βHe travels abroadβ makes the argument valid.
Answer: C
60. A man running a race course noted that the sum of the distances from the two flag posts
from him is always 10 meters and the distance between the flag posts is 8 meters. What is
the equation of the path traced by man? (Take flag posts to be on the x-axis with origin at
their midpoint)
A) π₯2
9+
π¦2
25= 1
B) π₯2
25+
π¦2
9= 1
C) π₯2
100+
π¦2
64= 1
D) π₯2
64+
π¦2
100= 1
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Solution: If a point moves in a plane in such a way that the sum of its distances from
two fixed points is constant, then the path is an ellipse. The constant value is equal to the
length of the major axis of the ellipse, 10 meters. That is 2π = 10 or π = 5. The Distance
between the foci is 2π = 8 which is π = 4, and the distance π of the minor axis satisfies
π2 = π2 β π2 = 25 β 16 = 9 .Then, π = 3. Then the equation of the ellipse is π₯2
π2 +π¦2
π2 =
1 becomes π₯2
52+
π¦2
32= 1. That is,
π₯2
25+
π¦2
9= 1.
Answer: B
61. What is the value of β«πβπ₯
βπ₯
9
1ππ₯ ?
A) π3 β π
B) π3
3β π
C) π (π2 β1
3)
D) 2(π3 β π)
Solution: Let = βπ₯ . Then 2ππ’ =ππ₯
βπ₯. Further, if π₯ = 1, π’ = 1 and π₯ = 9, π’ = 3. Then
β«πβπ₯
βπ₯
9
1ππ₯ = 2 β« ππ’3
1ππ’ = ππ’|π’ = 3
π’ = 1= π3 β π.
Answer: A
62. What is the value of β« π₯β1 β π₯2 ππ₯?
A) β1
3(1 β π₯2)
3
2 + π
B) β2π₯β1 β π₯2 + π
C) (1 β π₯2)3
2 + π
D) 1
2β1 β π₯2 + π
Solution: Let = 1 β π₯2 . Then ππ’ = β2π₯ππ₯ βΉ βππ’
2= π₯ππ₯ , then
β« π₯β1 β π₯2 ππ₯ = β1
2β« βπ’ ππ’ = β
1
2 π’
12
+1
1
2+1
+ π , where c is any constant
= β1
2
π’32
3
2
+ π = β1
3 π’
32
3+ π = β
1
3 (1 β π₯2 )
3
2 + π
Answer: A
63. Which one of the following is not true about the function π(π₯) = 3π₯4 β 4π₯3.
A) (0 ,0) is the point of inflection of π
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B) 0 and 1 are critical numbers of f
C) π is concave upward on (0 ,2
3) and concave downward on (ββ, 0) and on (
2
3, β)
D) π is decreasing on (ββ, 1) and increasing on (1, β)
Solution: π(π₯) = 3π₯4 β 4π₯3 β πβ²(π₯) = 12π₯3 β 12π₯2 = 12π₯2(π₯ β 1)
The critical points satisfy the equation 12π₯2(π₯ β 1) = 0. Then, = 0 , π₯ = 1 are the
critical points.
factors (ββ ,0) π₯ = 0 (0 ,1) π₯ = 1 (1, β)
12π₯2 + 0 + + +
π₯ β 1 - - - 0 +
12π₯2(π₯ β 1) - 0 - 0 +
From the table, π is decreasing on (ββ ,0) βͺ (0 ,1) and increasing on (1, β)
Since πβ²(π₯) = 12π₯3 β 12π₯2, πβ²β²(π₯) = 36π₯2 β 24π₯ = 6π₯(6π₯ β 4).
Setting πβ²β²(π₯) = 6π₯(6π₯ β 4) = 0, we have π₯ = 0, and π₯ =2
3.
Then the function is Concave upward on (ββ ,0) βͺ (2
3, β), downward on (0 ,
2
3), and
(0 ,0) is the inflection point of π.
Therefor the answer is the function f is concave upward on (0 ,2
3) and concave downward
on (ββ ,0) βͺ (2
3, β).
Answer: C
64. What is the value of limπ₯β01
π₯2 π ππ2 (π₯
2) ?
factor (ββ ,0) π₯ = 0 (0 ,
2
3) π₯ =
2
3 (
2
3, β)
6π₯ - 0 + + +
(6π₯ β 4) - - - 0 +
6π₯(6π₯ β 4) + 0 - 0 +
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A) 1
2
B) 1
4
C) 2
D) 4
Solution: Since limπ₯β0 π ππ2 (π₯
2) = 0 = limπ₯β0 π₯2, the limit is an indeterminate form of
0
0.
Applying Lβhopitals rule, we
getlimπ₯β01
π₯2 π ππ2 (π₯
2) = limπ₯β0
(π ππ2(π₯
2))β²
(π₯2)β² = limπ₯β0
2 sin(π₯
2) cos(
π₯
2)
1
2
2π₯== limπ₯β0
sin(π₯
2) cos(
π₯
2)
2π₯
which is also an indeterminate form of 0
0.
limπ₯β01
π₯2 π ππ2 (π₯
2) = limπ₯β0
sin(π₯
2) cos(
π₯
2)
2π₯= limπ₯β0
cos(π₯
2) cos(
π₯
2)
1
2βsin(
π₯
2) sin(
π₯
2)
1
2
2=
cos(0) cos( 0)1
2βsin(0) sin( 0)
1
2
2=
1
4.
Answer: B
65. Which one of the following is equal to π
ππ₯log2 β6π₯?
A) 3π₯
2 ln(2) B)
3
2π₯ ln(2) C)
1
6π₯ ln(2) D)
1
2π₯ ln (2)
Solution: Recalling that π
ππ’(logπ π’) =
1
π’ ππ (π) and noting that π(π₯) = log2 β6π₯ = π(β(π₯))
where π(π’) = log2 π’, π’ = β(π₯) = β6π₯, the derivative can be obtained by using chain rule. π
ππ₯π(π₯) =
π
ππ’π(π’).
π
ππ₯β(π₯) =
1
π’ ππ (2)Γ β6
1
2βπ₯ where π’ = β6π₯.
=1
β6π₯ Γ ln(2)Γ β6
1
2βπ₯=
1
2π₯ ππ (2)
Answer: D
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