Archival unit 1
ELBA RADO ⟦illegible⟧ RADO C
IJA 2550
Exams, Shamash Secondary School
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These are exam materials from the Shamash Secondary School in Baghdad. They include fourth year exams in arithmetic, trigonometry, and algebra, 1954 and 1957-1970, with handwritten answer keys. There are handwritten notes for one undated English exam. Included are final, mid-year, monthly, quarterly, and conditional exams as well as some monthly quizzes.
AI English Translation, Pages 1-25
Archival unit 2
⟦illegible⟧ ELBA
Archival unit 3
⟦illegible⟧ the exam ⟦illegible⟧ ⟦diagram with numerical values: 20, 55, 60, 75, 32.5, θ⟧ ⟦illegible⟧ = (20)² + (32.5)² - 2 x 20 x 32.5 cos ⟦illegible⟧ ⟦illegible⟧ = 400 + 1056.25 - 1300 x ⟦illegible⟧ = 1424 - 640 = 784 V = 28 ⟦illegible⟧ 20 / sin θ = 28 / sin 60 sin θ = 20 x 0.866 / 28 = 0.6186 θ = 38° 12' or 38° 13' 75 - 38° 12' = 36° 48' E
Archival unit 4
⟦illegible⟧ in Mathematics for the ⟦illegible⟧ grade 10/⟦illegible⟧/97 For the fourth general No 1 Let the man has £x stock 4x / 100 his profit x / 100 * 114 = £ 114x / 100 114x / 100 * 1/3 = £ 38x / 100 the first part 114x / 100 * 2/3 = £ 76x / 100 second part 38x / 100 * 100 / 95 * 5 / 100 = 2x / 100 profit 76x / 100 * 100 / 100 * 3 / 400 = 19x / 1200 profit 2x / 100 + 19x / 1200 = 43x / 1200 total profit 4x / 100 - 43x / 1200 = 25 / 2 (1200) <del>12x</del> 48x - 43x = ⟦illegible⟧ 5x = ⟦illegible⟧ x = ⟦illegible⟧ Stock ⟦illegible diagram/sketch⟧
Archival unit 5
2 ⟦illegible⟧ Mathematics solutions for the fourth general year ⟦illegible⟧ Final exam T A 22 34 B 200 yd C 5. AB = 200 Cos 34 = 200 x 0.829 = 165.8 yd. AT = 165.8 x tan 22 = 165.8 x 0.404 = 66.9832 yd. AC = 200 Sin 34 = 200 x 0.5592 = 111.84 tan θ = 66.9832 / 111.84 = 0.5985 ∴ θ = 30° 48' 5 6 5 5 ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
Archival unit 6
SHAMASH SECONDARY SCHOOL FINAL EXAMINATION - MAY 1970 Subject: Mathematics Date: 27/5/1970 Class: 4th year Secondary (scientific section) Time: 2 hours Q1- A chord AB of a circle is produced to T. From T a line TC is drawn to touch the circle at C . If BT = 9 cm. and TC = 12 cm. Find the length of and the ratio of the areas of the triangles BTC and ATC and then prove BC² : AC² = BT : AT . Q2- Two ships leave the same port at the same time and steam at 10 and 16 m.p.h. respectively in the direction N. 55 W. and S. 75 W. Find their distance apart after 2 hours and the bearing of the first ship to the second . Q3- PQ , CD are parallel chords of a circle , the tangent at D cuts PQ produ- ced at T , B is a point of contact of the other tangent from T ; prove that BC bisect PQ . Q4- A man has a certain amount of 4 % stock . He sells it at 114 and invests one - third the proceed in 5% at 95 , and the rest in 2¼ % stock at 108. He then finds that his annual income is reduced by £ 12 10s. . Find the amount of the original stock had he . Q5- ABC is a triangle in a horizotal plane $ BC = 120 yd. , angle BAC = 90 and angle ABC = 34 . At A is a vertical pole AT , the angle of elevation of T from B = 22 , calculate the height of AT , and the angle of elevation of T from C , ⟦sketch⟧ (TA) (TB) = (TC)² 9 (TA) = 144 TA = 144 / 9 = 16 AB = 16 - 9 = 7 cm. Δ ACT Δ BCT
Archival unit 7
⟦illegible⟧ ⟦illegible⟧ the roots of the ⟦illegible⟧ equation 2x⟦illegible⟧ y = 0 then x = -1 or x = ⟦illegible⟧ the function y ⟦illegible⟧ values x < -1 ⟦illegible⟧ the root of the equation 2x² - x - ⟦illegible⟧ the equation 2x² - x - 3 = -2 Draw the ⟦illegible⟧ ⟦illegible⟧ x = -1 and x = 1 ⟦illegible⟧ ⟦illegible⟧ the points of intersection of the graph of ⟦illegible⟧ y 3 2 1 -2 -1 -1/2 1/4 1/2 1 3/2 2 x -1 -2 1/2 - 3
Archival unit 8
SHAMASH SECONDARY SCHOOL ⟦Add⟧itional Examination, September, 1969. Subject: Algebra Class: 4th Year, Secondary Date: 5/9/1969. Time: 8:00-11:00 a.m. Attempt all questions: 1. (i) Find the values of 'a' and 'b' if 3x³ - ax + b is exactly divisible by (x+1)(x-2). If 'a' and 'b' have these values, factor the expre- ssion completely. (10 marks). (ii) If x³ + 3x + 5 = x(x+1)(x+2) + Ax(x+1) + Bx + C for all values of x, find the values of the constants A, B, and C. (10 marks). 2. (i) The 15th term of an arithmetic progression is 25 and the sum of the first 10 terms is 60. Find the first term of the progression, the common difference and the sum of the first 16 terms. (10 marks). (ii) p+ 3, p + 8, and p + 18, are the 3rd, 4th, and 5th terms of a geometric progression. Find the value of p. Find also the common ratio and the 9th term of the progression. (10 marks). 3. A can walk a mile in 2 minutes less time than B would take. In a walki⟦ng⟧ race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks). 4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0.. (10 marks). (ii) Use logarithms to compute the value of the following expression : ⁷√((0.002013)²(Sin 15° 12')³ / (4.004)³(Cos 42° 13')²) (10 marks). 5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch a⟦s⟧ one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 ⟦=⟧ 2x². ⟦4⟧ marks). (iii) Find the values of x for which the function 2x² - x ⟦- 3⟧ is always positive. ⟦(⟧ 4 marks). (iv) By drawing another stra⟦igh⟧t line on your diagram, fin⟦d⟧ the roots of the equation 2x² - x ⟦-⟧ 1 = 0. ⟦(⟧ 4 marks). ⟦line⟧
Archival unit 9
SHAMASH SECONDARY SCHOOL Conditional Examination, September, 1969. Subject: Algebra Date: 5/9/1969. Class: 4th Year, Secondary Time: 8:00-11:00 a⟦...⟧ ⟦line⟧ Attempt all questions: 1. (i) Find the values of 'a' and 'b' if 3x³ - ax + b is exactly divisible by (x+1)(x-2). If 'a' and 'b' have these values, factor the expre- ssion completely. (10 marks). (ii) If x³ + 3x + 5 = x(x+1)(x+2) + Ax(x+1) + Bx + C for all values of x, find the values of the constants A, B, and C. (10 marks). 2. (i) The 15th term of an arithmetic progression is 25 and the sum of the first 10 terms is 60. Find the first term of the progression, the common difference and the sum of the first 16 terms. (10 marks). (ii) p+ 3, p + 8, and p + 18, are the 3rd, 4th, and 5th terms of a geometric progression. Find the value of p. Find also the common ratio and the 9th term of the progression. (10 marks). 3. A can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks). 4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0. (10 marks). (ii) Use logarithms to compute the value of the following expression : 7 / (0.002013)²(Sin 15° 12')³ √ ⟦line⟧ (10 marks). (4.004)³(Cos 42° 13')² 5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch as one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 = 2x². (4 marks). (iii) Find the values of x for which the function 2x² - x - 3 is always positive. (4 marks). (iv) By drawing another straight line on your diagram, find the roots of the equation 2x² - x - 1 = 0. (4 marks). ⟦line⟧
Archival unit 10
Shamash Secondary School Final Examination May 1969 Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all questions: 1. (i) In the expression x³+Ax²+31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) (log x)² = log x⁷ - 10, finding two 2. (i) Solve the equation (log x)² = log x⟦⁷⟧ - 10, ⟦finding⟧ two values for x. (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ⟦line⟧ (1) x² -xy+y² -7=0 ⟦line⟧ (2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50) (10 marks) (ii) Compute the value of ⁷√((0.5002)² sin³ 14° 25') / ((4.003)³ cos² 15° 27') (10 marks) 4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) bounce (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous ⟦bounce⟧. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2).. (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50)
Archival unit 11
(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ⟦line⟧ 5. (i) Plot the curve of the function 3+2x-x² for values of x from x=-2 to x=4, choosing one half of an inch for each unit on the axis of x and on the axis of y. (4 marks) (ii) From your graph, find the roots of the equation x²-3=2x. (3 marks) (iii) Find the values of x for which the function 3+2x-x² is always positive. ( 3 marks) (iv) Find from your diagram the value of x at which the function 3+2x-x² is greatest and state the maximum value. (3 marks) (v) By plotting another curve on the same diagram, find the values of x for which 3+2x-x² > x/2 + 2. (4 marks) (vi) From your last diagram, find the roots of the equation 3+2x-x² = x/2 + 2. (3 marks) ⟦line⟧
Archival unit 12
Shamash Secondary School Final Examination May 1969 Solution to Algebra Questions, 4th Year 14/5/1969. 1. (i) Divide by (x-2) | x-2 | x³ + Ax² + 31x + B | x² + (A+2)x + 2A + 35 x³ - 2x² ⟦line⟧ (A+2)x² + 31x + B (A+2)x² - 2(A+2)x ⟦line⟧ or (2A + 4 + 31)x + B (2A + 35)x + B (2A + 35)x - 4A - 70 ⟦line⟧ 4A + B + 70 = 0 ∴ B = -4A - 70 .... ① Now Divide the Quotient by (x-3) ∴ x-3 | x² + (A+2)x + 2A + 35 | x + A + 5 x² - 3x ⟦line⟧ (A+5)x + 2A + 35 (A+5)x - 3A - 15 ⟦line⟧ 5A + 50 ∴ 5A + 50 = 0 .... ② ∴ 5A + 50 = 0 ∴ A = -50/5 = -10 ∴ the 3rd Factor is = x + A + 5 = ∴ B = -4(-10) - 70 = -30 = x - 10 + 5 = x - 5 ∴ A = -10 } Ans. | and the expression is: x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } 3rd factor = x-5 } An alternative method: <del>By</del> the remainder theorem, when x=2 then 2³ + 2²A + 2x31 + B = 0 ∴ 4A + B = -70 ... ① also when x=3, then also 3³ + 3²A + 3x31 + B = 0 ∴ 9A + B = -120 ... ② 5A = -50 ∴ A = -10 ∴ from ①: 4(-10) + B = -70 ∴ B = -30 ∴ A = -10 } Ans. Factoring, we get x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } Ans. 2 ⟦line⟧ (ii) 12 men + 7 boys = £ 9 13s. also 3 men = 4 boys + 8s. Let x shillings be the wages of one boy y " " " " " " man ∴ 12y + 7x = 9 x 20 + 13 or 12y + 7x = 193 ... ① also 3y = 4x + 8 or 3y - 4x = 8 ... ② 12y + 7x = 193 ... ① ∴ 23x = 161 ∴ x = 161/23 = 7 shillings 12y - 16x = 32 ... ② ∴ y = 4x+8/3 = 28+8/3 = 36/3 = 12s. ∴ wages of one boy = x shill. = 7s. } Ans. " " " man = y shill = 12s. }
Archival unit 13
Solution of Algebra ⟦illegible⟧ 14/5/1969 page 2 2 (i) solve the equation: (log x)² = log x⁷ - 10 ∴ (log x)² - 7 log x + 10 = 0 ∴ (log x - 2) (log x - 5) = 0 ∴ log x - 2 = 0 or log x = 2 ∴ x = 10² = 100 Ans. 1 or log x - 5 = 0 or log x = 5 ∴ x = 10⁵ = 100000 Ans. 2 (ii) 3x² + xy - 2y² + 7 = 0 ⟦line⟧ ① x² - xy + y² - 7 = 0 ⟦line⟧ ② adding, we get: 4x² - y² = 0 ∴ (2x + y) (2x - y) = 0 ∴ either y = 2x ⟦line⟧ ③ from ② + ③ : x² - x(2x) + (2x)² - 7 = 0 or y = -2x ⟦line⟧ ④ or x² - 2x² + 4x² = 7 or 3x² = 7 ∴ x = ± √7/3 ∴ y = 2x ∴ y = ± 2√7/3 From ② + ④ : x² - x(-2x) + (-2x)² - 7 = 0 or x² + 2x² + 4x² = 7 or 7x² = 7 ∴ x² = 1 ∴ x = ± 1 ∴ y = -2x ∴ y = -2 (± 1) ∴ y = ∓ 2 ∴ x = √7/3 } Ans. 1 x = -√7/3 } Ans. 2 y = 2√7/3 y = -2√7/3 x = 1 } Ans. 3 x = -1 } Ans. 4 y = -2 y = 2 3 (i) (5) (4³ˣ⁻¹) (√8)¹⁻ˣ = (√2)ˣ (√50) 5 x 2²⁽³ˣ⁻¹⁾ x 2³⁄₂⁽¹⁻ˣ⁾ = 2ˣ⁄₂ x 5 x 2¹⁄₂ ∴ 2⁶ˣ⁻² x 2³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 2⁶ˣ⁻²⁺³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 6x - 2 + 3/2 - 3x/2 = x+1/2 ∴ 9x/2 - 1/2 = x/2 + 1/2 ∴ 4x = 1 ∴ x = 1/4 Ans.
Archival unit 14
Solution to Algebra 4th Year Final Exam, 14/5/1969
page 3
3 (ii) x = ⁷√((0.5002)² sin³ 14° 25' / (4.003)³ cos² 15° 27')
log 0.5002 = 1̄.6992 | 2 log 0.5002 = 1̄.3984 | 3 log 4.003 = 1.8072
log sin 14° 25' = 1̄.3962 | 3 log sin 14° 25' = 2̄.1886 | 2 log cos 15° 27' = 1̄.9680
log 4.003 = 0.6024 | log Num = 3̄.5870 | log Den. = 1.7752
log cos 15° 27' = 1̄.9840 | log Den = 1.7752 |
7 log x = 5̄.8118
log x = 1̄.4017
x = 0.2522 or 2.522 x 10⁻¹
Ans.
(i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in A.P., prove that a², b², c² are
also in A.P.
By hypothesis: (a+b)⁻¹ - (c+a)⁻¹ = (c+a)⁻¹ - (b+c)⁻¹ or
1/(a+b) - 1/(c+a) = 1/(c+a) - 1/(b+c) or
(c+a-a-b)/((a+b)(c+a)) = (b+c-c-a)/((c+a)(b+c)) or (c-b)/((a+b)(c+a)) = (b-a)/((c+a)(b+c))
∴ (c-b)/(a+b) = (b-a)/(b+c) or (c-b)(c+b) = (b-a)(b+a) or
c² - b² = b² - a² which makes a², b², c² in A.P. by definition of A.P.
Q.E.D.
(ii)
⟦diagram of a bouncing ball with vertical arrows and labels 10, 5, 2.5⟧ + soon.
② total distance it travels before coming to rest =
= S_{n→∞} = 10 + (10 + 5 + 5/2 + ... to infinity)
= 10 + (10 / (1 - 1/2)) = 10 + 10 / (1/2) = 10 + 20 = 30 ft.
Ans. 2
① when the ball has struck the ground for the first time it has travelled 10 ft. When
it strikes the ground for the 2nd time it will have gone upward 5 ft & downward 5 ft also
∴ it will have travelled another 10 ft, the next distance will be 5 ft + so on
∴ the total distance travelled by the end of the 10th time it strikes the ground
will be 10 + 10 + 5 + 5/2 + 5/4 + ... to ten terms = 10 + (10 + 5 + 5/2 + 5/4 + ... to 9 terms)
∴ total distance = 10 + (10(1 - (1/2)⁹)) / (1 - 1/2) = 10 + (10(1 - 1/512)) / (1/2) = 10 + 20 [1 - 1/512] = 10 + 20 (511/512)
= 10 + 10220 / 512 = 10 + 19 123/128 = 29 123/128 ft Ans. 1 [See above →]
Archival unit 15
5 (i) Let y = 3 + 2x - x² x | y -2 | -5 -1 | 0 0 | 3 1 | 4 2 | 3 3 | 0 4 | -5 ⟦Graph showing parabola and straight line⟧ y = x/2 + 2 (1, 4) A(2, 3) B(-1/2, 1 3/4) (ii) The roots of the equation x² - 3 = 2x are the same as the roots of the equation 3 + 2x - x² = 0 ∴ the roots are the values of x when y = 0 or the roots are x = -1 } Ans. 2 they are the abscissas of the pts. of intersection with the and x = 3 } x-axis. (iii) the function 3 + 2x - x² is always positive for all points on the curve above the x-axis i.e. 3 + 2x - x² > 0 when -1 < x < 3 Ans. 3 (iv) the function 3 + 2x - x² is greatest when x = 1 + y_max = 4 Ans. 4 (v) we plot the st. line y = x/2 + 2 of which two points are: (-4, 0) + (0, 2) we join these points + we get the st. line AB which intersects the curve at A(2, 3) + B(-1/2, 1 3/4) ∴ for all values of x between -1/2 and 2, the curve of 3 + 2x - x² lies above the st. line x/2 + 2 ∴ 3 + 2x - x² > x/2 + 2 when -1/2 < x < 2. (Ans. 5) (vi) The roots of the equation 3 + 2x - x² = x/2 + 2 are the values of x which will make the ordinate of the curve 3 + 2x - x² equal to the ordinate of the st. line x/2 + 2. But the ordinates of the curve + the line are equal at the pts. of intersection A + B. ∴ the roots are the abscissas of A + B or x = -1/2 and x = 2 Ans. 6
Archival unit 16
⟦illegible⟧ Shamash Secondary School Final Examination May 1969 ⟦line⟧ Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. Answer all questions: 1. (i) In the expression x³ +Ax² +31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) 2. (i) Solve the equation (log x)² = log x ⟦-10, finding two values for x.⟧ (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ........(1) x² -xy+y² -7=0 .........(2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√¹⁻ˣ 8) = (√ˣ 2)(√ 50) (10 marks) (ii)Compute the value of ⁷√ (0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27' (10 marks) 4. (i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) bounce (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous ⟦bounce⟧. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2)..
Archival unit 17
(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ⟦line⟧ 5. (i) Plot the curve of the function 3+2x-x² for values of x from x=-2 to x=4, choosing one half of an inch for each unit on the axis of x and on the axis of y. (4 marks) (ii) From your graph, find the roots of the equation x²-3=2x. (3 marks) (iii) Find the values of x for which the function 3+2x-x² is always positive. ( 3 marks) (iv) Find from your diagram the value of x at which the function 3+2x-x² is greatest and state the maximum value. (3 marks) (v) By plotting another curve on the same diagram, find the values of x for which 3+2x-x² > x/2 + 2. (4 marks) (vi) From your last diagram, find the roots of the equation 3+2x-x² = x/2 + 2. (3 marks) ⟦line⟧
Archival unit 18
Shamash Secondary School Final Examination May 1969 Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. Answer all questions: 1. (i) In the expression x³+Ax²+31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) 2. (i) Solve the equation (log x)² = log x⁷ - ⟦10⟧, finding two values for x. (10 marks) ii) Solve the two simultaneous equations: 3x²+xy-2y²+7=0 ⟦line⟧(1) x²-xy+y²-7=0 ⟦line⟧(2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√ 8¹⁻ˣ) = (√ 2ˣ) (√ 50) (10 marks) (ii)Compute the value of ⁷√[(0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27'] (10 marks) 4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous bounce. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2)..
Archival unit 19
(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ⟦line⟧ 5. (i) Plot the curve of the function 3+2x-x² for values of x from x=-2 to x=4, choosing one half of an inch for each unit on the axis of x and on the axis of y. (4 marks) (ii) From your graph, find the roots of the equation x²-3=2x. (3 marks) (iii) Find the values of x for which the function 3+2x-x² is always positive. ( 3 marks) (iv) Find from your diagram the value of x at which the function 3+2x-x² is greatest and state the maximum value. (3 marks) (v) By plotting another curve on the same diagram, find the values of x for which 3+2x-x² > x/2 + 2. (4 marks) (vi) From your last diagram, find the roots of the equation 3+2x-x² = x/2 + 2. (3 marks) ⟦line⟧
Archival unit 20
Solution to the 3rd + 4th Quarter Exam. in Algebra 7/4/1969 1. (i) The square root of (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² is equal: √[(a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)²] = √[(a³ - 1/a³) - 3(a - 1/a)]² = a³ - 1/a³ - 3(a - 1/a) = a³ - 3a + 3/a - 1/a³ also ³√[a³ - 3a + 3/a - 1/a³] = ³√(a - 1/a)³ = a - 1/a Ans. (ii) Prove the identity: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) L.H.S. = bc(b-c) + ac² - a²c + a²b - ab² = bc(b-c) + a²(b-c) - a(b²-c²) = (b-c)[bc + a² - a(b+c)] = (b-c)(bc + a² - ab - ac) = (b-c)[a(a-b) - c(a-b)] = (b-c)(a-b)(a-c) = -(b-c)(c-a)(a-b) Q.E.D. 2. (i) solve: (x-1)/(√x - 1) = 3 + (√x + 1)/2 ∴ 2(x-1) = 6(√x - 1) + (√x - 1)(√x + 1) ∴ 2x - 2 = 6√x - 6 + x - 1 ∴ 6√x = x + 5 ∴ 36x = x² + 10x + 25 ∴ x² - 26x + 25 = 0 ∴ (x-1)(x-25) = 0 ∴ x = 1 and x = 25 } Ans. but x = 1 does not satisfy the original equation + should be rejected. Hence x = 25 Ans. 3 (i) x = ⁷√[(0.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')²] log 0.002001 = 3.3012 | 3 log 0.002001 = 9.9036 | 5 log 1.003 = 0.0060 log sin 16° 23' = 1.4504 | 2 log sin 16° 23' = 2.9008 | 2 log tan 41° 16' = 1.9466 log 1.003 = 0.0012 | log Num. = 10.8044 | log Den = 1.9526 log tan 41° 16' = 1.9433 | log Den. = 1.9526 | | 7 log x = 10.8518 | | log x = 2.7017 0.05031 | | x = 0.07975 | | or x = 7.975 X 10⁻² } Ans.
Archival unit 21
Solution to 3rd + 4th Quarter Exam - Cert. in Algebra 7/4/69
Page 2
√1+x + √1-x / √1+x - √1-x = (√1+x + √1-x)² / 1+x - (1-x) = 1+x + 1-x + 2√1-x² / 2x = 2 + 2√1-x² / 2x =
= 2(1 + √1-x²) / 2x = 1 + √1-x² / x = 1 + √1 - 4b² / (b²+1)² / 2b / b²+1 = 1 + √((b²+1)² - 4b²) / b²+1 / 2b / b²+1
= b²+1 + √b⁴+2b²+1-4b² / 2b = b²+1 + √(b²-1)² / 2b = b²+1 + b²-1 / 2b = 2b² / 2b = b Ans.
3 (ii) solve for x :
2(log x)² - 5(log x) + 2 = 0 ∴ [2(log x) - 1][log x - 2] = 0
∴ 2 log x = 1 or log x = 1/2 ∴ x = 10^(1/2) or x = √10 = 3.162 Correct to 3 dec. pl.
or log x = 2 or x = 10² or x = 100 } Ans.
4.(i) a = 3 } Prove S_2n = 4 S_n
d = 6 }
S_2n = 2n/2 {2a + (2n-1)d} = n {6 + (2n-1)(6)} = n {6 + 12n - 6} = 12 n²
S_n = n/2 {6 + (n-1)(6)} = n/2 {6n} = 3 n²
but 12 n² = 4 (3 n²) ∴ S_2n = 4 S_n Q.E.D.
(ii) t_3 / t_6 = 11 / 26 and S_4 = 34 , S_8 = ?
∴ a + 2d / a + 5d = 11 / 26 or 26a + 52d = 11a + 55d or 15a - 3d = 0 .... ①
also 34 = 4/2 {2a + 3d} or 34 = 2(2a + 3d) or 17 = 2a + 3d .... ②
∴ 15a - 3d = 0 .... ① } or <del>20 a + 7 d = 9</del> } <del>23 d = 170</del> <del>d = ⟦illegible⟧</del>
2a + 3d = 17 .... ② } <del>20 a + 30 d = 170</del> }
17a = 17 ∴ a = 1 ∴ 15 = 3d ∴ d = 5 ∴ a = 1 } Ans. 1
d = 5 }
+ the progression is 1, 6, 11, 16, ...
S_8 = 8/2 {2 x 1 + 7 x 5} or S_8 = 4(2 + 35) = 4 x 37 = 148 Ans. 2
Archival unit 22
Shamash Secondary School 3rd Quarter + 4th Quarter Examination Page. 1 Subject: Algebra Date: 7/4/1969 Class: 4th Year, scientific Time: 8:30 - 10:30 a.m. Answer all questions: 1. (i) By first taking the square root and then the cube root, find the sixth root of: (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² (12 marks) (ii) Prove that the left hand side is always equal to the right hand side in the following equation: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) (13 marks) 2. (i) solve the equation : (x-1)/(√x -1) = 3 + (√x +1)/2 (12 marks) (ii) Rationalise the denominator and then find the value of: (√1+x + √1-x) / (√1+x - √1-x) , when x = 2b / (b²+1) (13 marks) 3. (i) Compute : ⁷√[ (6.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')² ] (12 marks) (ii) Solve the following equation for x: 2(log x)² - 5(log x) + 2 = 0 (13 marks) 4. (i) In an arithmetic Progression the first term is 3 and the common difference is 6 . Show that the sum of 2n terms is always equal to four times the sum of n terms . (12 marks) (ii) In an A.P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34 . Find the progression and the sum of the first 8 terms . (13 marks)
Archival unit 23
SHAMASH SECONDARY SCHOOL 3rd & 4th quarter Examination. Subject: Algebra Class: 4th Year Scientific Date: 7/4/1969. Time: 8:30-10:30 a.m. ⟦line⟧ Answer all Questions : 1. (i) By first taking the square root and then the cube root, find the sixth root of : ( a³ - 1/a³ )² - 6( a - 1/a )( a³ - 1/a³ ) + 9( a - 1/a )². (12 marks). (ii) Prove that the left-hand side is always equal to the right-hand side in the following equation : bc(b - c) + ca(c - a) + ab(a - b) = - (b - c)(c - a)(a - b). (13 marks). 2. (i) Solve the equation : (x - 1)/(√x - 1) = 3 + (√x + 1)/2. (12 marks). (ii) Rationalise the denominator and then find the value of : (√(1 + x) + √(1 - x)) / (√(1 + x) - √(1 - x)) , when x = 2b / (b² + 1) . (13 marks). 3. (i) Compute by logarithms : ⁷√[ (0.⟦002001⟧)³(sin16° 23')² / (1.003)⁵(tan41° 16')² ] (12 marks). (ii) Solve the following equation for x : 2(log x)² - 5(log x) + 2 = 0 . (13 marks). 4. (i) In an Arithmetic Progression the first term is 3 and the common difference is 6. Show that the sum of 2n terms is always equal to four times the sum of n terms. (12 marks). (ii) In an A. P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34. Find the progression and the sum of the first 8 terms. (13 marks). ⟦line⟧
Archival unit 24
Solutions to Mid-year Exam., in Algebra 4th Secondary year, February, 1969. 1 1. Let the time now be x minutes after 5 o'clock. then x = 25 + x/12 + 30 (ACB = 30) ∴ x - x/12 = 55 ∴ 11x/12 = 55 ∴ x = 60 minutes after 5 ∴ the time now is exactly 6 o'clock Ans. ⟦Diagram of a clock with labels: B at 12, A at 6, C at 9, and an arrow at 5 labeled x min.⟧ 2. (i) solve 6 x³+19 x²+x-6 = 0 By trial + error we discover that x = -3 satisfies the equation. Hence by the remainder + factor theorems, (x+3) is a factor. Factoring, we get (x+3) (6x² + x - 2) = 0 or (x+3) (3x+2) (2x-1) = 0 ∴ x = -3 x = - 2/3 } Ans. x = 1/2 (ii) (3x-7)/(x-2) + (2x-5)/(x-3) = (3x+7)/(x+2) + (2x+5)/(x+3) ∴ (3(x-2)-1)/(x-2) + (2(x-3)+1)/(x-3) = (3(x+2)+1)/(x+2) + (2(x+3)-1)/(x+3) ∴ 3 - 1/(x-2) + 2 + 1/(x-3) = 3 + 1/(x+2) + 2 - 1/(x+3) 1/(x-3) - 1/(x-2) = 1/(x+2) - 1/(x+3) ∴ (x-2-x+3)/((x-2)(x-3)) = (x+3-x-2)/((x+2)(x+3)) ∴ 1/((x-2)(x-3)) = 1/((x+2)(x+3)) ∴ (x+2)(x+3) = (x-2)(x-3) ∴ x²+5x+6 = x²-5x+6 ∴ 10x = 0 ∴ x = 0 Ans.
Archival unit 25
2 3. (i) A (x² - 2x) + B (x + 4) + C = 3x² + x + 25 ∴ Ax² - 2Ax + Bx + 4B + C = 3x² + x + 25 ∴ Ax² + (B - 2A) x + 4B + C = 3x² + x + 25 . Equating coefficients of like terms, ∴ A = 3 , B - 2A = 1 ∴ B - 6 = 1 or B = 7 4B + C = 25 ∴ 4 x 7 + C = 25 or C = -3 ∴ A = 3 , B = 7 and C = -3 Ans. Another method : In the original identity, let x = 0 , then 4B + C = 25 ... ① again let x = 2 , then 6B + C = 12 + 2 + 25 = 39 ... ② 6B + C = 39 ... ② 4B + C = 25 ... ① } subtract 2B = 14 ∴ B = 7 and C = -3 now let x = 1 , then A (1 - 2) + B (1 + 4) + C = 3 + 1 + 25 or -A + 5 x 7 - 3 = 29 ∴ A = 32 - 29 ∴ A = 3 , B = 7 and C = -3 Ans. (ii) Solve : x² + xy + 2y² = 8 ... ① 2x² - 2xy - 3y² = 1 ... ② Let y = mx ∴ x² + mx² + 2m²x² = 8 from ① ..... (1.a) 2x² - 2m x² - 3m²x² = 1 from ② ..... (2.a) x² (1 + m + 2m²) = 8 } dividing, (1 + m + 2m²) / (2 - 2m - 3m²) = 8 / 1 x² (2 - 2m - 3m²) = 1 } ∴ 1 + m + 2m² = 16 - 16m - 24m² ∴ 26m² + 17m - 15 = 0 ∴ (13m + 15) (2m - 1) = 0 ∴ m = 1/2 or m = -15/13 when m = 1/2 , from (1.a) , we get: x² + x²/2 + x²/2 = 8 or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 when x = 2 , y = mx = 1/2 x 2 = 1 and when x = -2 , y = 1/2 (-2) = -1 x = 2 } Ans. 1 x = -2 } Ans. 2 y = 1 } y = -1 } when m = -15/13 then x² (1 - 15/13 + 2 x 225/169) = 8 or x² ( (169 - 13 x 15 + 450) / 169 ) = 8 ∴ x² ( (169 - 195 + 450) / 169 ) = 8 ∴ x² ( 424 / 169 ) = 8 ∴ x² = (8 x 169) / 424 = 169 / 53 ∴ x = ± 13 / √53 when x = 13 / √53 , y = -15/13 . 13 / √53 = -15 / √53 and when x = -13 / √53 , y = (-15/13) (-13 / √53) = 15 / √53 x = 13 / √53 } Ans. 3 x = -13 / √53 } Ans. 4 y = -15 / √53 } y = 15 / √53 }