AI English Translation, Pages 51-75
⟦illegible⟧ Algebra Exam
Mid-Year Exam
(i) (x+2)x - 3(x+2) = (x+2)⟦illegible⟧ = (x+2)(x-3) Ans.
(ii) 201x² - 99x - 102 = 3(67x² - 33x - 34) = 3(67x+34)(x-1) Ans.
(iii) x⁹ - 64x³ - x⁶ + 64 = (x⁹ - x⁶ - 64x³ + 64) = x⁶(x³-1) - 64(x³-1)
= (x⁶-64)(x³-1) = (x³-8)(x³+8)(x³-1)
= (x-2)(x²+2x+4)(x+2)(x²-2x+4)(x-1)(x²+x+1) Ans.
(b) Let x and y be factors of A and B in F, and let A = aF and B = bF.
Then ⟦illegible⟧ F ⊆ (a+b)F if F is also a common factor
of ⟦illegible⟧
2(a) (A+B)x² + (B+C)x + (A+C) = 5x² + 3x + 4 is true for all values
of x. Letting x=0, we get A+C=4; x=1, A+B+C=5 ... ①
Letting x=-1, A-B+C=3 ... ②
From ① & ② by subtraction, 2B = 2 or B = 1
From ①, A+1+C=5 or A+C=4
From ②, A-1+C=3 or A+C=4
⟦illegible⟧
A=2, B=1, C=2 Ans.
⟦illegible⟧ x² + ax + b = (x-3)Q + 0 ∴ (x-3) is a factor
⟦illegible⟧ (x-2) is a factor
⟦illegible⟧
A=2, B=3 Ans.
⟦illegible⟧ Secondary School
Final Examination, May 1968
Subject: Mathematics
Date: 14/5/1968
Class: 4th Scientific Year
Time: 8:00-11:00 a.m.
⟦line⟧
All questions are to be attempted.
1. (i) Resolve into four factors: x⁹+x³y⁶-8x⁶y³-8y⁹
(6 marks)
(ii) Simplify: { (a⁴-x⁴)/(a²-2ax+x²) ÷ (a²+ax)/(a-x) } x { (a⁵-a³x²)/(a³+x³) ÷ (a⁴-2a³x+a²x²)/(a²-ax+x²) }
(7 marks)
(iii) Find the value of B if 2x⁴+2x^(7/2)-5x³-x^(5/2)+3x²-x^(3/2)+x^(1/2)+x+3 +B is
exactly divisible by x+x^(1/2)-2
(7 marks)
2. (i) Compute by logarithms the value of x, if x= ⁷√[ (0.1023)³.Cos²41°28' / (1.007)²(tan⁵47°51') ]
arranging your work neatly.
(10 marks)
(ii) Solve for x the equation: (31.01)^(3x-1) = 104(2.003)^(2x+1)
(10 marks)
3. I rode one third of a journey at 10 miles an hour, one third more
at 9, and the rest at 8 miles an hour. If I had ridden half the
journey at 10, and the other half at 8 miles per hour, I should
have been half a minute longer on the way. What distance did I
ride ?
(20 marks)
4. (i) The sum of n terms of a series is 1/3n(4n²-1). Find the first two
terms.
(6 marks)
(ii) In boring a well 400 ft deep the cost is 2s. 3d. for the first
foot and an additional penny for each subsequent foot. What is
the cost of boring the last foot, and also of boring the entire
well ?
(7 marks)
(iii) The third term of a geometric series, in which all the terms
are positive, is 18 and the fifth term is 40.5. Find the first
term and the sum of the first six terms.
(7 marks)
5. (i) Draw the graph of y=1/4(3x²-5x-4) for values of x from -2 to +3,
using a scale of 1 inch to 1 unit on each axis.
(7 marks)
(ii) Use your graph to find the least value of 3x²-5x-4.
(6 marks)
(iii) By drawing the appropriate straight line on your graph solve
the equation 3x²-5x-6=0.
(7 marks)
⟦line⟧
y = 4x - 3
y = x² - 4x - 12
x | y
-3 | 9
-2 | 0
-1 | -7
0 | -12
1 | -15
2 | -16
3 | -15
4 | -12
5 | -7
6 | 0
7 | 9
1. The ⟦illegible⟧
2. ⟦illegible⟧
⟦illegible⟧
⟦illegible⟧ (2, -16) ⟦illegible⟧
⟦illegible⟧
x = -2, x = 6
⟦illegible⟧
⟦illegible⟧ x² - 8x - 9 < 0 ⟦illegible⟧
(2, -16)
Shamash Secondary School
3rd Quarter Examination, March, 1968
Subject: Algebra
Date: Sunday 24/3/1968
Class: 4th Year Scientific
Time: 10:15 - 11:45
⟦line⟧
1. Draw on the same diagram the graphs of the function 4x-3 and
of the function 4x²-4x-15, taking ½ inch as one unit on the
x-axis and one tenth of an inch as one unit on the y-axis.
(20 marks)
2. From your diagram, find the roots of the two simultaneous
equations:
y = 4x-3 ⟦line⟧ (1)
y = 4x²-4x-15 ⟦line⟧ (2) (20 marks)
3. From ⟦the⟧ graph of the function 4x²-4x-15, find the roots of the
equation 4x² = 4x + 15 (20 marks)
4. From your diagram find also the least value of 4x²-4x-15, and the
value of x corresponding to the least value of the function.
(20 marks)
5. By drawing an additional graph, find the values of x for which the
expression 4x²-4x-15 is always less than 9.
(20 marks).
⟦line⟧
SHAMASH SECONDARY SCHOOL
2nd Quarter Examination
Subject: Algebra
Date: 31/12/1967.
Class: 4th Year, Scientific Section.
Time: 10:15-11:45 a.m.
⟦line⟧
Attempt all questions:
12 12
1. (i) Resove into five factors : X - Y (10 marks)
(ii) Find the value of 'A' which will make the expression
3 2
6x + Ax + x - 6 divisible by (x + 3) and find the other factors.
(10 marks)
2. Find the square root of :
9 6 5 74 4 61 3 62 2 2 1
- x - 2x + --x - --x + --x - -x + - . (20 marks)
4 45 30 75 5 4
a - b a² - b²
1 + ----- 1 + -------
a + b a² + b²
3. (i). Reduce to simplest form : --------- ÷ ----------- (10 marks)
a - b a² - b²
1 - ----- 1 - -------
a + b a² + b²
5x - 8 6x - 44 10x - 8 x - 8
(ii). Solve the equation : ------ + ------- - ------- = ----- (10 marks)
x - 2 x - 7 x - 7 x - 6
4. Find the values of x, y, and z from the following equations :
3x - 2y + 4z = 3y - 2x + 7 = 7x + 2z - 2 = 11 . (20 marks)
5. A basket of eggs is emptied by one person taking half of them and
one more, a second person taking half of the remainder and one more, and a
third person taking half of the remainder and six more. How many did the
basket contain at first? (20 marks)
⟦line⟧
⟦illegible⟧ (x - y) (x + y) (x² + xy + y²) (x² - xy + y²)
⟦illegible⟧ (x - y) (x² + xy + y²) (x + y) (x² - xy + y²)
⟦illegible⟧ (x³ - y³) (x³ + y³)
⟦illegible⟧ (x⁶ - y⁶)
⟦illegible⟧ the divisibility ⟦illegible⟧
⟦illegible⟧ x⁴ + Ax³ + Bx² + x - 6 ⟦illegible⟧ x² + x - 2
⟦illegible⟧ (x⁴ + x³ - 2x²)
⟦illegible⟧ (A - 1) x³ + (B + 2) x² + x - 6
⟦illegible⟧ (A - 1) x³ + (A - 1) x² - 2(A - 1) x
⟦illegible⟧ (B - A + 3) x² + (2A - 1) x - 6
⟦illegible⟧ (B - A + 3) x² + (B - A + 3) x - 2(B - A + 3)
⟦illegible⟧ (3A - B - 4) x + 2B - 2A ⟦illegible⟧
⟦illegible⟧ 3A - B - 4 = 0 ⟦illegible⟧ 2B - 2A - 6 = 0
⟦illegible⟧ 3A - B = 4 ⟦illegible⟧ B - A = 3
⟦illegible⟧ 2A = 7 ⟦illegible⟧ A = 3.5 ⟦illegible⟧ B = 6.5
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ ⟦...⟧ ⟦illegible⟧
⟦illegible⟧ x - ⟦illegible⟧
⟦illegible⟧ x^2 + ⟦illegible⟧ x
⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧
⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧
⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧
⟦illegible⟧
31/12/1967
⟦illegible⟧ (1) ⟦illegible⟧ (1 + (a-b)/(a+b)) / (1 + (a^2-b^2)/(a^2+b^2)) = (a+b+a-b)/(a+b) / (a^2+b^2+a^2-b^2)/(a^2+b^2)
= 2a/(a+b) / 2a^2/(a^2+b^2) = a/(a+b) * (a^2+b^2)/a^2 = (a^2+b^2)/(a(a+b))
(2) (5x-2)/(x-2) + (6x-44)/(x-7) - (10x-8)/(x-1) = (x-5)/(x-6)
(5(x-2)+8)/(x-2) + (6(x-7)-2)/(x-7) - (10(x-1)+2)/(x-1) = (x-6+1)/(x-6)
5 + 8/(x-2) + 6 - 2/(x-7) - 10 - 2/(x-1) = 1 - 2/(x-6)
8/(x-2) - 2/(x-7) = 2/(x-1) - 2/(x-6) or 1/(x-2) - 1/(x-7) = 1/(x-1) - 1/(x-6)
(x-7-(x-2))/((x-2)(x-7)) = (x-6-(x-1))/((x-1)(x-6)) or -5/((x-2)(x-7)) = -5/((x-1)(x-6))
1/((x-2)(x-7)) = 1/((x-1)(x-6)) or (x-2)(x-7) = (x-1)(x-6)
x^2 - 9x + 14 = x^2 - 7x + 6 or 2x = 8 or x = 4 Ans.
(10 marks)
⟦line⟧
3x - 2y + 4z = 11 ; 2y + 4x + 8z = 24 ; 7x + 2y - z = 11
3x - 2y + 4z = 11 ---- (1) ; 4x + 4y + 8z = 24
-7x + 2y - z = 11 ---- (2) ; -6x + 3y = 12
⟦illegible⟧ + 8z = 24
2y - 2x + 7 = 11 ⟦illegible⟧ 24
7x + 2y = 13 ⟦illegible⟧
⟦illegible⟧ = 54
25x + 8z = 34 ⟦illegible⟧
21x + 4z = 22 ⟦illegible⟧
⟦illegible⟧ = 174
⟦illegible⟧ = 2
from eq (2) 14 + 2y - z = 11 or 2y - z = -3
from eq (1) -6x + 6(1) = 12 or -6x = 6 or x = -1
⟦illegible⟧
31/12/1948
5. ⟦illegible⟧
⟦illegible⟧ takes (x/2 + 1) ⟦illegible⟧ Remainder (x/2 - 1) ⟦illegible⟧
and ⟦illegible⟧ takes ⟦illegible⟧ + 1 ⟦illegible⟧ Remainder = (x/2 - 1) - (x/4 + 1/2 + 1) = x/4 - 5/2
⟦illegible⟧ = x/4 - 5/2 - (x/8 - 5/4 + 1) = x/8 - 9/4
⟦illegible⟧ who ⟦illegible⟧ + 6 = x - 6 + 42 = x + 42
⟦illegible⟧
x/2 + x/4 + x/8 + x/16 + ⟦illegible⟧ = x + 42
x = 54 ⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
1. (i) Factorise ⟦illegible⟧
(ii) Find the value of k which will make the expression 6x³ + 8x² + x - k
divisible by ⟦illegible⟧ the other factors. (10 marks)
2. Find ⟦illegible⟧
⟦illegible⟧ x⁶ - 2x⁵ + ⟦illegible⟧ x⁴ - ⟦illegible⟧ x³ + ⟦illegible⟧ x² - ⟦illegible⟧ x + ⟦illegible⟧
3. (i) Reduce to simplest form : ⟦illegible⟧
(ii) Solve the equation : (5x - 9) / (x - 2) + (4x - 11) / (x - 3) = (10x - 8) / (x - 1) - (x - 2) / (x - 4) (10 marks)
4. Find the values of x, y and z from the following equations:
3x - 2y + 4z = 2y - 3x + 7 = 7x + 2z - 2 = 11
5. A basket of eggs is emptied by one person taking half of them and one more,
a second person taking half of the remainder and one more, and a third person
taking half of the remainder + one more. How many did the basket
contain at first? (10 marks)
Solution to Monthly Quiz 12/11/1967
4th year secondary, 1967
II: (a) k = 20 ab / (4a + 5b) (i) 20ab = 4ak + 5bk ∴ 20ab - 4ak = 5bk
∴ 4a(5b - k) = 5bk ∴ a = 5bk / 4(5b - k) Ans. 1
5
(ii) 20ab - 5bk = 4ak ∴ 5b(4a - k) = 4ak
∴ b = 4ak / 5(4a - k) Ans. 2 5
√((k - 4a) / (k - 5b)) = √((20ab / (4a + 5b) - 4a) / (20ab / (4a + 5b) - 5b)) = √((20ab - 16a² - 20ab) / (20ab - 20ab - 25b²))
= √(16a² / 25b²) = 4a / 5b Ans. 3. 5
(b) a = 0, b = 1, c = -2, d = 2
10
(3abc - abcd) ∛(a³bc - c³bd + 3) = [(0) - (0)] ∛(0 - (-2)³(1)(2) + 3)
= 12 ∛(24 + 3) = 12 ∛(27) Ans. = 36 Ans.
(c) 3/2 x² - ax - 2/3 a²
3/4 x² - 1/2 ax + 1/3 a²
x ⟦line⟧
9/8 x⁴ - 3/4 ax³ - 1/2 a²x² 15
- 3/4 ax³ + 1/2 a²x² + 1/3 a³x
1/2 a²x² - 1/3 a³x - 2/9 a⁴
⟦line⟧
9/8 x⁴ - 3/2 ax³ + 1/2 a²x² - 2/9 a⁴
40
Number:
Shamash Secondary School
Monthly Quiz.
Name:
Subject: Algebra
Date: 12/11/1967
Class: 4th Year Secondary
Time: 1:15- 11:45 a.m.
I. Give the English equivalent to the following and fill in the blanks in
this sheet, handing it back with your answer book:
1- In the expression 3x, the coefficient of x is
1.
2- We move the terms from one side to the other side of the equation and combine like terms
2.
3- We eliminate fractions
3.
4- We unify the denominators of the fractions with the simplest common denominator
4.
5- The value of the expression 450384 to the nearest four significant figures is
5.
6- The terms of a fraction are its numerator and its denominator
6.
7- We measure the length of a straight line and find it equals 5.11 cm while its exact length is 6.0 cm. In this
case, we say that the absolute error is and the relative error is
and the percentage error is
7.
8- The equation: 2x² - 5y² + 4z² = 7 is a second-degree equation with three unknowns.
8.
9- In every division process there is a dividend, a divisor, and a quotient, and in some cases a remainder
9.
10- Reciprocal of the number - complementary angles - supplementary angles - perimeter of the polygon
10.
(Continued p. 2) . .
Number ::
Name ::
4th Year - Algebra -p.2- Monthly Quiz.
II. (a) If k = 20ab / (4a+5b) find (i) "a" in terms of "b" and "k"
(ii) "b" in terms of "a" and "k".
Find also the value of √((k - 4a) / (k - 5b)) in terms of "a" and "b".
(b) If a = 0, b=1, c=-2, d=3, find the value of
(3abc - 2bcd) ∛(a³bc - c³bd+3)
(c) Find the product of 3/2 x² - ax - 2/3 a² and 3/4 x² - 1/2 ax + 1/3 a²
⟦line⟧
Shamash Secondary School
Conditional Examination, Sept. 1967
Subject: Algebra
Date: 8/9/1967
Class: 4th Year Secondary
Time: 8:00 - 11:00 a.m.
⟦line⟧
Attempt all questions:
1. (i) Find the value of k if the expression 6x³-13x²+18x+k is exactly
divisible by 2x²-3x+4
(10 marks)
(ii) If the n th term of a series is (2n+1)/(2n+3) write down the first three
terms and express the difference between the n th and (n+1)th terms
as a single fraction in its simplest form.
(10 marks)
2. (i) If 3x²-4x+5 = a(x-b)²+c for all values of x, find the values of
a, b, and c.
(10 marks)
Hence, or otherwise, find the least value of 3x²-4x+5.
(ii) Find the lapse of time in minutes between the two instants when
the two hands of a watch are at right angles for the 1st and the 2nd
time between four O'clock and five O'clock.
(10 marks)
3. (i) Compute by logarithms the following expression :
⁹√[ (Sin² 15° 04' x Cos³ 31° 31') / ((510.7)² x (4.007)³) ]
(10 marks)
(ii) Find the value of x from the following equation correct to four
significant figures:
32^(2x-1) = 64^x * 40
(10 marks)
4. (i) Three times the third term of an arithmetic progression is twice
the sixth term. The sum of the first, third and fifth terms is 9.
Find:
(a) the ratio of the ninth term to the sixth term,
(b) the sum of the first thirteen terms of the progression.
(10 marks)
(ii) The third term of a geometric progression, in which all the
terms are positive, is 2/3 and the sum of the first two terms is 2½.
Find the first term, the common ratio and the fourth term of the
progression.
(10 marks)
5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the
y-axis draw the graphs of y = 4-x² and 4y = 5x + 4 for values
of x from -3 to +3.
( 8 marks)
(ii) From your graph, find:
a- the range of values of x for which 4-x² is greater than 5/4x+1,
(4 marks)
b- the values of x for which 4-x²=2.5,
(4 marks)
c- the square root of 5.6.
(4 marks)
⟦line⟧
Conditional examination, September 1967
1
subject: Algebra
Date: 8/9/1967
class: 4th Year Secondary
Time: 8:00 - 11:00 a.m.
Attempt all questions:
1. (i) Find the value of k if the expression 6x³-13x²+12x+k is exactly
divisible by 2x²-3x+4 (10 marks)
(ii) If the n'th term of a series is ⟦(2n+1)/(2n+1)⟧ write down the first three
terms and express the difference between the n'th and (n+1)th term
as a single fraction in its simplest form. (10 marks)
2. (i) If 3x²-4x+5 = a(x-1)² + ⟦b(x-1)⟧ + c for all values of x, find the values of 'a',
'b', and 'c'. (10 marks)
Hence, or otherwise, find the least value of 3x²-4x+5.
(ii) Find the lapse of time in minutes between the two instants
when the two hands of a watch are at right angles for
the 1st and the 2nd time between <del>four o'clock and five o'clock</del>
four o'clock and five o'clock. (10 marks)
3. (i) Compute by logarithms the following expression:
⟦√[ (sin 15° 04' * cos³ 21° 31') / ((5.127)² * (4.007)³) ]⟧ (10 marks)
(ii) Find the value of x from the following equation correct to
four significant figures.
<del>⟦illegible⟧</del> 3^(2x-1) = 6.4 * 40 (10 marks)
4. (i) Three times the third term of an arithmetic progression is twice the
sixth term. The sum of the first, third and fifth terms is 9.
Find:
(a) the ratio of the ninth term to the sixth term,
(b) the sum of the first thirteen terms of the progression. (10 marks)
(ii) The third term of a geometric progression, in which all the terms are positive,
is 2 and the ⟦illegible⟧
⟦Conditional⟧ exams ⟦illegible⟧
4th year secondary
2
5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis
draw the graphs of y = 4 - x² and 4y = 5x + 4 for values of x from -3
to +3. (8 marks)
(ii) From your graph find:
(a) the range of values of x for which 4 - x² is greater than
5/4 x + 1, (4 marks)
(b) the values of x for which 4 - x² = 2.5, (4 marks)
(c) the square root of 3.6. (4 marks)
⟦line⟧
⟦illegible⟧
⟦Conditional exam July 1957⟧
4th year secondary :
2
5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis
draw the graphs of y = 4 - x² and 4y = 5x + 4 for values of x from -3
to +3. (8 marks)
(ii) From your graph find:
(a) the range of values of x for which 4 - x² is greater than
5/4 x + 1, (4 marks)
(b) the values of x for which 4 - x² = 2.5, (4 marks)
(c) the square root of 3.6. (4 marks)
⟦line⟧
Solution to Conditional examination in Algebra , Sept., 1967, 4th year
①
1. (i) 2x² - 3x + 4 | 6x³ - 13x² + 18x + k | 3x - 2
| 6x³ - 9x² + 12x
⟦line⟧
- 4x² + 6x + k
| - 4x² + 6x - 8
⟦line⟧
8 + k = 0 ∴ k = -8 Ans.
(ii) nth term = 2n+1 / 2n+3 ∴ 1st term = 2+1 / 2+3 = 3/5 , 2nd term = 2x2+1 / 2x2+3 = 5/7
3rd term = 2x3+1 / 2x3+3 = 7/9
(n+1)th term - nth term = 2(n+1)+1 / 2(n+1)+3 - 2n+1 / 2n+3 = 2n+3 / 2n+5 - 2n+1 / 2n+3 =
= (2n+3)² - (2n+1)(2n+5) / (2n+5)(2n+3) = 4n² + 12n + 9 - (4n² + 12n + 5) / (2n+5)(2n+3)
= 4 / (2n+5)(2n+3) Ans.
2 (i) 3x² - 4x + 5 = a(x - b)² + c ∴ 3x² - 4x + 5 = ax² - 2abx + ab² + c
∴ a = 3 , ∴ -2ab = -4 or 3b = 2 or b = 2/3
∴ ab² + c = 5 or 3 x 4/9 + c = 5 or c = 5 - 4/3 or c = 11/3 = 3 2/3
∴ a = 3 , b = 2/3 , c = 11/3 = 3 2/3 Ans.
Hence 3x² - 4x + 5 = 3(x - 2/3)² + 11/3 and since (x - 2/3)² is always
positive its least value will be zero when x = 2/3
∴ therefore the least value of 3x² - 4x + 5 is 11/3 when x = 2/3
(ii) ⟦Let the⟧ ⟦illegible⟧ ⟦where x minutes past 3⟧
∴ x = 15 + ⟦illegible⟧ + x/12 or x = x/12 + 15
∴ 12x = x + 180 ⟦illegible⟧ 11x = 180 ∴ x = 180/11 = 16 4/11
∴ 11x = 360 ∴ x = 360/11 = 32 8/11 minutes
∴ x = 32 8/11 minutes past 3 ⟦illegible⟧
⟦Diagram of a clock face showing hands between 3 and 4⟧
Solution to Conditional Exam in Algebra, Sept. 1967 Cont.
2
3 (i) 9 √ [ (sin² 15° 04' × Cos³ 31° 31') / (510.7)² × (4.007)³ ] = x
log sin 15° 04' = Ī.4148 | 2 log sin 15° 04' = Ī.8296
log cos 31° 31' = Ī.9307 | 3 log cos 31° 31' = Ī.7921
log 510.7 = 2.7082 | log Num. = Ī.6217
log 4.007 = 0.6029 | log Den. = 7.2251
⟦line⟧
2 log 510.7 = 5.4164 | 9 log x = Ī.3966
3 log 4.007 = 1.8087 | log x = Ī.04407
log Den. = 7.2251 | = Ī.0441
x = 0.1107 Ans.
(ii) 32^(2x-1) = 64^x · 40 or 2^(5(2x-1)) = 2^(6x) · 40 or 2^(5(2x-1)-6x) = 40
<del>⟦illegible⟧</del>
<del>∴ (4x+5) log 2 = log 40</del>
<del>or 2^(4x+5) = 2² × 10 or 2^(4x+5-2) = 10 or 2^(4x+3) = 10</del>
<del>(4x+3) log 2 = log 10 ∴ (4x+3) log 2 = 1 ∴ 4x log 2 + 3 log 2 = 1</del>
<del>∴ 4x log 2 = 1 - 3 log 2 ∴ x = (1 - 3 log 2) / (4 log 2) = (1 - 3 × 0.3010) / (4 × 0.3010)</del>
<del>∴ x = (1 - 0.9030) / 1.2040 = 0.0970 / 1.2040 = 97 / 1204 = 0.0805647...</del>
<del>i.e. x = 0.08056 Correct to 4 significant figures.</del>
or 2^(4x-5) = 2² × 10 or 2^(4x-5-2) = 10 or 2^(4x-7) = 10
∴ (4x-7) log 2 = 1 or 4x log 2 - 7 log 2 = 1 or 4x log 2 = 1 + 7 log 2
or x = (1 + 7 log 2) / (4 log 2) = (1 + 7 × 0.3010) / (4 × 0.3010) = (1 + 2.1070) / 1.2040 = 3.1070 / 1.2040 = 3.107 / 1.204
∴ x = 2.5772... = 2.577 Correct to 4 significant figures
Ans.
(3)
4. (i) Let a = 1st term , d = common difference
" 3(a+2d) = 2(a+5d) or 3a+6d = 2a+10d or a-4d = 0 .... ①
also a + (a+2d) + (a+4d) = 9 or 3a+6d = 9 or a+2d = 3 .... ②
∴ 6d = 3 ∴ d = 1/2 ∴ a = 2
(a) ∴ 9th term / 6th term = a+8d / a+5d = 2+4 / 2+2 1/2 = 6 / 4 1/2 = 12 / 9 = 4/3 Ans. 1
(b) S_13 = n/2 {2a + (n-1)d} = 13/2 {2x2 + 12x1/2} = 13/2 {10} = 65 Ans. 2
(ii) l_3 = 2/3 , l_1 + l_2 = 2 1/2 , a = ? , r = ? , l_4 = ?
ar^2 = 2/3 } or ar^2 = 2/3 .... ① } by Division r^2 / 1+r = 2/3 x 2/5 or r^2 / 1+r = 4/15
a + ar = 5/2 } a(1+r) = 5/2 .... ② }
∴ 15r^2 = 4+4r or 15r^2 - 4r - 4 = 0 or (5r+2)(3r-2) = 0
or r = 2/3 and r = -2/5 {the latter value is to be discarded since}
{all the terms of the progression are positive, given}
∴ from eq. (1) a(2/3)^2 = 2/3 ∴ 4/9 a = 2/3 ∴ a = 3/2 ∴ l_4 = ar^3 = 3/2 (2/3)^3 = 4/9
a = 3/2 , r = 2/3 , l_4 = 4/9 Ans.
⟦line⟧
⟦illegible⟧
x | y
-3 | -5
-2 | 0
-1 | 3
0 | 4
1 | 3
2 | 0
3 | -5
y₁ = 4 - x²
y₂ = 5/4 x + 1
C(-1.22, 2.5)
D(1.22, 2.5)
1.23
-2.47
y = -1.6
x = ± 2.37
(i) From the table above, the graph of
y₁ = 4 - x² is plotted as shown
also the st. line y₂ = 5/4 x + 1
(ii) The two graphs intersect at two
(a) points whose abscissae are:
x = -2.47 and x = 1.23 Ans.
∴ 4 - x² > 5/4 x + 1 between
-2.47 and 1.23 since between these
two values of x the curve of 4 - x² lies above
the st. line 5/4 x + 1
(ii) (b) We draw the line y = 2.5
This line cuts the graph at two points ⟦illegible⟧
C(-1.22, 2.5) and D(1.22, 2.5)
∴ 4 - x² = 2.5 at x = -1.22 and x = 1.22
Ans.
(ii) (c) From y = 4 - x², we get x² = 4 - y
or x = ± √4 - y . Now let 4 - y = 5.6
∴ y = -1.6 ∴ when y = -1.6, x = ± √4 - (-1.6) or
x = ± √5.6 and from the graph we find that
when y = -1.6, x = ± 2.37 Ans.
⟦illegible⟧ Exam. paper in Algebra Page 1
4th year , 18/5/1967
x t = at + b ∴ t(x-a) = b ∴ t = b / (x-a)
y = b + a ( b / (x-a) ) = ( b(x-a) + ab ) / (x-a) = ( bx - ab + ab ) / (x-a) = bx / (x-a)
y = bx / (x-a) Ans.
(ii) 2^(x+y) = 32 ∴ 2^(x+y) = 2^5 ∴ x+y = 5
3^(x-y) = 9 ∴ 3^(x-y) = 3^2 ∴ x-y = 2
x+y = 5
x-y = 2
2x = 7 ∴ x = 3.5 Ans (7 marks)
y = 1.5 Ans
(iii) m/n = 5/4 , p/q = 3/4 , (3m+5p)/(n+q) = ?
m = 5n/4 , p = 3q/4 ∴ (3m+5p)/(n+q) = ( 15n/4 + 15q/4 ) / (n+q) = ( 15/4 (n+q) ) / (n+q)
= 15/4 = 3 3/4 Ans. (7 marks)
2. (a) When n tables are made for table costs: £ (300 + 8n) / n
or 1 table costs: £ (300/n + 8)
(b) when (50 + n) tables are made, 1 table costs: £ [300 + 8(50+n)] / (50+n)
or 1 table costs: £ ( 300 / (50+n) + 8 )
∴ 300 / (50+n) + 8 + 1 = 300/n + 8
or 300/n - 300 / (50+n) = 1 or 300(50+n) - 300n = n(50+n)
or 15000 + 300n - 300n = n² + 50n or n² + 50n - 15000 = 0
∴ (n+150)(n-100) = 0 ∴ n = -150 (to be discarded)
n = 100 Ans.
∴ the cost of one table in case (a) is £ (300/n + 8) = £ (3+8) = £ 11
" " " " " (b) is £ ( 300 / (50+n) + 8 ) = £ (2+8) = £ 10 Ans.
(20 marks)
Solution to Algebra paper Cont. (Final Exam 18/5/1967)
Page 2
7 ⟦illegible⟧ x = ⟦illegible⟧ Cos³ 42° 17' x tan⁵ 27° 34'
1.009³ x 90.04⁵
log cos 42° 17' = 1.8691 | 3 log cos 42° 17' = 1.6073
log tan 27° 34' = 1.7177 | 5 log tan 27° 34' = 2.5885
log 1.009 = 0.0037 | log Num. = 2.1958
log 90.04 = 1.9544 | log Den. = 9.7831
⟦line⟧
3 log 1.009 = 0.0111 | 7 log x = 12.4127
5 log 90.04 = 9.7720 | log x = 2.34467
log Den. = 9.7831 | or log x = 2.3447
| x = 0.02212
| or x = 2.212 x 10⁻²
| Ans.
| (10 marks)
(ii) N = 2⁴.¹³⁶ and 10⁰.³⁰¹ = 2 , log₁₀ N = ?
log₁₀ N = 4.136 log₁₀ 2 but log₁₀ 2 = 0.301
∴ log₁₀ N = 4.136 x 0.301 = 1.244936 = 1.2449 (Correct to
Ans. 5 significant
figures
(10 marks)
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
Solution to Algebra paper cont. Final Exam 18/5/1967
page 3
⟦4(i)⟧ is 4, 12, 20, 28, ...
In this progression a=4 , d=8
∴ S = n/2 {2x4 + (n-1)x8} or S = n/2 {8 + 8n - 8} or
S = n/2 (8n) ∴ S = 4n² or S = (2n)² Ans.
∴ Whatever the value of n is, the sum S is always a perfect square
Q.E.D.
(10 marks)
⟦7(ii)⟧ 12, x, y, 4 from the statement of the question:
12, x, y are in A.P. ∴ x-12 = y-x ..... ①
also x, y, 4 are in G.P. ∴ y/x = 4/y ..... ②
from eq. ① : y = 2x-12 ..... ③ ∴ y² = (2x-12)² = 4(x-6)²
from eq. ② : y² = 4x ..... ④ or y² = 4(x²-12x+36)
∴ 4(x²-12x+36) = 4x ∴ x²-13x+36 = 0
∴ (x-4)(x-9) = 0 ∴ x=4 or x=9
when x=4 , y = 2(x-6) = 2(4-6) = -4
when x=9 , y = 2(x-6) = 2(9-6) = 6
∴ x=4 } Ans. 1 x=9 } Ans. 2
y=-4 y=6
(10 marks)
∴ the terms are either 12, 4, -4, 4
or 12, 9, 6, 4
4th year Final Exam 18/5/1961 Algebra.
x | y₁ | y₂ | y₃
-2 | -8 | -4 | -8
-1 | -1 | -1 | -3
0 | 0 | 2 | 2
1 | 1 | 5 | ⟦illegible⟧
2 | 8 | 8 | 12
3 | 27 | |
(6 marks) (i) The curve y = x³ is plotted as shown below
(7 marks) (ii) From the equation x³ = 3x + 2 we get the same
roots for x as the roots obtained from solving
the two simultaneous equations:
y₁ = x³ and y₂ = 3x + 2
the graph of y₂ = 3x + 2 is the st. line shown in the diagram.
This line touches the curve y₁ = x³ at B(-1, -1) and
intersects it at C(2, 8)
∴ the roots of the original equation
are x = -1 and
x = 2 Ans.
y₁ = x³
y₃ = 5x + 2
y₂ = 3x + 2
D(2.4, 14)
C(2, 8)
B(-1, -1)
E(-0.4, -0.05)
A(-2, -8)
Again the roots of the equation x³ - 5x - 2 = 0
are the same as the roots of the two simultaneous
equations y₁ = x³ and y₃ = 5x + 2
since x³ = 5x + 2
Thus the graph of y₃ = 5x + 2 is another
st. line which intersects the curve y₁ = x³
at A(-2, -8) and at D(2.4, 14) and at
E(-0.4, -0.05) ∴ the roots of x³ - 5x - 2 = 0 are
x = -2 and
x = 2.4
x = -0.4 Ans.
(7 marks)
(iii) From the diagram,
since the st. line
y₂ = 3x + 2 lies above the
curve y₁ = x³ between x = -1 and
x = 2 and also between x = -∞ and
x = -1 ∴ 3x + 2 > x³ when x < -1
and also when -1 < x < 2 Ans.
Shamash Secondary School
Final Examination, May, 1967
Subject: Algebra
Date: 18/5/1967
Class: 4th Year Secondary
Time: 8:00 - 11:00 a.m.
⟦line⟧
Answer all questions:
1. (i) If x = a + b/t and y = b + at, find an expression for y in terms
of a, b, x.
(6 marks)
(ii) If 2^(x-y) = 8 and 3^(x-2y) = 9 find the values of x and y.
(7 marks)
(iii) If m/n = 5/4 and p/q = 3/4, find the value of (3m+5p)/(n+q)
(7 marks)
2. A manufacturer calculated that, allowing for the initial outlay on
equipment, the cost of production of n tables was £(300+8n). Write
down the cost of production of 1 table when
(a) n tables are made,
(b) (50+n) tables are made.
If the cost of production of one table decreases by £1 when the extra
50 tables are made, find the value of n and the cost of production of
a table in each case.
(20 marks)
3. (i) Compute by logarithms the following expression, arranging your
work neatly:
7√ (Cos³ 42° 17' x tan⁵ 27° 34') / (1.009³ x 90.04⁵)
(10 marks)
(ii) If N = 2^4.136 and 10^0.301 = 2, find, without the use of tables,
the logarithm of N to the base 10 correct to five significant
figures.
(10 marks)
4. (i) Prove that the sum of any number of terms of the progression
4, 12, 20, 28, ... is a perfect square.
(ii) The first three of the four terms, 12, x, y, 4, are in arithmetical
progression and the last three are in geometrical progression.
Find x and y.
(p.2)..