AI Transcription, Pages 26-50
3
An alternative method
x² + xy + 2y² = 8 ...... ①
2x² - 2xy - 3y² = 1 ...... ②
multiply Eq (2) by 8 :
Subtract { 16x² - 16xy - 24y² = 8 ...... ③
{ x² + xy + 2y² = 8 ...... ①
15x² - 17xy - 26y² = 0 ...... ④
∴ (x - 2y) (15x + 13y) = 0
∴ y = 1/2 x or y = - 15/13 x
When y = x/2 from Eq. ① : x² + x(x/2) + 2(x/2)² = 8 or x² + x²/2 + x²/2 = 8
or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 ∴ y = x/2 = ± 2/2 = ± 1
∴ x = 2 } Ans. 1 x = -2 } Ans. 2
y = 1 } y = -1 }
When y = - 15/13 x from Eq. ① : x² + x(- 15/13 x) + 2(- 15/13 x)² = 8 or
x² - 15/13 x² + 450/169 x² = 8 ∴ 169x² - 15x13x² + 450x² = 8 x 169
∴ 169x² - 195x² + 450x² = 1352 ∴ 424x² = 1352 ∴ x² = 1352/424
∴ x² = 169/53 ∴ x = ± 13/√53
When x = 13/√53 ∴ y = - 15/13 x or y = - 15/13 (13/√53) = - 15/√53
and when x = - 13/√53 ∴ y = - 15/13 x or y = - 15/13 (- 13/√53) = 15/√53
∴ x = 13/√53 } Ans. 3 x = - 13/√53 } Ans. 4
y = - 15/√53 } y = 15/√53 }
4
4 (i) x³ + y³ + 1/x³ + 1/y³ = (x³ + 1/x³) + (y³ + 1/y³) = [(x + 1/x)³ - 3(x + 1/x)] + [(y + 1/y)³ - 3(y + 1/y)]
= a³ - 3a + b³ - 3b = a³ + b³ - 3(a + b)
= (a + b)(a² - ab + b² - 3) Ans. 1
= (1 + 2)(1² - 1x2 + 2² - 3)
= (3)(1 - 2 + 4 - 3) = 3 x zero = 0 Ans. 2
(ii) a³ + a - 8b³ - 2b + c + 6abc + c³ = a³ - 8b³ + c³ + 6abc + a - 2b + c
= [a³ + (-2b)³ + c³ - 3a(-2b)c] + [a - 2b + c]
= (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc) + (a - 2b + c)
= (a - 2b + c) [(a² + 4b² + c² + 2ab - ac + 2bc) + 1]
= (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc + 1) Ans.
5. after the first replacement, there are
x/2 gall. of Brandy in Cask P and
(50 - x/2) gall. " " " Q
At the beginning of the 2nd operation:
100 gall water P
50 gall Brandy Q
(x²/200) gall. of Brandy are removed from Cask P and
x(50 - x/2) / 50 gall. " " " " Q
Mixture Brandy
100 x/2 gall. ? = x * x/2 / 100
x ? = x²/200 gall.
50 gall. (50 - x/2) gall.
x ? ? = x(50 - x/2) / 50
x²/200 + x(50 - x/2) / 50 / 2 gall. of Brandy are deposited in P after 2nd replacement.
∴ (x/2 - x²/200) + (x²/200 + x(50 - x/2) / 50) / 2 = 17 or
100x - x² / 200 + x² + 4x(50 - x/2) / 2 = 17 ∴ 100x - x² / 200 + x² + 200x - 2x² / 2 = 17
∴ 100x - x² / 200 + 200x - x² / 400 = 17 ∴ 200x - 2x² + 200x - x² = 6800
∴ 3x² - 400x + 6800 = 0 ∴ (3x - 340)(x - 20) = 0
∴ x = 340/3 = 113 1/3 inadmissible
x = 20 gallons Ans.
45
6. speed of train A = 22.5 mi/h = 22.5 x 22/15 ft/sec = 45/2 x 22/15 = 33 ft/sec.
speed " " B = 15 mi/h = 15 x 22/15 ft/sec = 22 ft/sec.
(i) when the two trains are travelling in
opposite directions (see Fig. I), points C and D
are separating at the rate of (33+22) ft/sec
= 55 ft/sec.
When the rear cars A and B just clear away
from each other, (see Fig. II) points C and D
have already separated by a distance
= (240+200) ft = 440 ft.
Time taken = total distance / rate of separation = 440 / 55 = 8 sec. Ans. 1
v₁ = 33 ft/sec. C
A ————————→
240 ft v₂ = 22 ft/sec
Fig. I ←————— B
200 ft
A ————————→ C
D ←————— B
Fig. II
(ii) When the two trains are travelling in the
same direction, (See Fig. III), points C and D
are separating at the rate of (33-22) ft/sec
= 11 ft/sec.
When the rear car A of the faster train
and the front car D of the slower train
just clear away from each other,
(see Fig. IV), points C and D have already
separated by a distance of 240 ft which
is the length of the faster train.
240 ft
A ————————→ C v₁ = 33 ft/sec
200 ft
B —————→ D v₂ = 22 ft/sec
Fig. III
A 240 ft
———————→ C
B 200 ft
—————→ D
Fig. IV
∴ Time taken = distance / rate of separation = 240 / 11 = 21 9/11 sec. Ans. 2
SHAMASH SECONDARY SCHOOL
Mid-Year Examination, February, 1969.
Subject: Algebra
Date: 17/2/1969.
Class: 4th Year, Secondary
Time: 8:30-11:30 a.m.
Five questions only are to be attempted.
1. The time now is x minutes after five and the two hands of the watch
stand in a straight line on opposite sides of the centre of the dial of the
watch. Find x and state, in words, the correct time. (20 marks)
2. (i) Find the value of x from the following equation :
6 x³ + 19 x² + x - 6 = 0. (10 marks)
(ii) Solve the following equation, using the shortest possible method,
by first reducing each fraction to a simpler form :
3 x - 7 2 x - 5 3 x + 7 2 x + 5
------- + ------- = ------- + ------- . (10 marks)
x - 2 x - 3 x + 2 x + 3
3. (i) In the following equation, A, B, and C are constants and the
equation is true for all values of x. Find the values of A, B and C.
A(x² - 2x) + B(x + 4) + C = 3x² + x + 25. (10 marks)
(ii) Solve the following equations simultaneously :
x² + xy + 2y² = 8 ...............(1)
2x² - 2xy - 3y² = 1 ...............(2) (10 marks)
4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression
(x³ + y³ + 1/x³ + 1/y³) in terms of "a" and "b". Hence or otherwise find the
value of (x³ + y³ + 1/x³ + 1/y³) if a = 1 and b = 2. (10 marks)
(ii) Resolve the expression a³ + a - 8 b³ - 2b + c + 6abc + c³ into two
factors one of which is (a - 2b + c). (10 marks)
5. A cask P is filled with 100 gallons of water, and a cask Q with
50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced;
and the same operation is repeated. Find x when there are 17 gallons of
brandy in P after the second replacement. (20 marks)
6. Two trains A and B are travelling on two railway tracks which are
parallel to each other. Train A is 240 ft long and it is travelling at 22.5
miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles
per hour. Find the length of time in seconds from the instant when the heads
of the front cars of the two trains are together, to the instant when the
P. T. O.
- 2 -
Mid-Year Exam. Cont., in Algebra ; 4th Year, Secondary, 17/2/1969.
⟦line⟧
two trains just clear away from each other in the two cases :
(i) When the two trains are travelling in opposite directions.
(ii) when the two trains are travelling in the same direction.
(20 marks).
⟦line⟧
3. (b) In the following equation, A, B, and C are constants and the
equation is true for all values of x. Find the values of A, B and C.
A(x² - 2x) + B(x + 4) + C = 3x² + x + 25 . (10 marks)
(ii) Solve the following equations simultaneously :
x² + xy + 2y² = 8 ........... (1)
2x² - 2xy - 3y² = 1 ........... (2) (10 marks)
4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression
(x² + 1/x² + 2)(y² + 1/y² - 2) in terms of "a" and "b". Hence or otherwise find the
value of (x + 1/x + y + 1/y) if a = 1 and b = 2 . (10 marks)
(ii) Resolve the expression a² + b² - c² - d² + 2ab + 2cd into two
factors one of which is (a + b - c + d). (10 marks)
5. A cask P is filled with 100 gallons of water, and a cask Q with
50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced,
and the same operation is repeated. Find x when there are 17 gallons of
brandy in P after the second replacement. (20 marks)
6. Two trains A and B are travelling on two railway tracks which are
parallel to each other. Train A is 240 ft long and it is travelling at 22.5
miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles
per hour. Find the length of time in seconds from the instant when the ⟦illegible⟧
of the front cars of the two trains are together, to the instant when the
P. T. O
B01 · header / latin
Solutions to Algebra Exam. (2nd quart.) 7/2/1969 page 1
B02 · paragraph / latin
1. 2x³ + Ax² + Bx - 4 | x - 2
2x³ - 4x² | 2x² + (4+A)x + (B+2A+8) | x+2
(4+A)x² + Bx - 4 | 2x² + 4x | 2x+A
(4+A)x² - 2(4+A)x | [(4+A)-4]x + B+2A+8
[2(4+A)+B]x - 4 or Bx + 2A+B+8
(B+2A+8)x - 2(B+2A+8) | Ax + 2A
2(B+2A+8) - 4 = 0 B+8 = 0 ----②
∴ 4+12B = -12 ∴ B = -8 from ① 2A-8 = -6 ∴ A = 1
or 2A + B = -6 ----① B = -8 Ans. 1
∴ the remaining factor is the last quotient, namely: 2x+A or
2x+1 Ans. 2
B03 · paragraph / latin
an alternative method: the factors of x²-4 are (x-2) & (x+2). By the
remainder theorem, when x=2 the expression 2x³+Ax²+Bx-4 becomes zero
∴ 2x8 + 4A + 2B - 4 = 0 or 4A + 2B = -12 or 2A + B = -6 ----①
also, when x = -2, then 2(-2)³ + 4A - 2B - 4 = 0 ∴ 4A - 2B = 20 ∴ 2A - B = 10 ----②
Now 2A + B = -6 ----① ∴ 4A = 4 ∴ A = 1 } Ans. 1
2A - B = 10 ----② ∴ B = -8
∴ 2x³ + x² - 8x - 4 = (x²-4)(2x+1) ∴ (2x+1) is the remaining factor Ans. 2
B04 · paragraph / latin
2. 9/4 x⁶ - 3x⁵ + 4x⁴ - 5x³ + 5/3 x² - 2/3 x + 1/9 (arranging according to descending powers)
9/4 x⁶ - 3x⁵ + 4x⁴ - 3x³ + 5/3 x² - 2/3 x + 1/9 | 3/2 x³ - x² + x - 1/3
9/4 x⁶
3x³ - x² | -3x⁵ + 4x⁴ - 5x³
x | -3x⁵ + x⁴
3x³ - 2x² + x | +3x⁴ - 5x³ + 5/3 x²
x | +3x⁴ - 2x³ + x²
3x³ - 2x² + 2x - 1/3 | -x³ + 2/3 x² - 2/3 x + 1/9
| -x³ + 2/3 x² - 2/3 x + 1/9
∴ the square root is;
3/2 x³ - x² + x - 1/3 Ans.
B05 · marginalia / latin
(20 marks)
(20 marks)
B06 · paragraph / latin
3. ⟦illegible⟧
⟦illegible⟧ = 1 Ans.
B01 · paragraph / latin
3. (a) 4(x²-1) + 2(x+3) = 2 + 2x(1+2x)
(b) x(6x+1) = 2x+1
(c) x(x+2) = 2(x-2)
B02 · paragraph / latin
6 Simplifying (a), we get: 4x²-4+2x+6 = 2+2x+4x² or
4x²+2x+2 = 4x²+2x+2 (always true, being an identity)
B03 · paragraph / latin
8 from (b): 6x²+x = 2x+1 ∴ 6x²-x-1 = 0 or
(3x+1)(2x-1) = 0 ∴ x = -1/3 and x = 1/2 (conditional equation)
and the values of x are x = -1/3 } Ans.
x = 1/2
B04 · paragraph / latin
6 from (c):
x²+2x = 2x-4 or x² = -4 (never true, x being real)
B05 · paragraph / latin
4 (i) 1/x + 1/y + 1/z = 2 1/2 .... ① Dividing eq. ② by 2 & subtracting with eq. ①, we get:
2/x + 2/y + 2/z = 5 .... ② { 1/x + 1/y + 1/z = 2 1/2 .... ①
3/x - 5/y + 7/z = 2 5/6 .... ③ { 1/x + 3/y - 3/z = 1 .... ⟦illegible⟧
∴ 2/x + 3/y = 3 1/2 .... ④ from ① multiplying by 7: 7/x + 7/y + 7/z = 17 1/2
from ③ (⟦illegible⟧) 3/x - 5/y + 7/z = 8 1/2
25/y = 12 1/2 ∴ y = 25 / 12 1/2 = 2 -2/x + 22/y = 9 .... ⑦
y = 2 from ④ 2/x + 3/2 = 3 1/2 ∴ 2/x = 2 ∴ x = 1
from ① 1/1 + 1/2 + 1/z = 2 1/2 or 1/z = 1 ∴ z = 1
∴ x = 1
y = 2 } Ans.
z = 1
B06 · marginalia / latin
(10 marks)
B07 · paragraph / latin
(ii) (x/y² + y/x² - 1) / (x²/y² + y²/x² + 1) . (1 + y/x) / (x - y) ÷ (1 + y/x) / (x² - y²) = (x³+y³-xy²) / (x²+x²y+xy²) . (x+y) / (x²-xy) ÷ (x³+y³) / (x⁵-x²y³)
= (y(y²+x²-xy)) / (x(y²+x²+xy)) . (x+y) / (x(x-y)) . (x²(x³-y³)) / (y(x³+y³)) = (y(x+y)(x²+xy+y²)(x/y)(x²+xy+y²)) / (x²y(x-y)(x²+xy+y²)(x+y)(x²-xy+y²))
= 1 Ans.
B08 · marginalia / latin
(10 marks)
⟦line⟧
(ii) \frac{\frac{x}{y} + \frac{y}{x} - 1}{\frac{x^2}{y^2} + \frac{y^2}{x^2} + 1} \cdot \frac{x + y}{x - y} \div \frac{x^3 + y^3}{x^4 - y^4} = \frac{x^2 + y^2 - xy}{x^2 y^2} \cdot \frac{x^2 y^2}{x^4 + x^2 y^2 + y^4} \cdot \frac{x + y}{x - y} \div \frac{x^3 + y^3}{x^4 - y^4}
= y(y^2+x^2-xy) / x(y^2+x^2+xy) " (x+y) / x(x-y) " x^2(x^2-y^2) / y(x^3+y^3) = x^2y(x+y)(x^2-xy+y^2)(x-y)(x+y) / xy^2(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)
= 1 Ans.
Make-up Exam. 2nd Quarter ⟦الصف الرابع⟧
17/1/1969
I. (i) Find the value of x² - 1/y when x = -1/2 & y = -3 (6 marks)
(ii) Solve the equation x/3 + (x-1)/2 = 7 (7 marks)
(iii) Find x if 2x + y = 4 and 3y + 4 = 6x (7 marks)
II (i) If t = ∛((x² + 4y) / 2yz) , find z in terms of x, y and t. (10 marks)
(ii) If F = av - b/v² and if F = 4 when v = 5 and F = 36 when
v = 10, find the values of "a" and "b" and the value of F when v = 20
(10 marks)
III. a man can cycle at x m.p.h. in still air. His speed increases
y m.p.h. when he cycles with the wind, and decreases y m.p.h. when
he cycles against the wind. The difference in his time to cycle one mile
with the wind and one mile against the wind is z hours. Find a
formula for z in terms of x and y, and find x if y = 2, z = 1/3.
(20 marks)
IV. In how many days will "a" horses eat 1/n th of the corn of a field
the whole of which can be eaten by "b" horses in "c" days.
(20 marks)
V. Find the square root of:
16x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1
showing your steps neatly. (20 marks)
SHAMASH SECONDARY SCHOOL
Subject: Algebra
Class: 4th Year Secondary
Date: 7/1/1969
Time: 10:15-11:45
Attempt all questions:
1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x²-4.
Find the values of A and B and find the remaining factor.
(20 marks)
2. Find the square root of:
4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9
(20 marks)
3. Which of the following equations is always true, which is sometimes
true and which is never true? Find the values of x which satisfy the
equatioh which is sometimes true.
(a) 4(x²-1) + 2(x + 3) = 2 + 2x(1 + 2x)
(b) x(6x + 1) = 2x + 1
(c) x(x + 2) = 2(x - 2)
(20 marks)
4. (i) Solve simultaneously the following equations:
1/x + 1/y + 3/z = 2½ ....................(1)
2/x + 4/y - 6/z = 2 ....................(2)
3/x + 5/y + 7/z = 2 5/6 ....................(3)
(ii) Simplify the following expression to simplest form:
x/y + y/x - 1 / (x²/y² + x/y + 1) . (1 + y/x) / (x - y) ÷ (1 + y³/x³) / (x²/y - y²/x)
(20 marks)
5. A man bought "A" lbs of coffee for a certain sum of money. He
kept "B" lbs to himself and sold the remainder at "C" shillings a pound
more than he paid for it. He found that he received for this portion
an amount equal to the original sum of money which he paid for the whole.
Find the original sum of money which he paid for the whole. (20 marks)
B01 · header / latin
SHAMASH SECONDARY SCHOOL
Subject: Algebra
Date: 7/1/1969
Class : 4th Year Secondary
Time: 10:15-11:45
B02 · paragraph / latin
Attempt all questions:
B03 · paragraph / latin
1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x² - 4.
Find the values of A and B and find the remaining factor.
(20 marks)
B04 · paragraph / latin
2. Find the square root of:
4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9
(20 marks)
B05 · paragraph / latin
3. Which of the following equations is always true, which is sometimes
true and which is never true? Find the values of x which satisfy the
equation which is sometimes true.
(a) 4(x² - 1) + 2(x + 3) = 2 + 2x(1 + 2x)
(b) x(6x + 1) = 2x + 1
(c) x(x + 2) = 2(x - 2)
(20 marks)
B06 · paragraph / latin
(i) Solve simultaneously the following equations:
1/x + 1/y + 3/z = 2½ ⟦line⟧ (1)
2/x + 4/y - 6/z = 2 ⟦line⟧ (2)
3/x - 5/y + 7/z = 2 5/6 ⟦line⟧ (3)
B07 · paragraph / latin
(ii) Simplify the following expression to simplest form:
((x/y + y/x - 1) / (x²/y² + x/y + 1)) . ((1 + y/x) / (x - y)) ÷ ((1 + y³/x³) / (x²/y - y²/x))
(20 marks)
B08 · paragraph / latin
5. A man bought "A" lbs of coffee for a certain sum of money. He
kept "B" lbs to himself and sold the remainder at "C" shillings a pound
more than he paid for it. He found that he received for this portion
an amount equal to the original sum of money which he paid for the whole.
Find the original sum of money which he paid for the whole.
(20 marks)
SHAMASH SECONDARY SCHOOL
Monthly Examination, November 1968
Subject: General Mathematics Date: 18/11/1968.
Class : 4th Year Secondary Time: 8:30 - 10:00 a.m.
الرقم:
الاسم:
1. Give the English Equivalent of the following, filling the blanks in
this sheet and hand it over with your examination book.
| Numerals = figures | ١- ارقام |
| Digits | ٢- مراتب |
| Subtraction | ٣- الطرح |
| Factors | ٤- العوامل |
| The index or exponent of the power | ٥- اس القوة |
| Multiple | ٦- مضاعف |
| Consecutive even numbers | ٧- اعداد زوجية متتالية |
| ⟦Consecutive⟧ odd ⟦numbers⟧ | ٨- اعداد فردية متتالية |
| The integral part of a number | ٩- الجزء الصحيح من العدد |
| Prime numbers | ١٠- اعداد اولية |
| The least common Denominator | ١١- المقام المشترك الاصغر |
| An improper fraction | ١٢- كسر لفظي |
| The reciprocal of a number | ١٣- مقلوب العدد |
| Terminating decimals | ١٤- الكسور العشرية المنتهية |
| Recurring or Repeating decimals | ١٥- الكسور العشرية الدورية |
| The percentage error | ١٦- الخطأ المئوي |
| Ratio + Proportion | ١٧- النسبة والتناسب |
| The mean proportional between two numbers | ١٨- الوسط المتناسب بين عددين |
| The Dividend | ١٩- ربح المساهم ( ربح حامل الاسهم ) |
| Axiom | ٢٠- البديهية |
| Postulate | ٢١- الموضوعة |
| an acute angle | ٢٢- زاوية حادة |
| an obtuse ⟦angle⟧ | ٢٣- زاوية منفرجة |
| a Reflex ⟦angle⟧ | ٢٤- زاوية منعكسة |
| a segment of a circle | ٢٥- قطعة دائرة |
| a Sector ⟦of a circle⟧ | ٢٦- قطاع دائرة |
| The Data | ٢٧- المعاليم |
| The unknowns | ٢٨- المجاهيل |
| Two Complementary angles | ٢٩- زاويتان متتامتان |
| ⟦Two⟧ Supplementary ⟦angles⟧ | ٣٠- زاويتان متكاملتان |
| an equilateral polygon | ٣١- مضلع متساوي الاضلاع |
| an isosceles triangle | ٣٢- مثلث متساوي الساقين |
| The rhombus | ٣٣- المعين |
| The Locus | ٣٤- المحل الهندسي |
| The secant to a circle | ٣٥- المستقيم القاطع للدائرة |
| The removal + insertion of brackets | ٣٦- ازالة وادخال الاقواس |
| Transposition from one side of an equation to the other | ٣٧- نقل حدود المعادلة من جهة الى الجهة الاخرى |
| Identity | ٣٨- متطابقة |
| Inequality | ٣٩- متباينة |
- يتبع -
- ٢ -
الرقم:
الاسم:
٤٠- مقدار جبرى متجانس ⟦line⟧ 1 mark
A homogeneous algebraic expression
٤١- درجة المقدار الجبرى ⟦line⟧ "
The degree or the dimension of an algebraic expression
٤٢- المعامل الحرفى ⟦line⟧ "
The literal coefficient
٤٣- مقدار جبرى من الدرجة الثانية ⟦line⟧ "
An algebraic expression of the second degree
or a quadratic expression
٤٤- ان حدى الكسر هما بسطه ومقامه ⟦line⟧ 2 marks
The two terms of a fraction are its numerator
and denominator
٤٥- في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة . 5 marks
In every process of division there is a dividend, a divisor, a quotient + in some
cases a remainder
٤٦- ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ⟦line⟧ "
The perpendicular bisectors of the sides of a triangle meet at the centre
of the circumscribed circle.
٤٧- ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الرأس "
وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث.
The medians of a triangle meet at a point which divides each of them
two thirds from the vertex and one third from the base. This point is
called the centroid of the triangle.
٤٨- نقيس طول مستقيم فنجد انه يساوى ٦١,٥ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم . "
وفي هذه الحالة نقول ان الخطأ المطلق هو ⟦line⟧
والخطأ النسبي هو ⟦line⟧ والخطأ المئوى هو ⟦line⟧
We measure the length of a st. line + we find that it is equal 61.5 cms. We then find that
its exact length is 60 cms. In this case we say that the absolute error is 1.5 cm, the relative error is 1.5/60
and the percentage error is 2.5%
٤٩- ان قيمة المقدار ٥٣٠٩,٧٢ لاقرب اربعة ارقام معنوية هي ⟦line⟧ 5 marks
The value of 5309.72 correct to 4 significant figures is 5310.00
٥٠- ان المعادلة ٣س٢ - ٢س ص + ص = ٥ع - ٣ع هي معادلة من الدرجة ⟦line⟧ في ⟦line⟧ "
مجاهيل ⟦line⟧
The equation 3x² - 2xy + y = 5z - 3z is a quadratic equation in three unknowns.
(75 marks)
(II) Fill in the blanks in the following equations:-
| (2.5 marks) | 1. | one furlong = | ( 10 ) | chains= | ( 1/8 ) | mile |
| " | 2. | one chain = | ( 22 ) | yards = | ( 100 ) | links |
| " | 3. | one statute mile = | ( 1760 ) | yds. = | ( 5280 ) | ft. |
| " | 4. | one nautical mile = | ( 6080 ) | ft. |
| " | 5. | one sq. chain = | ( 484 ) | sq. yds. |
| " | 6. | one acre = | ( 10 ) | sq. ch. = | ( 4840 ) | sq. yds. |
| " | 7. | one gallon = | ( 8 ) | pints |
| " | 8. | one bushel = | ( 8 ) | gallons = | ( 4 ) | pecks |
| " | 9. | one English ton = | ( 2240 ) | lbs. = | ( 1016 ) | kilograms |
| " | 10. | one English ton = | ( 20 ) | cwt. = | ( 80 ) | qr. = | ( 160 ) | stones. |
[Marginalia] 1 quarter = 1/4 of one cwt = 28 lbs = 2 stones
[Marginalia] 1 stone = 14 lbs
(25 marks).
SHAMASH SECONDARY SCHOOL
Monthly Examination, November 1968
الرقم::
الاسم::
Subject:: General Mathematics
Date:: 18/11/1968.
Class :: 4th Year Secondary
Time:: 8:30 - 10:00 a.m.
1. Give the English Equivalent of the followinh, filling the blanks in
this sheet and hand it over with your examination book.
١- ارقام
٢- مراتب
٣- الطرح
٤- العوامل
٥- اس القوة
٦- مضاعف
٧- اعداد زوجية متتالية
٨- اعداد فردية متتالية
٩- الجزء الصحيح من العدد
١٠- اعداد اولية
١١- المقام المشترك الاصغر
١٢- كسر لفظي
١٣- مقلوب العدد
١٤- الكسور العشرية المنتهية
١٥- <del>الكسور</del> العشرية ⟦الدورية⟧
١٦- الخطأ المئوي
١٧- النسبة والتناسب
١٨- الوسط المتناسب بين عددين
١٩- ربح المساهم ( ربح حامل الاسهم )
٢٠- البديهية
٢١- الموضوعة
٢٢- زاوية حادة
٢٣- زاوية منفرجة
٢٤- زاوية منعكسة
٢٥- قطعة دائرة
٢٦- قطاع دائرة
٢٧- المعاليم
٢٨- المجاهيل
٢٩- زاويتان متتامتان
٣٠- زاويتان متكاملتان
٣١- مضلع متساوي الاضلاع
٣٢- مثلث متساوي الساقين
٣٣- المعين
٣٤- المحل الهندسي
٣٥- المستقيم القاطع للدائرة
٣٦- ازالة وادخال الاقواس
٣٧- نقل حدود المعادلة من جهة الى الجهة الاخرى
٣٨- متطابقة
٣٩- متباينة
- يتبع -
- ٢ -
الرقم::
الاسم::
٤٠ - مقدار جبري متجانس
٤١ - درجة المقدار الجبري
٤٢ - المعامل الحرفي
٤٣ - مقدار جبري من الدرجة الثانية
٤٤ - ان حدي الكسر هما بسطه ومقامه
٤٥ - في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة .
٤٦ - ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ...........
٤٧ - ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الراس
وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث.
٤٨ - نقيس طول مستقيم فنجد انه يساوي ٦١,٥ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم .
وفي هذه الحالة نقول ان الخطأ المطلق هو ...........
والخطأ النسبي هو ........... والخطأ المئوي هو ...
٤٩ - ان قيمة المقدار ٥٣,٠٧٢ لاقرب اربعة ارقام معنوية هي ...........
٥٠ - ان المعادلة ٣س٢ - ٢س ص + ص٢ = ٤٥ - ٤ع هي معادلة من الدرجة ........ في
...... مجاهيل .
(75 marks)
(II) Fill in the blanks in the following equations:-
1. one furling = ( ) chains = ( ) mile
2. one chain = ( ) yards = ( ) links
3. one statute mile = ( ) yds. = ( )
4. one nautical mile = ( ) ft.
5. one sq. chain = ( ) sq. yds.
6. one acre = ( ) sq. ch. = ( ) sq. yds.
7. one gallon = ( ) pints
8. one bushel = ( ) gallons = ( ) pecks
9. one English ton = ( ) lbs. = ( ) kilograms
10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.
SHAMASH SECONDARY SCHOOL:
الرقم ::
الاسم ::
Monthly Examination, November 1968:
Subject: General Mathematics:
Date: 18/11/1968.:
Class : 4th Year Secondary:
Time: 8:30 - 10:00 a.m.:
1. Give the English Equivalent of the followinh, filling the blanks in
this sheet and hand it over with your examination book.
١- ارقام
٢- مراتب
٣- الطرح
٤- العوامل
٥- اس القوة
٦- مضاعف
٧- اعداد زوجية متتالية
٨- اعداد فردية متتالية
٩- الجزء الصحيح من العدد
١٠- اعداد اولية
١١- المقام المشترك الاصغر
١٢- كسر لفظي
١٣- مقلوب العدد
١٤- الكسور العشرية المنتهية
١٥- <del>الكسور العشرية الدورية</del>
١٦- الخطأ المئوي
١٧- النسبة والتناسب
١٨- الوسط المتناسب بين عددين
١٩- ربح المساهم ( ربح حامل الاسهم )
٢٠- البديهية
٢١- الموضوعة
٢٢- زاوية حادة
٢٣- زاوية منفرجة
٢٤- زاوية منعكسة
٢٥- قطعة دائرة
٢٦- قطاع دائرة
٢٧- المعاليم
٢٨- المجاهيل
٢٩- زاويتان متتامتان
٣٠- زاويتان متكاملتان
٣١- مضلع متساوي الاضلاع
٣٢- مثلث متساوي الساقين
٣٣- المعين
٣٤- المحل الهندسي
٣٥- المستقيم القاطع للدائرة
٣٦- ازالة وادخال الاقواس
٣٧- نقل حدود المعادلة من جهة الى الجهة الاخرى
٣٨- متطابقة
٣٩- متباينة
يتبع
- ٢ -
الرقم ::
الاسم ::
٤٠- مقدار جبري متجانس
٤١- درجة المقدار الجبري
٤٢- المعامل الحرفي
٤٣- مقدار جبري من الدرجة الثانية
٤٤- ان حدي الكسر هما بسطه ومقامه
٤٥- في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة .
٤٦- ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ...........
٤٧- ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الرأس وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث.
٤٨- نقيس طول مستقيم فنجد انه يساوي ٥ر٦١ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم .
وفي هذه الحالة نقول ان الخطأ المطلق هو ...........
والخطأ النسبي هو ........... والخطأ المئوي هو ...
٤٩- ان قيمة المقدار ٧٢ر٥٣٠٩ لاقرب اربعة ارقام معنوية هي ...........
٥٠- ان المعادلة ٣س٢ - ٢س ص + ص٢ = ٤٥ - س ع هي معادلة من الدرجة ....... في ....... مجاهيل .
(75 marks)
(II) Fill in the blanks in the following equations:-
1. one furlong = ( ) chains= ( ) mile
2. one chain = ( ) yards = ( ) links
3. one statute mile = ( ) yds. = ( ) ft.
4. one nautical mile = ( ) ft.
5. one sq. chain = ( ) sq. yds.
6. one acre =( ) sq. ch. = ( ) sq. yds.
7. one gallon = ( ) pints
8. one bushel = ( ) gallons = ( ) pecks
9. one English ton = ( ) lbs. ⟦=⟧ ( ) kilograms
10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.
(25 marks).
B01 · header / mixed
— ٧ —
١٤٤٩ : ⟦illegible⟧
١٧ : ⟦illegible⟧
B02 · paragraph / arabic
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
B03 · paragraph / arabic
٥ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
B04 · paragraph / arabic
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧
B05 · paragraph / arabic
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
B06 · paragraph / arabic
٨ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ . ٣٥ —
⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧
⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
B07 · paragraph / arabic
٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧
B08 · paragraph / arabic
٥ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧
⟦illegible⟧ ⟦illegible⟧ ⟦line⟧
B09 · header / latin
(25 marks)
(II) Fill in the blanks in the following equations :-
B10 · form / latin
1. one furlong = ( ) chains = ( ) mile
2. one chain = ( ) yards = ( ) links
3. one statute mile = ( ) yds. = ( ) ft.
4. one nautical mile = ( ) ft.
5. one sq. chain = ( ) sq. yds.
6. one acre = ( ) sq. ch. = ( ) sq. yds.
7. one gallon = ( ) pints
8. one bushel = ( ) gallons = ( ) pecks
9. one English ton = ( ) lbs. = ( ) kilograms
10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones
B11 · footer / latin
(25 marks)
B01 · header / latin
y = 1/4(3x² - 5x - 4)
Final Exam in Algebra
14/5/1962
B02 · table / latin
x | y
-2 | 4 1/2
-1 | 1
0 | -1
1 | -1 1/2
2 | -1/2
3 | 2
4 | 6
B03 · other / latin
y = 1/4(3x² - 5x - 4)
(-0.81, 1/2)
(2.48, 1/2)
y = 1/2
(5/6, -1.5)
B04 · paragraph / latin
(ii) From the curve the value of x for
of y (i.e. for the root of the eq.)
1/4(3x² - 5x - 4) is (-0.54)
is the least value of (3x² - 5x - 4)
(i.e.) the least value of y is =
1/4(-1.54) = -6.08 or -6 1/2
Ans. II
B05 · paragraph / latin
(iii) The equation 3x² - 5x + 6 = 0
is identical with the equation
3x² - 5x - 4 = 2 ≡ 1/4(3x² - 5x - 4) = 1/2
∴ the roots of 3x² - 5x - 6 = 0
are the same as the roots of
1/4(3x² - 5x - 4) = 1/2 .
Draw the curve y = 1/2, this line will
intersect the original curve at (2.48, 1/2)
and (-0.81, 1/2) ; the roots are:
x = 2.48
x = -0.81
Ans.
B01 · header / latin
⟦illegible⟧ Exam
14/5/1968
B02 · paragraph / latin
Let distance ridden be x miles also ½ minute = 1/120 hours.
x/3x10 + x/3x8 + x/3x8 = x/2x10 + x/2x8 - 1/120
x/30 + x/24 + x/24 = x/20 + x/16 - 1/120 the least common denominator is 2³x3x5
or x/2x3x5 + x/2³x3 + x/2³x3 = x/2²x5 + x/2⁴ - 1/2³x3x5 multiplying all the equation by ⟦illegible⟧
72x + 90x + 90x = 108x + 135x - 18
x = 18 Ans.
B03 · paragraph / latin
4(i) the sum Sₙ = ⅓ n (4n²-1) when n=1 S₁ = ⅓(1)(4-1) = ⅓(3) = 1 = 1st term
when n=2 S₂ = ⅓ ⋅ 2 (4x4-1) = 2/3 x 15 = 10 = sum of 1st + 2nd terms
∴ 1st term = 1 Ans.
2nd " = 10 - 1 = 9
B04 · paragraph / latin
(ii) 2s. 3d. = 27d. ∴ the boring of the 1st foot costs 27d.
" " 2nd " " 28d. + 29d.
∴ we have an A.P. in which a = 27, d = 1, n = 400
∴ l = a + (n-1)d = 27 + 399 = 426 d. = £1. 15s. 6d. cost of boring the last foot
S = n/2 (a + l) = 400/2 (27 + 426) = 200 x 453 = 90600 d. (be used)
∴ S = 90600 / 240 = £377 ½ = £377. 10s. cost of boring the entire well
Ans.
B05 · paragraph / latin
(iii) l₃ = 18 a = ? 40.5 = ar⁴ ∴ r² = 81/2 / 18 = 81/36
l₅ = 40.5 S₆ = ? 18 = ar²
∴ r = ± 9/6 = ± 3/2
Since all the terms of the G.P. are positive ∴ r = 3/2
∴ 18 = a(3/2)² ∴ a = 18x4 / 9 = 8 Ans. 1
S₆ = a(r⁶-1) / r-1 = 8[(3/2)⁶-1] / 3/2-1 = 8[729/64 - 1] / 1/2 = 16(729-64 / 64)
∴ S₆ = 16 x 665 / 64 = 665 / 4 = 166 ¼ Ans. 2
B06 · paragraph / latin
5.(i) Draw the graph of y = ¼(3x²-5x-4) from x = -2 to x = 4 using
1 inch = 1 unit on both axes
(ii) Find the least value of 3x²-5x-4
(iii) solve the equation 3x²-5x-6 = 0
B07 · table / latin
x | y
-2 | 4.5
-1 | 1
0 | -1
1 | -1.5
2 | -0.5
3 | 2
4 | 6
B01 · header / latin
Final Examination in Algebra, ⟦illegible⟧ year
14/5/1968
B02 · paragraph / latin
x⁹y³ - xy⁹ = x³(x⁶ - y⁶) - xy³(x⁶ - y⁶)
(x³ + y³)(x³ - xy³) = (x³ + y³)(x² - x²y² + y⁴)(x - 2y)(x² + 2xy + 4y²)
Ans.
B03 · paragraph / latin
⟦illegible⟧ ÷ ⟦illegible⟧ { ⟦illegible⟧ ÷ ⟦illegible⟧ }
= ⟦illegible⟧
= ⟦illegible⟧ = x²/y³ Ans.
B04 · paragraph / latin
2x⁴ + 2x³ - 5x² - x + 3 + x + x + 3x² + B | x + x² - 2
⟦line⟧ | 2x³ - x² + x + 3
Remainder = B + 6 = 0 ∴ B = -6 Ans.
B05 · table / latin
x = ⁷√ (0.1023)³ cos 41° 28' / (1.007)⁴ tan 47° 51'
log 0.1023 = 1.0098 | 3 log 0.1023 = 3.0294
log cos 41° 28' = 1.8747 | log cos 41° 28' = 1.7494
log 1.007 = 0.0029 | log Num. = 4.7788
log tan 47° 51' = 0.0433 | log Denom. = 0.2223
log x = 4.5565
log x = 1.50807 = 1.5081
x = 0.3222 Ans.
B06 · paragraph / latin
(ii) (31.01)³ˣ = 104(2.003)²ˣ⁺¹ ∴ (3x-1) log 31.01 = log 104 + (2x+1) log 2.003
3x log 31.01 - log 31.01 = log 104 + 2x log 2.003 + log 2.003
x(3 log 31.01 - 2 log 2.003) = log 104 + log 2.003 + log 31.01
x = (log 104 + log 2.003 + log 31.01) / (3 log 31.01 - 2 log 2.003)
B07 · table / latin
log 104 = 2.0170 | x = (2.0170 + 0.3016 + 1.4915) / (4.4745 - 0.6032)
log 2.003 = 0.3016 | x = 3.8101 / 3.8713 = 0.98419 = 0.9842
log 31.01 = 1.4915
B01 · header / latin
⟦illegible⟧ Examination ⟦illegible⟧
⟦illegible⟧ 1958
B02 · paragraph / latin
1. (i) Find the value of ⟦illegible⟧
(ii) Find the values of a and b which will make the expression
x⁴ + ax³ + bx - 6 divisible by (x-2) and (x+3), and find the ⟦illegible⟧
⟦line⟧
B03 · paragraph / latin
2. Find the square root of:
x⁶/25 - 2/5 x⁵ + 244/225 x⁴ - 122/105 x³ + 136/49 x² - 2/3 x + 1/4 (20 marks)
B04 · paragraph / latin
3. (i) Reduce to simplest form:
x - 4/x
⟦line⟧
x + 2 - 4x / (1 + x/ (2x-1 / (1 + 1/x-1))) (10 marks)
B05 · paragraph / latin
(ii) Solve the equation:
1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x (10 marks)
B06 · paragraph / latin
4. Find the values of x, y and z from the following equations:
4x - y + 2z = 15; y + 2z = 3x - 2; y + 4z = ⟦illegible⟧ (20 marks)
B07 · paragraph / latin
5. A football match a charge of 2s is made for sitting in the ground,
and an extra charge of 2s 6d for a reserved seat. If N people are
admitted to the ground and x people take seats, show that the
total amount received, £P, is given by 40P = 4N + 5x.
Write the formula so that x is the subject. Then find the
number of people taking seats if 2000 enter the ground and
the amount received is £600. (20 marks)
⟦line⟧
Shamash Secondary School
Mid-Year Examination, Feb. 1968
Subject: Algebra
Date: 7/2/1968
Class: 4th Year, Scientific Section.
Time: 8:30 - 11:00 a.m.
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Attempt all questions:
1. (a) Resolve into factors:
(i) (2a+b)x² - (a-b)x - (a + 2b) (4 marks)
(ii) 201x² - 99x - 102 (4 " )
(iii) x⁹ - 64x³ - x⁶ + 64 (4 " )
(b) Show that any common factor of A and B is also a factor of mA & nB.
(8 ma⟦rk⟧
2. (a) The following equation is true for all values of x:
(2x-3)²-c = 2Ax² - 4Bx. Find the values of A,B and C. (8 " )
(b) Of the following three equations, one is always true, one is
sometimes true and one is never true. Find which is which,
giving your reasons:
(i) 3x(x-4)+x = 5(x²-1)+13-11x (4 marks)
(ii) x²(2x-5)+3(x-1) = 2x³-x(5x-3)-3 (4 " )
(iii) x(x²-1)+2(1+x)(1-x) = 0 (4 " )
3. (i) Solve the equation: 3x³-14x²+32 = 0 (10 " )
(ii) Solve the two simultaneous equations:
x+y+2xy+x²+y² = 0 ...........(1)
x-y-2xy+x²+y² = 6 ...........(2) (10 " )
4. (i) Running separately, two taps can fill a bath with water in "a"
and "b" minutes respectively. Prove that they take ab/a+b minutes
to fill it when running together. (10 marks)
(ii) If, when they are running separately, the first tap can fill the
bath in 7 minutes less time than the second, and when they are
running together they fill it in 12 minutes, find the values
of "a" and "b". (10 m⟦a⟧
5. (i) In an examination taken by both boys and girls, 41 candidates out
of every 68 pass. Five boys out of every 8 pass and 7 girls out
of every 12 pass. Find the ratio of boy candidates to girl
candidates.
(ii) If 168 girls passed the examination, find the total number of
candidates.
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B01 · paragraph / latin
(4) (i) The 1st tap fills the bath in "a" minutes ∴ it fills 1/a of the bath in one minute
The 2nd " " " " " "b" " ∴ it " 1/b " " " " " "
∴ both taps together fill (1/a + 1/b) of the bath in one minute
" " " " " (a+b/ab) " " " " " "
" " " " " (1/ab) " " " in (1/(a+b/ab)) minutes
" " " " " (ab/ab) " " " i.e. all the bath in (ab/a+b) minutes. Ans.
B02 · paragraph / latin
(ii) When running separately, the 1st tap fills the bath in 7 minutes less than
the 2nd tap ⟦...⟧ i.e. b - a = 7 ⟦line⟧ ①
also, when running together, they both fill the bath in (ab/a+b) minutes i.e. ab/a+b = 12 ⟦line⟧ ②
from ① b = a + 7 and from ② ab = 12a + 12b
∴ a(a + 7) = 12a + 12(a + 7) or a² + 7a = 12a + 12a + 84
a² - 17a - 84 = 0 ∴ (a + 4)(a - 21) = 0 ∴ a = -4 to be discarded
or a = 21 minutes
∴ b = a + 7 = 21 + 7 = 28 minutes or [and b = 28 minutes] Ans.
B03 · paragraph / latin
5(i) Let the number of boy candidates in any case be = b and the
" " " girl " " " " " be = g
5b/8 + 7g/12 = 41/68 (b + g) or 5x17x3b + 7x17x2g = 6x41(b + g)
255b + 238g = 246b + 246g or 9b = 8g ∴ b/g = 8/9
No. of boy candidates = 8 : 9 Ans.
No. of girl candidates
B04 · paragraph / latin
(ii) When the no. of girl candidates who passed the examination is
equal to 168, then in the above relation we get: 7/12 g = 168
∴ g = 168 x 12 / 7 = 24 x 12 = 288
b = 8/9 g = 8/9 x 288 = 8 x 32 = 256
∴ b + g = 256 + 288 = 544
∴ Total number of candidates = 544 Ans.
B01 · paragraph / latin
(i) 3x(x-1) + 2 = 5(x²-4) + 13 - 11x
∴ 3x² - 3x + 2 = 5x² - 20 + 13 - 11x or 2x² - 8x - 9 = 0 or x² - 4x - 4.5 = 0
⟦illegible⟧
(ii) x²(x-5) + 3(x-1) = 2x³ - x(5x-3) - 3
∴ x³ - 5x² + 3x - 3 = 2x³ - 5x² + 3x - 3
which is an identity and ⟦illegible⟧ true for all values of x
(iii) x(x²-1) + 2(1+x)(1-x) = 0
or x(x+1)(x-1) + 2(x+1)(1-x) = 0
∴ (x+1) [x(x-1) + 2(1-x)] = 0 or (x+1) [x(x-1) - 2(x-1)] = 0
∴ (x+1)(x-1) [x-2] = 0 ∴ x = 1, -1, 2
This is a conditional equation which is true for these values only. In other words it is not true.
3. (i) 2x³ - 14x² + 22x = 0 By trial x = 2 satisfies the equation
since 2(2)³ - 14(2)² + 22(2) = 16 - 56 + 44 = 4 ∴ x = 2 is a root
and hence (x-2) is a factor
∴ 2x³ - 14x² + 22x = (x-2)(2x² - 10x - 2) = 0
2(x-2)(x² - 5x - 1) = 0 ∴ x = 2
x = ⟦illegible⟧
x + y + 2xy + x² + y² = 0 ... ① ∴ (x+y) + (x+y)² = 0 ... (1a)
x - y + 2xy + x² + y² = 6 ... ② ∴ (x-y) + (x-y)² = 6 ... (2a)
from (1a) (x+y)[1+x+y] = 0 ∴ x+y = 0 or y = -x ... ③
also 1+x+y = 0 or y = -x-1 ... ④
substitute from ③ into (2a), since (x+x) + (x+x)² = 6 or 2x + 4x² - 6 = 0
∴ 2x² + x - 3 = 0 ∴ (2x+3)(x-1) = 0 ∴ x = 1 from ③ ∴ y = -1
x = -3/2 y = 3/2
Ans. 1 x = 1, y = -1 Ans. 2 x = -3/2, y = 3/2
substitute from ④ into (2a), since (x+x+1) + (x+x+1)² = 6 ∴ 2x+1 + (2x+1)² = 6
∴ 4x² + 6x - 4 = 0 or 2x² + 3x - 2 = 0 ∴ (2x-1)(x+2) = 0 or x = 1/2 or x = -2
x = 1/2 y = -3/2 Ans. 3 x = -2, y = 1 Ans. 4
x = -2, y = 1