Mathematics (Core), 2007, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2007

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

41 - 50 of 67 Questions

#	Question	Ans
41.	Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean A. 9 B. 8 C. 7 D. 6 Show Content Detailed Solution Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 mean = \(\frac{39}{9}\) = 4.33 then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\) = 9	A
42.	Q is 32 km away from P on a bearing 042^o and R is 25km from P on a bearing of 132^o. Calculate the bearing of R from Q. A. 122^o B. 184^o C. 190^o D. 226^o	B
43.	The bearing of a point P from another point Q is 310^o. If \|PQ\| = 200m, how far west of Q is P? A. 128.6m B. 153.2m C. 167.8m D. 187.9m Show Content Detailed Solution cos 40 = \(\frac{x}{200}\) x - 200 x cosx = 200 x 0.7660 x = 153.2m	B
44.	Find the value of a if log10 a + log10a2 = 0.9030 A. 4.0 B. 2.0 C. 1.6 D. 0.0 Show Content Detailed Solution Log a + log a2 = 0.9030 log (a x a2) = 0.9030 log a3 = 0.9030 a3 = 10^.9030(antilogarithm table) a3 = 7.998 a = 3\(\sqrt{7.998}\) a = 1.6	C
45.	A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime? A. 2 B. 3 C. 4 D. 6 Show Content Detailed Solution Let normal working hour = x let overtime = y after 10 hours = N31.00 1 x 2.50x + 4y = 31 4 x x + y = 10 2.50x + 4y = 31 - 4x + 4y = 30 ---------------- -1.5x = -9 x = 6; x + y = 10 6 + y = 10 y = 10 - 4 y = 4hrs for over time	C
46.	At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years? A. 4% B. 3\(\frac{1}{2}\)% C. 3% D. 2\(\frac{1}{2}\)% Show Content Detailed Solution SI = \(\frac{PRT}{100}\) 39 = \(\frac{520 \times R \times 3}{100}\) 1560R = 3900 R = \(\frac{3900}{1560}\) = 2.5% R = 2\(\frac{1}{2}\)%	D
47.	PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height A. 1.7cm B. 3.0cm C. 3.2cm D. 3.9cm Show Content Detailed Solution By Pythagoras theorem h² = (2\(\sqrt{3})^2 - (\sqrt{3})^2\) = 2²(\(\sqrt{3})^2 - 3\) 4(3) - 3 = 112 - 3 h² = 9 h = \(\sqrt{9}\) = 3cm	B
48.	Which of the following is a factor of 2 - x - x²? A. 1 - x B. 1 + x C. x - 1 D. 2 - x Show Content Detailed Solution 2 - x - x²; (2 - 2x) + (x - x²) = 2(1 - x) + x(1 - x) = (1 - x)(2 + x) = 1 - x	A
49.	Make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g A. \(\frac{a + bc - fg}{dg}\) B. \(\frac{a - bc + fg}{dg}\) C. \(\frac{a + bc - f}{dg}\) D. \(\frac{a + bc - dg}{dg}\) Show Content Detailed Solution \(\frac{a + bc}{wd + f}\) = g(cross multiply) a = bc + wdg + fg wdg = a + bc - fg w = \(\frac{a + bc - fg}{dg}\)	A
50.	For what range of values of x is 4x - 3(2x - 1) > 1? A. x > -1 B. x > 1 C. x < 1 D. x < -1 Show Content Detailed Solution 4x - 3(2x - 1) > 1 4x - 6x + 3 > 1 -2x > 1 - 3; 2x > -2 x < \(\frac{-2}{-2}\) = x < 1	C

41.	Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean A. 9 B. 8 C. 7 D. 6 Show Content Detailed Solution Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 mean = \(\frac{39}{9}\) = 4.33 then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\) = 9	A
42.	Q is 32 km away from P on a bearing 042^o and R is 25km from P on a bearing of 132^o. Calculate the bearing of R from Q. A. 122^o B. 184^o C. 190^o D. 226^o	B
43.	The bearing of a point P from another point Q is 310^o. If \|PQ\| = 200m, how far west of Q is P? A. 128.6m B. 153.2m C. 167.8m D. 187.9m Show Content Detailed Solution cos 40 = \(\frac{x}{200}\) x - 200 x cosx = 200 x 0.7660 x = 153.2m	B
44.	Find the value of a if log10 a + log10a2 = 0.9030 A. 4.0 B. 2.0 C. 1.6 D. 0.0 Show Content Detailed Solution Log a + log a2 = 0.9030 log (a x a2) = 0.9030 log a3 = 0.9030 a3 = 10^.9030(antilogarithm table) a3 = 7.998 a = 3\(\sqrt{7.998}\) a = 1.6	C
45.	A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime? A. 2 B. 3 C. 4 D. 6 Show Content Detailed Solution Let normal working hour = x let overtime = y after 10 hours = N31.00 1 x 2.50x + 4y = 31 4 x x + y = 10 2.50x + 4y = 31 - 4x + 4y = 30 ---------------- -1.5x = -9 x = 6; x + y = 10 6 + y = 10 y = 10 - 4 y = 4hrs for over time	C

46.	At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years? A. 4% B. 3\(\frac{1}{2}\)% C. 3% D. 2\(\frac{1}{2}\)% Show Content Detailed Solution SI = \(\frac{PRT}{100}\) 39 = \(\frac{520 \times R \times 3}{100}\) 1560R = 3900 R = \(\frac{3900}{1560}\) = 2.5% R = 2\(\frac{1}{2}\)%	D
47.	PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height A. 1.7cm B. 3.0cm C. 3.2cm D. 3.9cm Show Content Detailed Solution By Pythagoras theorem h² = (2\(\sqrt{3})^2 - (\sqrt{3})^2\) = 2²(\(\sqrt{3})^2 - 3\) 4(3) - 3 = 112 - 3 h² = 9 h = \(\sqrt{9}\) = 3cm	B
48.	Which of the following is a factor of 2 - x - x²? A. 1 - x B. 1 + x C. x - 1 D. 2 - x Show Content Detailed Solution 2 - x - x²; (2 - 2x) + (x - x²) = 2(1 - x) + x(1 - x) = (1 - x)(2 + x) = 1 - x	A
49.	Make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g A. \(\frac{a + bc - fg}{dg}\) B. \(\frac{a - bc + fg}{dg}\) C. \(\frac{a + bc - f}{dg}\) D. \(\frac{a + bc - dg}{dg}\) Show Content Detailed Solution \(\frac{a + bc}{wd + f}\) = g(cross multiply) a = bc + wdg + fg wdg = a + bc - fg w = \(\frac{a + bc - fg}{dg}\)	A
50.	For what range of values of x is 4x - 3(2x - 1) > 1? A. x > -1 B. x > 1 C. x < 1 D. x < -1 Show Content Detailed Solution 4x - 3(2x - 1) > 1 4x - 6x + 3 > 1 -2x > 1 - 3; 2x > -2 x < \(\frac{-2}{-2}\) = x < 1	C

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