Mathematics (Core), 1999, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1999

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 45 Questions

#	Question	Ans
1.	A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group? A. 25 B. 19 C. 18 D. 17 Show Content Detailed Solution Let the three items be M, Y and P. n{M ∩ Y} only = 4-3 = 1 n{M ∩ P) only = 5-3 = 2 n{ Y ∩ P} only = 2 n{M} only = 12-(1+3+2) = 6 n{Y} only = 10-(1+2+3) = 4 n{P} only = 14-(2+3+2) = 7 n{M∩P∩Y} = 3 Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.	A
2.	If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X. A. \(\frac{1}{2}\)X B. X-\(\frac{1}{4}\) C. X-\(\frac{1}{3}\) D. X-\(\frac{1}{2}\) Show Content Detailed Solution \(log_810\) = X = \(log_8{2 x 5}\) \(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks	C
3.	Find the value of X if \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) A. 3√2+4 B. 3√2-4 C. 3-2√2 D. 4+2√2 Show Content Detailed Solution Start your solution by cross-multiplying, \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - √2] √2 = x + √2 then collect like terms x√2 - x = √2 + 2 and factorize accordingly to get the unknown. x(√2 - 1) = √2 + 2 x = {√2 + 2} / {√2 - 1} rationalize x = {√2 + 2} / {√2 - 1} * {	A
4.	If \(\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}\) What is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10 Show Content Detailed Solution Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at p = \(\frac{5}{2}\), q = -\(\frac{25}{4}\) and r = -\(\frac{17}{4}\) NUMERATOR: a\(^{\frac{2}{1}}{\frac{3}{4}}\) b\(^{\frac{-3}{1}}{\frac{3}{4}}\) c\(^{\frac{1}{1}}*{\frac{3}{4}}\) a\(\frac{3}{2}\) b\(\frac{-9}{4}\) c\(\frac{3}{4}\) Using index method for numerator & denominator : a\(^{\frac{3}{2}}-{\frac{-1}{1}}\) b\(^{\frac{-9}{4}}-{\frac{4}{1}}\) c\(^{\frac{3}{4}}-{\frac{5}{1}}\) : a\(\frac{5}{2}\) b\(\frac{-25}{4}\) c\(\frac{-17}{4}\) Then p+2q will giv	D
5.	If {(a²b^-3c)^3/4/a^-1b⁴c⁵} = a^pb^qc^r; what is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10 Show Content Detailed Solution Hint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = -25/4 and r = -9/2. Then substitute the values of p and q into p+2q to get -10.	D
6.	A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss. A. 30% gain B. 25% gain C. 30% loss D. 25% loss Show Content Detailed Solution Cost price (cp) = (100/5) x N1.20 =N24.00 Selling price (sp) = 100-20 = 80 (80/4) x N1.50 = N30.00 Gain = SP-CP = N30.00-N24.00 = N6.00 Gain% = (gain/CP) x 100 = 25%	B
7.	What is the answer when 2434\(_6\) is divided by 42\(_6\)? A. 23₆ B. 35₆ C. 52₆ D. 55₆ Show Content Detailed Solution Hint: Two methods can be used; 1. Direct division (if you know division in number bases) 2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356 2434\(_6\) to base 10 2 \(\times\) 6\(^3\) + 4 \(\times\) 6\(^2\) + 3 \(\times\) 6\(^1\) + 4 \(\times\) 6\(^0\) 2 \(\times\) 216 + 4 \(\times\) 36 + 3 \(\times\) 6 + 4 \(\times\) 1 432 + 144 + 18 + 4 = 598 And: 42\(_6\) to base 10 4 \(\times\) 6\(^1\) + 2 \(\times\) 6\(^0\) 4 \(\times\) 6 + 2 \(\times\) 1 24 + 2 = 26 Then divide 598 by 26 = 23 Then convert 23\(_{10}\) to base 6 23\(_{10}\) = 35\(_6\)	B
8.	If 2\(_9\) x (Y3)\(_9\) = 3\(_5\) x (Y3)\(_5\), find the value of Y. A. 4 B. 3 C. 2 D. 1 Show Content Detailed Solution The correct answer is 1. PROVE==> (2x9\(^0\)) (3x9\(^0\) + Yx9\(^1\)) = (3x5\(^0\))(3x5\(^0\) + Yx5\(^1\)). You multiply to get 2(3+9Y)=3(3+5Y). You further multiply to get 6+18Y=9+15Y. ==> Collect like terms 18Y-15Y=9-6. 3Y/3=3/3. Y=1 Explanation Credit: endowednuel	D
9.	Simplify \(\sqrt{\frac{(0.0023 \times 750)}{(0.00345 \times 1.25)}}\) A. 15 B. 20 C. 40 D. 75 Show Content Detailed Solution Multiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\) Dividing, => \(\sqrt{400} = 20\).	B
10.	If \(mn = (\frac{m}{n} - \frac{n}{m}\)) for m, n belong to R, evaluate -34 A. \(\frac{-25}{12}\) B. \(\frac{-7}{12}\) C. \(\frac{7}{12}\) D. \(\frac{25}{12}\) Show Content Detailed Solution \(m*n = \frac{m}{n}-\frac{n}{m} = \frac{-3}{4} - \frac{4}{-3} = \frac{-3}{4} + \frac{4}{3} = \frac{(-9+16)}{12} = \frac{7}{12}\)	C

1.	A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group? A. 25 B. 19 C. 18 D. 17 Show Content Detailed Solution Let the three items be M, Y and P. n{M ∩ Y} only = 4-3 = 1 n{M ∩ P) only = 5-3 = 2 n{ Y ∩ P} only = 2 n{M} only = 12-(1+3+2) = 6 n{Y} only = 10-(1+2+3) = 4 n{P} only = 14-(2+3+2) = 7 n{M∩P∩Y} = 3 Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.	A
2.	If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X. A. \(\frac{1}{2}\)X B. X-\(\frac{1}{4}\) C. X-\(\frac{1}{3}\) D. X-\(\frac{1}{2}\) Show Content Detailed Solution \(log_810\) = X = \(log_8{2 x 5}\) \(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks	C
3.	Find the value of X if \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) A. 3√2+4 B. 3√2-4 C. 3-2√2 D. 4+2√2 Show Content Detailed Solution Start your solution by cross-multiplying, \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - √2] √2 = x + √2 then collect like terms x√2 - x = √2 + 2 and factorize accordingly to get the unknown. x(√2 - 1) = √2 + 2 x = {√2 + 2} / {√2 - 1} rationalize x = {√2 + 2} / {√2 - 1} * {	A
4.	If \(\frac{(a^2 b^{-3}c)^{\frac{3}{4}}}{a^{-1}b^{4}c^{5}}=a^{p} b^{q} c^{r}\) What is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10 Show Content Detailed Solution Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at p = \(\frac{5}{2}\), q = -\(\frac{25}{4}\) and r = -\(\frac{17}{4}\) NUMERATOR: a\(^{\frac{2}{1}}{\frac{3}{4}}\) b\(^{\frac{-3}{1}}{\frac{3}{4}}\) c\(^{\frac{1}{1}}*{\frac{3}{4}}\) a\(\frac{3}{2}\) b\(\frac{-9}{4}\) c\(\frac{3}{4}\) Using index method for numerator & denominator : a\(^{\frac{3}{2}}-{\frac{-1}{1}}\) b\(^{\frac{-9}{4}}-{\frac{4}{1}}\) c\(^{\frac{3}{4}}-{\frac{5}{1}}\) : a\(\frac{5}{2}\) b\(\frac{-25}{4}\) c\(\frac{-17}{4}\) Then p+2q will giv	D
5.	If {(a²b^-3c)^3/4/a^-1b⁴c⁵} = a^pb^qc^r; what is the value of p+2q? A. (5/2) B. -(5/4) C. -(25/4) D. -10 Show Content Detailed Solution Hint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = -25/4 and r = -9/2. Then substitute the values of p and q into p+2q to get -10.	D

6.	A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss. A. 30% gain B. 25% gain C. 30% loss D. 25% loss Show Content Detailed Solution Cost price (cp) = (100/5) x N1.20 =N24.00 Selling price (sp) = 100-20 = 80 (80/4) x N1.50 = N30.00 Gain = SP-CP = N30.00-N24.00 = N6.00 Gain% = (gain/CP) x 100 = 25%	B
7.	What is the answer when 2434\(_6\) is divided by 42\(_6\)? A. 23₆ B. 35₆ C. 52₆ D. 55₆ Show Content Detailed Solution Hint: Two methods can be used; 1. Direct division (if you know division in number bases) 2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356 2434\(_6\) to base 10 2 \(\times\) 6\(^3\) + 4 \(\times\) 6\(^2\) + 3 \(\times\) 6\(^1\) + 4 \(\times\) 6\(^0\) 2 \(\times\) 216 + 4 \(\times\) 36 + 3 \(\times\) 6 + 4 \(\times\) 1 432 + 144 + 18 + 4 = 598 And: 42\(_6\) to base 10 4 \(\times\) 6\(^1\) + 2 \(\times\) 6\(^0\) 4 \(\times\) 6 + 2 \(\times\) 1 24 + 2 = 26 Then divide 598 by 26 = 23 Then convert 23\(_{10}\) to base 6 23\(_{10}\) = 35\(_6\)	B
8.	If 2\(_9\) x (Y3)\(_9\) = 3\(_5\) x (Y3)\(_5\), find the value of Y. A. 4 B. 3 C. 2 D. 1 Show Content Detailed Solution The correct answer is 1. PROVE==> (2x9\(^0\)) (3x9\(^0\) + Yx9\(^1\)) = (3x5\(^0\))(3x5\(^0\) + Yx5\(^1\)). You multiply to get 2(3+9Y)=3(3+5Y). You further multiply to get 6+18Y=9+15Y. ==> Collect like terms 18Y-15Y=9-6. 3Y/3=3/3. Y=1 Explanation Credit: endowednuel	D
9.	Simplify \(\sqrt{\frac{(0.0023 \times 750)}{(0.00345 \times 1.25)}}\) A. 15 B. 20 C. 40 D. 75 Show Content Detailed Solution Multiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\) Dividing, => \(\sqrt{400} = 20\).	B
10.	If \(mn = (\frac{m}{n} - \frac{n}{m}\)) for m, n belong to R, evaluate -34 A. \(\frac{-25}{12}\) B. \(\frac{-7}{12}\) C. \(\frac{7}{12}\) D. \(\frac{25}{12}\) Show Content Detailed Solution \(m*n = \frac{m}{n}-\frac{n}{m} = \frac{-3}{4} - \frac{4}{-3} = \frac{-3}{4} + \frac{4}{3} = \frac{(-9+16)}{12} = \frac{7}{12}\)	C