Mathematics 2004 Jamb Examination

\[ \begin{array}{rrrr}
& 4 & 2 & 4 & 3 \\
- & 1 & 3 & 4 & 4 \\
\hline
& y & 3 & 4 & 4 \\
\end{array}\] Column 4:  3 - 4 = 4 after borrow ⇒ OK
Column 3: 4 - x = 4 ⇒ x = 4
Column 2: (2 - 1 = 1), then borrow ⇒ 6 - 3 = 3 OK
Column 1: 4 - 1 = 2 = y\[\boxed{x = 4,\ y = 2}\]  

The correct answer is C. 116₇

Given 451₆ – p₇ = 305₆
First we collect the like base
451₆ – 305₆ = p₇

Next we convert the base 6 to base 10

(4*6² + 5*6¹ + 1*6⁰) - (3*6² + 0*6¹ + 5*6⁰) = p₇
(144 + 30 +1) - (108 + 0 + 5) = p₇
175 - 113 = p₇
p₇ = 62₁₀

Next convert 62₁₀ to base 7

7   |   62
7   |   8    |   6
7   |   1    |   1
7   |   0    |   1

Thus  p₇ = 116₇

\[\dfrac{\dfrac{1}{10}×\dfrac{2}{3} + \dfrac{1}{4}}{\dfrac{1}{2} ÷ \dfrac{3}{5} - \dfrac{1}{4}}\]\[ \text{Numerator: } \dfrac{1}{10}\times\dfrac{2}{3}=\dfrac{2}{30}=\frac{1}{15} \] \[ \dfrac{1}{15}+\dfrac{1}{4} =\dfrac{4}{60}+\dfrac{15}{60} =\dfrac{19}{60} \] \[ \text{Denominator: } \dfrac{1}{2}\div\dfrac{3}{5} =\dfrac{1}{2}\times\dfrac{5}{3} =\dfrac{5}{6} \] \[ \dfrac{5}{6}-\dfrac{1}{4} =\dfrac{10}{12}-\dfrac{3}{12} =\frac{7}{12} \] \[ \dfrac{\dfrac{19}{60}}{\dfrac{7}{12}} =\dfrac{19}{60}\times\frac{12}{7} =\frac{228}{420} =\frac{19}{35} \] 

To find the ratio of the number of grains sown to the number of grains harvested, follow these steps:

1. Determine the total number of grains harvested
   - The farmer harvested 5000 cobs, each bearing 500 grains.
   - The total number of grains harvested is:
     \[      5000 \times 500 = 2,500,000 \text{ grains}
     \]

2. Set up the ratio of grains sown to grains harvested:
   - The number of grains sown was 5000.
   - The number of grains harvested is 2,500,000.
   - The ratio is:
     \[      \dfrac{\text{Number of grains sown}}{\text{Number of grains harvested}} = \dfrac{5000}{2,500,000}
     \]

3. Simplify the ratio:
   - Simplify \( \dfrac{5000}{2,500,000} \) by dividing both the numerator and the denominator by 5000:
     \[      \dfrac{5000 \div 5000}{2,500,000 \div 5000} = \dfrac{1}{500}
     \]

The ratio of the number of grains sown to the number of grains harvested is 1:500

Final Answer:
The correct answer is A. 1:500

To determine the fraction of the chalk that the third teacher received, follow these steps:

1. Calculate the fraction of the chalk remaining after the first teacher received their share:The first teacher received \( \dfrac{2}{5} \) of the total chalk. Thus, the remaining chalk is:
   \[    1 - \dfrac{2}{5} = \dfrac{5}{5} - \dfrac{2}{5} = \dfrac{3}{5}
   \]

2. Calculate the fraction of the remaining chalk that the second teacher received: The second teacher received \( \dfrac{2}{15} \) of the remaining chalk. To find out how much this is, multiply \( \dfrac{2}{15} \) by the remaining \( \dfrac{3}{5} \):\[\dfrac{2}{15} \times \dfrac{3}{5} = \dfrac{6}{75} = \dfrac{2}{25}\]

3. Calculate the fraction of the chalk that the third teacher received: The third teacher received the remainder of the chalk. Subtract the amounts received by the first and second teachers from the total: \[\text{Amount received by third teacher} = 1 - \left(\dfrac{2}{5} + \dfrac{2}{25}\right)\]To combine \( \dfrac{2}{5} \) and \( \dfrac{2}{25} \), find a common denominator. The least common denominator is 25:\[\dfrac{2}{5} = \dfrac{10}{25}\]\[\dfrac{2}{5} + \dfrac{2}{25} = \dfrac{10}{25} + \dfrac{2}{25} = \dfrac{12}{25}\]Thus, the fraction of chalk left for the third teacher is: \[1 - \dfrac{12}{25} = \dfrac{25}{25} - \dfrac{12}{25} = \dfrac{13}{25}\]

So, the third teacher received \( \dfrac{13}{25} \) of the chalk, which corresponds to option C.

The correct answer is B. 3

Given the equation \( \sqrt[3]{4^{2x}} = 16 \), let's solve for \( x \):

1. Cube both sides to eliminate the cubic root:
   \[    4^{2x} = 16^3
   \]

2. Simplify \( 16^3 \):
   \[    16^3 = (2^4)^3 = 2^{12}
   \]

   So, the equation becomes:
   \[    4^{2x} = 2^{12}
   \]

3. Express \( 4 \) as \( 2^2 \):
   \[    (2^2)^{2x} = 2^{12}
   \]

   Simplifying further:
   \[    2^{4x} = 2^{12}
   \]

4. Since the bases are the same, equate the exponents:
   \[    4x = 12
   \]

5. Solve for \( x \):
   \[    x = \dfrac{12}{4} = 3
   \]

\[ \frac{1}{\sqrt{3} + 2} * \frac{\sqrt{3} - 2}{\sqrt{3} - 2} \]\[= \frac{\sqrt{3} - 2}{(\sqrt{3} + 2)(\sqrt{3} - 2)} \]\[= \frac{\sqrt{3} - 2}{3 - 4} \]\[= \frac{\sqrt{3} - 2}{-1} = 2 - \sqrt{3} \]

6log_x2 - 3log_x3 = 3log₅0.2
= logx²⁶ - 3logx³³ = log5^(0.2)3
= logx(⁶⁴/₂₇) = log5(¹/₅)³
logx(⁶⁴/₂₇) = log5(¹/₁₂₅)
let logx(⁶⁴/₂₇) = y
∴x^y = ⁶⁴/₂₇
and log5(¹/₁₂₅) = y
∴ 5^y = ¹/₁₂₅
5^y = 125^-1
5^y = 5^-3
∴ y = -3
substitute y = -3 in x^y = ⁶⁴/₂₇
implies x^-3 = ⁶⁴/₂₇
¹/_x³ = ⁶⁴/₂₇
64x³ = 27
x³ = ²⁷/₆₄
x = ³√(²⁷/₆₄)
x = ³/₄

D. P^c ∩ (Q ∩ R)

Let
N = Total number of students.
P = Number of students that offers Physics.
P' = Number of students offering physics only.
C = Number of students that offers Chemistry.
X = Number of students that offers both.

P = 3X; X = P/3
C = 2P 
P + C  - X = N
3X + 2(3X) - X = 40
X = 5

P = 3X = 3(5) = 15
P' = P - X 
P' = 15 - 5 = 10.