Mathematics 2013 Jamb Examination

To convert a number from base ten to another base, we perform successive division by the target base and record the remainders. Here, we divide 27 by 3:
27 ÷ 3 = 9 remainder 0
9 ÷ 3 = 3 remainder 0
3 ÷ 3 = 1 remainder 0
1 ÷ 3 = 0 remainder 1
Reading the remainders from bottom to top, we get 1000. Therefore: \[ 27_{10} = 1000_{3} \] This is expected because \(3^3 = 27\), which in base three is represented by 1 followed by three zeros.

In the ratio 5:3:2, the highest share corresponds to the largest part of the ratio, which is 5. We are told this share equals 40 apples. We can set up a relationship to find the value of a single 'part' in the ratio:
Let the constant of proportionality be \(k\).
Highest share = \(5k = 40\)
Dividing both sides by 5: \[ k = \frac{40}{5} = 8 \] The smallest share corresponds to the smallest part of the ratio, which is 2. Using the value of \(k\) we just found: \[ \text{Smallest share} = 2k = 2 \times 8 = 16 \] Thus, the girl with the smallest portion receives 16 apples.

To evaluate this expression, we can simplify the decimal division first. Let's rewrite the expression: \[ \frac{1.25 \times 0.025}{0.05} \] We can multiply the numerator and denominator by 100 to clear the decimal in the divisor: \[ = \frac{1.25 \times 2.5}{5} \] Now, simplify by dividing 1.25 by 5 or 2.5 by 5. Dividing 2.5 by 5 gives 0.5: \[ = 1.25 \times 0.5 = 0.625 \] The question asks for the answer correct to 1 decimal place. Looking at the second decimal digit (2), which is less than 5, we do not round up the first decimal digit: \[ 0.625 \approx 0.6 \] 

The formula for Simple Interest is given by: \[ I = \frac{P \times R \times T}{100} \] Where:
\(I\) = Interest (₦600)
\(P\) = Principal (₦3000)
\(R\) = Rate (8%)
\(T\) = Time (in years)
Substituting the values into the formula: \[ 600 = \frac{3000 \times 8 \times T}{100} \] \[ 600 = 30 \times 8 \times T \] \[ 600 = 240T \] To find \(T\), we divide 600 by 240: \[ T = \frac{600}{240} = \frac{60}{24} = \frac{5}{2} = 2.5 \; \text{years} \] In fractional form, 2.5 years is equal to 2\(\frac{1}{2}\) years.

\[\frac{3^{-5n}}{9^{1-n}} \times 27^{\,n+1}\]

To evaluate the expression, we use the power rule of logarithms, which states that \(\log_b(a^n) = n \log_b a\). Applying this rule to the given problem: \[\log_{10} 4^{1/3} = \frac{1}{3} \log_{10} 4\]We are given the value of \(\log_{10} 4\) as 0.6021. Substituting this value into the equation: \[= \frac{1}{3} (0.6021)\]Performing the division: \[0.6021 \div 3 = 0.2007\]The logarithm of the cube root of a number is exactly one-third of the logarithm of that number.

We start by simplifying the square roots inside the parentheses by looking for perfect square factors:
\(\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
Substitute these back into the numerator: \[\sqrt{5}(7\sqrt{3} - 2\sqrt{3}) = \sqrt{5}(5\sqrt{3})\]The numerator becomes \(5\sqrt{15}\). Now, divide this by the denominator, \(\sqrt{15}\): \[\frac{5\sqrt{15}}{\sqrt{15}} = 5\]The radical terms cancel out, leaving a whole number result of 5.

First, we list the members of each set within the specified ranges:
P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} (Odd numbers between -1 and 20)
Q = {2, 3, 5, 7, 11, 13, 17, 19, 23} (Prime numbers between -2 and 25)
The intersection \(P \cap Q\) consists of elements that appear in both sets. Since all prime numbers except 2 are odd, we look for odd primes common to both lists:
3, 5, 7, 11, 13, 17, and 19 are present in both sets. 2 is prime but not odd, and 1 is odd but not prime. Therefore, the intersection set contains all common odd primes up to 19.

Observe the expression under the square root: \(t^2 - 4t + 4\). This is a perfect square trinomial, which factors as \((t - 2)^2\).
So the equation is: \[S = \sqrt{(t - 2)^2}\]Taking the square root of a squared term gives the absolute value, but typically in these problems, we consider the principal root: \[S = t - 2\]To solve for \(t\) in terms of \(S\), we add 2 to both sides of the equation: \[t = S + 2\]Thus, \(t\) is the sum of \(S\) and 2.

According to the Factor Theorem, if \((x - a)\) is a factor of a polynomial \(f(x)\), then \(f(a) = 0\).
Here, the factor is \(x - 4\), so \(a = 4\). Let \(f(x) = x^2 - x - k\).
Setting \(f(4) = 0\): \[(4)^2 - (4) - k = 0\]\[16 - 4 - k = 0\]\[12 - k = 0\]\[k = 12\]When \(k\) is 12, the quadratic expression becomes \(x^2 - x - 12\), which factors into \((x - 4)(x + 3)\).