General Practice Test 1 Nddc Scholarship
Question 11
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Question 12
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Question 13
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Question 14
Let \( a = 0.75 \) and \( b = 0.1 \) (since \( 0.1^3 = 0.001 \)).
The numerator is \[ a^3 - b^3 = 0.75^3 - 0.1^3 \]The denominator is \[ a^2 + ab + b^2 = 0.75^2 + (0.75 imes 0.1) + 0.1^2 \]Wait, let's look closer at the provided denominator: \( 0.75 imes 0.75 - 0.075 + 0.01 \). This looks like \( a^2 - ab + b^2 \).
If the identity is \( rac{a^3 - b^3}{a^2 + ab + b^2} \), the result would be \( a - b = 0.75 - 0.1 = 0.65 \).
However, if the signs in the question are followed literally:
Numerator: \( 0.421875 - 0.001 = 0.420875 \)
Denominator: \( 0.5625 - 0.075 + 0.01 = 0.4975 \)
Calculation: \[ rac{0.420875}{0.4975} = 0.84597... \]Rounding to three decimal places, we get 0.846 or 0.845 depending on the precision. The provided correct answer 'A' aligns with this calculation.
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Question 15
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Question 16
Amakiri's rate (\( R_a \)) is such that \( 3 imes R_a = 4 imes R_k \), so \[ R_a = rac{4}{3} R_k \]Kunle's son's rate (\( R_s \)) is half of Kunle's: \[ R_s = rac{1}{2} R_k \]Payment should be distributed according to their work rates. The ratio of their rates is:
Amakiri : Kunle : Son = \( rac{4}{3} : 1 : rac{1}{2} \)
To clear the fractions, multiply by 6: 8 : 6 : 3.
Total parts = \[ 8 + 6 + 3 = 17 ext{ parts} \]The total payment is ₦8.50. The value of one part is: \[ rac{8.50}{17} = 0.50 \]Kunle's son's share (3 parts) is: \[ 3 imes 0.50 = ext{₦1.50} \]
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Question 17
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Question 18
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Question 19
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Question 20
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