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Identify the longest chain containing the double bond: 5 carbons (pent), with double bond at C2 (pent-2-ene).
Number from the end that gives the double bond the lower locant. A methyl group appears at C3 and Br at C4.
IUPAC name: 4-Bromo-3-methylpent-2-ene
This is a 5-carbon chain ($n = 5$: pent) with double bond at position 2 (C2=C3), methyl at C3, and Br at C4.
The recorded value $3.50 \text{ cm}$ has a precision of $0.01 \text{ cm}$. We must find an instrument with a least count (LC) of $0.01 \text{ cm}$.
- Option A: $LC = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm}$
- Option B: $LC = \frac{1 \text{ mm}}{50} = 0.02 \text{ mm} = 0.002 \text{ cm}$
- Option C: $LC = 0.1 \text{ cm}$
- Option D: $1 \text{ MSD} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm}$. Since $10 \text{ VSD} = 9 \text{ MSD}$, $1 \text{ VSD} = 0.9 \text{ MSD}$. $LC = 1 \text{ MSD} - 1 \text{ VSD} = 0.1 \text{ MSD} = 0.01 \text{ cm}$.
Thus, Option D matches the required precision.
Financial data: Profit from operations $59,800; Finance costs $12,000; Profit for year $47,800; Ordinary share capital $700,000; Retained earnings $72,500; 10% Debentures $120,000. What was the ROCE?
Option C (6.70%) is correct.
Capital Employed = ($700,000 + $72,500) + $120,000 = $892,500
ROCE = $59,800 ÷ $892,500 × 100 = 6.70%
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