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Motor vehicles costing $530,000 were depreciated at 10% straight-line instead of the correct 25%. After incorrect entries: Motor vehicles at cost $530,000; Provision for depreciation $53,000. Which journal entries will correct the error?
Option C is correct.
Correct depreciation = $530,000 × 25% = $132,500
Already charged = $530,000 × 10% = $53,000
Additional needed = $132,500 − $53,000 = $79,500
Entry: Debit Statement of profit or loss $79,500 | Credit Provision for depreciation $79,500
Let $T_n$ denote the total number of triangles that can be formed by joining the vertices of a regular polygon with $n$ sides. If $T_{n+1} - T_n = 10$, find the value of $n$.
A triangle is formed by choosing any 3 vertices from $n$ vertices, so $T_n = \binom{n}{3}$.
$T_{n+1} - T_n = \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$
Setting $\binom{n}{2} = 10$: $\dfrac{n(n-1)}{2} = 10 \Rightarrow n(n-1) = 20 \Rightarrow n = 5$... wait: $5 \times 4 = 20$ ✓. But checking $n=5$: $T_6 - T_5 = \binom{6}{3}-\binom{5}{3}=20-10=10$ ✓. So $n = 5$.
Actually re-checking: $\binom{n}{2}=10 \Rightarrow n=5$. Answer: $n=\mathbf{5}$, but the JEE 2013 answer is $n=5$. Wait — options show 7 as (A). Let me verify: $T_{n+1}-T_n=\binom{n}{2}=10 \Rightarrow n(n-1)=20 \Rightarrow n=5$. The correct answer is $\mathbf{5}$ (option B, index 1).
Size decreases as nuclear charge increases (more protons pull same electrons closer):
- \(\text{Ca}^{2+}\): Z=20
- \(\text{Ar}\): Z=18
- \(\text{K}^+\): Z=19
- \(\text{Cl}^-\): Z=17
- \(\text{S}^{2-}\): Z=16
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