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Sia ? 6 years, 6 months ago
Let n = 6q + 5, where q is a positive integer.
We know that any positive integer is of the form 3k or, 3k +1 or, 3k + 2.
{tex}\therefore{/tex} q = 3k or, 3k + 1 or, 3k + 2.
If q = 3k, then,we have,
n = 6q + 5 = 18k + 5 = 3 (6k + 1) + 2 = 3m + 2, where m = 6k + 1
If q = 3k + 1, then
n = 6q + 5 = 6 (3k + 1) + 5 = 3 (6k + 3) + 2 = 3m + 2, where m = 6k + 3
If q = 3k + 2, then
n = 6q + 5 = 6 (3k + 2) + 5 = 3 (6k + 5) + 2 = 3m + 2, where m = 6k + 5.
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Sia ? 6 years, 5 months ago
Let the original list price of the toy be Rs. x.
{tex}\therefore{/tex} Number of toys can be bought for Rs 360 = {tex}\frac{360}{x}{/tex} toys
Now, Reduced list price of the toy = Rs(x - 2)
{tex}\therefore{/tex} Number of toys can be bought with new reduced list price for Rs 360 = {tex}\frac{360}{x-2}{/tex} toys
According to the question:
{tex}\frac{\displaystyle360}{\displaystyle x-2}=2+\frac{\displaystyle360}{\displaystyle x}{/tex} (2 extra toys can be bought if price reduces by 2 rupees)
{tex}\therefore \frac{360}{x - 2 } - \frac{360}{x }{/tex} = 2
{tex}\Rightarrow \frac{ 360x - 360(x - 2 )}{x (x - 2)}{/tex} = 2
{tex}\Rightarrow \frac{360x - 360x + 720}{x^2 - 2x }{/tex} = 2
{tex}\Rightarrow{/tex} 720 = 2(x2 - 2x)
{tex}\Rightarrow x^2 - 2x = \frac{720 }{2}{/tex}
{tex}\Rightarrow{/tex} x2 - 2x = 360
{tex}\Rightarrow{/tex} x2 - 2x - 360 = 0
{tex}\Rightarrow{/tex} x2 - 20x + 18x - 360 = 0
{tex}\Rightarrow{/tex} x(x - 20) + 18(x - 20) = 0
{tex}\Rightarrow{/tex} (x - 20)(x + 18) = 0
{tex}\Rightarrow{/tex} x - 20 = 0 [{tex}{/tex} Since, Cost cannot be negative. {tex}\therefore{/tex},x + 18 {tex}\neq{/tex} 0]
{tex}\Rightarrow{/tex} x = 20
Hence, the original list price of the toy is x = Rs 20.
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