Class 12th

Factorised from = 2⁴ * 3⁴ * 5² * 7²

Total no. of factor = 5* 5 * 3 * 5

= 225 factors

113^rd factor = $\sqrt[]{N}$

 i.e. 7 * 5 * 3² * 2² = 1260

150/- wax; 8.5/- thread; 12.5/- coloured paper; 9/- silver powder

Profit = 40%.

Net input = 150 + 8.5 + 12.5 + 9 = 180/-

Net profit = 180 * .4 = 72/-

$$ $\frac{1}{2}$  the candles = $\frac{}{}$ $\frac{1}{2}$   (180) = 90/-

At 50% profit = 45.

90 + 45 = 135/-

Let the remaining $\frac{1}{2}$ $\frac{}{}$ = x

Net price = (72 + 180) = 135 + x

x = 117

Profit on second $\frac{1}{2}$ $\frac{}{}$ = 117 – 90 = 27

$\frac{}{}$ $\frac{27}{90}*100$  = 30%

P (1.05)2 − P = 369 or P = 3600

$\frac{x}{5}+\frac{x}{6}$ = CP (x oranges of each type) = $\frac{11x}{30}$  and SP =  $\frac{2x}{7}$

On selling 2x oranges, loss = $\frac{x}{5}+\frac{x}{6}-\frac{2x}{7}=\frac{17x}{210}$

If loss is 34, Total oranges =  $\frac{2x*34}{\left(\frac{17x}{210}\right)}=840$

(c):Double in 4 years, four times in 8 years, eight times in 12 years.

(b):Notice that the difference between the divisors and the remainder is constant. Therefore take the LCM of the divisors or any multiple of the LCM and subtract.

The difference between the divisor and the remainder

which is 2 in this case.

Therefore, the smallest 3 digit number

= 15 * 7? 2 = 105 – 2 = 103

If Capital = X, then

$\left(\frac{X}{2}\right)*\left(3*\frac{1}{100}\right)+\frac{X}{3}*\left(5*\frac{1}{100}\right)+\left(\frac{X}{6}\right)\left(6*\frac{1}{100}\right)$

= 250

X = Rs. 6000

(d): $\frac{{(1956-1)}^{475}}{4}=\frac{{(-1)}^{475}}{4}=-1or3$

(b):Rise of milk-level = volume of 3 spheres / Base area of cylinder

^{$=\frac{\{\frac{4}{3}\pi {(5)}^3*3\}}{\{\pi {(5)}^{2}cm\}}=20cm$}

After x hours $\frac{x}{12}$ $\frac{}{}$ of the first candle is burnt out = $\frac{x}{8}$ $\frac{}{}$ of the second is burnt out.

Remaining parts are $\left(1-\frac{x}{12}\right)and\left(1-\frac{x}{8}\right)$ $$

$Now,\left(1-\frac{x}{12}\right)=2\left(1-\frac{x}{8}\right)orx=6hrs.$