Hi, Let there are x ballsso black balls = x4now red balls = x4+x9= 13x36blue balls = 7xwhite balls = 56 so here according to the question x = x4+13x36+7x+56x= 22x36+7x+56x-22x36-56= 7x14x36-56= 7xx18-8= xx-144= 18xsquaring both side x2+1442-288x= 324xx2-612x+20736= 0 using quadratic formula x = 612±6122-4×207362=612±5402= 576, 36so both can be number of balls

Question 15

Hi, Let there are x balls so black balls = \frac{x}{4} now red balls = \frac{x}{4} + \frac{x}{9} = \frac{13 x}{36} blue balls = 7 \sqrt{x} white balls = 56 so here according to the question x = \frac{x}{4} + \frac{13 x}{36} + 7 \sqrt{x} + 56 x = \frac{22 x}{36} + 7 \sqrt{x} + 56 x - \frac{22 x}{36} - 56 = 7 \sqrt{x} \frac{14 x}{36} - 56 = 7 \sqrt{x} \frac{x}{18} - 8 = \sqrt{x} x - 144 = 18 \sqrt{x} squaring both side x^{2} + 144^{2} - 288 x = 324 x x^{2} - 612 x + 20736 = 0 using quadratic formula x = \frac{612 \pm \sqrt{612^{2} - 4 \times 20736}}{2} = \frac{612 \pm 540}{2} = 576, 36 so both can be number of balls

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