Question

when faridabibi put oranges in some boxes, she observed that if she would put 20
oranges
in each box, 1 more box would be required. Forming simultameous equations, let us calculate
more in each box, then 3 boxes would be less required. But if she would put 5 oranges less
how many boxes and oranges did faridabidi have?​

Answer 1

Answer:

$\huge{\underline{\mathtt{\red{T}\pink{O}\green{\:F}\blue{I}\purple{N}\orange{D}}}}$

$\sf Total \: no. \:of\: boxes$

$\sf Total\: no.\: of \:oranges$

$\huge{\underline{\mathtt{\red{S}\pink{O}\green{L}\blue{U}\purple{T}\orange{I}\red{O}\pink{N}}}}$

$\sf let\: the \:number\: of\: oranges\: be\: x \:and \:number \:of \:boxes\: be\: y$

$\sf so,\: total \:no.\: of \:oranges= xy$

$\sf in \:case - 1$

$\sf no.\:of \:oranges \:in \:each\: box = x+20$

$\sf no.\: of \:boxes = y-3$

$\sf\blue{Point \:to \:be \:noted:}$

$\sf\ Total \:Number\: of\: Boxes\:$

$\ = \dfrac{Total \:Number\: of \:Oranges}{Number\: of \:oranges\: in\: each\: box}$

$\ ⟹(y-3)= \dfrac{xy}{x+20}$

$\sf ⟹(x+20)(y-3) = (xy)$

$\sf ⟹xy-3x+20y-60=xy$

$\sf ⟹xy-xy-3x+20y=60$

$\sf ⟹-3x+20y=60 -----equation(i)$

$\sf in case -2$

$\sf no.\: of \:oranges\: in\: each\: case = x-5$

$\sf no. \:of\: boxes = y+1$

$\sf Total\: Number\: of \:Boxes$

$\ = \dfrac{Total \:Number \:of \:Oranges}{Number\: of\: oranges\: in\: each\: box}$

$\ (y+1) = \dfrac{xy}{x-5}$

$\sf ⟹(x-5)(y+1)= xy$

$\sf ⟹xy+x-5y-5=xy$

$\sf ⟹xy-xy+x-5y-5=0$

$\sf ⟹x-5y=5 ------ equation (ii)$

$\sf from\: both \:equations\: (i) \:and\: (ii)$

$\sf multiplying \:equation \:(ii)\: by \:3$

$\sf -3x+20y=60$

$\sf 3x-15y= 15$

____________ $\sf (by \:adding)$

$\sf 0x+5y=75$

$\sf ⟹y=15$

$\sf putting \:the \:value\: of\: y\: in\: equation\: (ii) ,we \:get$

$\sf x-5×15=5$

$\sf ⟹x=80$

$\sf therefore, \:no. \:of\: oranges \:in\: each\: box\:= 80 \:and \:number\: of\: boxes = 15$