Question

100 pencils are to be kept inside 3 types of boxes a,b,c if 5 boxes of type a and 3 boxes of type b,2 boxes of type c are used 6 pencils are left out if 3 boxes of type a,5 boxes of type b 2 boxes of type c 2 pencils are left out if2 boxes of type a,4 boxes of types,4 boxes oftype c there is space for 4 more pencil find the no of pencils that each box can hold ​

Answer 1

$\text{Let x, y and z be respectively the number of pencils}$

$\text{that each box A,B and C can hold}$

$\text{As per given data,}$

$\text{we have the equations}$

$5\,x+3\,y+2\,z=94\;$......(1)

$3\,x+5\,y+2\,z=98\;$......(2)

$2\,x+4\,y+4\,z=104\;$......(3)

$\text{Subract (2) from (1)}$

$2\,x-2\,y=-4$

$x-y=-2\;$......(4)

$\bf(2){\times}2-(3)\implies$

$6\,x+10\,y+4\,z=196$

$2\,x+4\,y+4\,z=104$

$4\,x+6\,y=92$

$2\,x+3\,y=46\;$.....(5)

$\bf(5)+3{\times}(4)\implies$

$2\,x+3\,y=46$

$3\,x-3\,y=-6$

$5\,x=40$

$x=\dfrac{40}{5}$

$\implies\boxed{\bf\,x=8}$

$\text{Put x=8 in (4) we get}$

$8-y=-2$

$-y=-10$

$\implies\boxed{\bf\,y=10}$

$\text{Put x=8 and y=10 in (1), we get}$

$5(8)+3(10)+2\,z=94$

$40+30+2\,z=94$

$2\,z=94-70$

$2\,z=24$

$\implies\,z=\dfrac{24}{2}$

$\implies\boxed{\bf\,z=12}$

$\therefore$

$\textbf{Box A can hold 8 pencils}$

$\textbf{Box B can hold 10 pencils}$

$\textbf{Box C can hold 12 pencils}$

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