A student needs at least three notebooks and three pencils. Notebooks cost #60 and pencil #36 and the student has #360 to spend. The student decides to spend as much as possible of his #360. (a) How many ways can he spend his money (b) Does any of the ways give him change ??? if so, how much ?


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Question Number 10294 by Tawakalitu ayo mi last updated on 02/Feb/17

$A student needs at least three notebooks and$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $student decides to spend as much as possible$ $You can't use 'macro parameter character #' in math mode$ $(a) How many ways can he spend his money$ $(b) Does any of the ways give him change ? ? ?$ $if so, how much ?$
	Answered by mrW1 last updated on 02/Feb/17

	$f o r 3 n o t e b o o k s a n d 3 p e n c i l s :$ $3 \times 60 + 3 \times 36 = 288$ $m o n e y l e f t :$ $360 - 288 = 72$ $t h e r e a r e 2 w a y s t o s p e n d t h i s m o n e y :$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$
	Commented by Tawakalitu ayo mi last updated on 02/Feb/17

	$wow, God bless you sir .$
	Answered by sandy_suhendra last updated on 02/Feb/17

	$let the notebooks he gets = x$ $and the pencils he gets = y$ $the linear programs are :$ $x ⩾ 3, y ⩾ 3$ $60 x + 36 y ⩽ 360 \Rightarrow 5 x + 3 y ⩽ 30$ $if x = 3 \Rightarrow 15 + 3 y ⩽ 30$ $he gets y = 5 (no change)$ $if x = 4 \Rightarrow 20 + 3 y ⩽ 30$ $he only gets y = 3$ $You can't use 'macro parameter character #' in math mode$ $if x = 5 \Rightarrow 25 + 3 y ⩽ 30 \Rightarrow y < 3$
	Commented by Tawakalitu ayo mi last updated on 02/Feb/17

	$wow, God bless you sir .$
	Answered by arge last updated on 03/Feb/17

	$a = N o o f n o t e b o o k s = 3$ $b = N o o f p e n c i l s = 3$ $x = c o s t o f e a c h n o t e b o o k s = 60$ $y = c o s t o f e a c h p e n c i l = 36$ $z = m o n e y o f s t u d e n t s = 360$ $a x + b y = 360$ $3! x + 3! y = 360$ $3 \times 2 \times 1 x + 3 \times 2 \times 1 y = 360$ $3 x + 3 y$ $3 x + 2 y$ $3 x + y$ $2 x + 3 y$ $2 x + 2 y$ $2 x + y$ $x + 3 y$ $x + 2 y$ $x + y$ $c o u l d b e 9 w a y s, b u t :$ $360 - (2 x + 3 y) = 132, n o t$ $R t a : 8 w a y s$
	Commented by Tawakalitu ayo mi last updated on 03/Feb/17

	$God bless you sir .$
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