<p>4:Convert the word problem into numbers:
what: x (a variable)
percent: (__/100) (divided by 100)
of: * (multiplication or times)
(q + 1): (q + 1) (exactly what it says)
is: = (equals)
r:r (exactly what it says)</p>
<p>so you get:
(x/100)*(q+1) = r</p>
<p>now solve for x:
(x/100) = (r/(q+1))
x = (100r/(q+1))</p>
<p>wow. very nice sgk</p>
<p>That doesn't answer my question sgk..</p>
<p>least common D ^</p>
<p>the first three problems are GRID IN problems</p>
<p>For the first problem:
What I did was.....
x=total no. of students. s=staplers r=rulers g=glue bottles
So, s=x/2 r=x/3 g=x/4
r+s+g=65
x/2+x/3+x/4=65
solving,
x=60</p>
<p>As for how,2s=3r=4g
x=2s or x=3r or x=4g
So,2s=3r=4g</p>
<p>Gluttony, I'm depressed you didn't ask me :(
OK , the questions are not well-framed but still...</p>
<hr>
<p>1)in an art class,there was just enough staplers,rulers and glue bottles.if every 2 student share a stapler,every 3 student share a ruler and every 4 students share a glue bottle and the total number of staplers,rulers and glue bottles were 65 then how many student was there?</p>
<p>Let number of staplers, rules, glue bottles = x,y,z respectively.
Number of students = 2x = 3y = 4z
x + y + z = 65
x + 2x/3 + x/2 = 65
13x / 6 = 65
x = 30
And number of students = 2x = 60 students </p>
<hr>
<p>2)how many positive integers less than 1000 are multiples of 5 and are equal to 3 times an even integer?</p>
<p>The number should be of the type: 5k, as well as 6n
for 5k = 6n the numbers should be multiples of 30.
Between 1 to 999, inclusive, the number of multiples of 30 can be found by:
Last multiple = 990 = 30 + (n-1)30
(this is by using the formula of mth term of an arithmetic progression)
So n = 33
So there are 33 positive integers satisfying the given conditions.</p>
<hr>
<p>3)if a and b are positive integers and a+b<1000 and a/b=0.625 then what is the biggest value of b can be?</p>
<p>a/b = 0.625 = 625/1000 = 5/8</p>
<p>a + b < 1000
b + 5b/8 < 1000
13b < 8000
b < = 615
but 5b/8 should also be an integer.
so b should be a multiple of 8 .
if we divide 615/8, we get a remainder of 7... which means the max multiple of 8 acceptable is 615 - 7 = 608
So the answer is 608.</p>
<hr>
<p>4) if q and r are positive integers, what percent of (q+1) is r?</p>
<p>[r / (q+1)] x 100 %
OR
[(100r)/(q+1)] %</p>
<p>Spidey to the rescue ;)</p>
<p>spidey.....i solved them by myself......just giving out this ques so others can get a bite.....thnx for ur detailed solutions though..... :D</p>
<p>The questions have been answered before in much the same way. I fail to see how spidey's explanations were any clearer than the others.</p>
<p>I do like how he solved number 3 though.</p>
<p>i wrote detailed,not clearer..all the solutions are good,it's just that everyone has their different way</p>