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[1989hemantrajput]

The question is ,

In a group, each boy's quota of match stick to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total no. of stick assigned to him. Which of the following is the possible no. of stick assigned to each boy?

a) 200 b)150 c)125 d)175


Q2. Alford got an order from a garment manufacturer for 480 denim shirts. He bought 12 sewing machines and appointed some expert tailors to do the job. However many didn't report for duty. As a result, each of those who did, had to stitch 32 more shirts than originally planned by Alford, with equal distribution of work. How many tailors had been appointed earlier and how many had not reported for work?

a) 12,4 b)10,3 c)10,4 d) none of these

[jnelson0612]

Welcome to the forum!

Two things:
1) Will you please list the original source for each of these questions?
2) Will you also please post each question in a separate post?

Please see our forum guidelines for more information: read-before-you-post-general-math-folder-guidelines-t2717.html

Thanks!

[krishnan.anju1987]

Hi,

Are the answers b and c respectively. Please let me know.

This is how I solved it.

1) Let the initial number of boxes be y and sticks per box be x.
Then, after reducing the number of sticks in each box, sticks per box would be x-25 and number of boxes would be y+3

now (x-25)(y+3)=xy

as the number of sticks are constant.

hence, I arrived at the equation

3x-25y=75--------------1)

Now the answer choices are 200, 150, 125 and 175.

I worked backwards to fine the factors of these numbers
200= 2,2,5,10,50,4,10,20,200
150=3,5,10,2,15,50,6,100,30,60,150,25
125=5,5,5,25,125
175=5,5,7,35,25,175

Only 150 has factors 50 and 3 that satisfy the equation 1)

2) for the second problem, let the number of tailors appointed be x and the number of shirts that tailors were supposed to stitch be y. Then x.y=480

However, since many tailors did not turn up, the actual number of tailors became x1 and the number of shirts they had to stitch became y+32.

thus x1.(y+32)=480

now we have the solutions given as answer. Reinserting the values in the question and finding the value of y in both equations as same would tell us the answer.

a) x=12, x1=8
y=40 and y=28

b) x=10 and x1=7
y=48 and y=some fractional value

c) x=10, x=6
y=48 and y=(80-32)=48

Hence, I arrived at c.

Please let me know whether my approach is correct.

Thanks

[tim]

we need a source on these before we can discuss them further..