A carpenter has to build 71 wooden boxes in one week. He can build as

Question

A carpenter has to build 71 wooden boxes in one week (7 days). He can build as many per day as he wants but he has decided that the number of boxes he builds on any one day should be within 4 off the number he builds on any other day. What is the least number of boxes that he could have build on Saturday?

A. 4
B. 5
C. 6
D. 7
E. 8

A. 4
B. 5
C. 6
D. 7
E. 8

yashikaaggarwal · Answer

Let The least no. On Saturday be X
Then the wooden box remaining to build in 6 days = 71-x

Now, we have to find the least possible no. So that even if the Carpenter made equal no. Of boxes on rest 6 days.
The difference between the no. Of boxes On 6 days & on Saturday = 4 at max.

Let's put value of X as options one by one.
1) Let the Carpenter made 4 box on Saturday
Rest 71-4 = 67 boxes are to be made in 6 days with an average of 11 per day.
So 11-4 > 4 {Incorrect}

2) Let the Carpenter made 5 box on Saturday
Rest 71-5 = 66 boxes are to be made in 6 days with an average of 11 per day.
So 11-5 > 4 {Incorrect}

3) Let the Carpenter made 6 box on Saturday
Rest 71-6 = 65 boxes are to be made in 6 days with an average of 11 per day.
So 11-6 > 4 {Incorrect}

4) Let the Carpenter made 7 box on Saturday
Rest 71-7 = 63 boxes are to be made in 6 days with an average of 10.5 per day.
So 10.5-7 < 4 {Correct}

So the least no. Of boxes can be build on Saturday = 7

Answer is D

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Akp880 · Answer

A carpenter has to build 71 wooden boxes in one week (7 days). He can build as many per day as he wants but he has decided that the number of boxes he builds on any one day should be within 4 off the number he builds on any other day. What is the least number of boxes that he could have build on Saturday?

A. 4
B. 5
C. 6
D. 7
E. 8

A. 4
B. 5
C. 6
D. 7
E. 8

Explanation:

Let us do the backsolving in this question. ..assume that he makes 6 boxes on last day of week

then on remaining 6 days he has to make 71-6=65 boxes
but he can make max 10 boxes only per day now( read the ques max-min=4) 
so then total count 10*6+6=66<71

thus we should move towards higher number

assume that he makes 7 boxes on last day of week 
then on remaining 6 days he has to make 71-7=64 boxes
but he can make max 11 boxes olt per day now( read the ques max-min=4) 
so then total count 11*6+67= 73>71

Thus he makes at least 7 boxes on last day of week

Hence D is the correct answer.

IanStewart · Answer

The carpenter needs to build, on average, more than 10 boxes per day. So on at least one day, he must build 11 or more boxes, because you can't have an average greater than 10 unless your set contains values greater than 10. So the carpenter needs to build at least 7 boxes on his least productive day. We can then confirm 7 is possible -- the remaining six days need to sum to 64, which is possible within the constraints if he builds 11 boxes on four days, and 10 boxes on two.

sambitspm · Answer

Let's assume that on Saturday the carpenter builds x boxes. 
On any day the maximum boxes he can build is x+4

So Let's say on every other day the number of boxes is x+4 so as to minimize x

Hence 6*(x+4) + x = 71
7x = 47
Now definitely x has to be numeric, which means that if x = 6 then the equation will not hold.

He needs to make 7 boxes on Saturday, so that equation holds.

D

adstudy · Answer

As we have 7 days, then if we try to divide 71 by 7 we get remainder as 1 and quotient as 10. This implies any one day, except Saturday, to have count as 11 and other days as 10. So

Mon = 10 boxes
Tue = 10 boxes
Wed = 10 boxes
Thu = 10 boxes
Fri = 10 boxes
Sat = 10 boxes
Sun = 11 boxes

Now as it is given we can have maximum difference between any two days as less than or equal to 4, it would be best to reduce boxes from Saturday one-by-one and then distribute among maximum possible days to maximize our case.

So we notice, at max, we can reduce the count of Saturday to 7 and increasing three boxes by 1 each. (As below)

Mon = 10 boxes
Tue = 10 boxes
Wed = 11 boxes
Thu = 11 boxes
Fri = 11 boxes
Sat = 7 boxes
Sun = 11 boxes

Note that we cannot make it 6 or below, as then difference between 6 and 11 would be 5 and would break the condition of difference being less than or equal to 4

Hence Answer is    D

ScottTargetTestPrep · Answer

Solution:

We can check the answer choices starting from choice A since we want the least number of boxes for Saturday.

A. If the number of boxes built on Saturday is 4, then he will have 71 - 4 = 67 boxes to be built for the other 6 days. The average for those 6 days would be 67/6 = 11 ⅙. We see that 11 (or 12) boxes exceeds the daily limit of 4 + 4 = 8 boxes. Eliminate Choice A.

B. If the number of boxes built on Saturday is 5, then he will have 71 - 5 = 66 boxes to be built for the other 6 days. The average for those 6 days would be 66/6 = 11. We see that 11 boxes exceeds the daily limit of 5 + 4 = 9 boxes. Eliminate Choice B.

C. If the number of boxes built on Saturday is 6, then he will have 71 - 6 = 65 boxes to be built for the other 6 days. The average for those 6 days would be 65/6 = 10 5/6. We see that on more than one of the days he would have to build 11 boxes, and this would exceed the daily limit of 4 + 6 = 10 boxes. Eliminate Choice C.

D. If the number of boxes built on Saturday is 7, then he will have 71 - 7 = 64 boxes to be built for the other 6 days. The average for those 6 days would be 64/6 = 10 2/3. We see that on more than one of the days he would have to build 11 boxes, but this would be acceptable because 11 boxes does not exceed the daily limit of 4 + 7 = 11 boxes. Choice D is correct.

Alternate Solution:

Let x be the number of boxes built on Saturday. To minimize x, we should maximize the number of boxes built on every other day. According to the question, the number of boxes built in one day cannot exceed the number of boxes built in any other day by more than 4; therefore, let’s assume that x + 4 boxes were built on every day besides Saturday. Then, the number of boxes built in one week is:

(x + 4) + (x + 4) + (x + 4) + (x + 4) + (x + 4) + x + (x + 4)

7x + 24

We want this number to equal 71, so we get the following equation:

7x + 24 = 71

7x = 47

x ≈ 6.7

We see that if x is 6 or less, then the carpenter won’t be able to build 71 boxes in one week. Therefore, the minimum value of x is 7.

Answer: D

Regor60 · Answer

Call the least value X and the sum of the differences from X across the other 6 days Y.

So then:

7X+Y = 71

We're told that the maximum difference in any one day over the minimum is 4 and since there are 6 days at or above the minimum:

Y<= 24

So that means:

7X = 71-Y and

7X >= 47 or

X>=7

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JuvenalUrbino · Answer

I was confused by the phrasing of this question. It says "the number of boxes he builds on any one day should be  within 4  off the number he builds on any other day". Shouldn't we interpret this as not including 4? That is, if he made 6 boxes on one day, he can only make max. 9 boxes on any other day? If the question makers had intended that in this example he could have made 10 boxes on any other day, surely they would have written "within and including 4", or am I missing something here?

Bunuel · Answer

"Within 4 off" means the difference can be at most 4, including 4 itself. The phrase "within X" generally includes the boundary. For example, "within 10 miles" means up to and including 10 miles. Similarly, here, "within 4 off" means the difference can be up to and including 4. So if one day has 6 boxes, another day can have up to 10 boxes.

Dereno · Answer

Given that we need to build 71 wooden 📦 boxes within 7 days, which on average is 10.xx per day. Sufficiently greater than 10 boxes =  approx 11 boxes per day .

Constraint : the difference between the number of boxes made between any two days should be within 4. So maximum difference can be 4.

Asked is LEAST NUMBER OF BOXES THAT CAN BE MADE WITHIN 7 days.

Suppose all days make 11 boxes = 77 > 71. Not possible 
6 days make 11 boxes and the rest in the last day = 6*11+5=71 not possible. As the difference between 
11 and 6 exceeds 4.

Case 2: all 6 days make 10 boxes = 60 boxes. On the 7th day, make 11 boxes to get 71. But, 11 is not the least number, even though the constraint is fulfilled.

Case 3: 11 boxes 📦 for 5 days , 10 📦 boxes on 1 day and on the last day we make 6 boxes. But, the constraint is not fulfilled.

Case 4: 11 boxes on 4 days, and 10 boxes on 2 days = 64 boxes on 6 days. The last day we make 7 boxes 📦. And the constraint of 4 📦 boxes is also fulfilled.  Correct answer

Case 5: 11 boxes for 3 days, and 10 boxes for 3 days, will result in 63 boxes and we need 8 boxes on the last day to fulfill the contract of 71 boxes. But, 8 is not the LEAST number. So not correct .

A carpenter has to build 71 wooden boxes in one week. He can build as

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