A factory is working on a production order using identical machines.

Question

A factory is working on a production order using identical machines. The production begins with a certain number of machines operating in the first hour. At the start of the second hour, and at the start of each hour thereafter, the same fixed number of additional machines is added, thus increasing the units produced in each subsequent hour. By the end of 8 hours, the machines have produced exactly 220 units.

Select for 1st hour the number of units produced in the first hour, and select for 8th hour the number of units produced in the eighth hour that would be jointly consistent with the given information. Make only two selections, one in each column.

Bunuel · Accepted Answer

In 8 hours, the machines produced a total of 220 units.

So the average production per hour is 220/8 = 27.5 units.

Because the output increases each hour,  the production in the first hour must be less than the average . Among the choices, the only value below 27.5 is  17 .

Thus, in the first hour, 17 units were produced.

Since the same number of additional machines is added each hour, the hourly outputs form an arithmetic sequence. For such a sequence, the average equals (first hour + eighth hour)/2.

So:

(17 + eighth-hour output)/2 = 27.5

17 + eighth-hour output = 55

Eighth-hour output =  38

I29-170

ODST117 · Answer

I tried to solve this using work and rate. That did not work. It was so simple once I understood we can assign variables to the number of units produced. 
Say there were  N  machines in the first hour. They produce  x  units in one hour.
Say  M  machines were added each hour and they produce  y  units on their own.
In the first hour,  N  machines produce  x  units.
In the second hour,  N+M  machines produce  x+y  units.
In the third hour,  N+2M  machines produce  x+2y  units and so on.
In the eighth hour,  N+7M  machines produce  x+7y  units.
Adding all the units ->  (x)+(x+y)+(x+2y)+........+(x+7y) = 220

8x+28y = 220

2x+7y=55

y=\frac{55-2x}{7}

Now plug in values for  x

1.   x=17

y = \frac{55-34}{7}

y=3

Units produced in the eighth hour  = x+7y = 17+7(3) = 17+21 = 38

First column ->  17 
Second column ->  38

Aryaa25 · Answer

Given Some set of machines will run during 1st hour , let it be a.

and for every Consequent hour same no. machines are added , let it be b .

So for example : for 2nd hour , total no. of machines operating will be "a+b"

for 3rd hour : a+2b, Similarly for nth hour it will be a+(n-1)b

as per the question , for 8th hour no. of machines operating will be a+7b.

it is given that machines have produced 220 units all together.

so here we can apply sum of the terms in AP formula, n/2 (First term + last term) =sum of terms.

8/2(a+a+7b)=220 , which when simplified gives us 2a+7b=55.

or a= (55-7b)/2

here we can plug in the values in table and check which values are satisfying:

for a= 17 , a+7b will be 38 ,Hence the combination is Satisfied

a=17 (1st hour) 
 a+7b=38 (8th hour)

Souradeep17 · Answer

if the machine produces N units in first hour then it will produce n+x , n+2x, n+3x .. and so on upto n+7x

The total of it is 8n+28x=220

x=(220-8n)/28
55-2x /7

now if n=17 
then x is 55-34/7=3

if first hour is 17 then 8th hour will be n+7x =17+7*3=38

So 17 and 38

Dereno · Answer

Let the number of machine working at the first hour be n.

With addition of fixed number of machine at each hour, let the fixed. Machine be “a”.

Let’s see the number of machines operating subsequently from the second hour, they are:

n+a, n+2a, n+3a, n+4a, n+5a, n+6a , n+7a.

If we add all the number of machines, we get (8n + 28a).

(8n + 28a) machines made 220 units.

(2n + 7a) = 55.

Putting a = 7, n = 3,

If a = 5, n = 10

If  a = 3, n = 17 . ( this value matches with the options given).

If a = 1, n = 24.

Therefore : (n+7a) = (17+ (7*3))= 17+21= 38 .

First hour : 17

Eighth hour : 38

AbhishekP220108 · Answer

Initially I thought it was work rate time type questions but after giving some time, i thought to change the approach and recognised that it is an arithematic progression type. here is how I approached later.

Lets say X machines started doing the work and they Produced X units
later on after 1st hour, at the start of 2nd hour A machines joined the X machines and both made X+A units at the end of 2nd hour, then at the start of every hour from 2nd it was given A machines will be added, I noticed the pattern, X, X+A, X+2A.......X+7A(at the 8th hour)

Sum of from 1st to 8th is given as 220 unit

X+X+A+X+2A+.......+X+7A=220
8X+28A=220
2X+7A=55

Start substituting the minimum value of X from table as  17 , it gives A as 3, hencce we got out answer for 1st hour
for 8th hour substitute the value of X and A in X+7A which will give 17+7*3= 38

Adit_ · Answer

Easiest approach:
In an AP n/2(first+last)=sum

8/2(first+last)=220
first+last=55
find a combo adding upto 55.

Answer 17,38.

egmat · Answer

This is an  arithmetic sequence  problem. Here's why:
- Identical machines produce identical units per hour
- A  fixed number  of machines is added each hour
- So production increases by a  constant amount  each hour

Setting Up: 
Let:
-  a  = units produced in the 1st hour
-  d  = additional units produced each subsequent hour (due to added machines)

The production pattern looks like:
- Hour 1:  a 
- Hour 2:  a + d 
- Hour 3:  a + 2d 
- ...
- Hour 8:  a + 7d

Key Formula: 
 Sum of an arithmetic sequence  = (number of terms / 2) × (first term + last term)

Total =  (8/2) × (a + a + 7d) = 220

Simplifying:
 4 × (2a + 7d) = 220 
 2a + 7d = 55

Testing Answer Choices:

If 1st hour = 17: 
2(17) + 7d = 55
34 + 7d = 55
7d = 21
d = 3

8th hour = 17 + 7(3) =  38  ✓ (This is in our choices!)

(As such, once you find an answer that satisfies both parts, you can stop, but showing other options for demonstration)

If 1st hour = 27.5: 
2(27.5) + 7d = 55
d = 0

This would mean NO machines are added each hour - contradicts the question!

If 1st hour = 32 or higher: 
This gives negative d (production decreasing) -  invalid

Verification: 
With a =  17  and d =  3 :
- Hours 1-8 produce: 17, 20, 23, 26, 29, 32, 35, 38
- Sum =  220  ✓

Answer: 1st hour = 17, 8th hour = 38

pappal · Answer

lets start with n machines and let a machines are added at the beginning of each hour and suppose each unit produces 1 unit per hour.
so the no. of units operating each hour=> n, n+a, n+2a, n+3a,........n+7a
given total of units produced in 8 hours =220
i.e. 8n+28a=220
a=(55-2n)/7
checking for the values of n, all the given values will either result in 0 or a negative value of a (not possible) except  n=17 
which results in a=3
so n+7a (in the 8th hour)= 38 
so 17,38

saurabhbajpai · Answer

1st hour units - 1 X A (initial machines)

Let B machines get added every hour.
After 8 hours, the number of units produced will be

8A + B (1+2+3+4+5+6+7) = 220
8A + 28B = 220
2A + 7B = 55

Let A be 17, we will get B as 3

Number of units produced in the 8th hour will be A + 7B = 17 + 21 = 38

We get our answers -
1st hour units = 17
8th hour units = 38

A factory is working on a production order using identical machines.

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A factory is working on a production order using identical machines.

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