A toy manufacturer runs his plant for 30 days in a month such that his

Question

A toy manufacturer runs his plant for 30 days in a month such that his daily output is constant during the month. He undertakes the maintenance of his plant every 6 months for 5 days in a month, during which period no toys are produced. As he has to undertake the maintenance of the plant this month, he increases his constant daily output of toys by 10 percent compared to the last month’s constant daily output. If the manufacturer wants to keep his monthly revenue unchanged as compared to last month, by approximately what percentage should be increase the price per toy?

A. 7%
B. 8%
C. 9%
D. 10%
E. 11%

A. 7%
B. 8%
C. 9%
D. 10%
E. 11%

yashikaaggarwal · Answer

Let the manufacturer monthly output be 30
The month in which maintenance has to be took place. The daily output increased by 10%
Daily output is 1
Manufacturer take 5 days for maintainance that means for the certain month, the manufacturer would only be able to produce 25 output if he hadn't increased output by 10%
Due to 10% increase in daily output. Output is equal to 
25*110/100 = 27.5
Therefore the manufacturer is able to produce 27.5 output.
Let the cost per output be 1.
For previous month manufacturer was been able to get 30$ 
But for this month he will only be able to get 27.5
For same level of revenue.

The cost at which one output is sold in that particular month is 30/27.5 = 1.09$
The increase in price 
= (new price-old price/old price)*100
=> (1.09-1/1)*100
=> 0.09*100
=> 9%

Answer should be C

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NitishJain · Answer

IMO  C

Let the output for 1 day = x
price of 1 item = y
Total price for the products produced in 1 day = xy
Total price for the products produced in 30 days = 30xy

Fro 25 days of production:
output for 1 day = 1.1x
Let the new price be p
Total price for the products produced in 25 days = 25*(1.1)xp
Since monthly revenue has to be same::
30xy = 25*(1.1)xp
p =  \frac{12}{11}  *y ~ 1.09y
 ==> 9% increase in price is needed.

hacorus · Answer

We have a factory working 30days-5Days this month or 25days.
Production is uped by 10% so equivalent to 27.5days of normal output
To generate the same revenue, we would need an equivalence of 30 days output.
So we are missing 2.5 days.

2.5/27.5 = 25/275 = 0.0909... so 9%
If you had noticed that 27.5=11*2.5 then 1/11 may have been more familiar to you for calculation.

Answer C 9%

ScottTargetTestPrep · Answer

Solution:

Let’s assume that the daily production is 10 toys last month and each toy is $1. Thus, the production last month is 10 x 30 = 300 and the total revenue last month is $300. For this month, the daily production would be 10 x 1.1 = 11 toys, and since there are only 30 - 5 = 25 production days, the production this month is 11 x 25 = 275. To have the same revenue as last month, each toy must be sold for 300/275 ≈ $1.09, which is a 9% increase from last month’s price of $1.

Answer: C

ODST117 · Answer

Let  x  be the number of toys manufactured in a normal month.
Daily output of toys in a normal month  = x/30 
If their selling price per toy is  S , then revenue in a normal month is  Sx

During the maintenance month, they increase the daily output by  10 % and work only for 25 days -> 
Daily output in maintenance month  = 1.1(\frac{x}{30}) 
Total toys manufactured in maintenance month  = 25(1.1(\frac{x}{30})) = \frac{55x}{60} 
Let  S_n  be the new selling price, then revenue in maintenance month  = S_n(\frac{55x}{60})

They want the revenue to be the same for both months -> 
 S_n(\frac{55x}{60}) = Sx  
 S_n = 1.09S

Even though the above equation tells us the % increase we need, optional step ->  \frac{S_n-S}{S}*100 = \frac{1.09S-S}{S}*100 = 0.09*100 = 9 %

This means they need to increase the selling price by  9 %.

A toy manufacturer runs his plant for 30 days in a month such that his

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