A furniture retailer bought 100 identical chairs at a cost of $20 each

Question

A furniture retailer bought 100 identical chairs at a cost of $20 each. He sold some of the chairs within the first week for $35 each. In the second week, he sold all the remaining chairs for $30 each. If his total profit on the sale of the chairs was greater in the second week than in the first week, which of the following can be the number of chairs sold by the retailer in the second week?

I. 30
II. 60
III. 80

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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Show Answer

Official Answer

C

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rohit8865 · Answer

1) second week 30 
means first week=70 chairs sol
CP of 70 chairs = 20*70= 1400
SP of 70 chairs = 35*70=2450
Profit of 1st week = 2450-1400=$1050
2nd week CP of chairs =20*30 =600
SP of 30 chairs = 30*30 =900
Profit of 2nd week = 900-600=$300

clearly profit 1> profit 2
thus this option is incorrect.

2) for 1st week 40 chairs profit = $600
 for 2nd week 60 chair profit=$600
Profit1 = Profit2...incorrect

3) for 1st week 20 chairs profit = $300
 for 2nd week 60 chair profit=$800
Profit1 < Profit2...correct

only option 3 is correct

Ans C

EgmatQuantExpert · Answer

Hey, PFB the official solution - Given : • Number of chairs bought by retailer = 100 • Cost price of 1 chair = $ 20 • First Week o Let the number of chairs sold in the first week be z . o Selling price per chair = $ 35 o So, profit per chair = $ 35 - $ 20 = $ 15 o So, total profit on selling z chairs = $ 15z • Second Week o The number of chairs sold =100 – z o Selling price per chair = $ 30 o So, profit per chair = $ 30 - $ 20 = $ 10 o So, total profit on selling 100 - z chairs = $ 10(100 – z) = $ (1000-10z) • Profit in 2nd week > Profit in 1st week To find : • Which of the 3 values is/are possible for (100 – z) ? Approach : 1. To answer the question, we need to find the possible values of z 2. We’ve already inferred above the profit of 1st and 2nd weeks in terms of z. We’re also given a relation between the profit of 1st and 2nd weeks. We’ll use this relation to find the possible values of z. Working Out : • Finding possible values of z o Profit in 2nd week > Profit in 1st week o 1000 - 10z > 15z o So, 1000 > 25z o Or z < \frac{1000}{25} o That is, z < 40 • Finding possible values of 100 – z o Since z < 40 , -z > -40 o So, 100-z > 100 – 40 o That is, number of chairs sold in the 2nd week > 60 • Evaluating the 3 given values o Of the 3 given values, the only value that is greater than 60 is the 3rd value (80) Looking at the answer choices, we see that the correct answer is Option C Thanks, Saquib Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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RastogiSarthak99 · Answer

I did it slightly different from others and so i'm sharing my method

total chairs = 100; cost 20 per chair; total cost = 2000

1st week. Assume x chairs were sold. 
Cost = 20x
Revenue = 35x
Profit = 15x

2nd week. 100-x chairs sold
Cost = 20(100-x)
Revenue = 30(100-x)
Profit = 1000-10x

We are told 2nd week profit is greater than 1st week

1000-10x>15x

40>x

Question asks which of the following can be the number of chairs sold in second week.

chairs sold 2nd week is 100-x

since 40>x only value from the options applicable is 80. (not 60 because 40 is not equal to x, 40 is GREATER than x, x will always be below 40 eliminating 30 and 60.

C!

G-Man · Answer

Let x be the number of chairs sold in the second week

Profit in the first week = 35-20 = 15
Profit in the second week = 30-20 = 10
Total profit = number of chair sold * profit per chair
Then the question states that 15*(100-x) < 10*x
1500 < 25x
x> 60
Only option III is greater than 60

Option C

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A furniture retailer bought 100 identical chairs at a cost of $20 each

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