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A certain manufacturer sells its products to stores in 113 different r
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Bunuel
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A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  03 Jul 2016, 10:05
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A certain manufacturer sells its products to stores in 113 different regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of manufacturer’s product sold worldwide last year?

A) 10^6
B) 10^7
C) 10^8
D) 10^9
E) 10^10
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warriorguy
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  17 Oct 2016, 03:35
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Total Number of Stores = 181 * 113

Total Number of Products sold = 181 * 113 * 51752

= 1.81 * 10^2 * 1.13 * 10^2 * 5.1752 * 10^4
=1.8*1.1*5.17 * 10^8

Approx = 2*1*5 * 10^8 = 10*10^8= 10^9
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  03 Jul 2016, 10:22
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Number of different regions = 113
Average of stores per region = 181
Average units sold per store = 51752
Total number of units of manufacturer’s product sold worldwide last year = 113 * 181 * 51752
We should use estimation here
110 * 180 * 50,000
= 198 * 10^2 * 50,000
= 2 * 10^4 * 5* 10^4
= 10 * 10^8
=10^9
Answer D
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  03 Jul 2016, 23:28
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D

51752*181*113 or
51000*180*113 or
50000*200*100
=10^9

hope it helps
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Abhishek009
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  17 Oct 2016, 07:02
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Bunuel wrote:
A certain manufacturer sells its products to stores in 113 different regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of manufacturer’s product sold worldwide last year?

A) 10^6
B) 10^7
C) 10^8
D) 10^9
E) 10^10


Use approximation ab initio

113 ~ 100
181 ~ 200
51,752 ~ 50000

So, 113 *181 *51,752 ~ 100*200*50000 =>20000*50000 = 1000000000 =10^9

Hence answer will be (D) 10^9
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  18 Oct 2016, 08:18
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Bunuel wrote:
A certain manufacturer sells its products to stores in 113 different regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of manufacturer’s product sold worldwide last year?

A) 10^6
B) 10^7
C) 10^8
D) 10^9
E) 10^10


We are given that a certain manufacturer sells its products to stores in 113 different regions, with an average (arithmetic mean) of 181 stores per region and 51,752 units per store.

Since we are asked, which of the following is closest to the total number of units of manufacturer’s product sold worldwide last year, we can use estimation.

113 ≈ 100

181 ≈ 200

51,752 ≈ 50,000

Thus, the approximate number of units sold worldwide last year was:
100 x 200 x 50,000 = 1,000,000,000 = 10^9

Answer: D
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  28 Dec 2016, 06:55
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Excellent Question.

Testing out Knowledge on approximation.

Total Sold goods => 181*113*51752 => 200*100*52000 (Approx.) -> 1040000000 => 104 Crores => 100 Crores Approx => 10^9

Hence D
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  25 Aug 2017, 21:44
1. 181 Stores x 113 regions = 20,453
2. 20,453 x 51,752 (I worked this out the long way but scientific notation is faster)
3. 2 x 10^4 x 5 x 10^4 = 10 x 10^8 (just add exponents since base and exponent is the same)
4. 10^9

Answer (D)
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  02 Sep 2018, 18:02
As from question statement:
181*131*51,752
For approximation
180*(100+13)*50,000
180*(100+10)*50,000
(18000+1800)*50,000
~20,000*50,000
10^9
Key is to approximate at each step to make calculations more easier.

Answer D

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Amlu95
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  09 Dec 2019, 16:13
how do we know what is the correct level of approximation ?

I tried

180 * 110 * 52000 = which gives a 10^8
why cant this be the answer ?
Bunuel can you please tell me how much of an approximation is correct on the gmat ?
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Harry8014
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Re: A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  16 Oct 2020, 03:37
113 ~ 100
181 ~ 200
51,752 ~ 50000

so 10^2 X (2*10^2) X (5*10^4)

Here 2 and 5 adds 1 to 10 power

add 1 + 2 + 2 + 4 = 10^9
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A certain manufacturer sells its products to stores in 113 different r [#permalink] New post  25 Nov 2020, 10:34
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Bunuel wrote:
A certain manufacturer sells its products to stores in 113 different regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of manufacturer’s product sold worldwide last year?

A) 10^6
B) 10^7
C) 10^8
D) 10^9
E) 10^10



\(113*181*51,752\)

\(Approximating:\)

\(=100*200*50,000\)

\(=10^2*2*100*5*10,000\)

\(=10^2*10^2*10^4*5*2\)

\(=10^2*10^2*10^4*10^1\)

\(=10^9\)

The answer is D
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