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

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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D

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

hope it helps

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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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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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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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)

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

Posted from my mobile device
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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 ?

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

\(113*181*51,752\)

\(Approximating:\)

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

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

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

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

\(=10^9\)

The answer is D
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