For the years 1972-2007, Total World Credit Market Debt (TWCMD), as me

Question

https://gmatclub.com/forum/download/file.php?id=73142 For the years 1972-2007, Total World Credit Market Debt (TWCMD), as measured in trillions of US dollars, is accurately modeled by the equation y = N • 2^{k(t - 1972)} , whose graph is given. Here, N and k are positive constants and t denotes the year. From each drop-down menu, select the option that creates the most accurate statement based on the information provided. The constant N is approximately equal to GRAPH_1_PLACEHOLDER If the model continues to be accurate beyond 2007, the TWCMD will equal approximately double the 2007 value in the year GRAPH_2_PLACEHOLDER 111.jpg

GMATCoachBen · Accepted Answer

Short video solution here (2:00):

https://www.youtube.com/watch?v=qsfKCrBHoWQ

MartyMurray · Answer

For the years 1972-2007, Total World Credit Market Debt (TWCMD), as measured in trillions of US dollars, is accurately modeled by the equation  y = N • 2^{k(t - 1972)} , whose graph is given. Here, N and k are positive constants, and t denotes the year.

From each drop-down menu, select the option that creates the most accurate statement based on the information provided.

The constant N is approximately equal to

We can determine N by looking at the value of TWCMD in 1972 since, in 1972,  t - 1972 = 0 , so  2^{k(t - 1972)} = 1 . Thus, in 1972,  y = N .

In 1972, TWCMD appears to be around 3.

Select  3 .

If the model continues to be accurate beyond 2007, the TWCMD will equal approximately double the 2007 value in the year

Since TWCMD  = y = N • 2^{k(t - 1972)} , TWCMD will double each time  k(t - 1972)  increases by  1 . After all, each time  k(t - 1972)  increases by  1 ,  2^{k(t - 1972)}  increases by a power of 2, and thus  y  doubles.

Looking at the graph, we see that y doubles from 5 to 10 and from 10 to 20 in approximately 9 years in each case. So, it takes 9 years for  k(t - 1972)  to increase by  1 .

If the model continues to be accurate, we can expect this pattern to hold after 2007.

So, it will take another approximately 9 years for TWCMD to double from its level in 2007.

Select  2016 .

Correct Answer   3, 2016

Fish181 · Answer

Easier way to do it that involves approximation/number sense:

For question 1, when the year is 1972, t - 1972 = 0, this simplifies n * 2k^(1972−1972) to n * 2k^0 = n * 1 = n. This tells us that n = whatever the value on the graph in 1972 is, which is 3.

For question 2, take a step back and realize this graph is growing exponentially. The answer choices of years are really spread out. This means that the graph most likely doubles the earliest possible year of the answer choices which is 2016. If you have a hard time imagining that just think about how rapid exponential growth is.

xyz02611 · Answer

I didn ́t understand

"TWCMD will double each time k(t−1972)k(t−1972) increases by 1. After all, each time k(t−1972)k(t−1972) increases by 1, 2k(t−1972)2k(t−1972) increases by a power of 2, and thus y doubles."

Could you provide an example maybe?

KarishmaB · Answer

The constant N is approximately equal to _____ To get the value of N, plug in the value of y from 1972 so that k disappears. Put t = 1972 3 = N * 2^0 N = 3 If the model continues to be accurate beyond 2007, the TWCMD will equal approximately double the 2007 value in the year _______ When given a graph (line or curve) consider the pattern it shows. You may not need to do any calculations. From 1972 to 1982, the value of y goes from 3 to 6 - doubles From 1997 to 2007, the value of y more than doubles. We can expect the value of y to double again in less than 10 years. The given options are 2016, 2025, 2034. Answer must be 2016 Discussion on Graphs: https://youtu.be/ilMxPjHNeic

harshvardhanindia · Answer

i got the first part wrong and know how the answer is 3 now but i cant find the flaw in my earlier approach . I saw that the  1997  both y axis and x axis were definitely integers where yaxis = 20 and x axis = 1997 so, i applied the given formula y=N•2^k(t−1972) 
so 20= N.2^k(25)
 2^2 . 5^1= N. 2^25k 
so i arrived at the conclusion that N has to be a multiple of 5 and in the options only 10 was a multiple of 5 so it had to be 10. where am i going wrong ?

KarishmaB · Answer

k can take any positive value such as k = 0.1. In that case, N is about 3.5. or k could be 0.2 in which case N = 0.625 etc. Here the exponents are not required to be integers so N doesn't need to be a multiple of 5. A discussion on exponents: https://youtu.be/ibDqnatAMG8

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=84490 A slightly lengthy and math-y approach, in case it helps deepen understanding (1) Finding N is the easiest part. When t = 1972 (in the formula), y = 3. This gives N = 3 . (2) To find k, we can use 2 data points from the graph that show doubling , and use the formula to get k. I have used 1997 (y = 20) and 2006 (y = 40) for this. We can get k = 1/9. (3) For 2007, y = 43. This is 3 x 2^(35k) . To double this means multiplying by 2. So, y (t) = 3 x 2 ^ (k x (t - 1972)) = 3 x 2^(35k+1) This means that -> k (t - 1972) = 35k+1 => t - 1972 = 35 + 1/k = 35 + 9 = 44 => t = 1972 + 44 = 2016 --- Harsha GMAT-Club-Forum-8hbdd9j1.png

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For the years 1972-2007, Total World Credit Market Debt (TWCMD), as me

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