Over a seven-year period, from 2002 to 2009, the number of babies born

Question

Over a seven-year period, from 2002 to 2009, the number of babies born to married couples increased despite a decrease in marriages from 17,000 marriages in 2002.

In the given expression, B and M represent the percent change in the babies and marriages, respectively. I represents the number of babies per married couple in 2002. The percent change in a quantity is calculated by the formula:

(\frac{X_{new}-X_{old}}{X_{old}})*100

In the table below select for (I) The expression that represents the number of babies born in 2002 and (II) The expression of the number of babies born per family in 2009. Select one value in each column.

pintukr · Answer

In the table below select for :-

(I) The expression that represents the number of babies born in 2002

Marriages in 2002 = 17,000

'I' represents babies born per married couple

So,    the number of babies born in 2002 = 17000 * I

(II) The expression of the number of babies born per family in 2009. Select one value in each column..

Change in babies born (%) = B (from 2002 to 2009)
Change in familes, or marriages (%) = m (from 2002 to 2009)

Since, 'I' represents babies born per married couple

I' (ie, number of babies born per family in 2009) =   * I 
 =   * I

Hereil · Answer

Hello, can someone help me understand how answer to q1 is 17000/I ? Thank you

Sajjad1994 · Answer

Recall the information given in the question.

We know that in 2002 there were 17000 marriages, and I represent  \frac{Babies}{Marriage.}

Using the information, make a mathematical expression:

17,000 Marriages x babies/marriage = 17000 marriages x I = 17000I

Deepa444 · Answer

Sajjad1994  can u plz explain the 2nd solution?

Sajjad1994 · Answer

Use the formula given

(\frac{X_{new}-X_{old}}{X_{old}})*100

This part of the problem is asking for a ratio, but the variables that are given are percents.

Therefore, the 100’s are there because M and B are percent quantitie s.  They are both added because M is a negative number.

So the number in the numerator is the percentage of babies that exist in 2009 compared to 2002 (a number that will be less than 100). The denominator is the percentage of the 2002 marriages that exists in 2009 (a number greater than 100).

Thus the solution is,

×I

Gargi7777 · Answer

Hello, Can you please help me understand this. "  They are both added because M is a negative number."- Where have they mentioned M is negative?
 If M is not negative was the answer supposed to be =>  I ?

Bunuel · Answer

(I) The number of babies born in 2002 equals the number of marriages that year multiplied by the number of babies per marriage. Since there were 17,000 marriages and the average was I babies per marriage, the total number of babies born in 2002 is  17000 * I .

(II) To find the number of babies born per family in 2009, we need to consider how both the total number of babies and the total number of marriages changed over time.

The number of babies increased by B%, so the total number of babies in 2009 is  17000 * I * \frac{100 + B}{100} . 
 The number of marriages changed by M%, so the number of marriages in 2009 is  17000 * \frac{100 + M}{100} .

The number of babies per family in 2009 is therefore:

\frac{17000 * I * \frac{100 + B}{100}}{17000 * \frac{100 + M}{100}}

The 17000 and 100 terms cancel out, leaving:

\frac{100 + B}{100 + M} * I

Bunuel · Answer

The stem says:  “the number of babies born to married couples increased  despite a decrease in marriages from 17,000 marriages in 2002. ”
 
This means marriages  decreased , so M is a  negative  percent change.

(I)	(II)
		\(\frac{17000}{I}\)
		\(17000*I\)
		\(\frac{100+B}{100+m}*I\)
		\(\frac{100-B}{100+m}*I\)
		\(\frac{100+m}{100+B}*I\)
		\(\frac{100+m}{100-B}*I\)

Over a seven-year period, from 2002 to 2009, the number of babies born

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Over a seven-year period, from 2002 to 2009, the number of babies born

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

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Perfect 805 Score Debrief by Julia

My Rewards