Two-Part Analysis Data Insights Practice Questions

To solve this logic problem, first list the constraints:

- 12 hours = total hours available per day
- 4 hours = maximum hours for walking per day
- Minimum 4 art or architecture activities during the 2 days
- Maximum 1 art museum per day
- Minimum of 1 beach activity during the trip
- Minimum of 1 shopping activity each day

Now evaluate the activities the family has already planned to see which constraints are satisfied and which are not.

**Day 1:** The family has planned 11 hours of activities with 1 hour of walking. Therefore, only 1 hour remains, and a walking activity could be chosen if desired.

**Day 2:** The family has planned 10 hours of activities (3 + 2 + 4 + 1 = 10) with 4 hours of walking. This means that 2 hours remain for additional activities, but none can involve walking since the 4-hour walking limit has been reached.

Given these constraints, the only activities possible for each day are:

- **Day 1:** Mirador De Colon, Montserrat, or La Pedrera
- **Day 2:** Montserrat

Therefore, the family must choose the sightseeing trip to Montserrat on Day 2, leaving only Mirador De Colon and La Pedrera as options for Day 1.

Now consider the family preferences. Mom already has shopping on each day (Las Ramblas and Barri Gotico), and Little Brother has a beach activity on Day 2 (Nova Icària). However, Big Sister only has 3 of her required 4 art or architecture activities. Both Mirador De Colon and La Pedrera fit this category, but Dad will not go to more than 1 art exhibit on a single day, and the family will already visit Park Güell on Day 1. This means they cannot also visit La Pedrera. Therefore, the family will visit Mirador De Colon on Day 1.

To solve this fractions problem, we must find the ratio of the number of total boxes packed by the first-shift to the total number of boxes packed by both shifts together. From the information given, we know that there were 2/3 as many first-shift workers as second-shift workers and, inverting the second fraction, we know that each first-shift worker packed 3/4 as many boxes as each individual second-shift worker. From here, multiplying the ratio of workers by the ratio of work per individual gives the fraction of total first-shift boxes relative to the second shift. This is done as (2/3) × (3/4) = 6/12 = 1/2. Thus, the first shift packs half as many boxes as the second shift. We can compute the first-shift boxes relative to the total by:
(first-shift fraction) / (first-shift fraction + second-shift fraction) = 1 / (1+2) = 1/3.
Thus the first shift does 1/3 of the total work. We must look for two numbers in the table that are related by a factor of 3. The only two numbers are 12 and 36, meaning that the first shift packed 12 boxes, and the total number of boxes packed by both shifts was 3 × 12 = 36.

Alternatively, one could use numbers to establish the relationship between the number of total boxes packed by the first shift and the number of total boxes packed by the two shifts together. We use our fractional ratios to choose smart numbers and assign 2 workers to the first shift, 3 workers to the second shift, 3 boxes per individual on the first shift, and 4 boxes per individual on the second shift. This gives:
Total First-shift Boxes = (2 workers) × (3 boxes per worker) = 6 boxes
Total Second-shift Boxes = (3 workers) × (4 boxes per worker) = 12 boxes
Total Boxes Overall = 12 boxes + 6 boxes = 18 boxes.
Again, from this we can derive that the ratio of first-shift boxes to the total boxes packed is 6/18 = 1/3.
Column 1: The correct answer is C.
Column 2: The correct answer is E.

This short problem is quite difficult because of the subject matter. The first sentence describes a value that is increasing steadily "by an unknown multiplier" or growth factor. So there is some multiplier that could be used to calculate the size of the population at the end of every hour. For example, if you know that a population starts out with 2 bacteria and the multiplier is 5, then at the end of the first hour, there will be 2 * 5 = 10 bacteria; at the end of the second hour, there will be 10 * 5 = 50 bacteria, and so on.

The second sentence then provides information to calculate that unknown multiplier, or growth factor; let's call that multiplier r. At t = 1, the population is p. At t = 5, the population is p². You are told that the growth factor corresponds to hourly growth, so there are 4 one-hour intervals in the timeframe described. So the question becomes this: what value of r, multiplied by itself 4 times (for the 4 one-hour intervals), would take you from p to p²?

Multiplying r by itself 4 times is the same thing as raising r to the 4th power: (r)(r)(r)(r) = r⁴. In the four hours described, the bacteria grew from p to p² (or p times p) so the bacteria grew by a factor of p. Therefore, p = r⁴, or p^(1/4) = r. ("p to the one-fourth power" is the same thing as the fourth root of p.)

The hourly growth factor, r, is p^(1/4), or ⁴√p.

Here's a way to look at the growth, step by step:

t = | given | calculation
1 | p | (given)
2 | | p × p^(1/4) = p^(5/4)
3 | | p^(5/4) × p^(1/4) = p^(6/4) = p^(3/2)
4 | | p^(3/2) × p^(1/4) = p^(7/4)
5 | p² | p^(7/4) × p^(1/4) = p^(8/4) = p²

You can now use the growth factor, p^(1/4), to calculate the population size at t = 0. If you were going from hour 1 to hour 2, you would multiply the hour 1 population by the growth factor in order to get to the hour 2 population (as in the above table). Because you are working backwards, from hour 1 to hour 0, you instead divide hour 1's population by the growth factor: p / p^(1/4):

p / p^(1/4) = p^(1 - 1/4) = p^(3/4)

"p to the 3/4 power" is the fourth root of p³, or ⁴√(p³).

Column 1: The correct answer is A.
Column 2: The correct answer is D.