DAT Question of the Day

Correct Answer: A. 5:4

We need to figure out which numbers are primes and which ones are composite. The following shows primes in blue and composites in red: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Note that 1 is neither prime nor composite, it only has one factor. We have 5 reds and 4 blues. The correct answer is A.

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Correct Answer: E. (2√2)/π

There are several ways you can compute the area of the octagon. No matter which way you choose, all of them will be easier if you draw the picture.

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Correct Answer: B. √x.

It’s commonly known that logarithmic growth and s-curves can be used to represent populations, but these aren’t the only functions. In our question we have two key features, population growth and an upper limit. We can quickly rule out choice D and E. They will not cap out, in fact they will grow even faster never reaching an upper limit. Choice C is the negative absolute value. This is not growing but decreasing. The cosine is also bounded by a certain amplitude, in this case it has an amplitude of 1. It is possible to use this function to model a bounded population, but the cosine function is a periodic function. This means there are places where the function grows and there are places where the function decreases. The square root, on the other hand, is always growing even when it reaches its upper limit. This makes the square root function, choice B, the correct answer.

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Correct Answer: C. 128 cells

If the bacteria double every 15 minutes, then that means over the course of 2 hours, they will double 8 times. 500 x 2⁸ = 128,000 cells on the surface. Then we apply the anti bacterial wash, which kills 99.9% of cells, meaning 0.1%, or 0.001 of the cells remain. 128,000 x 0.001 = 128 cells.

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Correct Answer: C. 4

After the first ball is picked, the second may not be the same color. After two balls picked, the third may still be a different color than the first two (since there are a total of 3 colors). But when the fourth ball is picked, it must match one of the prior colors. Therefore, picking 4 balls will ensure that you have at least one set of 2 balls with the same color.

The numbers 5, 7, 8 do not matter here (they could matter if these numbers were lower, or the number of balls picked were higher).

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Correct Answer: D. 13

If we denote the hypotenuse of the triangle as c and the two other sides as a and b, then the values of the area and perimeter can be written as (1/2)ab = 30 and a + b + c = 30. We have three unknowns but two equations so far. However, a right triangle also follows the relation a² + b² = c². We can now solve these three equations and obtain c = 13.

However, the “spirit” of the problem is not checking whether you can solve this system of 3 equations (which is pretty complicated!) – instead, this problem is trying to see if you can come up with a “shortcut” to get to the right answer. Since both the perimeter and area are both nice, whole numbers, we can conclude that the lengths of the sides are also nice, whole numbers, most likely from one of the “special” triangles that we know about (eg. 3-4-5, 5-12-13) or a similar triangle. We can immediately see that the sides of the 5-12-13 triangle add up to 30. The area of the 5-12-13 triangle is 30 as well, and we can thus get to the correct answer without doing any solving at all. The work is shown below for your convenience, however.

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Correct Answer: B. 20√2

As can be seen from the figure below, the diagonal of the square is equal to the diameter of the circle, which is 5×2=10 cm. If the length of one side of the square is x, the diagonal is given by √2 x. Therefore, √2(x) = 10 -> x = 5√2 and the perimeter is 4 × 5√2 = 20√2.

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Correct Answer: C. 79

For a grade of B, either the condition 85 × 0.4 + x × 0.6 ≥ 82 needs to be met under the first policy, or the condition 85 × 0.5 + x × 0.5 ≥ 82 needs to be met under the second policy. These two conditions yield x ≥ 80 and x ≥ 79, respectively. Since the minimum value of x is to be computed, we pick the lower of these two bounds, which is 79.

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Correct Answer: D. {60, 61, 59}

This can be done by computing the variance of each of these expressions. Also, by inspection, it can be seen that the numbers in D vary by the least amount from each other.

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