I am taking the Praxis II test and I can't seem to answer certain styles of questions. Please help!!!!!!

  1. In a class of 29 children, each of 20 children has a dog and each of 15 has a cat. How many of the children have both a dog and a cat?

  2. If there are exactly 5 times as many children as adults at a show, which of the following CANNOT be the number of people at the show? a. 102 b. 80 c. 36 d. 30

  3. In order to estimate the population of snails in a certain woodland, a biologist captured and marked 84 snails there were then released back into the woodland. Fifteen days later the biologist captured 90 snails from the woodland, 12 of which bore the markings of the previously captured snails. If all of the marked snails were still active in the woodland when the second group of snails were captured, what should the biologist estimate the snail population to be based on the probabilities suggested by this experiment? a. 630 b. 1,010 c. 1040 d. 1080.

Hi,

In a class of 29 children, each of 20 children has a dog and each of 15 has a cat. How many of the children have both a dog and a cat? I think there is not enough information to narrow this question done to a single answer. Here's why: 20+15=35 which is greater than 29, so there must be overlap in children who have dogs and children who have cats. But, we need more information to determine how much overlap.

For example, suppose we assume that 26 (an arbitrary choice between 20 and 29) children have at least a dog or a cat. Then the number of children with both a dog and a cat is (20+15)-26=9. To see that this is true, consider the following quantities:

  1. Number of children with a dog and a cat=9

  2. Number of children with a dog and not a cat=20-9=11

  3. Number of children with a cat and not a dog=15-9=6
So, 9+11+6=26 is the number of children with at least a dog or a cat.

Another possibility is to assume that 25 children have at least a dog or a cat. Then the number of children with both a dog and a cat is (20+15)-25=10.

We can compute several possibilities:

for 29 children with at least one, #with both=6
for 28 children with at least one, #with both=7
for 27 children with at least one, #with both=8
for 26 children with at least one, #with both=9
for 25 children with at least one, #with both=10
for 24 children with at least one, #with both=11
for 23 children with at least one, #with both=12
for 22 children with at least one, #with both=13
for 21 children with at least one, #with both=14
for 20 children with at least one, #with both=15
 (can't go lower than 20 since 20 have a dog, and can't go higher than 29 since that is the class size)

I SPECULATE that the question is implicitly assuming that every child in the class has either a dog of a cat, so that there are 6 chikdren with both a dog and a cat. OR, it could be that the question should specify a range of possibilities OR, this is a faulty question.

If there are exactly 5 times as many children as adults at a show, which of the following CANNOT be the number of people at the show?

a. 102 b. 80 c. 36 d. 30

This is a ratio question:

Before giving an answer, let's try an example: Suppose there are 4 adults Then there are 5x4 = 20 children So, the number of people is 4+20=24

The ratio adults:children:total = 1:5:6 Equivalent ratios are: 2:10:12, 3:15:18, 4:20:24 (see above), 5:25:30, 6:30:36, etc. The last number in the ratio is the answer we are looking for. Notice that the last 2 ratios take care of d. and c. respectively. We can continue the list of ratios or we can notice that the last number in every ratio is divisible by 6 (adult+child=1+5=6. Now, 102/6 = 17 and 80/6=13.333 Therefore, the answer is b.

Another way:

Let n=number of adults. Then #children=5n So, #people=n+5n=6n (note that n must be a whole number) Try a. 6n=102 gives n=17 Try b. 6n=80 gives n=13.3333 (doesn't work) Note that c works: 6n=36 gives n=6 Note that d works: 6n=30 gives n=5 Therefore, the answer is c.

In order to estimate the population of snails in a certain woodland, a biologist captured and marked 84 snails there were then released back into the woodland. Fifteen days later the biologist captured 90 snails from the woodland, 12 of which bore the markings of the previously captured snails. If all of the marked snails were still active in the woodland when the second group of snails were captured, what should the biologist estimate the snail population to be based on the probabilities suggested by this experiment?

a. 630 b. 1,010 c. 1040 d. 1080.

This can be done as a ratio question: Let n be the snail population in the woodland. For the second sample, marked:total = 12:90 For the whole woodland, marked:total = 84:n We can estimate n by equating the two ratios above: 12:90 = 84:n

Rewriting as fractions:
12/90 = 84/n

Cross multiplying:

12n = 90x84
12n = 7560
n=7560/12
n=630

Therefore, the woodland population can be estimated as 630 snails.

This question can also be done using probabilities: Let n be the snail population in the woodland. The probability of a snail in the woodland being marked is 84/n. Using the second sample, we can estimate the probability of a snail in the woodland being marked as 12/90. The two probabilities can be equated: 12/90=84/n The same as before: n=630.

Paul

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