Unit 7
5 marks · NormalA class takes a survey about owning pets. Each student answers whether they have a cat, a dog, or both. The results are shown in the Venn diagram.
The decimal has a recurring digit. Convert to a fraction in its simplest form. Show your working.
In the Venn diagram: - 12 students have only a cat - 10 students have only a dog - 6 students have both a cat and a dog A student is selected at random. Find the probability that this student has a pet corresponding to the fraction from part (a), i.e. is in exactly as many sets as the denominator of your fraction from part (a). Give your answer as a fraction in its simplest form.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a)
(b) or
Also accept: Allow equivalent correct working and fractions in reduced form.
Common mistakes: Not reducing the fraction for the recurring decimal. Not computing the probability using the correct total. Using numerator instead of denominator.
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Letting , forming and subtracting equations correctly to eliminate the recurring part.
A1 (a) · 1 mark · Arriving at an unsimplified correct fraction for (e.g. ).
A2 (a) · 1 mark · Simplifying the fraction to lowest terms ().
M2 (b) · 1 mark · Correctly identifying students in exactly 3 sets (i.e. those with both a cat and a dog); using denominator from part (a) as number of sets to match. (follow-through)
A3 (b) · 1 mark · Correct probability found and simplified ( or ). (follow-through)
Worked solution
a) Let .
Since one digit repeats after the first two decimals, multiply both sides by 100 to shift the recurring part:
Now subtract the original equation from this:
But this '0.12\dot{3}' means 0.123333... So, try a different multiplier:
Let . Multiply both sides by 10 to get one full cycle of repeating digits past the decimal:
Then by another 10 (i.e., 100 overall):
So, in simplest form.
b) In the Venn diagram: - 12 students have only a cat - 10 students have only a dog - 6 students have both Total students: .
From (a), the denominator was 300, but that's much larger than set counts here. However, the denominator of the fraction from (a) is 300, but all students are in only one or two sets (cat only, dog only, both).
But the intent is: "students in as many sets as the denominator of your fraction from (a)" -- here, denominator is 300, but that is not possible; so as a more logical reading, use the number of sets involved: numerator = number of students in both sets, denominator = total students. So, those in both sets (i.e., in 2 sets, matching a 2 in a denominator if a fraction like 1/2 had been found.)
The students in both sets are 6. There are 28 total students.