Junior Cycle Maths Higher free practice questions with answers

Question 1 (10 marks) [Probability and outcomes]

Suggested maximum time: 5 minutes. A game consists of spinning a fair spinner and rolling a fair 4-sided

die. The spinner has 4 equal coloured sectors: black (K), white (W), grey (Y) and pink (P).

(a) Complete the table below to show all 16 possible outcomes of the game. Two have already been done for you. For example, K1 means black on the spinner and 1 on the die.

(a) Answer: All 16 outcomes: K1 K2 K3 K4; W1 W2 W3 W4; Y1 Y2 Y3 Y4; P1 P2 P3 P4

• Every colour on the spinner can be paired with every number on the die, so there are 4 × 4 = 16 outcomes.

• Write one row for each colour: the colour letter followed by each die number from 1 to 4.

• Rows: K: K1 K2 K3 K4; W: W1 W2 W3 W4; Y: Y1 Y2 Y3 Y4; P: P1 P2 P3 P4.

• Check the count: 4 rows of 4 outcomes = 16.

(b) Each outcome in the table is equally likely. The game is played once. Find the probability that the outcome is grey on the spinner.

(b) Answer:

• There are 16 outcomes in total, and each is equally likely.

• The outcomes with grey on the spinner are the whole Y row: 4 outcomes.

• Probability = favourable outcomes ÷ total outcomes =

(c) Find the probability that the outcome is grey on the spinner and an even number on the die.

(c) Answer:

• Look at the Y row and pick out only the outcomes where the die shows an even number.

• The numbers on the die that fit are 2, 4, so there are 2 favourable outcomes.

• Probability = 1 4 4 16 = 1 4 . 1 8 2 16 = 1 8 . Question 2 (25 marks) [Statistics: tables, histograms, means and pie charts]

Suggested maximum time: 10 minutes. 45 students were asked how much time they spent on their phone on a particular day. The results are shown in the frequency table below. [Note: 2 − 4 means 2 hours or more but less than 4 hours hours, and so on.]

(a) Draw a histogram to show the information in the table. Use the axes below. 0 2 4 6 8 10 12 14 16 18 0 2 4 6 8 10 Time (hours) Number of students

(a) Answer: A histogram with bars of equal width and heights (0–2: 14, 2–4: 14, 4–6: 8, 6–8: 5, 8–10: 4).

• The class intervals all have the same width (2 hours), so the height of each bar is simply the frequency.

• Draw one bar for each class, with no gaps, from the lower to the upper class boundary on the horizontal axis.

• Bar heights: 0–2: 14, 2–4: 14, 4–6: 8, 6–8: 5, 8–10: 4.

• Label both axes and check that the total of the heights is 45.

(b) Find the minimum number of students who could have given an answer of less than 7 hours.

(b) Answer: 36

• The value 7 hours lies inside the class 6 − 8, so we do not know exactly how many in that class are below it.

• Everyone in the classes below that one is definitely under 7 hours: 14 + 14 + 8 = 36.

• Minimum: none of the 5 in the class 6 − 8 need be below it, so the answer is 36.

(c) Use mid-interval values to estimate the mean for the 45 students. Give your answer in hours, correct to 1 decimal place.

(c) Answer: 3.7 hours

• Write the mid-interval value of each class: 1, 3, 5, 7, 9.

• Multiply each mid-interval value by its frequency and add: (1 × 14) + (3 × 14) + (5 × 8) + (7 × 5) + (9 × 4).

• Sum = 167.

• Mean = 167 ÷ 45 = 3.7 hours (1 d.p.).