A Level Mathematics 9709 Paper 51, May/June 2025

Question paper and mark scheme50 marks6 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Download PDF Mark scheme

100%

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9709/51 · May/June 2025

Download PDF

The masses of the bags of rice made by a company are normally distributed with mean $μ kg$ and standard deviation $0.14 kg$ . The probability that the mass of a randomly chosen bag of this rice is less than $1.48 kg$ is $0.22$ . Find the value of $μ$ .

[3]

Find the number of different arrangements of the 8 letters in the word KANGAROO in which the two As are together and the two Os are not together.

[3]

A fair 8-sided dice has faces labelled K, A, N, G, A, R, O, O. The dice is rolled repeatedly. Find the probability that fewer than 6 rolls of this dice are required to obtain an A.

[2]

Find the probability that the second A is obtained on the 6th roll of the dice.

[2]

Last Sunday, teams of runners took part in a charity event. The time taken, in seconds, to run $50 m$ was recorded, correct to 1 decimal place, for each runner. The times recorded for 11 runners from each of the Gulls and the Herons are shown in the table.

Gulls	$7.9$	$8.2$	$8.3$	$8.6$	$8.6$	$8.8$	$9.2$	$9.7$	$9.8$	$10.0$	$10.4$
Herons	$9.5$	$9.9$	$8.5$	$8.1$	$9.2$	$10.8$	$8.3$	$9.7$	$9.3$	$9.9$	$8.7$

Draw a back-to-back stem-and-leaf diagram to represent this information, with Gulls on the left-hand side.

[4]

Find the median and the interquartile range of the times of the runners from the Gulls.

[3]

Two other teams of runners, the Eagles and the Swifts, also took part in the event. The recorded times in seconds for 20 runners from the Eagles and 30 runners from the Swifts are denoted by $x$ and $y$ respectively. It is given that $Σ x = 175.0$ and that the mean of $y$ is $8.4$ . Find the mean of the times taken by all 50 runners.

[2]

It is given that $Σ x^{2} = 1823.0$ . It is also known that the standard deviation of the times taken by all 50 runners is $1.38$ seconds. Find the value of $Σ y^{2}$ , correct to 1 decimal place.

[3]

Every Saturday, a particular community holds a ‘Puzzle’ event to raise money for a new Leisure Centre. Competitors attempt to solve a puzzle as quickly as possible. Last Saturday, 600 competitors took part. The times taken to complete the puzzle were normally distributed with mean 32.4 minutes and standard deviation 2.5 minutes.

How many competitors would you expect to have times within 1.2 minutes of the mean time?

[4]

In this Saturday’s event, 60% of the competitors had times less than 36.0 minutes. 9 competitors who took part in this Saturday’s event are selected at random. Find the probability that at least 2 and fewer than 8 of these competitors had times less than 36.0 minutes.

[3]

80 competitors who took part in this Saturday’s event are selected at random. Use a suitable approximation to find the probability that more than 50 of these competitors had times less than 36.0 minutes.

[5]

In a group of 20 musicians, there are 9 guitarists, 6 pianists and 5 drummers. 6 musicians are selected from these 20 to perform at a concert.

Find the number of different ways in which the 6 musicians can be selected if there must be at least 3 guitarists, at most 2 pianists and exactly 1 drummer.

[4]

Three bands will be selected from the original group of 20 musicians. Each band will consist of 3 guitarists, 1 pianist and 1 drummer. No musician can be in more than one band. The first band selected will play at a concert in France, the second band selected will play in Italy and the third band selected will play in Spain. Find the number of different ways in which these three bands can be selected.

[3]

A bag contains 10 marbles, of which 4 are red and 6 are blue. Four marbles are selected from the bag at random, without replacement. The random variable $X$ denotes the number of blue marbles selected.

Show that $P (X = 2) = \frac{3}{7}$ .

[2]

Draw up the probability distribution table for $X$ .

[4]

Find the probability that at least 2 of the marbles chosen are blue, given that at least 1 red marble and at least 1 blue marble are chosen.

[3]

Mark scheme

9709/51 · May/June 2025

Download PDF

1 · 3 marks

B1
P( $Z < \frac{1.48 - μ}{0.14}$ ) = 0.22
M1
$\frac{1.48 - μ}{0.14} = - 0.772$
A1
$μ = 1.59$

2a · 3 marks

M1
$\frac{7!}{2!} - 6!$ [Method 1] OR $5! \times \frac{6 \times 5}{2}$ [Method 2]
M1
$n - 6!$ [Method 1] OR $q \times \frac{6 \times 5}{2}$ [Method 2]
A1
1800

2b · 2 marks

M1
$1 - (\frac{3}{4})^{5}$ [Method 1] OR $\frac{1}{4} + (\frac{1}{4}) (\frac{3}{4}) + (\frac{1}{4}) (\frac{3}{4})^{2} + (\frac{1}{4}) (\frac{3}{4})^{3} + (\frac{1}{4}) (\frac{3}{4})^{4}$ [Method 2]
A1
$= 0.763, \frac{781}{1024}$

2c · 2 marks

M1
$(\frac{1}{4})^{2} (\frac{3}{4})^{4} \times 5 = \frac{405}{4096}$
A1
= 0.0989

3a · 4 marks

B1
Correct stem, ignore extra values (not in reverse, not split).
B1
Correct Gulls labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no commas or other punctuation.
B1
Correct Herons labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation.
B1
Correct key for their diagram, need both clubs labelled and ‘sec’ or ‘s’ stated at least once here, or in leaf headings or title.

3b · 3 marks

B1
Median = 8.8 (seconds)
M1
9.8 – 8.3
A1
1.5

3c · 2 marks

M1
Mean = $\frac{175.0 + 30 \times 8.4}{50}$ or $\frac{175.0 + 252.0}{50}$
A1
= 8.54

3d · 3 marks

M1
${1.38}^{2} = \frac{1823.0 + \sum y^{2}}{50} - their (\frac{427.0}{50})^{2}$
M1
Solve to find $\sum y^{2}$
A1
$\sum y^{2} = 1918.8$

4a · 4 marks

M1
$P (\frac{- 1.2}{2.5} < Z < \frac{1.2}{2.5})$
M1
$2 \times 0.6844 - 1$ or $2 \times (0.6844 - 0.5)$ or $0.6844 - 0.3156$
A1
0.3688
B1
Expected number = $0.3688 \times 600 = 221.28$ so 221 or 222

4b · 3 marks

M1
One term of the form $^{9} C_{x} (p)^{x} (1 - p)^{9 - x}, 0 < p < 1, x \neq 0$ or 9
A1
Correct un-simplified expression.
B1
0.926

4c · 5 marks

B1
Mean = $80 \times 0.6 = 48$ , Variance = $80 \times 0.6 \times 0.4 = 19.2$
M1
Substituting their 48 and their 19.2 into the ±standardising formula (any number for 50.5), allow $σ^{2}$ or $\sqrt{σ}$ .
M1
Use continuity correction 49.5 or 50.5 in their standardisation formula.
M1
Appropriate probability area, from final process, must be a probability. $[P (Z > 0.571) = 1 - Φ (0.571) =] 1 - 0.7160 =$
A1
0.284

5a · 4 marks

B1
Correct outcome/value for either the 1st or 2nd scenario clearly identified, accept un-simplified, WWW.
M1
2 correct outcomes/values obtained, accept un-simplified
M1
Sum of 3 correct scenarios, may be identified by un-simplified expression.
A1
Total: 10710

5b · 3 marks

M1
$^{9} C_{3} \times^{6} C_{1} \times^{5} C_{1}$ or $^{9} C_{3} \times 6 \times 5$ seen, condone $\times 3$ or $\times 3!$ .
M1
their $2520 \times$ their $400 \times$ their 12 seen, accept un-simplified, condone $\times 3$ or $\times 3!$ .
A1
12096000

6a · 2 marks

M1
$\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times k$ or $\frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} \times \frac{5}{7} \times k$ for $k$ an integer, $k > 1$ [Method 1] OR $\frac{^{6} C_{2} \times^{4} C_{2}}{^{10} C_{4}}$ seen as the numerator of a fraction [Method 2]
A1
$\frac{3}{7}$

6b · 4 marks

B1
Table with correct values of $x$ and at least one further non-zero probability correct.
B1
Third probability correct.
B1
Fourth probability correct.
B1
Fifth probability correct.

6c · 3 marks

M1
$\frac{3}{7} + their \frac{8}{21}$ seen as the numerator of a fraction.
B1
$their \frac{4}{35} + \frac{3}{7} + their \frac{8}{21}$ seen as the denominator of a fraction.
A1
$= \frac{170}{194}, \frac{85}{97}, 0.876$

Questions

1
Question 13 marks
Mark scheme ›
2
Question 27 marks
Mark scheme ›
3
Question 312 marks
Mark scheme ›
4
Question 412 marks
Mark scheme ›
5
Question 57 marks
Mark scheme ›
6
Question 69 marks
Mark scheme ›