Calculate the mean mark of the distribution (correct to one decimal place). Find the median mark. State the modal mark.

The mean mark is $\boxed{\text{34.2}}$.

b) Find the median mark.

The total number of students is $N=50$. Since $N$ is even, the median is the average of the $\frac{N}{2}$-th and $(\frac{N}{2}+1)$-th values.

The median is the average of the 25th and 26th marks.

• Marks 1-5 are 10.
• Marks 6-13 are 20.
• Marks 14-25 are 30.
• Marks 26-40 are 40. The 25th mark is 30. The 26th mark is 40.

The median mark is $\boxed{\text{35}}$.

c) State the modal mark.

From the table, the highest frequency is 15, which corresponds to the mark of 40.
The modal mark is $\boxed{\text{40}}$.

Question 5

a) Form a pair of simultaneous equations to represent the information.

Let $A$ be the cost of one pen of type A and $B$ be the cost of one pen of type B.
From the first statement: "Four pens of type A and three pens of type B cost N2,100."

From the second statement: "Two pens of type A and five pens of type B cost N1,750."

The pair of simultaneous equations is:

b) Solve the equations to find the cost of one pen of type A and one pen of type B.

The cost of one pen of type B is N200.

The cost of one pen of type A is N375.
The cost of one pen of type A is $\boxed{\text{N375}}$ and one pen of type B is $\boxed{\text{N200}}$.

c) If a customer buys 6 pens of type A and 2 pens of type B, how much will he pay?

Cost of 6 pens of type A = $6 \times A = 6 \times 375$
Cost of 2 pens of type B = $2 \times B = 2 \times 200$

The customer will pay $\boxed{\text{N2,650}}$.