Question

(a) Candy bars were distributed equally among students of a class. The number of candy bars each student got was one-eighth of the number of students. Had the number of students been half, each student would have got 16 candy bars. What is the total number of candy bars that were distributed?

(b) Reduce to simplest form.

(i) $\frac{6}{3}+\left(\frac{-1}{6}\right)$

(ii) $\frac{-3}{2}-\frac{3}{8}$

(iii) $\frac{-9}{6}+\left(\frac{-3}{5}\right)$

(iv) $\frac{-1}{3}-\left(\frac{-3}{5}\right)$

[4 MARKS]

Answer 1

a) Steps: 1 Mark
Answer: 1 Mark
b) Each sub-part: 0.5 Marks


(a) Let the total number of students $=x$.

Then, number of candy bars each student has $=\frac{x}{8}$

Now, no. of student has been half $=\frac{x}{2}$,

Number of candies each student would have got = 16

$\frac{x}{2}\times 16=x\times \frac{x}{8}$

Or No. of initial students = 64 ; $\frac{x}{2}=32.$

So, number of candies distributed = 32 x 16 = 512


(b)
(i) $\frac{6}{3}+\frac{-1}{6}=2-\frac{1}{6}=\frac{11}{6}$

(ii) $\frac{-3}{2}-\frac{3}{8}=\frac{-12-3}{8}=\frac{-15}{8}$

(iii) $\frac{-9}{6}+\frac{-3}{5}=\frac{-3}{2}-\frac{3}{5}$

$\frac{-15-6}{10}=\frac{-21}{10}$

(iv) $\frac{-1}{3}-\left(\frac{-3}{5}\right)=\frac{-1}{3}+\frac{3}{5}$

$\frac{-5+9}{15}=\frac{4}{15}$