There’s multiple methods you can use to solve both parts (a) and (b)

For every parcel delivered on time = + $8

For every late parcel = – $2

(a)

Method 1: Algebra

Let the number of parcels delivered on time be x. Thus, the number of late parcels will be (700 – x)

Form an equation with the information provided in the question:

8 (x) + (-2) (700 – x) >= 0 (8 dollars per parcel on time, -2 per late parcel, total sum must be more than or equal to 0 to avoid losing money)

8 x – 1400 + 2 x >= 0

10 x >= 1400

x >= 140

Hence, the least number of parcels which needs to be delivered on time is 140.

Method 2: Ratio?

8 / 2 = 4 (For every 1 parcel delivered on time, 4 parcels can be delivered late without any loss)

5 u = 700

u = 700 / 5

=140

Hence, the least number of parcels which needs to be delivered on time is 140.

In this case, Method 2 seems to be easier, but it’s also because the numbers are “nice” in a sense where 8 is easily divisible by 2. If the numbers get larger or aren’t going to be divisible, Method 1 should be the go-to method, but I’m not sure if P5s would have learnt Algebra.

(b)

Method 1: Guess-and-Check

This should be the method which most schools teach. Find a random number and start working from there. In this case, parcels on time should be >500, since 500 x 8  is 4000, which is less than the profit in the question, so we can start with a number like 550. Subsequently, compare the total profits and adjust the numbers accordingly.

By guess-and-check, 542 parcels were delivered on time.

Method 2: Big-brained method

This method was taught to me when I was in Primary School and I loved it, and never had to resort to guessing-and-checking, which would sometimes be time-consuming. This method requires a bit of understanding to work, so if you can’t grasp it then you can just rely on guess-and-check.

If you take a look at the last three rows of our guess-and-check, you can see that by adding one parcel on time and removing on late parcel, we’re shifting the total profit by increments of 10. This number actually comes from 8 – (-2) which is the profit difference between a parcel on time and a late parcel.

Keeping this in mind, we can start by assuming that all the parcels were delivered on time.

Total profit would be 8 x 700 = 5600.

Now, we know that the actual profit isn’t 5600, because for every late parcel, we have to deduct $10 from this total. (You lose the $8 you would’ve gotten, plus you incur a $2 penalty).

By using 5600-4020, we can see how much we’re deducting from the total.

5600-4020 = 1580

1580 / 10 = 158 (This means that we’re taking away $10 158 times from the maximum total, which also means there were 158 late parcels)

Hence, total parcels delivered on time = 700 – 158 = 542//

These type of questions are really popular; sometimes they might say there’s a total of x legs on a farm which has cows and chickens, then you need to find out how many of each type there are etc. so if you can understand Method 2 well and practice it proficiently, then it’s an advantage you will have in your exams, because it’s less time-consuming than trying out various numbers.

Hi,

Thank you so much and appreciated that you take time to explain and provide me with few methods. 

It really helps alot for me to explain to my DS.

Thanks again.