# Category Archives: Problem of the Week

## 2010 4th Year Math Challenge – National Level with answer key

Below are the 2010 MTAP 4th Year Math Challenge – National Level questions and answers. Solutions will be posted later. More reviewers can be found on the Past Tests and All Posts pages.

Easy

1.) Ricky has (4a + 1) ₧100 bills and Emil has (2a – 2) Php 50 bills. How much more money has Ricky than Emil?

2.) The graphs of $y = 10 - x^2$ and $y + 1 = 0$ intersect at two points. Find the distance between these points

3.) If r and s are the roots of $x^2 = x + 30$, what is ∣r – s∣?

4.) A triangle has sides with integral length. If its perimeter is 12 cm, what is the longest possible side of the triangle?

5.) If the length of a rectangle is increased by 10% and the area is increased by 120%, by how many percent is the width increased?

## Problem of the Week 8

This is the 8th problem of the week of MTAP Reviewer.

Problem

The diagonal of a square is 6cm. Find the area of the square that is twice its size.

Solution

We can divide a square into several ways, one of which is using its diagonal.

Click on the image to enlarge

Now, using the diagonal of the square as side of the larger square, we can create a larger square as shown above. As seen, the diagonal of the square cuts the square into half. This half region makes a fourth of the larger square (Can you see why?).

Since the length of the side of the larger square is the diagonal of the smaller square 6, the area of the larger square is 6× 6 = 36.

## Problem of the Week 8 – Shaded Part

This is the eighth problem in our Problem of the Week.

Problem

Two semi-circles are placed inside a square as shown below. What is the area of the shaded part?

Solution 1

To get the area of the shaded part, we have to get the area of the square and then subtract the area of the two semicircles.  Continue reading

## Problem of the Week 7

My apologies for not sustaining the Problem of the Week section of this site. Anyway, I’ll do my best to make Problem of the Week a really weekly post. This week, we are going to tackle two problems since they are quite easy. These problem are classified under the third Grade and were taken from Grade 3 Sample Problem Set 1, so these are actually very easy.

Problem 1

My uncle’s age is 7 more than twice my eldest brother’s age. My eldest brother is 19 years old. How old is my uncle?

Solution

The eldest brother’s age is 19. The uncle’s age is 7 more than twice the brother’s age. Now, twice 19 is equal to 38. Since the uncle is 7 more than twice, we add 7 to 38. So, the uncle is 38 + 7 = 45 years old. Continue reading

## Problem of the Week 6 – Divisibility by 6

Problem

The number 25a where a is a digit is divisible by 6. What is the largest possible value of a?

Solution

A number is divisible by 6 if it is both divisible by 2 and divisible by 3.

First, a number if divisible by 2 if it is even. Now, for 25a to be even, a must be even. Since a is even, it could be one of the following numbers: 0, 2, 4, 6 or 8.

Second, a number is divisible by 3 if the sum of its digit is divisible by 3. So, we substitute the possible numbers above, add the digits, and see if they are divisible by 3.

For a = 0

2 + 5 + 0 = 7  Continue reading