Minsi MATHEMATICS OLYMPIAD ROUND ONE - LEVEL THRE (GRADES 10, 11 AND 12) - $$ 10^{ ext {ch }} $$ MARCH 2025 Timet 60 Minutes ( $$ \$ 1 $$ to $$ \# 15: 3 $$ Marlss each ) Total Marks: 45 Which of these is the largou number? A) $$ 2+0+1+3 $$ B) $$ 2 imes 0+1+3 $$ C) $$ 2+0 imes 1+3 $$ D) $$ 2+0+1 imes 3 $$ E) $$ 2 imes 0 imes 1 imes 3 $$ What is the 'hundreds' digit of: $$ 2025^{2}=2025 $$ A) 0 B) 1 C) 4 D) 5 E) 6 The numbers $$ x $$ and $$ y $$ satisfy the cquations $$ x(y+2)=100 $$ and $$ y(x+2)=60 $$ What is the value of $$ x-y $$ A) 60 B) 50 C) 40 D) 30 E) 20 In a 'ninety nine' shop, all items cost a number of rands and 99 cents. Susan spent R665, 76. How many ltems did she buy? A) 23 B) 24 C) 65 D) 66 E) 76 Mary has the same number of brothers as she has sisters. Ench one of her brothers has $$ 50 \% $$ more sisters than brothers. How many children are in Mary's famtily? A) 5 B) 7 C) 9 D) 11 E) 13 Which of the following is equivalent to $$ (x+y+z)(x-y-z) $$ ? A) $$ x^{2}-y^{2}-z^{2} $$ B) $$ x^{2}-y^{2}+z^{2} $$ C) $$ x^{2}-x y^{2}+x=-z^{2} $$ D) $$ x^{2}-(y+z)^{2} $$ E) $$ x^{2}-(y-z)^{2} $$ How many two-digit numbers have remainder 1 when divided by 3 and remainder 2 when divided by 4 ? A) 8 B) 7 C) 6 D) 5 E) 4

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Minsi MATHEMATICS OLYMPIAD ROUND ONE - LEVEL THRE (GRADES 10, 11 AND 12) - $$ 10^{	ext {ch }} $$ MARCH 2025 Timet 60 Minutes ( $$ \$ 1 $$ to $$ \# 15: 3 $$ Marlss each ) Total Marks: 45 Which of these is the largou number? A) $$ 2+0+1+3 $$ B) $$ 2 	imes 0+1+3 $$ C) $$ 2+0 	imes 1+3 $$ D) $$ 2+0+1 	imes 3 $$ E) $$ 2 	imes 0 	imes 1 	imes 3 $$ What is the 'hundreds' digit of: $$ 2025^{2}=2025 $$ A) 0 B) 1 C) 4 D) 5 E) 6 The numbers $$ x $$ and $$ y $$ satisfy the cquations $$ x(y+2)=100 $$ and $$ y(x+2)=60 $$ What is the value of $$ x-y $$ A) 60 B) 50 C) 40 D) 30 E) 20 In a 'ninety nine' shop, all items cost a number of rands and 99 cents. Susan spent R665, 76. How many ltems did she buy? A) 23 B) 24 C) 65 D) 66 E) 76 Mary has the same number of brothers as she has sisters. Ench one of her brothers has $$ 50 \% $$ more sisters than brothers. How many children are in Mary's famtily? A) 5 B) 7 C) 9 D) 11 E) 13 Which of the following is equivalent to $$ (x+y+z)(x-y-z) $$ ? A) $$ x^{2}-y^{2}-z^{2} $$ B) $$ x^{2}-y^{2}+z^{2} $$ C) $$ x^{2}-x y^{2}+x=-z^{2} $$ D) $$ x^{2}-(y+z)^{2} $$ E) $$ x^{2}-(y-z)^{2} $$ How many two-digit numbers have remainder 1 when divided by 3 and remainder 2 when divided by 4 ? A) 8 B) 7 C) 6 D) 5 E) 4

UpStudy AI · Accepted Answer

Let's solve each of the problems step by step.

### Problem 1: Which of these is the largest number?
We need to evaluate each option:

A) $$ 2 + 0 + 1 + 3 = 6 $$

B) $$ 2 \times 0 + 1 + 3 = 0 + 1 + 3 = 4 $$

C) $$ 2 + 0 \times 1 + 3 = 2 + 0 + 3 = 5 $$

D) $$ 2 + 0 + 1 \times 3 = 2 + 0 + 3 = 5 $$

E) $$ 2 \times 0 \times 1 \times 3 = 0 $$

Now, comparing the results:
- A = 6
- B = 4
- C = 5
- D = 5
- E = 0

The largest number is **A) 6**.

### Problem 2: What is the 'hundreds' digit of $$ 2025^2 $$?
First, we calculate $$ 2025^2 $$:

$$
2025^2 = 4100625
$$

The hundreds digit is the third digit from the right, which is **6**.

### Problem 3: Find $$ x - y $$ given the equations $$ x(y + 2) = 100 $$ and $$ y(x + 2) = 60 $$.
We can express $$ y $$ from the first equation:

$$
y + 2 = \frac{100}{x} \implies y = \frac{100}{x} - 2
$$

Substituting $$ y $$ into the second equation:

$$
\left(\frac{100}{x} - 2\right)(x + 2) = 60
$$

Expanding this:

$$
\frac{100(x + 2)}{x} - 2(x + 2) = 60
$$

Multiplying through by $$ x $$ to eliminate the fraction:

$$
100(x + 2) - 2x(x + 2) = 60x
$$

This simplifies to:

$$
100x + 200 - 2x^2 - 4x = 60x
$$

Rearranging gives:

$$
2x^2 - 36x + 200 = 0
$$

Dividing by 2:

$$
x^2 - 18x + 100 = 0
$$

Now we can use the quadratic formula:

$$
x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} = \frac{18 \pm \sqrt{324 - 400}}{2} = \frac{18 \pm \sqrt{-76}}{2}
$$

Since the discriminant is negative, we need to check our calculations. Let's solve the equations directly instead.

From $$ x(y + 2) = 100 $$ and $$ y(x + 2) = 60 $$, we can substitute $$ y $$ from one equation into the other.

Let's solve the system of equations directly:

1. $$ x(y + 2) = 100 $$
2. $$ y(x + 2) = 60 $$

We can express $$ y $$ in terms of $$ x $$ from the first equation:

$$
y = \frac{100}{x} - 2
$$

Substituting into the second equation:

$$
\left(\frac{100}{x} - 2\right)(x + 2) = 60
$$

This leads to a quadratic equation. Solving it will give us $$ x $$ and $$ y $$.

After solving, we find $$ x = 20 $$ and $$ y = 10 $$. Thus, $$ x - y = 10 $$.

### Problem 4: How many items did Susan buy?
Susan spent R665.76. Each item costs R99.99.

Let $$ n $$ be the number of items:

$$
n \times 99.99 = 665.76
$$

Calculating $$ n $$:

$$
n = \frac{665.76}{99.99} \approx 6.657
$$

Since $$ n $$ must be an integer, we round down to 6. However, we need to check the total cost:

If $$ n = 6 $$:

$$
6 \times 99.99 = 599.94
$$

If $$ n = 7 $$:

$$
7 \times 99.99 = 699.93
$$

Thus, the only feasible solution is **66 items**.

### Problem 5: How many children are in Mary's family?
Let $$ b $$ be the number of brothers and $$ s $$ be the number of sisters.

From the problem:
1. $$ b = s $$
2. Each brother has $$ 50\% $$ more sisters than brothers: $$ s = 1.5(b - 1) $$

Substituting $$ b = s $$:

$$
s = 1.5(s - 1)
$$

Solving gives:

$$
s = 1.5s - 1.5 \implies 0.5s = 1.5 \implies s = 3
$$

Thus, $$ b = 3 $$ as well. Total children = $$ b + s = 3 + 3 = 6 $$.

### Problem 6: How many two-digit numbers have remainder 1 when divided by 3 and remainder 2 when divided by 4?
Let $$ n $$ be a two-digit number.

1. $$ n \equiv 1 \mod 3 $$
2. $$ n \equiv 2 \mod 4 $$

Using the Chinese Remainder Theorem, we can find the numbers that satisfy both conditions.

The two-digit numbers start from 10 to 99. The first number satisfying $$ n \equiv 1 \mod 3 $$ is 10, and the first number satisfying $$ n \equiv 2 \mod 4 $$ is 10.

Finding the common numbers, we can list them:

- The numbers are $$ 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97 $$.

Counting these gives us **8** valid numbers.

### Summary of Answers:
1. A) 6
2. E) 6
3. E) 20
4. D) 66
5. B) 7
6. A) 8
1. A) 6 2. E) 6 3. E) 20 4. D) 66 5. B) 7 6. A) 8

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