Identify the original value and the ratio for increase.

13) We can increase or decrease numbers in a given ratio. Increase 690 in the ratio 4:3

Step 1: Identify the original value and the ratio for increase.
Original value = 690.
Ratio for increase = $\frac{\text{new part}}{\text{old part}} = \frac{4}{3}$.

Step 2: Multiply the original value by the ratio.

$$\text{Increased value} = 690 \times \frac{4}{3}$$ $$\text{Increased value} = \frac{690 \times 4}{3}$$ $$\text{Increased value} = \frac{2760}{3}$$ $$\text{Increased value} = 920$$

The increased value is $\boxed{920}$.

14) In a class of 75 pupils, there are 13 more girls than boys. Find the number of boys in that class.

Step 1: Let $B$ be the number of boys and $G$ be the number of girls.
We are given two pieces of information:

1. Total pupils: $B + G = 75$
2. Girls are 13 more than boys: $G = B + 13$

Step 2: Substitute the second equation into the first equation.

$$B + (B + 13) = 75$$ $$2B + 13 = 75$$

Step 3: Solve for $B$.

$$2B = 75 - 13$$ $$2B = 62$$ $$B = \frac{62}{2}$$ $$B = 31$$

The number of boys in the class is $\boxed{31}$.

15) The population of a school increased from 8000 pupils to 10000 pupils. What is the percentage increase?

Step 1: Calculate the absolute increase in population.

$$\text{Increase} = \text{New population} - \text{Old population}$$ $$\text{Increase} = 10000 - 8000 = 2000 \text{ pupils}$$

Step 2: Calculate the percentage increase using the formula: $\frac{\text{Increase}}{\text{Old population}} \times 100\%$.

$$\text{Percentage increase} = \frac{2000}{8000} \times 100\%$$ $$\text{Percentage increase} = \frac{1}{4} \times 100\%$$ $$\text{Percentage increase} = 25\%$$

The percentage increase is $\boxed{25\%}$.

16) The radius of a circle is 14cm. Calculate its perimeter (circumference).

Step 1: Recall the formula for the circumference of a circle: $C = 2\pi r$.
Given radius $r = 14 \text{ cm}$. Use $\pi = \frac{22}{7}$.

Step 2: Substitute the values into the formula.

$$C = 2 \times \frac{22}{7} \times 14$$ $$C = 2 \times 22 \times \frac{14}{7}$$ $$C = 2 \times 22 \times 2$$ $$C = 44 \times 2$$ $$C = 88 \text{ cm}$$

The perimeter (circumference) of the circle is $\boxed{88 \text{ cm}}$.

17) When money is borrowed interest is paid. Find the simple interest on 150,000frs for 2 years at $5 \frac{1}{2}\%$

Step 1: Identify the principal ($P$), rate ($R$), and time ($T$).
$P = 150,000 \text{ frs}$
$R = 5 \frac{1}{2}\% = 5.5\%$
$T = 2 \text{ years}$

Step 2: Use the simple interest formula: $I = \frac{P \times R \times T}{100}$.

$$I = \frac{150000 \times 5.5 \times 2}{100}$$ $$I = 1500 \times 5.5 \times 2$$ $$I = 1500 \times 11$$ $$I = 16500 \text{ frs}$$

The simple interest is $\boxed{16,500 \text{ frs}}$.

18) $1/2y = 1/4$. Calculate the value of y.

Step 1: Write the equation.

$$\frac{1}{2}y = \frac{1}{4}$$

Step 2: To solve for $y$, multiply both sides of the equation by 2.

$$2 \times \frac{1}{2}y = 2 \times \frac{1}{4}$$ $$y = \frac{2}{4}$$ $$y = \frac{1}{2}$$

The value of $y$ is $\boxed{\frac{1}{2}}$.

19) The sum of angles of a triangle is 180°. Two angles are 45° and 33° respectively. What is the third angle?

Step 1: Let the three angles of the triangle be $A$, $B$, and $C$.
We know $A + B + C = 180^\circ$.
Given $A = 45^\circ$ and $B = 33^\circ$.

Step 2: Substitute the known angles into the sum equation.

$$45^\circ + 33^\circ + C = 180^\circ$$ $$78^\circ + C = 180^\circ$$

Step 3: Solve for $C$.

$$C = 180^\circ - 78^\circ$$ $$C = 102^\circ$$

The third angle is $\boxed{102^\circ}$.

20) What is the place value of 7 in 70465?

Step 1: Identify the position of the digit 7 in the number 70465.
The number can be broken down by place value:

• 5 is in the ones place.
• 6 is in the tens place.
• 4 is in the hundreds place.
• 0 is in the thousands place.
• 7 is in the ten thousands place. The place value of 7 in 70465 is $\boxed{\text{Ten Thousands}}$.

21) The distance round a square plot is 240m. How long is each side?

Step 1: The distance round a square plot is its perimeter.
Perimeter of a square = $4 \times \text{side}$.
Given Perimeter = $240 \text{ m}$.

Step 2: Set up the equation and solve for the side.

$$4 \times \text{side} = 240$$ $$\text{side} = \frac{240}{4}$$ $$\text{side} = 60 \text{ m}$$

Each side of the square plot is $\boxed{60 \text{ m}}$.

22) The average of 3 numbers is 107. What is the sum of the 3 numbers?

Step 1: Recall the formula for average: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Count of numbers}}$.
Given Average = 107, Count = 3.

Step 2: Rearrange the formula to find the sum.

$$\text{Sum of numbers} = \text{Average} \times \text{Count of numbers}$$ $$\text{Sum of numbers} = 107 \times 3$$ $$\text{Sum of numbers} = 321$$

The sum of the 3 numbers is $\boxed{321}$.

23) Three tickets for a dance cost 2,400frs. Find the cost of 10 such tickets.

Step 1: Find the cost of one ticket.

$$\text{Cost per ticket} = \frac{\text{Total cost for 3 tickets}}{\text{Number of tickets}}$$ $$\text{Cost per ticket} = \frac{2400 \text{ frs}}{3}$$ $$\text{Cost per ticket} = 800 \text{ frs}$$

Step 2: Calculate the cost of 10 tickets.

$$\text{Cost for 10 tickets} = \text{Cost per ticket} \times 10$$ $$\text{Cost for 10 tickets} = 800 \text{ frs} \times 10$$ $$\text{Cost for 10 tickets} = 8000 \text{ frs}$$

The cost of 10 tickets is $\boxed{8,000 \text{ frs}}$.

24) Find the LCM of 5 and 7.

Step 1: Identify the numbers. 5 and 7 are both prime numbers.

Step 2: The Least Common Multiple (LCM) of two prime numbers is their product.

$$\text{LCM}(5, 7) = 5 \times 7$$ $$\text{LCM}(5, 7) = 35$$

The LCM of 5 and 7 is $\boxed{35}$.

25) Numbers can be added or subtracted. Subtract 3479 from 8001.

Step 1: Perform the subtraction.

$$8001 - 3479$$ $$8001$$ $$\underline{-3479}$$ $$4522$$

The result of the subtraction is $\boxed{4522}$.

26) Find the average of 3, 4, 5, 6 and 7.

Step 1: Sum the numbers.

$$\text{Sum} = 3 + 4 + 5 + 6 + 7 = 25$$

Step 2: Count the number of values. There are 5 numbers.

Step 3: Calculate the average: $\text{Average} = \frac{\text{Sum}}{\text{Count}}$.

$$\text{Average} = \frac{25}{5}$$ $$\text{Average} = 5$$

The average of the numbers is $\boxed{5}$.

27) Twelve years ago, Blaise was 15 years old. How old is he now?

Step 1: To find Blaise's current age, add the number of years that have passed to his age 12 years ago.

$$\text{Current age} = \text{Age 12 years ago} + 12 \text{ years}$$ $$\text{Current age} = 15 + 12$$ $$\text{Current age} = 27 \text{ years old}$$

Blaise is $\boxed{27 \text{ years old}}$.

28) What distance will a car cover in 20 minutes if it is running at the speed of 75km/hr?

Step 1: Identify the speed and time.
Speed = $75 \text{ km/hr}$
Time = $20 \text{ minutes}$

Step 2: Convert the time to hours, as the speed is given in km/hr.

$$20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours}$$

Step 3: Use the formula Distance = Speed $\times$ Time.

$$\text{Distance} = 75 \text{ km/hr} \times \frac{1}{3} \text{ hours}$$ $$\text{Distance} = \frac{75}{3} \text{ km}$$ $$\text{Distance} = 25 \text{ km}$$

The car will cover a distance of $\boxed{25 \text{ km}}$.

29) The product of a score and a gross divided by a dozen is ____.

Step 1: Define the terms:

• A score is 20 units.
• A gross is 144 units (12 dozen).
• A dozen is 12 units.

Step 2: Set up the expression based on the problem statement.
Product of a score and a gross: $20 \times 144$.
Divided by a dozen: $\frac{20 \times 144}{12}$.

Step 3: Perform the calculation.

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