Solve the following problems: 6a, 6b, 7a, 7b, 8a, and 8b.

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for problems 6a, 6b, 7a, 7b, 8a, and 8b.

:

Let the unknown number be $N$ .
Translate the problem into an equation
"3.5% of a number is 3.5".
$0.035 \times N = 3.5$
To find $N$ , divide both sides by 0.035
$N = \frac{3.5}{0.035}$
Simplify the division
$N = \frac{35}{0.35} = \frac{3500}{35} = 100$
The number is $\boxed{100}$ .

:
Evaluate the numerator
$0.01 \times 0.04 = 0.0004$
Rewrite the expression with the evaluated numerator
$\frac{0.0004}{0.0002}$
Perform the division
$\frac{0.0004}{0.0002} = 2$
Express the answer in standard form (scientific notation).
Standard form requires a number between 1 and 10 multiplied by a power of 10.
$2 = 2 \times 10^0$
The value is $\boxed{2 \times 10^0}$ .

:

Identify the given information

- Number of girls = 375
- Ratio of boys to girls = 4:5

Let the number of boys be $B$ .
Set up a proportion based on the ratio:
$\frac{\text{Number of boys}}{\text{Number of girls}} = \frac{4}{5}$ $\frac{B}{375} = \frac{4}{5}$
Solve for $B$ by multiplying both sides by 375
$B = \frac{4}{5} \times 375$
Perform the multiplication
$B = 4 \times \frac{375}{5} = 4 \times 75$ $B = 300$
The number of boys in the school is $\boxed{300}$ .

:

Identify the given values for simple interest calculation

- Principal (P) = ₦650.00
- Time (T) = 2 years
- Rate (R) = 6% per annum = 0.06

Use the formula for simple interest
$\text{Simple Interest (SI)} = \frac{P \times R \times T}{100}$
Substitute the values into the formula
$\text{SI} = \frac{650 \times 6 \times 2}{100}$
Perform the multiplication in the numerator
$\text{SI} = \frac{650 \times 12}{100}$ $\text{SI} = \frac{7800}{100}$
Perform the division
$\text{SI} = \text{₦}78.00$
The simple interest is $\boxed{\text{₦}78.00}$ .

:
This is an inverse proportion problem.
Let $M$ be the number of men and $D$ be the number of days. The total work is constant, so $M_1 D_1 = M_2 D_2$ .

Identify the given values

- $M_1 = 6$ men
- $D_1 = 4$ days
- $M_2 = 8$ men
- $D_2 = ?$ days

Substitute the values into the formula
$6 \times 4 = 8 \times D_2$ $24 = 8 D_2$
Solve for $D_2$ by dividing both sides by 8
$D_2 = \frac{24}{8}$ $D_2 = 3$
It will take 8 men $\boxed{3 \text{ days}}$ to finish the same work.

:
Find the prime factorization of 60.
$60 = 2 \times 30 = 2 \times 2 \times 15 = 2^2 \times 3 \times 5$
For a number to be a perfect square, all the exponents in its prime factorization must be even.
In the prime factorization of 60 ( $2^2 \times 3^1 \times 5^1$ ), the exponent of 2 is even (2), but the exponents of 3 and 5 are odd (1).
To make the exponents of 3 and 5 even, we need to multiply by $3^1$ and $5^1$ .
The smallest number to multiply by is $3 \times 5 = 15$ .
- $60 \times 15 = (2^2 \times 3 \times 5) \times (3 \times 5) = 2^2 \times 3^2 \times 5^2 = (2 \times 3 \times 5)^2 = 30^2 = 900$, which is a perfect square.
The smallest number by which 60 must be multiplied to be a perfect square is $\boxed{15}$ .

Solve the following problems: 6a, 6b, 7a, 7b, 8a, and 8b.

Step-by-step solution

Let the unknown number be $N$ .

Translate the problem into an equation

To find $N$ , divide both sides by 0.035

Simplify the division

Evaluate the numerator

Rewrite the expression with the evaluated numerator

Perform the division

Express the answer in standard form (scientific notation).

Identify the given information

Let the number of boys be $B$ .

Solve for $B$ by multiplying both sides by 375

Perform the multiplication

Identify the given values for simple interest calculation

Use the formula for simple interest

Substitute the values into the formula

Perform the multiplication in the numerator

Perform the division

This is an inverse proportion problem.

Identify the given values

Substitute the values into the formula

Solve for $D_2$ by dividing both sides by 8

Find the prime factorization of 60.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

In the prime factorization of 60 ( $2^2 \times 3^1 \times 5^1$ ), the exponent of 2 is even (2), but the exponents of 3 and 5 are odd (1).

To make the exponents of 3 and 5 even, we need to multiply by $3^1$ and $5^1$ .

The smallest number to multiply by is $3 \times 5 = 15$ .

Still stuck on this one?

More Mathematics Questions

Solve the following problems: 6a, 6b, 7a, 7b, 8a, and 8b.

Step-by-step solution

Let the unknown number be NNN.

Translate the problem into an equation

To find NNN, divide both sides by 0.035

Simplify the division

Evaluate the numerator

Rewrite the expression with the evaluated numerator

Perform the division

Express the answer in standard form (scientific notation).

Identify the given information

Let the number of boys be BBB.

Solve for BBB by multiplying both sides by 375

Perform the multiplication

Identify the given values for simple interest calculation

Use the formula for simple interest

Substitute the values into the formula

Perform the multiplication in the numerator

Perform the division

This is an inverse proportion problem.

Identify the given values

Substitute the values into the formula

Solve for D2D_2D2​ by dividing both sides by 8

Find the prime factorization of 60.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

In the prime factorization of 60 (22×31×512^2 \times 3^1 \times 5^122×31×51), the exponent of 2 is even (2), but the exponents of 3 and 5 are odd (1).

To make the exponents of 3 and 5 even, we need to multiply by 313^131 and 515^151.

The smallest number to multiply by is 3×5=153 \times 5 = 153×5=15.

Still stuck on this one?

More Mathematics Questions

Let the unknown number be $N$ .

To find $N$ , divide both sides by 0.035

Let the number of boys be $B$ .

Solve for $B$ by multiplying both sides by 375

Solve for $D_2$ by dividing both sides by 8

In the prime factorization of 60 ( $2^2 \times 3^1 \times 5^1$ ), the exponent of 2 is even (2), but the exponents of 3 and 5 are odd (1).

To make the exponents of 3 and 5 even, we need to multiply by $3^1$ and $5^1$ .

The smallest number to multiply by is $3 \times 5 = 15$ .