# Translating a sentence into a multi-step equation Online Quiz

#### Differential Equations

46 Lectures 2.5 hours

#### Linear Equations in Two Variables

9 Lectures 2.5 hours

Following quiz provides Multiple Choice Questions (MCQs) related to Translating a sentence into a multi-step equation. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz. Q 1 - A certain number was multiplied by 9. Then, this product was divided by 33. Finally, 8 was subtracted from this quotient, resulting in a difference of -5. What was this number?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\frac{9x}{33} −8 = −5$

Step 2:

Adding 8 to both sides

$\frac{9x}{33} −8 + 8 = −5 + 8; \: \frac{9x}{33} = 3$

Step 3:

Cross multiplying and simplifying

$x = \frac{99}{9} =11$

So, required number is 11

Q 2 - Some number was divided by 7. After which, the quotient is added to 2. Next, the sum is multiplied by 6, which resulted in 12. Given this product, find the initial number.

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\left ( \frac{x}{7} + 2 \right )6 = 12$

Step 2:

Dividing both sides by 6

$\frac{6\left ( \frac{x}{7} + 2 \right )}{6} = \frac{12}{6}; \: \left ( \frac{x}{7} + 2 \right ) = 2$

Step 3:

Subtracting 2 from both sides

$\frac{x}{7} + 2 − 2 = 2 − 2 = 0; \: \frac{x}{7} = 0$

Cross multiplying

$x = 7 \times 0 = 0$

So, required number is 0

Q 3 - 8 was multiplied by a number. Then, the product was divided by 6. Finally, 3 was added to this quotient, giving 7. State the initial number.

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\frac{8x}{6} + 3 = 7$

Step 2:

Subtracting 3 from both sides

$\frac{8x}{6} − 3 + 3 = −3 + 7; \: \frac{8x}{6} = 4$

Step 3:

Cross multiplying and simplifying

$x = \frac{24}{8} = 3$

So, required number is 3

Q 4 - A certain number was multiplied by 10. Then, this product was divided by 26. Finally, 8 was subtracted from this quotient, resulting in a difference of −3. What was this number?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\frac{10x}{26} − 8 = −3$

Step 2:

Adding 8 to both sides

$\frac{10x}{26} − 8 + 8 = −3 + 8; \: \frac{10x}{26} = 5$

Step 3:

Cross multiplying and simplifying

$x = \frac{130}{10} = 13$

So, required number is 13

Q 5 - First, 120 was divided by some number. The resulting quotient was then multiplied by 3. Following this, 5 was subtracted from the product, giving 40. What was the initial divisor?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$3(\frac{120}{x}) − 5 = 40$

Step 2:

Adding 5 to both sides

$\frac{360}{x} − 5 + 5 = 40 + 5; \: \frac{360}{x} = 45$

Step 3:

Cross multiplying and simplifying

$x = \frac{360}{45} = 8$

So, required number is 8

Q 6 - 17 was divided by a number. This quotient was then multiplied by 6, and 5 was taken from that product. If the previous operation resulted in 1, find the initial number

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$6(\frac{17}{x}) − 5 = 1$

Step 2:

Adding 5 to both sides

$\frac{102}{x} − 5 + 5 = 1 + 5; \: \frac{102}{x} = 6$

Step 3:

Cross multiplying and simplifying

$x = \frac{102}{6} = 17$

So, required number is 17

Q 7 - Some number was divided by 6. After which, the quotient is added to 2. Next, the sum is multiplied by 6, which resulted in 12. Given this product, find the initial number.

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\left ( \frac{x}{6} + 2 \right )6 = 12$

Step 2:

Dividing both sides by 6

$\frac{\left [ \left ( \frac{x}{6} + 2\right )6 \right ]}{6} = \frac{12}{6}; \: \left ( \frac{x}{6} + 2\right ) = 2$

Subtracting 2 from both sides

$\frac{x}{6} − 2 + 2 = 2 − 2 = 0; \: \frac{x}{6} = 0$

Step 3:

Cross multiplying and simplifying

$x = 6 \times 0 = 0$

So, required number is 0

Q 8 - A number was subtracted from 14 and that difference was then divided by 4. After which, the quotient was multiplied by 3. This product was 9. What was the number?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$[\left ( \frac{14 - x}{4} \right )]3 = 9$

Step 2:

Dividing both sides by 3

$\frac{\left [ [\left ( \frac{14 - x}{4} \right )]3 \right ]}{3}= \frac{9}{3}; \: \frac{\left ( 14 - x \right )}{4} = 3$

Cross multiplying and simplifying

$14 − x = 4 \times 3 = 12$

Step 3:

Subtracting 14 from both sides

$14 − x − 14 = 12 − 14 = − 2; \: -x = −2; \: x = 2$

So, required number is 2

Q 9 - A certain number was multiplied by 7. Then, this product was divided by 2.8. Finally, 6.8 was subtracted from this quotient, resulting in a difference of −30.8. What was this number?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$\frac{7x}{2.8} − 6.8 = −30.8$

Step 2:

Adding 6.8 to both sides

$\frac{7x}{2.8} − 6.8 + 6.8 = −30.8 + 6.8; \: \frac{7x}{2.8} = −24.8$

Step 3:

Cross multiplying and simplifying

$x = −24.8 \times \frac{2.8}{7} = −9.6$

So, required number is −9.6

Q 10 - 6 was added to a certain number. This sum was then divided by 3. Finally, the quotient was multiplied by 5. This product came out to 10. What was the number?

### Explanation

Step 1:

Let the required number be = x

Reducing the problem to an equation we get

$[\left ( \frac{6 + x}{3} \right )]5 = 10$

Step 2:

Dividing both sides by 5

$\frac{\left [ [\left ( \frac{6 + x}{3} \right )]5 \right ]}{5}= \frac{10}{5}; \: \left [ \frac{\left ( 6 + x \right )}{3} \right ] = 2$

Cross multiplying and simplifying

$6 + x = 3 \times 2 = 6$

Step 3:

Subtracting 6 from both sides

$6 + x − 6 = 6 − 6 = 0$

So, required number is 0

translating_sentence_into_multistep_equation.htm