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## MTEL General Curriculum Mathematics Practice

Question 1 |

#### Use the expression below to answer the question that follows.

#### \(\large \dfrac{\left( 155 \right)\times \left( 6,124 \right)}{977}\)

#### Which of the following is the best estimate of the expression above?

## 100Hint: 6124/977 is approximately 6. | |

## 200Hint: 6124/977 is approximately 6. | |

## 1,000Hint: 6124/977 is approximately 6. 155 is approximately 150, and \( 6 \times 150 = 3 \times 300 = 900\), so this answer is closest. | |

## 2,000Hint: 6124/977 is approximately 6. |

Question 2 |

#### A family has four children. What is the probability that two children are girls and two are boys? Assume the the probability of having a boy (or a girl) is 50%.

\( \large \dfrac{1}{2}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |

\( \large \dfrac{1}{4}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |

\( \large \dfrac{1}{5}\) Hint: Some configurations are more probable than others -- i.e. it's more likely to have two boys and two girls than all boys. Be sure you are weighting properly. | |

\( \large \dfrac{3}{8}\) Hint: There are two possibilities for each child, so there are \(2 \times 2 \times 2 \times 2 =16\) different configurations, e.g. from oldest to youngest BBBG, BGGB, GBBB, etc. Of these configurations, there are 6 with two boys and two girls (this is the combination \(_{4}C_{2}\) or "4 choose 2"): BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG. Thus the probability is 6/16=3/8. |

Question 3 |

#### Which of the following is equal to one million three hundred thousand?

\(\large1.3\times {{10}^{6}}\)
| |

\(\large1.3\times {{10}^{9}}\)
Hint: That's one billion three hundred million. | |

\(\large1.03\times {{10}^{6}}\)
Hint: That's one million thirty thousand. | |

\(\large1.03\times {{10}^{9}}\) Hint: That's one billion thirty million |

Question 4 |

#### The equation \( \large F=\frac{9}{5}C+32\) is used to convert a temperature measured in Celsius to the equivalent Farentheit temperature.

#### A patient€™s temperature increased by 1.5° Celcius. By how many degrees Fahrenheit did her temperature increase?

## 1.5°Hint: Celsius and Fahrenheit don't increase at the same rate. | |

## 1.8°Hint: That's how much the Fahrenheit temp increases when the Celsius temp goes up by 1 degree. | |

## 2.7°Hint: Each degree increase in Celsius corresponds to a \(\dfrac{9}{5}=1.8\) degree increase in Fahrenheit. Thus the increase is 1.8+0.9=2.7. | |

## Not enough information.Hint: A linear equation has constant slope, which means that every increase of the same amount in one variable, gives a constant increase in the other variable. It doesn't matter what temperature the patient started out at. |

Question 5 |

#### The chart below gives percentiles for the number of sit-ups that boys of various ages can do in 60 seconds (source , June 24, 2011)

#### Which of the following statements can be inferred from the above chart?

## 95% of 12 year old boys can do 56 sit-ups in 60 seconds.Hint: The 95th percentile means that 95% of scores are less than or equal to 56, and 5% are greater than or equal to 56. | |

## At most 25% of 7 year old boys can do 19 or more sit-ups in 60 seconds.Hint: The 25th percentile means that 25% of scores are less than or equal to 19, and 75% are greater than or equal to 19. | |

## Half of all 13 year old boys can do less than 41 sit-ups in 60 seconds and half can do more than 41 sit-ups in 60 seconds.Hint: Close, but not quite. There's no accounting for boys who can do exactly 41 sit ups. Look at these data: 10, 20, 41, 41, 41, 41, 50, 60, 90. The median is 41, but more than half can do 41 or more. | |

## At least 75% of 16 year old boys can only do 51 or fewer sit-ups in 60 seconds.Hint: The "at least" is necessary due to duplicates. Suppose the data were 10, 20, 51, 51. The 75th percentile is 51, but 100% of the boys can only do 51 or fewer situps. |

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