Number and Operations

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Texas 7th Grade Math › Number and Operations

Questions 1 - 10
1

The temperature at sunrise was 18.4°F. It dropped 2.6°F each hour for 3.5 hours, then rose 4.75°F around noon. A cold front then decreased the temperature by 1.2°F. What is the final temperature with correct units?

What is the final temperature?

12.85°F

33.45°F

14.05°F

9.30°F

Explanation

Model the situation with signed numbers and a rate: $18.4 - 2.6 \times 3.5 + 4.75 - 1.2$. Compute: $2.6 \times 3.5 = 9.1$, so $18.4 - 9.1 = 9.3$, $9.3 + 4.75 = 14.05$, and $14.05 - 1.2 = 12.85$. Final temperature: 12.85°F, which makes sense because overall there was more cooling than warming.

2

Which proportion correctly represents converting 3,750 grams to kilograms? Use 1 kg = 1000 g.

$\frac{1\ \text{kg}}{1000\ \text{g}}=\frac{3750\ \text{g}}{x\ \text{kg}}$

$\frac{1000\ \text{kg}}{1\ \text{g}}=\frac{x\ \text{kg}}{3750\ \text{g}}$

$\frac{1000\ \text{g}}{1\ \text{kg}}=\frac{x\ \text{g}}{3750\ \text{kg}}$

$\frac{1000\ \text{g}}{1\ \text{kg}}=\frac{3750\ \text{g}}{x\ \text{kg}}$

Explanation

Correct proportion: $\frac{1000\ \text{g}}{1\ \text{kg}}=\frac{3750\ \text{g}}{x\ \text{kg}}$. Cross-multiply: $1000\cdot x=1\cdot 3750\Rightarrow x=3.75$ kg. Unit rate: $3750\ \text{g}\times\frac{1\ \text{kg}}{1000\ \text{g}}=3.75\ \text{kg}$; g cancels. Reasonableness: 3750 g is a few thousand grams, so a few kilograms; 3.75 kg makes sense.

3

A train travels at a constant speed. At 2 hours it has gone 150 miles; at 5 hours it has gone 375 miles. Use $d = rt$.

What is the train's constant speed?

50 miles/hour

60 miles/hour

75 miles/hour

225 miles/hour

Explanation

A constant rate means the ratio distance/time stays the same. Compute $r = \frac{d}{t}$: $\frac{150}{2} = 75$ and $\frac{375}{5} = 75$. In $d = rt$, $r$ is 75, so the speed is 75 miles/hour.

4

The relationship between hours worked (x) and pay (y) is represented by the ordered pairs: (2, 26), (5, 65), (8, 104). What is the constant of proportionality $k$ in $y=kx$?

45670

5

13

8

Explanation

The constant of proportionality is the consistent ratio $k=\dfrac{y}{x}$. Using (2, 26): $26/2=13$; using (5, 65): $65/5=13$; using (8, 104): $104/8=13$. So $k=13$, the unit rate (slope) of dollars per hour.

5

A bakery packs 1,260 cookies in 7 hours. What is the unit rate in cookies per hour?

0.0056 cookies/hour

160 cookies/hour

180 cookies/hour

7 cookies/hour

Explanation

Divide total cookies by hours to get per 1 hour: $1260 \div 7 = 180$, so 180 cookies/hour. Unit rates are useful for comparing production speeds between different days or bakeries.

6

A school places 252 students into 9 buses. What is the unit rate in students per bus?

28 students/bus

0.036 bus/student

9 students/bus

27 students/bus

Explanation

Divide students by buses to get per 1 bus: $252 \div 9 = 28$, so 28 students/bus. Unit rates help compare how crowded different buses are.

7

A person weighs 150 pounds. About how many kilograms is that? Use 1 lb ≈ 0.45 kg. What is the measurement in the new units?

67.5 kg

333.3 kg

300 kg

68.5 kg

Explanation

Proportion method: $\frac{0.45\ \text{kg}}{1\ \text{lb}}=\frac{x\ \text{kg}}{150\ \text{lb}}$. Cross-multiply: $1\cdot x=0.45\cdot 150\Rightarrow x=67.5$ kg. Unit rate: $150\ \text{lb}\times\frac{0.45\ \text{kg}}{1\ \text{lb}}=67.5\ \text{kg}$; lb cancels. Reasonableness: kilograms should be less than pounds since 1 lb is less than 1 kg; 67.5 < 150.

8

Consider these four representations of distance over time.

Which representation shows a constant rate of change?

Table: time (h) 1, 2, 3; distance (mi) 52, 104, 156

Table: time (h) 1, 2, 3; distance (mi) 50, 110, 165

Equation: $d = 120t^2$

Verbal: A jogger speeds up, covering 0.9 mile in the first 10 minutes and 1.1 miles in the next 10 minutes

Explanation

A constant rate has a consistent ratio $\frac{d}{t}$. Choice A: $\frac{52}{1}=52$, $\frac{104}{2}=52$, $\frac{156}{3}=52$ (constant). B changes (50, 55, 55). C is quadratic ($t^2$) so the rate varies with $t$. D describes speeding up, not a constant rate.

9

A recipe that serves 6 people uses 2 2/3 cups of flour. How much flour is needed for 4 people, and how much extra flour would be used if you accidentally made enough for 9 people instead of 4?

What are the two amounts (for 4 people; and the extra if making for 9)?

$1 \tfrac{1}{3}$ cups; extra $1 \tfrac{2}{3}$ cups

$1 \tfrac{7}{9}$ cups; extra $2 \tfrac{2}{9}$ cups

$1 \tfrac{5}{9}$ cups; extra $2 \tfrac{4}{9}$ cups

$1 \tfrac{7}{9}$ cups; extra $4$ cups

Explanation

Total flour for 6 is $2\tfrac{2}{3}=\tfrac{8}{3}$ cups. Per person: $\tfrac{8}{3}\div 6=\tfrac{8}{18}=\tfrac{4}{9}$ cup. For 4 people: $4\cdot \tfrac{4}{9}=\tfrac{16}{9}=1\tfrac{7}{9}$ cups. For 9 people: $9\cdot \tfrac{4}{9}=4$ cups. Extra compared to the 4-person amount: $4-\tfrac{16}{9}=\tfrac{20}{9}=2\tfrac{2}{9}$ cups. These values are proportional to servings, so they are reasonable.

10

When $x$-values are 4, 6, 8, the corresponding $y$-values are 12, 18, 24. What is the value of $k$ in $y=kx$?

3

45660

2

4

Explanation

In a proportional relationship, $k=\dfrac{y}{x}$. Compute: $12/4=3$, $18/6=3$, and $24/8=3$. The ratio $y/x$ is the same for all pairs, so $k=3$ (the unit rate/slope).

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