2014 AMC amc10b 真题 答案 详解

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中英双语真题

2014 AMC 10B
Problem 1
Leah has 13 coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?

Leah有13枚硬币，它们都是1分或5分的硬币，假设她比现在再多一枚5分的硬币，那么她拥有的5分和1分硬币的数量就一样多了，问她现在所拥有的所有硬币总共价值多少？
(A) 33 (B) 35 (C) 37 (D) 39 (E) 41

Problem 2
What is ( \frac{2^3 + 2^3}{2^{-3} + 2^{-3}} )?
( \frac{2^3 + 2^3}{2^{-3} + 2^{-3}} ) 的值是多少？
(A) 16 (B) 24 (C) 32 (D) 48 (E) 64

Problem 3
Peter drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip?
Peter在石子路上行驶了他总路程的前三分之一，接下来的20英里走的是石板路，最后剩下的总路程的五分之一走的是泥路。问Peter的总路程是多少英里？
(A) 30 (B) ( \frac{400}{11} ) (C) ( \frac{75}{2} ) (D) 40 (E) ( \frac{300}{7} )

英文真题

2014 AMC 10B Problems
Problem 1
Leah has 13 coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies as nickels. In cents, how much are Leah's coins worth?
(A) 33 (B) 35 (C) 37 (D) 39 (E) 41

Problem 2
What is (\frac{2^3 + 2^{-3}}{2^{-3} + 2^{-3}}) ?
(A) 16 (B) 24 (C) 32 (D) 48 (E) 64

Problem 3
Peter drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip?
(A) 30 (B) (\frac{400}{11}) (C) (\frac{75}{2}) (D) 40 (E) (\frac{300}{7})

Problem 4
Susie pays for 4 muffins and 3 bananas. Calvin spends twice as much paying for 2 muffins and 16 bananas. A muffin is how many times as expensive as a banana?
(A) (\frac{3}{2}) (B) (\frac{5}{3}) (C) (\frac{7}{4}) (D) 2 (E) (\frac{13}{4})

Problem 5
Camden constructs a square window using 8 equal-size panes of glass, as shown. The ratio of the height to width for each pane is (5:2), and the borders around and between the panes are 2 inches wide. In inches, what is the side length of the square window?
(A) 26 (B) 28 (C) 30 (D) 32 (E) 34

Problem 6
Orvin went to the store with just enough money to buy 30 balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 1 balloon at the regular price and get a second at (\frac{1}{3}) off the regular price. What is the greatest number of balloons Orvin could buy?
(A) 33 (B) 34 (C) 36 (D) 38 (E) 39

Problem 7
Suppose (A > B > 0) and (A) is (x\%) greater than (B). What is (x)? (A) (100\left(\frac{A-B}{B}\right)) (B) (100\left(\frac{A+B}{B}\right)) (C) (100\left(\frac{A+B}{A}\right)) (D) (100\left(\frac{A-B}{A}\right)) (E) (100\left(\frac{A}{B}\right))

真题文字详解(英文)

Problem 1

Leah has 13 coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?

(A) 33 (B) 35 (C) 37 (D) 39 (E) 41

Solution

If Leah has 1 more nickel, she has 14 total coins. Because she has the same number of nickels and pennies, she has 7 nickels and 7 pennies. This is after the nickel has been added, so we must subtract 1 nickel to get 6 nickels and 7 pennies. Therefore, Leah has $6 \cdot 5 + 7 = \boxed{37}$ (C) cents.

Problem 2

What is $\frac{2^3+2^3}{2^{-3}+2^{-3}}$ ?

(A) 16 (B) 24 (C) 32 (D) 48 (E) 64

Solution 1

We can synchronously multiply $2^3$ to the expressions both above and below the fraction bar. Thus,

$\frac{2^3+2^3}{2^{-3}+2^{-3}}=\frac{2^6+2^6}{1+1}=2^6.$

Hence, the fraction equals to $\boxed{\text{(E)}\ 64}$.

Solution 2

We have

$\frac{2^3+2^3}{2^{-3}+2^{-3}}=\frac{8+8}{\frac18+\frac18}=\frac{16}{\frac14}=16\cdot4=64,$

so our answer is $\boxed{\text{(E)}\ 64}$.

Problem 3

Randy drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?

(A) 30 (B) $\frac{400}{11}$ (C) $\frac{75}{2}$ (D) 40 (E) $\frac{300}{7}$

Solution 1

Let the total distance be $x$. We have $\frac{x}{3}+20+\frac{x}{5}=x$, or $\frac{8x}{15}+20=x$. Subtracting $\frac{8x}{15}$ from both sides gives us $20=\frac{7x}{15}$. Multiplying by $\frac{15}{7}$ gives us $x=\boxed{\text{(E)}\ \frac{300}{7}}$.

Solution 2

The first third of his distance added to the last one-fifth of his distance equals $\frac{8}{15}$. Therefore, $\frac{7}{15}$ of his distance is 20. Let $x$ be his total distance, and solve for $x$. Therefore, $x$ is equal to $\frac{300}{7}$, or E.

Problem 4

Susie pays for 4 muffins and 3 bananas. Calvin spends twice as much paying for 2 muffins and 16 bananas. A muffin is how many times as expensive as a banana?

(A) $\frac{3}{2}$ (B) $\frac{5}{3}$ (C) $\frac{7}{4}$ (D) 2 (E) $\frac{13}{4}$

Solution

真题文字详解(中英双语)

Problem 1 Leah has 13 coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
(A) 33 (B) 35 (C) 37 (D) 39 (E) 41

Solution If Leah has 1 more nickel, she has 14 total coins. Because she has the same number of nickels and pennies, she has 7 nickels and 7 pennies. This is after the nickel has been added, so we must subtract 1 nickel to get 6 nickels and 7 pennies. Therefore, Leah has $6\cdot5+7=\boxed{37(C)}$ cents.

Problem 2 What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}?$
(A) 16 (B) 24 (C) 32 (D) 48 (E) 64

Solution 1 We can synchronously multiply $2^3$ to the expressions both above and below the fraction bar. Thus,
$\frac{2^3 + 2^3}{2^{-3} + 2^{-3}} = \frac{2^6 + 2^6}{1 + 1} = 2^6.$ Hence, the fraction equals to $\boxed{(E) 64}$.

Solution 2 We have
$\frac{2^3 + 2^3}{2^{-3} + 2^{-3}} = \frac{8 + 8}{\frac18 + \frac18} = \frac{16}{\frac14} = 16\cdot4 = 64,$
so our answer is $\boxed{(E) 64}$.

Problem 3 Randy drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?
(A) 30 (B) $\frac{400}{11}$ (C) $\frac{75}{2}$ (D) 40 (E) $\frac{300}{7}$

Solution 1 Let the total distance be $x$. We have $\frac{x}{3} + 20 + \frac{x}{5} = x$, or $\frac{8x}{15} + 20 = x$. Subtracting $\frac{8x}{15}$ from both sides gives us $20 = \frac{7x}{15}$. Multiplying by $\frac{15}{7}$ gives us $x = \boxed{\text{(E)} \frac{300}{7}}$.

Solution 2 The first third of his distance added to the last one-fifth of his distance equals $\frac{8}{15}$. Therefore, $\frac{7}{15}$ of his distance is 20. Let $x$ be his total distance, and solve for $x$. Therefore, $x$ is equal to $\frac{300}{7}$, or E.

Problem 4 Susie pays for 4 muffins and 3 bananas. Calvin spends twice as much paying for 2 muffins and 16 bananas. A muffin is how many times as expensive as a banana?
(A) $\frac{3}{2}$ (B) $\frac{5}{3}$ (C) $\frac{7}{4}$ (D) 2 (E) $\frac{13}{4}$

Solution