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Prep T1-DS-52

文章hwatai » 2007-11-02 01:51

If the average (arithmetic mean) of four different numbers is 30, how many of the numbers are greater than 30 ?

(1) None of the four numbers is greater than 60.

(2) Two of the four numbers are 9 and 10, respectively.

答案(c) 請教解題思路,感恩!
hwatai
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文章iven9 » 2007-11-02 06:33

如果四個數的算術平均數是30,那麼在四個數之中有幾個數大於30?
(1)四個數中沒有一個數大於60。
(2)四個數中的兩個數分別是9跟10。

乍看之下條件(1)沒什麼用,但是由條件(2)就可以知道另外兩個數加起來要有(30*4)-(9+10)=101
也就是說,當沒有一個數大於60的時候,另外兩個數,分別一定都大於30。

如果沒有條件(1)的限定,那麼,雖然另外兩個數加總要到101,但是可以是100跟1都可以,所以條件(1)+條件(2)充分,答案選(C)。
請參考指教。
即使被地獄的業火燒盡,我依然嚮往著天堂。
iven9
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文章vantreal » 2007-11-02 06:38

the average =30 and the sum of four numbers=120

(1) None of the four numbers is greater than 60.

if there are four 30s, then there's NO number greater than 30.
if there are three 10s, then there's 90 as the fourth number.
-> ONE number is greater than 30

---> insufficient

(2) Two of the four numbers are 9 and 10, respectively.
since the sum of four numbers=120,
and Two of the four numbers are 9 and 10,
and sum of the rest of two numbers must be 120-9-10=101

there are many possibilities, such as 1and 100, 50and 51.
we still cannot determine how many numbers are greater than 30.

---> insufficient

(1)+(2)
since none of the numbers is greater than 60.
we can assume that,
for (2) the sum of the rest of two numbers=101,
the greater number is 60, then the rest number is 51.
other possibilities include (59,52), (58,53), (57,54),(56,55),(55,56),(54,57),(53,58),(52,59),(51,60)

therefore, we can know there are TWO numbers greater than 30.

---> sufficient

Ans: (C)
vantreal
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文章hwatai » 2007-11-02 12:25

i see
thank you guys!
hwatai
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文章mikesimon » 2007-11-06 11:29

(1) None of the four numbers is greater than 60.

if there are four 30s, then there's NO number greater than 30.
if there are three 10s, then there's 90 as the fourth number.

-> ONE number is greater than 30(題目不是四個不同數的平均嗎??)

---> insufficient

(2) Two of the four numbers are 9 and 10, respectively.
since the sum of four numbers=120,
and Two of the four numbers are 9 and 10,
and sum of the rest of two numbers must be 120-9-10=101

there are many possibilities, such as 1and 100, 50and 51.
we still cannot determine how many numbers are greater than 30.

---> insufficient

(1)+(2)
since none of the numbers is greater than 60.
we can assume that,
for (2) the sum of the rest of two numbers=101,
the greater number is 60, then the rest number is 51.
other possibilities include (59,52), (58,53), (57,54),(56,55),(55,56),(54,57),(53,58),(52,59),(51,60)......上面的例子是不是要減十阿?因為這樣加起來,好像是111......還是我算錯了!!謝謝答覆....

therefore, we can know there are TWO numbers greater than 30.

---> sufficient

Ans: (C)[/quote]
mikesimon
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