下载文档原格式
1.Calculating,3×(7−5)−5=3×2−5=6−5=1.
Answer:(B) 2.Since x is less than−1and greater than−2,then the best estimate of the given choices is−1.3.
Answer:(B) 3.The shaded square has side length1so has area12=1.
The rectangle has dimensions3by5so has area3×5=15.
Thus,the fraction of the rectangle that is shaded is1
15.
Answer:(A)
4.Calculating,25−52=32−25=7.
Answer:(E) 5.In3hours,Leona earns$24.75,so she makes$24.75÷3=$8.25per hour.
Therefore,in a5hour shift,Leona earns5×$8.25=$41.25.
Answer:(E)
6.Calculating,√
64+
√
36
√
64+36
=
8+6
√
100
=
14
10
=
7
5
.
Answer:(A)
7.Solution1
In total Megan and Dan inherit$1010000.
Since each donates10%,then the total donated is10%of$1010000,or$101000.
Solution2
Megan donates10%of$1000000,or$100000.
Dan donates10%of$10000,or$1000.
In total,they donate$100000+$1000=$101000.
Answer:(A) 8.We think of BC as the base of ABC.Its length is12.
Since the y-coordinate of A is9,then the height of ABC from base BC is9.
Therefore,the area of ABC is1
2(12)(9)=54.
Answer:(B)
9.Calculating the given difference using a common denominator,we obtain5
8−1
16
=9
16
.
Since9
16is larger than each of1
2
=8
16
and7
16
,then neither(D)nor(E)is correct.
Since9
16is less than each of3
4
=12
16
,then(A)is not correct.
As a decimal,9=0.5625.
Since3
5=0.6and5
9
=0.5,then9
16
>5
9
,so(C)is the correct answer.
Answer:(C)
10.Since M=2007÷3,then M=669.
Since N=M÷3,then N=669÷3=223.
Since X=M−N,then X=669−223=446.
Answer:(E)
11.The mean of6,9and18is 6+9+18
3
=
33
3
=11.
Thus the mean of12and y is11,so the sum of12and y is2(11)=22,so y=10.
Answer:(C) 12.In P QR,since P R=RQ,then∠RP Q=∠P QR=48◦.
Since∠MP N and∠RP Q are opposite angles,then∠MP N=∠RP Q=48◦.
In P MN,P M=P N,so∠P MN=∠P NM.
Therefore,∠P MN=1
2(180◦−∠MP N)=1
2
(180◦−48◦)=1
2
(132◦)=66◦.
Answer:(D)
13.The prime numbers smaller than10are2,3,5,and7.
The two of these numbers which are different and add to10are3and7.
The product of3and7is3×7=21.
Answer:(B) 14.Since there were21males writing and the ratio of males to females writing is3:7,then there
are7
3×21=49females writing.
Therefore,the total number of students writing is49+21=70.
Answer:(D) 15.Solution1
Thefirst stack is made up of1+2+3+4+5=15blocks.
The second stack is made up of1+2+3+4+5+6=21blocks.
There are36blocks in total.
We start building the new stack from the top.
Since there are more than21blocks,we need at least6rows.
For7rows,1+2+3+4+5+6+7=28blocks are needed.
For8rows,1+2+3+4+5+6+7+8=36blocks are needed.
Therefore,Clara can build a stack with0blocks leftover.
Solution2
Since the new stack will be larger than the second stack shown,let us think about adding new rows to this second stack using the blocks from thefirst stack.
Thefirst stack contains1+2+3+4+5=15blocks in total.
Thefirst two rows that we would add to the bottom of the second stack would have7and8 blocks in them,for a total of15blocks.
This uses all of the blocks from thefirst stack,with none left over,and creates a similar stack.
Therefore,there are0blocks left over.
Answer:(A) 16.Solution1
The sum of the numbers in the second row is10+16+22=48,so the sum of the numbers in any row,column or diagonal is48.
In thefirst row,P+4+Q=48so P+Q=44.
In the third row,R+28+S=48so R+S=20.
Therefore,P+Q+R+S=44+20=64.
