2 Graeme, aged 57, is married to Catherine, aged 58. They work as medical consultants, and

(b) Explain the capital gains tax (CGT) and inheritance tax (IHT) implications of Graeme gifting his remaining ‘T’

ordinary shares at their current value either:

(i) to his wife, Catherine; or

(ii) to his son, Barry.

Your answer should be supported by relevant calculations and clearly identify the availability and effect of

any reliefs (other than the CGT annual exemption) that might be used to reduce or defer any tax liabilities

arising. (9 marks)

A.n

B.[log2n]-1

C.n／2

D.[log2n]+1

A.57

B.39

C.24

D.11

对n个元素的有序表A[1．．n]进行二分(折半)查找(除2取商时向下取整)，查找元素A[i](1≤i≤n)时，最多与A中的(57)个元素进行比较。

A．n

B．[log2n]-1

C．n／2

D．[log2n]+1

有9名学生到英语培训班学习，培训前后各进行了一次水平测验，成绩如下表：

假设测验成绩服从正态分布，试在显著性水平α=0. 05下，检验学生的培训效果是否显著

I always enjoy the drive into Marley.It is a good straight road with pleasant views of the countryside (1) either side.There are woods and hills,villages and farms and,about halfway,a large lake.And because there is rarely much traffic on the road, I can usually enjoy the view (2)I drive along. I was rather (3) the other morning,therefore,when a small green car began to drive very close behind me.I went a little (4)，hoping to get rid of it.But whenever I slowed down，the little car caught up (5) me.The driver,a middle aged man，was grinning and waving to me. (6)，I did not know him.

“Perhaps I am doing (7) wrong，”I thought.I checked my lights and my indicator,but they were both (8).And I certainly had not got a puncture.I did not slow down until I got to Marley.As I was crossing the railway bridge into Marley,the lights turned (9).I was waiting there when the little green car pulled up beside me.The driver got out and tapped on my window.I opened it.“Can I help you？”I asked rather (10).“I’m trying to help you!”the man said.“You see,your number plate fell off about fifteen miles back.I tried to attract your attention,but you didn’t take any notice!”

1. (A). of (B). by (C). on (D). with

2. (A). since (B). as (C). until (D). for

3. (A). annoyed (B). pleased (C). excited (D). disappointed

4. (A). slower (B). faster (C). earlier (D). later

5.(A). after (B). along (C). on (D). with

6. (A). Therefore (B). However (C). Thus (D). So

7. (A). everything (B). nothing (C). something (D). anything

8. (A). in order (B). out of order (C). in place (D). out of place

9. (A). on (B). off (C). green (D). red

10. (A). coldly (B). helpfully (C). earnestly (D). excitedly

阅读以下说明及C++程序代码，将应填入(n)处的语句写在对应栏内。

【说明】

本程序的功能是实现任意两个大整数的乘法运算，例如：

输入整数1：8934793850094505800243958034985058

输入整数2：234584950989689084095803583095820923

二者之积：

209596817742739508050978890737675662366433464256830959194834854876 8534

【C++代码】

include＜iostream.h＞

const int MAXINPUTBIT=100;

const int MAXRESULTBIT=500;

class LargeNumber{

int i,j;

int temp;

int one[MAXINPUTBIT+1];

int onebit; //one的位数

int two[MAXINPUTBIT+1];

int twobit; //two的位数

int result[MAXRESULTBIT+1];

public:

LargeNumber();

~LargeNumber();

int inputone(); //出错返叫0，否则返回1

int inputtwo(); //同上

void multiplication(); //乘

void clearresult(); //清零

void showresult(); //显示

};

LargeNumber∷LargeNumber()

{

for(i=0;i＜=MAXINPUTBIT;i++)

{

one[i]=0;

two[i]=0;

}

nebit=0;

twobit=0;

inputone();

inputtwo();

}

LargeNumber∷～LargeNumber()

{

}

int LargeNumber∷inputone()

{

char Number[MAXINPUTBIT+1];

cout＜＜"Please enter one:";

cin＞＞Number;

i=0;

j=MAXINPUTBIT;

while(Number[i]!='\0')

i++;

nebit=i;

for(i--;i＞=0;i--,j--)

{

if(int(Number[i])＞=48&&int(Number[i])＜=57)

(1); //由字符转换为数字

else

return 0;

}

return 1;

}

int LargeNumber∷inputtwo()

{

char Number[MAXINPUTBIT+1];

cout＜＜"Please enter two:";

cin＞＞Number;

i=0;

j=MAXINPUTBIT;

while(Number[i]!='\0')

i++;

twobit=i;

for(i--;i＞=0;i--,j--)

{

if(int(Number[i])＞=48&&int(Number[i])＜=57)

two[j]=int(Number[i]-48); //由字符转换为数字

else

return 0;

}

return 1;

}

void LargeNumber∷multiplication() //乘法

{

clearresult();

int m;

for(i=MAXINPUTBIT;i＞=0;i--)

{

temp=two[i];

for(j=(2),m=MAXINPUTBIT;m＞=0;m--,j--)

{

result[j]+=temp*one[m];

if(result[j]＞9)

{

result[j-1]+=result[j]/10;

(3);

}

}

&n

It can be concluded from Paragraph 2 that ______.

A．a population growth leads to a greater proportion of workforce

B．a large population does not necessarily mean a high proportion of workforce

C．the size of the aged determines the proportion of workforce

D．the proportion of the very young determines the size of workforce

Harry: I hate history.

Susan: I don't. (57) I'm reading about the Romans.