團影:
(29)育教 日三初月二十年巳己曆夏
i.
Mamo A, B and C.
1990 中學會考預習專欄
ii. The concentrations of urea and salt are higher in urine than in blood plasma. How is this brought about?
Homeostasis
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Biology (13)
報日僑華
2,,
The diagram below shows a section of the mammalian skin.
W. H. Leung
iii. Give two ways in which the
composition of sweat
May differ from that of urine.
FL
1. The diagram below shows part of the human. urinary System and the contents of blood plasma and urine in Card A respectively.
Contents of blood plasta in 9/1000 cm3
water
protein
glucose
salt
(sodium chloride!
910
24
}
iv. The average production of sweat is
500 cm3 per day in Man.:
in very hot surroundings as much 25 3000 cm3 may be lost in a day.
i.
(1) What changes would you expect in the urine as a result of this greatly increased loss of sweat?
Explain.
(2) New might a great loss of sweat 11 these conditions endanger the I fe of the person concerned?
Name the parts A to H,
G
ii. State two structures which indicate that this skin sample belongs to a maceral.
iii. Describe how our skin can help in regulating our body temperature.
(1) when we feel hot, and
(2) When the body is chilled by a sudden drop of the surrounding temperature.
iv. What does F secretes? State the
functions of this secretion,
V. What are the functions of G?
V.
Contents ut vrine in g/1600 or
water
960
Canfat why protein and glucose are present in blood plasma but absent in urine?
Solution
Prateln
0
glucose
C
1.
ured
sali
12
(sodium chlorides i
vi. State aree main ways in which
water is lost. from the human body.
1990 中學會考預習專欄
ORA WARJA, MILL & QALE PRESS
Mathematics (13)
7. The sun of first n terms of an A.P.
is n(n+3), Find the first term and the common difference.
8. The tenth term of an A.P. is -15 and the sum of the first six teras is -17.
A ureter
Brenal vein Crenal artery
(c) T(11)+1(32)+...+1(20):
5(20)-5(10)
-- 2870 - 385
= 2485
(Ans.)
d
六期星日十三月二十(九八九-)年八十七國民搴中
ii. The concentration of urea and salt are higher in urine than in blood plasma because of the re-absorption of water along the kidney tubules while urea and salt are only partly re-absorbed.
This leads to a relatively higher concentration of salt and urea.
iii. The concentration of urea and salt in sweat 15 less than that in urine.
iv. (1) A small volume of concentrated urine is produced. The lost of kater through Sweating decreases the osmotic potential of blood, thus a greater. proportion of water is being re-absorbed from the glomerular filtrate by kidney tubules.
Y.
(2) The decreases in volume of the plasma as a result of great loss of water in sweat decreases the ability of the circulatory system to transfer heat from the body core to the surface quickly enough to prevent overheating of body
cells.
Protein molecules are too large to pass the pores in the: capillary wall of glomerulus into the cavity of Bowman's capsule by ultra- filtration. Glucose is completely reabsorbed
into the blood
Capillaries' surrounding proximal convoluted tubule.
the
vi. Water is lost from human body mainly through urination, sweating and expiration.
.' n = -55
n. G
2.
A epidermis
B- dermis
C hair
0 erector muscle
blood capillaries
Ed
F
sebaceous gland
G
adipose tissue
H
-sweat gland
ii. The presence of hair and sweat
gland.
iii.(1) Our skin responses by vaso- dilation of blood capillaries.
to dissipate body heat through radiation and by sweating to lose heat through evaporation.
(2) Our skin responses by vaso- constriction of superficial blood capillaries, so that less blood flows to the body surface to reduce heat ioss by - radiation. The contraction of erector muscle result in erection of hair, thus trapping a thick layer of air to enable better insulation of heat.
iv. F produces sebum which prevents wetting of hair and its anti-septic nature kills bacteria and fungi.
v. G stores energy in the form of fat and acts as a good insulating Tayer.
(rejected)
[Ans.
(ii)
S(n) > -320
2
[24
+ (n - 1)d1 > -320
They will meet in 5 days."
(a) and the first term and the
demon difference of
the
progression.
3. (a) Let the first term be a and the
common difference be d.
Section B
6. (a) 3m-2)-(2m+4)
of
- 6
(5m-20)-(3m-2)
2m - 18
12-
(Ans.)
2m + 4 = 2(12) + 4 = 28
20(64
(a + 9d)
(1)
d = (3m 2) - (2m + 4).
6
9. (a)
a = 20 .....(2)
[B(28)4(20-1)(6)]
From (1), (2).
1700
(Ans.)
a = 4, d 2
[Ans.)
7,
5(n) - ntn + 3),
(b) T(n) = a + (n - 1)d
The first term
= 4+ (n - 1)(2)
= 2n + 2
= S(1)
1)(-2)] -320
n(4) > -320
n2 - 4n - 320 < 0·
(n = 20)(n+16) < 0
-16 (n (20
.. the greatest value of n
required is 19.
(Ans.)
(i) Monthly salary in the nth year
- 2000+ En - 1)(120)
= 120 + 1830
(ii) The total amount
{Ans,)
[2(2000) + (n-11(120)] × 12
{Ans.
+
W. K. Lo
Exercise 13
Arithmetic Progression and sequence
Section A
1. The term Tal or a sequence s
given by 1(n). . 2(n-1 17:
(a) find the 6th term.
(b) find the rth term.
2. The sum of the first n terms of a
sequence is given by
S(n) = (n + 1)(2n + 11.
If
1(0)
represents the nih tera, find
(a) T(1)+T(2)41 (3) + ... ... ... ... ... +T (10)
[b] T(1)T(2)+.....+T(20)
(c) T(11)+(12) + ..... T(20)
3. The oth term of an arithmetic pregression is denoted by Tin). If
2T (7) T(C) = 10 and T(14) = 5T(2), find
4
[a) the first term and the common
difference of the A.P., and
(b) the nth term.
4. Find the sum of 1-2-3-4-5-617-81....
to 2n terms
5. A and & set out to meet each other from two places 660 km apart. A travels 60 km the first day, 56 km the second, 52 km the third and so
@ travels 40 km the first day, 48 km the second, 56 km the third and so on. When will they nee!?
Section B
6. If the first three terms of an A.P. are Za 4, 37 - 2 and 5 - 20 respectively, find
(b) Let S(n) denotes the Sum first terms of the A.P.
(i) Fing n such that S(n)=-192, (ii) Find the greatest value of A such that Sin)>-320..
9. Mary and John begin, war together
Mary's initial punteley is $2000
after anc
each year of completed service, honthly salary is increased by 120, John's initial monthly salary is $1500 and after each year of completed
service, hasnthly salary is increased by $200
(a) If Mary Works in the company for
n years
(i) what is her monthly salary
in the nth year. (ii) what is the total
receives.
amount
(b) If works in the company for
(i) is his monthly salary mth year.
the total amount he
.*. 1(7) = a + 6d, 1(2) = a + d
T(10) =
.. T(14)
a + 3 10
a13d5(ed)
4. The sum of all positive terms
- 1+3+5+7+..... (2n-1)
(Ans.)
(201) + (n - 1)(2))
S(2) - S(1)
2(2+3) - 4
= (1)(1 + 3)
= 4
The 2nd term
T(2) = [T(1) + T(2)] [19]
(b)
= 6n(120h + 3880)
(i) Monthly salary
= 1500+ (m- 1) (200)
200m+1300
(ii) the total amount
[Ans.!
- [2(3500) + (m - 11(200)] × 12
= 2
= 6
The sum of all negative terms
.. Common difference
= 6m[2800 + 200m)
roce
= -2-4-6-8.....-2n
= 2
(Ans.)
(c)
(c) (i) In how many
years will John S monthly salary be more than a
::
[2(-2) + (n = 1}{-2}}
8, (a) Let
the flast teen and common differgiof the A.P. be
the
-nin
have
.. Sum of the series to 20 terms
= n2 + (-n2 + n)
a and d respectively.
15.....(1)
[2a + (61)d] = -12
2a + 5d-4 .....(2)
from (1) and (2)
a = 3, 4 = -2
(Ans.)
(b)
(i)
S(n) = -192
12a + (n - 1)d] = -192
.. y 0 or y y
1080 80
(6 + (n - 1)(-2)) = -192
n{4 - n) = -192
(1) Let the number of years be x.
120x + 1880 < 200x + 1300
BOX > 580
x > 7.25
.*. in 8 years' time,
salary will be move Mary's monthly salary.
John's that
(i) Let the number of years taken be
y.
.*. 6y[120y+3880) < 6y(2800+200y)
y(2800+200y)-3(120y+3880) > 0 y(80y-1080) > 0
(ii) At the end of how many
years wil eshn received more ty
Solutions
1. (a) 1(6) = 2(6 - 1)2
- 2 x 25
= SC
(b) T(r) = 2(r - 1)2
PUBL
5. Suppose 4 and 8 meat in n days.
Distance travelled by A in n days
(Ans.)
(Ans.)
=
2(60) + (n − 1)(-4)]
= 26(31 - n)
2. (a) 5(10) = I{1}+B{2}+...........+7(10)
10(10+ 1)(20+ 1:
= 335
Distance travelled by B in a days
=
[2140) + (0
-
1)(8)]
(Ans.)
- 4n(n = 9)
(b) 5(2) = 1(1)+7 ( 2 ) + . . . . . + T{20}
{Ans.)
y is rejected
1080
y> 13.5
.. for 14 years, John's total
income is more than Mary's total income.
correction doesn't
affect
trial balance agreement and will- not be done
suspense account.
thought
(a)
the value of m
(b) the sum of first 20 terms
the A.F.
1990 中學會考預習專欄
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MILL & DALE PRESS
Accounting (13)
DJ
-
20(20 - *)(40 + 1)
: 2870
[Ans.)
Sucn imbalance warrants detailed examination 30% owing to the time limits and requirements to present the fina account.s to reveal immediate financial situation, it may not be possible for us to find out the errors immediately. Therefore, we have tc establish a nominal account to make the two sides of the trial balance agreeing with each other. This account is called 'Suspense Account!.
Example:-
2n131 n) + 4n(n+9)= 660
2 49n - 330 - 0
(55)(-6)= 0
After opening this suspense account, the trial balance will appear as:
Trial Balance as on 31st December 1988
Dr. Cr. ($) (3)
Assets (Totals) &
Expenses (Totals) Capital
Liabilities (Totals) &
Revenues (Totals) Suspense
5.000
4,800
600 400 5,400 5,400
n2 - 4 - 192 = 0
(n = 16) (n + 12) = 0
-12
... n 16
Example:-
مود
n - 16
(Ans.)
(+) This
As in the previous example, the suspense account is having a debit balance $400 and we find the following errors:-
(1) Sales is undercasted $200.
(2) Purchases is overcasted $100.
(3) Expenses is undercasted $700.
(4) The purchases of motor van $500 is completed not recorded in both the assets account and bank account.
Effect to Net Profit
If some of the above errors were detected after the profit and loss account was prepared, it is necessary for us to report the correct net profit The following table may serve
amount.
as a guide:-
↗或九龍尖沙咀麼爲三十號錦档糊中心
PENEK) ABSE
4
B窣䍲* 于影像中心(五——二九七九年)
然選擇。請即到香港金鐘道統一中心 08 饭非常大眾化,桐和定成爲精明消費者的必
Trial Balance as on 31st December 1988
Dr.
Cr
(5)
($)
Affecting
Trial
M. T. Kwok
Balance
Assets (Totals) &
Expenses (Totals) Capital
5,000
Liabilities (Totals) & Revenues (Totals)
4,800
600
5,000 5,400
In that case, the trial balance agreed and we can proceed to prepare our final accounts and the suspense account will be treated in Balance Sheet as:-
(a) Current Assets (if the suspense
account is having a debit balance)
(b) Current
Correction:-
Suspense
Errors Agreement
If the total
entries dre nat
arount
ΓΕ1,
5
of debit
with corresponding total amount of credit entries obviously it indicates the existence of error and the imbalance of trial balance may mean the following mistakes:-
(1) Incorrect calculation (1.0. wrong additions or deduction ir any account)
(2) Single entry (ie. having debit or
credit entry only!
(3) Figures
01 the two sides not identical (.e. entering $50 on debit side but $510 en credit side)
This trial balance is obviously imbalanced and the credit,
side is greater than the debit side and the difference is $400. So, we need to insert $400 onto the debit side and this insertment is done by opening an account Suspense' and making a single entry on debit sides of the account:-
31/12 Difererce
Suspense
The Journal entries are required to record the above correction:-
However, after preparing the final accounts, we should look into the account entries carefully to find out the source of errors and correction should be made through the suspense account. After such correcting entries have been made, the suspense account Suspenses should have a zero balance. (If there
is any errors which would not affect
correction is done US under the previous issue "Errors Not Affecting Trial Balance Agreement".)
31/12 Difference 31/12 Sales
31/12 Purchases
Liabilities (if the Suspense account is having a credit balance)
400/31/12 Expenses 2001 100)
700:
700
700
Opening Shock
Closing Stock
decrease
Errors found on The error is to Action needed Account:~
understata or to Increase or overstate Account:- Overstated (+) Increase (+) Understated (-) Decrease. (-)
Profit-
Overstated Understated
Decrease
Increase
Purchases
Overstated Understated
Increase
Sales
The Journal
Other Revante
Dr.
($) ($)
Expenses
Suspense
Sales
200
Cverstated Understated Overstated Understated Overstated Understated
Decrease Decrease
Increase
Decrease
increase
-Increase
Cecrease
200
100
Purchases
100
the trial balance
agreement, the
Expenses
Suspense
700
700
* Other accounts:-
offects only amount on Balance Sheet.
Motor Van (*)
500
Bank (*)
500
END -