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,/7 means" before that, do this! "
Answer all questions.
Electronic Calculators Must Not Be Used In This Paper. Omission Of Essential Working Will Result In Loss Of Marks.
1 7(Jpic: 1 Numbers
(a) Evaluate 12 + 8 + (9 - 5).
(b) Evaluate 0.018+0.06.
Thinking Process
[ I]
[ I]
(a) Evaluate the bracket first. Observe BODMAS rules. (b) You may change each decimal to fraction first.
Solution
(a) 12+8+(9-5) =12+8+(4)
=12+.§. 4 = 12+ = 14 Ans.
(b) 0.018+0.
= 16~0 + 1~ =--x-^18 1000 6
=10=.)^ 0'..)^ Ans.
2 Topic: 1 Numhers
Tasnim records the temperature, in °e, at 6 a.m. every day for 10 days.
-6 -3 0 -2 -1 -7 -5 2 -\ -
(a) Find the difference between the highest and the lowest temperatures. [] (b) Find the median temperature. [I]
Thinking Process (a) j7 Subtract the lowest temperature from the highest temperature. (b) Find the middle temperature j7 Write the temperatu res in increasing order.
Solution
(a) 2-(-7) = 2 + 7 = 9 °e Ans. (b) Writing the temperatures in increasi
-7 -6 -5 -3 -3 -2 -\
medIan temperature.^ = -2--3-
= -~ = -2.5 °e
3 Topic: 3 Inequalities
4
[ (^) t .IS^ gIven.^ t ath^ "4^3 < 11 < s·^7
(a) Write down a decimal value of n
this inequality. (b) Write down a fractional value of 1 this inequality.
Thinking Process
(a) To find a decimal value of n j
fractions as decimals. (b) j7 You may change your answer to a fraction.
Solution
(a) "4 < 11 < S
n = 0.80 Ans.
(b) 11 = 54 Ans.
Topic: 1a Everyday Mathematics
Here is part of a bus timetable.
0956 1026 1056 I 1003 10 33 1103 I 10 17 1047 11 17 1 1028 1058 1128 1 1043 II 13 1143 1
time. [1]
Thinking Process (a) Use second column. Subtract 10 26 from 11 13 j/ Express 1113 as 1073. (b) Find the length of time from City Hall to airport. Subtract this time and 2 hours from 1405.
Solution
(a) Consider the second column. 1113= 1073-1026= bus takes 47 minutes from the bus station to the airport ADS.
(b) Time oftlight = 1405 check-in time at the airport = 1405 - 0200 = 1205 Time taken from the City Hall to Airport = 1043 -1003 = 40 minutes 1205 - 0040 = 1165 - 0040 = 1125 latest bus that Chris can take trom the City Hall is at 1103 ADS.
5 1l'pic: 2 Indices and Standard Form
(a) Express 0.0000852 in standard form. [1]
(b) Calculate (3 x 10 5 ) + (6 x 10-^1 ) , giving your answer in standard form. [1]
Thinking Process (a) j/ Recall that standard form is A x 10 n^ , where 1s:A<10. (b) j/ To find (ax10m)+(bx10n) in standard form
j/ use --- axiOm^ = - (a)^ xi Om-n and express - a bx10n b b in standard form to reach the final expression.
= 0.5 x 107 = 5.0 X 106 ADS.
6 Topic: I J Symmetry
(a) Complete the description of the
The pattern has rotational symm ............... and ............... lines of (b) Shade in two more small squares to make a pattern with exactly 2 symmetry.
Thinking Process (a) Understand the definition of line rotational symmetry. (b) j/ Remember that line of symme of reflection.
Solution
(a) The pattern has rotational symme
...... .3 ........ and ...... 0 ......... lines o
(b)
= ~O = 5 Ans.
11 Topic: 20 Sets and Venn Diagram
(a) On the Venn diagram, shade the set P' n (Qu R).
"&
(b) A group of 40 children are asked what pets
they own.
[1]
Of these children, 13 own a cat, 5 own both a cat and a dog and 15 own neither a cat nor a dog. U sing a Venn diagram, or otherwise, find the number of children who own a dog, but not a cat. [2]
Thinking Process (a) Note that set P is empty (P '). Therefore shade set Q and set R which is outside of set P. (b) P Draw a Venn diagram by using the given information and use it to solve the question.
Solution
(a)
x+5+8+15= x+28= x= Number of children who ow = 12 Ans.
12 Topic: 21 Matrices
A cafe sells hot drinks. On Monday it sells 80 teas, 60 coffe chocolates. On Tuesday it sells 70 teas, 90 coffe chocolates.
A cup of tea costs $0.80, a cup of c
and a cup of hot chocolate costs $1. This information can be represented b
M and N below.
M =(80 60 40)
70 90 50 N=
(a) Work out MN. (b) Explain what the numbers in you represent.
Thinking Process (a) P To find MN P perform matrix (b) To understand what the numbers consider the intermediate steps in tion of the product MN.
Solution with TEACHER's
(a) MN = (80 60 40)(Oi 8 ]
= (80 x 0.8) + (60 x 1) + (40 x 1 (70 x 0.8) + (90 x 1) + (50 x 1
= (64 + 60+ 48) 56+90+
= (172) 206 Ans.
(b) First row represents total amou Monday while the second row re amount earned on Tuesday.
13 Topic: 22 Functions
f(x)=2-3x
Find
(a) f( -5),
(b) f-I(x).
Thinking Process
(a) j/ Replace x by -5.
(b) Let y = f(x). Make x the subject of formula.
Solution (a) f(-5) = 2-3(-5)
= 2 + 15 = 17 Ans.
(b) f(x)=2-3x
Let y = f(x)
=> y= 2-3x
3x= 2- y
x=--2-y
=> C 1 (y)=2- y
14 Topic: 1a Everyday Mathematics
[1 ]
[2]
A rectangular garden has length 35 metres and width
25 metres.
These distances are measured correct to the nearest
metre.
(a) Write down the upper bound of the length of
the garden. [1]
(b) Work out the lower bound of the perimeter of
the garden. [2]
Thinking Process (a) Divide 1 m by 2. Add the value to the length of garden.
(b) To find the lower bound of the perimeter j/ sub-
tract 0.5 m from length and width.
Solution
(a) Upper bound of the length = 35 + 0.
= 35.5 m
(b) Lower bound of the perimeter
= 2(L+ B)
= 2[(35 - 0.5) + (25 - 0.5)]
= 118 mAns.
15 Topic: 3 Inequalities
Y
5,·············,············,·············,·············f ,
4*·············+·············+············,··········· .., ;
L
3~···················~··············-+··········· , •...............~~......•..
+---4---~--~--~---+~x o (^2 )
(a) Find the gradient of the line L. [I]
(b) The shaded region on the diagram is defined by
three inequalities.
One of these is x + y S; 4.
Write down the other two inequalities. [2]
Thinking Process
(a) j/ Apply formula, Gradient = Y2 - Y
x2 - x 1
(b) j/ Locate the given inequality first and then
write the equations of the other two. Convert the equations into inequalities.
Solution
(a) Taking two points (0, 1) and (4,3) on the line.
gradient = ~ =b 2 1 =4"=2" Ans.
(b) y-intercept of line L = 1
equation of line L is: y = tx + 1
equation of second line is: x = I
required inequalities are:
y~tx+1 and x~1 Ans.
16 Topic: Ia Everyday Mathematics
(a) Dwayne buys a camera for $90.
He sells the camera for $126.
Calculate his percentage profit. [I]
(b) The price of a computer was $375.
In a sale, the price was reduced by 15%.
Calculate the reduction in the price of the
computer. [1]
(c) The exchange rate between euros and dollars is
€I = $1..
(i) Convert €I80 to dollars. [I]
(ii) Convert $500 to euros. [I]
(c)
-7 -6 -5 -4 -3 -2 -1 o 2
18 Topic; 13 Angles and Circle Properties
(a) Find the size of the interior angle of a regular octagon. [I] (b) A regular octagon, an equilateral triangle and a regular n-sided polygon fit together at a point.
(i) An interior angle of the regular n-sided polygon is 0°.
Find a. [I]
(ii) Find the value of n. [2]
Thinking Process
(a) j/ Apply, one interior angle = (n - 2). n (b) (i) Sum of angles at a point is equal to 360°.
Solution
(a) Number of sides in an octogan, n = 8
.. Intenor angle.^ = (8-2)180° 8
= \080° = 1,,-0 8 ~ .",ns.•
(b) (i) a O^ + 60° + 135° = 360° (LS sum at a point)
a O^ = 360° - 60° -135°
(ii)
= 165° Ans.
({° = -'---'----(n-2)xI80° n 1650= (11-2)xI80° 11 165°11 = 180°11- 360° 15°n = 360° 11 = 24 Ans.
19 Topic; 2 Indices and Stamfard Form
(a) Evaluate (i) ifij6, 1 (ii) 16 2 -16°.
(b) Simplify ( 3a~b4)-~
12ab
Thinking Process (a) (i) j/ Express 216 as 6' (ii) j/ Rewrite 16 as base of 4 and simplify.
(b) Apply rules of indices:
Solution
(a) (i) ifil
1 =(216) 1 =(6x6x6) 1 =(6 3 )3 =6 Ans.
(b)
1 (ii) 16 1 -16° 1 =(4~)1- =4-
= 3 Ans.
( r
3a 2 b - 12ab 4
=(4~3 r
=( 4t
3 r
16b , 6 a-
Ans.
[I]
[I]
[2]
The car accelerates uniformly from a speed of
v mls to a speed of 3v mls in 50 seconds.
It then continues at a constant speed.
Speed
(mls)
3v - - - - - - - -.,..------.,
v
O+---------r-------~--~ o 50 100
Time (I seconds)
(a) Find, in terms of v, the acceleration of the car
in the first 50 seconds. [1]
(b) The car travels 2500 metres during the 100
seconds. Find v. [2]
(c) Find the speed of the car, in kilometres per
hour, when 1 = 75. [1]
Thinking Process (a) Acceleration = gradient of speed-time graph. Calculate gradient of line from t= 0 to t= 50. (b) Total distance travelled in a speed-time graph
= area under speed-time graph.
(c) To find the speed in kmlh f/ first find the
speed in m/s. Note that 1 km = 1000 m, and
1 hour = 3600 seconds.
Solution
(a) Acceleration in first 50 seconds = 3~O v
2v 50
= 25 V^ ml^ s^2 Aos.
(b) Distance travelled during 100 second
= area of trapezium + area of rectangle
=> 2500=1(50)(v+3v)+(50X3V)
2500 = 100v + 150v
2500= 250v
v = 10 mls ADS.
(c) When 1 = 75 seconds,
speed of the car = 3v mls
= 3(10) mls
= 30 mls
= (30 x 3600) kmlh
= 108 km/h ADS.
replacement.
(a) Complete the tree diagram to show
outcomes and their probabilities.
First pen
TO
Second
black
red
(b) Find, as a fraction in its lowest te
probability that
(i) Luis takes two black pens,
(ii) Luis takes two different colour
Thinking Process (a) To complete the tree diagram f/ co number of pens at each stage of th (b) (i) Find p(black) x p(black)
(Ii) Find p(black, red) + p(red, blac
Solution
(a)
First pen Second p
3 black
TO
.... .!.9.... red
(a) On the grid, draw a frequency polygon to represent this information for the girls and another frequency polygon for the boys. [3] (b) Write down the modal group for the girls. [I] (c) Make a comment compar- ing the distribution of the times spent by the girls with the times spent by the boys. [I]
Thinking Process
0< t :s; 0.5 0.5 < t :s; 1 I < t :s; 1.5 1.5 < t :s; 2 2 < t :s; 2.5 2.5 < t :s; 3
o 5 8 6 o
o 0.5 1.5 2 2. Time (t hours)
(a) To draw a frequency polygon .p plot each frequency against the mid-value of the class interva (b) .p Look for the class with highest frequency. (c) For comparison, consider the modal groups for the two graphs or the range of times spent.
Solution
(a)
o 0.5 1.5 2 Time (I hours)
(b) 1< t ~ 1.5 An
(e) Comparing the groups, we see modal group f
1.5 < t ~ 2 ho
than the moda girls. Therefor longer time do homework than Alternatively: Most girls spe 0.75 hours to 2 their homework boys times are spread between to 3 hours.
angle B = (3y + xt angle C = (2y + 10)° angle D = (3x + 5)°
(a) By finding the sum of the angles in the
quadrilateral, show that 7y + 5x = 345. [I]
(b) Given that angle A = 90° then 2y + x = 90.
Solve the simultaneous equations to find x and y.
7y + 5x = 345 2y + x = 90 [3]
(c) Find the size of the smallest angle in the quad-
rilateral. [I]
Thinking Process (a) To find x f/ apply, sum of angles in a quadri- lateral = 3600 • (b) f/ Substitute the value of x from 2nd equation
into the 1st equation and solve for y. Substitute
the value of y back into the expression of x
and solve for x.
Solution
(a) Sum of angles in a quadrilateral = 360°
=> (2y+x)+(3y+x)+(2y+1O)+(3x+5)= => 2y+x+3y+x+2y+JO+3x+5= => 7 y + 5x + 15 = 360 => 7 y + 5x = 345 Shown.
(b) 7y+5x=345 ............. (i)
2y + x = 90 ............. (2)
From (2): x = 90 - 2y ............. (3)
substitute (3) into (I):
7y + 5(90 - 2y) = 345
7y+ 450-IOy = 345
-3y=-
y = 1~5 = 35 .)
Substitute y = 35 into (3),
x = 90- 2(35)
x = 20, y = 35 Ans.
(c) Smallest angle = angle D
= (3x+ 5)°
= 65° ADS.
aircraft.
(ii) Given that the aircraft is east of the line AB, find, by measuring, its bearing from C.
Thinking Process (a) Measure AB. Using the given scale, convert the distances into kilometres. (c) Measure the bearing using protractor. (d) (i) Locus of I: Construct ..I. bisector of AC. Locus of II: Draw a circle with centre at B. J' convert 90 km into cm to find the radius o Look for the points of intersection where circle from B and perpendicular bisector of AC m (ii) To find the bearing J' measure the acute angle and subtract it from 360°.
Solution
(a) By measurement, AB = 7 em
Given scale is : 1 em -- 20 km
7 em - 20x 7 = 140 km
:. actual distance between A and B = 140 km. ADS.
(b) Refer to drawing.
(c) Bearing of C from B = 103° ADS. (d) (i) Refer to drawing. (ii) From drawing, the possible posit
aircraft due east of AB is at poin
.. Bearing of P from C
= 2490 ADS.
(a) /=-4-
(i) Find r when c = 8 and d = - 4.
(ii) Express c in terms of d and (.
(b) Solve 17-5x:S;2x+3.
[I]
[2]
[2]
(c) Factorise 9 - 25x 2. [I]
(d) Factorise completely 8px + 6qy - 3qx -16 pY. [2]
(e) Solve 5x 2 + 6x -13 = O.
Give your answers correct to two decimal
places. [4]
Thinking Process (a) (i) f/ Substitute the given values into the equa- tion to find f. (ii) f/ Make c the subject of the formula. (b) Arrange the inequality such that the unknowns are all on one side of the inequality. (c) f/ Apply, a^2 - b^2 = (a + b)(a - b). (d) Factorise by grouping. (e) f/ Apply quadratic formula and solve for x.
Solution
(a) (i) 6c^
2 -d
/=-4-
(b)
(c)
(ii)
= 388 = 97 Ans.
6c 2 -d
/=-4-
4/= 6c 2 -d 6c 2 =4/ +d
c^2 =--- 4/+d
c=± ~4/6+d Ans.
17-5x::> 2x+
-5x - 2x ::> 3 - 17
-7x::>-
x;;::-^ -
x;;:: 2 Ans.
9-25x 2
= (3)2 _ (5X)
= (3 + 5x)(3 - 5x) Ans.
= (x-2y)(8p-3q) Ans.
(e) 5x 2 +6x-13=
by quadratic formula,
-6 ± J'( 6--:::)2-_-4(-5)-(--13-) X=-~~-----
-6 ± ../36 + 260
-6±..J2%
IO
~ X= -6^ +^ ..J2%^ or x=---- -6^ -..J2%
IO IO
= I.I205 = -2.
x = I.I2 or - 2.32 (2dp) Ans.
Topic: 1a
(a) Mariam works in a shop.
She earns $5.20 per hour.
She also earns a bonus of 15% of the value o
the items she sells in a week.
(i) In one week she works for 32 hours and
sells items with a value of £2450.
Calculate Mariam's total earnings for the
week. [2]
(ii) In another week, Mariam worked for 28
hours and earned a total of $409.60.
Calculate the value of the items she sold
that week. [3]
(b) (i) Jack opens a bank account paying simple
interest.
He pays in $800 and leaves it in the
account for 4 years.
At the end of 4 years he closes the
account and receives $920.
Calculate the percentage rate of simple
interest paid per year. [2]
(ii) Jack uses some of the $920 to pay for a
holiday and a computer.
He saves the remainder.
The money is divided between the holiday,
computer and savings in the ratio 4 : 5 : 7.
Calculate the amount he saves. [2]
Thinking Process (a) (i) f/ Find the basic earning. Find the bonus. Add them together. (ii) Find the basic earnings for 28 hours and subtract it from $409.60 to find bonus. This bonus is equal to 15% of Mariam's weekly total sale. Thus, form an equation and solve it to find total sales.
5 Topic: 1-1 (d) Area of trapezium CDEF = ~ x 40 x (65 + 80)
= 2900 cm^2 Ans.
6 Topic· 23
(a) The first nve terms of a sequence are
~~~~~:t:r~~p:,D 17, 11, 5, -1, -7.
,;; Find, in terms of n, an expression for the nth
The diagram shows a framework ABCD supporting a
shop sign.
The framework is fixed to a vertical wall AB with
CD horizontal.
AC = 64 cm and CD = 80 cm.
BAC =35°, BCA = 90° and ACD = 125°.
(a) Calculate AB. [2]
(b) Calculate AD. [3]
(e) Calculate Aix. [3]
(d) On the sign CDEF, FE is parallel to CD fU1d
is 40 cm below it. FE = 65 cm.
Calculate the area of the sign CDEF. [2J
Thinking Process
(a) j7 Apply, cosB = ::~.
(b) j7 Apply cosine rule: 8 2 = b^2 + c^2 - 2bccosA
(c) j7 Apply sine rule to find L.ADe.
(d) j7 Apply, area of trapezium = ~h(SUm of II sides)
Solution
(a) In!1 ABC, cos.»~^ -0^ =-^64
AB
AB=~ cos 35° = 78.129 "<' 78.1 cm. Ans. (b) In!1ACD, using cosine rule,
AD = )(64)" + (80)2 - 2(64)(80)cosI25°
AD = .JI 0496 + 5873.
= 127.94"<' 128 cm (3sf) Ans,
(e) In !1ACD, using sine rule,
sinADC sinACD
AC AD
sinADC sin 125° ~ 64 127. ~ sin 125° ~ sinADC =---x
AOC=24.19°"<'24.2° (ldp) Ans.
term of this sequence. L2]
(b) The nth term, S/1' of a different sequence is
found using the formula S/1 = 11" + 311.
(i) Work out the first four terms of this sequence. [2]
(ii) The nth term T, of another sequence is
found using tl~e formula Tn = 5n - 12. There are two values of n for which S/1 =6.
Form and solve an equation in n to find
these two values. [4]
Thinking Process (a) To find the nth term j7 note that the difference between consecutive terms is decreasing by 6.
(b) (i) j7 Substitute n = 1, 2, 3, 4 into the given
formula. (ii) Substitute the formulas Tn and Sn into the given equation and solve for n.
Solution
(a) 7'" = 23 - 611 Ans.
(b) (i) S" = /1 2 + 311 SI = (1)2 + 3( I) = 4
S" = (2)" + 3(2) = 10
S3 = (3)2 + 3(3) = 18
S4 = (4)2 +3(4) = 28
first four terms are: 4. 10, 18. 28 Ans.
(ii) ~= 6
~^112 +^^311 = 6
~ 11-3=0 or n - 24 = 0
~ 17=3 or 11 = 24
.. (^) 11=3 and (^) 2'1 Ans,
7 1(,p'C: 19
(a) The pie chart summarises the results of a local election.
Candidate D Candidate A
(i) Candidate B received 1600 votes. Work out the total number of people who voted in the election. [2] (ii) What fraction of the vote did candidate D receive? Give your answer in its lowest terms. [I] (iii) How many more votes than candidate A did candidate C receive? [2] (b) The table summarises the ages of the members of a film club.
Thinking Process (a) (i) Note that 1600 votes are repres 60°. Find the number of votes re by 360°. (ii) To find the fraction 17 Find the an represented by Candidate D. (iii) 17 Find the number of votes rec candidate A and C.
(b) (i) Use mean = If; 17 Compute the
(x) of each interval. (ii) To draw histogram 17 first find the frequency density for each range. (iii) Estimate by referring to the histogra
Solution
(a) (i) (^) 60° represents - 1600 votes
360° represents - I~goo x 360°
= 9600 votes total number of people who vote = 9600 Ans.
(ii) Angle representing candidate D
= 360° - 144° - 60° - 90° (Ls arou = 66°
fraction of votes that candidate D rec
66° II
= 3600 = 60 Ans.
(iii) Votes received b\J candidate A = ~ J 36 = 24 r;:;iiJ~ii~~)115:S: a < 20 20:s: a < 30 30:s: a < 40 40:s: a < 60 60:s: a < 80 Votes received by candidate C = .!.:! 12 36 45 33^36
(i) Calculate an estimate of the mean age of the members. [3] (ii) On the grid below, draw a histogram to represent this data. [3]
E~t==te" t-t
I
~ H-
f-I-
++,. l++t± , 10 20 30 40 50
Age (a years)
±
, i· 1 --.-
.- -.
I-r
=F
t
difference. 3840 - 2400 = 1440 .. candidate C receives 1440 more
than candidate A Ans.
. - -
-i--
,=±t1$ml+1t+++;+
70 80
Thinking Process
(a)
(b)
-) -+ -+ (i) PO=OO-OP
(ii) (^) if AB = (~) then IABI = ~ a^2 + b 2
(iii) To find equation of line PO j/ use y = mx + c where m is the gradient and c is the y-intercept. (iv) Let R be (x, y). Find the midpoint of PO and equate it to the midpoint of OR. -+ -+ -) (i) (a) (^) AB= OB-OA ~ -> -> (b) AC=AB-BC (c) j/ Apply similarity concept and express --> -> CD as a ratio of AB (ii) (a) To find the ratio j/ consider the ratio of the linear dimensions of two triangles. (b) Apply concept of area of similar
triangles. ~ = (Ji) A2 ~
Solution
-) ~ --t (a) (i) PQ = OQ - OP
(ii) I iQI = ~(4)" + (_5)" =.J16+ = J4i units Ans.
(iii) We have. P(4. 2) and Q(8, -3)
gradient of PO = --^ -3-2^ =--^5
eC1uatIOn.^01 'PO' IS:^ J' = --^5 x + c
using ~ P(4.2). 2=_1(4)+(' 4 => c=
Y = --x+^5 7 Ans.
(iv) Let the coordinates of R be (.Y. )')
1111 'J POlllt.^0 t'PR.^ = (4+2'^ x^ -2-'2+^ l')
since Q is the miJpoint of PR.
=> (4;.. 2;Y)=(8. -3)
=> 4+.Y=^
2+ J' ~
2 -2-'^ =-.J
4+.Y = 16 2+y=-
.Y = 12 y=-
.. R is (12. 8) Ans.
---> ---> (b) (i) (a) AB=OB-OA = b-a Ans. -, (b) AC = AB+ BC = b-a+4a-b = 3a Ans.
(c) OC = OA + I1C
= a +3a = 4a OA I
OC 4
since />;OAB and />;OCD are similar OB AB OA -=-=-=-
OD CD OC 4
=> CD=4AB
= 4(b - a) Ans.
(ii) (a) perimeter of />;OI1B: perimeter of />;OCD 1:4 Ans.
9 1 ()pic: II!
(b) area of />;OI1B = (.!.)
area of />;OCD 4 16
therelore. area of />;OAB: area of trapezium ABDC I: 15 Ans.
IVolumeof acone=i1frZfll
I Curved surface area of acone=1frl]
The diagram shows a solid cone of height 15 cm and base radius 6 cm. (a) Show that the slant height of the cone is
16.2 cm, correct to one decimal place. [I]
(b) Calculate the total surface area of the cone. [3]
(c) Calculate the volume of the cone. (d) The cone is made from wood.
The mass of 1 m 3 of the wood is 560 kg.
[2]
Calculate the mass of the cone in grams. [2] (e) Another cone is made of the same material and is geometrically similar to the first. The mass of the second cone is double the mass of the first. (i) Calculate the height of the second cone. [2] (ii) Calculate the total surface area of the second cone. [2]
Thinking Process (a) f/ use Pythagoras' theorem. (b) Total surface area = curved surface area + area of circular base. (d) To find the mass f/ Express 1 m^3 into cm 3. Convert 560 kg to grams. (e) (i) Apply rule of similar figures:
mass A = (length A) mass B length B (ii) f/ Ratio of area of similar figures = ratio of squares of corresponding lengths.
Solution
(a) Using Pythagoras theorem.
slant height = J6" + 15 2
= .J26I
= 16.155 '" 16.2 cm Shown.
(b) Total surface area = lfr" + lfrl
= If(6)2 + If(6)(16.155)
= 132.93lf = 417.61", 418 cm" Ans.
(c) Volumeofcone=J.. lfr 2/
=J.. lf (6)2(15)
= 180lf = 565.487 '" 565 cm 3 (3sf) Ans.
(d) Volume of cone in m 3 = 565. 1000000 = 5.65487 X 10-^4 111
1m 3 --560kg 5.65487 x 10-4 111 3 -- 5.65487 x 10-^4 x 560 =0.3167 kg mass of cone in grams = 0.3167 x 1000 =316.7 g",317 g Ans.
(e) (i) mass^ of^ I^ st cone^ (height^ of^ I^ st cone^ ) mass of 2nd cone height of 2nd cone
I ( 15 )
"2 = height of 2nd cone
3fT 15 ~"2 = height of 2nd cone 15 0.7937 = ------ height of 2nd cone
IlClg. I lt^ of'2 nd cone = ---^15 ~ 0. = 18.899", 18.9 cm Ans.
(ii) area of 2nd cone
area of I st cone
( height^ of^ 2nd^ cone) height of 1st cone
e8;~99J
area of 2nd cone
area of 2nd cone = (^ ~18899)2 x 417.
= 662.93 '" 663 cm" Ans.
10 TopIC.· 7
Adil wants to fence off some land as an enclosure for his chickens. The enclosure will be a rectangle with an area of 50 m2.
x
(a) The enclosure is x m long.
Show that the total length of fencing, L m, required for the enclosure is given by 100 L=2x+-. x
[2]
(b) The table below shows some values of x and
the corresponding values of L, correct to one
decimal place where appropriate, for 100 L=2."+-. X
ll:.§ 54 .'J"^ 28.7^ 28.5^30 32.3^ 35.1^ 38.
Complete the table. (^) [2] (c) On the grid opposite draw a horizontal x-axis for 0::; x ::; 20 using a scale of I cm to represent 2 m and a vertical
L-axis for 0::; L ::; 60 using a scale of 2 em to
represent 10m. On the grid, plot the points given in the table and join them with a smooth curve. [3]
