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2019-Grade-10-Mathematics-Third-Term-Test-Paper-with-Answers-Western-province (1)

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wrh = 14 cm
 14  h
14. P(XY) =
23. (i)

C
AC fyda AB f¾Ldj, ,ïn iuÉfþolhg
1
wod, ia:dkh ,l=Kq lsÍug
1

50
B fldgi
01.
(i)
1
ka
(ii)
+
1
+
1
1

uq¿ uqo, = re' 16 000
1
165
(iii) fjk;a  30
wdrdê;hska  40
1
1
1
44
1
1

(ii) 4 .=Khla
2

(iii)
 14 + 28
2

(iii)
2

1

1
5
44 + 28
1
72 cm
1
f;dard .ekSu
(b) (i)
4
5
(ii)
1
f;dard fkd.ekSu
1
5

fkd
 14  14
2
1 260 m
re' 12
(ii) re' 20 000 
re' 2 400
(iii) re' 2 400  3 + 20 000
re' 7 200 + 20 000
re' 27 200
1
5
.e
1
(iii)
+
+
1
1
1+1
1
1

10

10

1
1

fkd
1

II m;%h
ñksia meh 8  12  5
ñksia meh 480
(ii) ñksia meh 10  12  3
ñksia meh 360
1
1
1
1
(iii)
1
01. (i)
1
1
1
1

re' 400  4
re' 1 600
1
1

(ii) re' 1 600 
1
re' 80 000
1
(b) (i)
fkd
4
5
1568 – 308
03. (a) (i)
1
4
5
4
5

.e
.e
1
5
(iv) 56  28 – r2
56  28 –

(ii)
 2r + 28
2

1
,CIHh u.ska ksrEmKhg

14 m


10

05. (a) (i)
1
1
1
1

1
(v) 110 

10
02. (i)
 360 fyda 360 – 3000
60
1
re' 400
1
 3600
(iv)
= re' 2 000
(iv)
1
(ii)
1
b;sß fldgi =
04. (i)
60
1
1
(iii)


10
(iv) b;sß jev m%udKh
Èkla ;=, wjika
l<hq;=jev m%udKh
= ñ'me' 120
1
ñ'me' 120
=
2
= ñ' me' 60
1
tla lïlrejl= Èklg
jevl< hq;= meh .Kk =
=
60
5
12



1
1

10
–3
1

(ii) (0 , –3)
2

(iii) y = x2 – 3
2

(iv) 0 ;a 1.8 ;a w;r fyda 0 < x < 1.8
2

02. (i)
(v) wkql%uKh
05.
=
=
=
1
1
wka;#LKavh =
–1
1
iólrKh" y =
x–1
1

10
03. (a) (i)
x = 2y
 (1) 1
x – 2y = 0
100x + 150y = 14 000  (2) 1

iqÿiq mßudKhla ,sùug
;srig 90 la ,l=Kq lsÍug
mßudKhg wkqj f.dvke.s,af,a Wi
fiùug
f.dvke.s,a, we£ug^mßudKhg&
40 fldaKh we£ug
30 fldaKh we£ug
lKqj iy fmd,j w;r 90 ,l=Kq
lsÍug
lKqj we| oelaùug
lKqfõ Wi uek ,sùug
lKqfõ ienE Wi fiùug
1
1
(500 – 550)
1
1
1
1
1
1
1
1
1
(ii) (1)  75,
75 x – 150y = 0
 (3) 1
x = 80
2
y = 40
2
06. (i)
(ii)
x
d
f
fxd
375
– 150
2
– 300
1
425
– 100
3
– 300
v – 2as = u
1
475
– 50
5
– 250
√
1

525
0
9
0
10
575
+ 50
4
+ 200
625
+ 100
4
+ 400
675
+ 150
3
+ 450
30
200
(b) v2 = u2 + 2as
2
2
= u
04. (a)
=
1
1
5
=
2x + 4
1
5–4
=
2x
1
=
x
1
(b) x (x + 4)
=
45
1
=
0
1
(x + 9) (x – 5) =
0
2
x = – 9 fyda x = 5
1
x2 + 4x – 45

x , d yd fd ;Srj,g

1
A

3
3
+
(
=
525 +
=
=
525 + 6.66
531.66
)
1
1
1

Èkl uOHjH wdodhu = re' 532
(iii) Èk 20 l wdodhu
os. yd m<, iDK úh fkdyel'
u,a md;a;sfha m<, = 5m
uOHkHh =

10

10
=
re' 532  20 1
=
re' 10 640
1
10 640 > 10 000
wdodhu m%udKj;a fõ'
1

10
- - www.mathspapers.info - -
07. (i)
Tn
=
T12 =
(ii) Sn
a + (n – 1) d
5 + 11  4
=
5 + 44
=
49
=
^
2ACB
1
=
^ (OA = OB)
OBA
1
^ + OAB
^ + OBA
^ = 180
AOB
1
^
AOB
^
180 – 2 OAB
1
1
^ = 180 – 2 OAB
^
túg 2 ACB
1
1
^
^
2 ACB
+ 2 OAB
1
^
10. AOB
1
^
OAB
1
(a + l)
S12 =
=
1
(5 + 49)
=
6  54
=
324

^
ADB
1

377
1

5 + (n – 1)  4
1
56 =
(n – 1)  4
1
14 =
n–1
15 =
n
(iii) S13 =
=
(iv) 61 =
324 + 53
1
=
=
^
OAB
=
180
(o;a;h)
^
^
2 ACB
+ 2 ADB
=
180
1
^ = ACB
^ jk ksid
;jo ADB
^ + 2 ACB
^
2 ACB
=
180
1
=
^
ACB

11. (i)
1
=
45
BD (o;a;h)
1
^
^
AEB
= DBE
(o;a;h)
08. (i)
(ii)
(iii)
(iv)
(v)
AB f¾Ldj we£ug
60 fldaKh ks¾udKhg
45 fldaKh ks¾udKhg
,ïnlh ks¾udKhg
AC ys ,ïn iuÉfþolh ks¾udKhg
O ,laIHh ,l=Kq lsÍug
jD;a;h we£ug
wrh uek ,sùug
1
1
1
2
2
1
1
1


10
1
BE (fmdÿ mdoh)
BE =



2 
 r
=
88
=
88
r
=
r
=
mßudj
=
^
(ii) túg ABE
=
^
BED fõ'
=
1
(iii) ;jo AB = DE (wkqrEm wx.)
1
AB =
 DE =
14 cm
 14  14  20
12 320 cm3
BC (o;a;h)
1
BC
1

10
12
1
anti log 0.3295
=
2.135

1
= 0.3295
18
.eyeKq
<uhs
1
3
=
50
1
= 1.8949 + 0.9694 – 2.5348
20
(ii) 12
w÷re lsÍug
(iii) AB’
(iv)
50
25
1

1
la fõ'
12. (i)
1

1
1

(b) lgx = lg 78.5 + lg 9.321 – lg 342.6
x
1
 AB // DE (taldka;r < )
 ABCD
r2h
=
1
 ABE   BDE  (md'flda'md')
;jo DE // BC fõ'
09. (a) 2r

10
ABE yd BDE  j,
AE =
10
1

10
r;=
nekshï
we¢ wh

fldKavh
3
f.d;d isá
<uhs
1
1
1




13 12 18
7
.eyeKq 4
<uhs

10
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