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1 0 . K s ; h - 1
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1 0 . K s ; h - 1
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1 0 . K s ; h - 1
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1 0 . K s ; h - 1
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1 0 . K s ; h - 1
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1 0 . K s ; h - 1
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පිළිතුරු පත්රය 10 ශ්රේණිය 2017 අවසාන වාර පරීක්ෂණය
ගණිතය 11
01. වසරකට වරිපනම් බද්ද =රු100000×15
100= රු15000 (ල. 01)
කුලියට දීශ්ෙන් ලැශ්බන මුදල = රු 7000×12 = රු 84000 (ල. 01)
නඩත්තතුව + වරිපනම් බද්ද = රු 15000 +රු 29000 = රු 44000 (ල. 01)
ඉතුරු මුදල =රු84000-44000 (ල. 01)
= රු 40000
වියදම් සඳහා ලබාගත්ත මුදල =රු 40000× 25
100 (ල. 01)
=රු 10000 (ල. 01)
බැැංකුශ්ේ තැන්පත්ත කළ මුදල = රු 30000
වසර 3 ශ්පාළිය = රු40800 - රු 30000
= රු 10800 (ල. 01)
වාර්ෂික ශ්පාළී අනුපාතය = 3600
30000× 100% (ල. 02)
= 12% (ල. 01)
02. a)
i. 𝑥 = 0 දී 𝑥 = 2 දී
𝑦 = 2 × 02 − 1 𝑦 = 2 × 22 − 1 (ල. 01)
𝑦 = 0 − 1 𝑦 = 8 − 1 (ල. 01)
𝑦 = −1 𝑌 = 7
ii. අක්ෂ ලකුණු කිරීෙ (ල. 01)
ලක්ෂය ලකුණු කිරීෙ (ල. 01)
ප්රසේාාරය ඳීමෙ ුමෙට බවට (ල. 01)
iii. අවෙ අගය =(-1) (ල. 01)
iv. 𝑦 = 2𝑥2 + 3 (ල. 02)
b) i. 𝑥 = .7 ∓ .1 𝑥 = −.7 ∓ .1 (ල. 02)
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03.
a) 𝑥
(𝑥−3)+
3
(3−𝑥)
𝑥
𝑥−3−
3
(𝑥−3) (ල. 01)
𝑥−3
𝑥−3= 1 (ල. 01)
b) i. 7𝑥 + 5𝑦 = 12900 (ල. 01)
3𝑥 + 10𝑦 = 12600 (ල. 01)
ii. ×2 14𝑥 + 10𝑦 = 25800 (ල. 01)
3𝑥 + 10𝑦 = 12600
- 11𝑥 = 13200
𝑥 =13200
11 (ල. 01)
𝑥 = 1200
𝑥 = 1200 ට ආශ්ද්ශශ්යන්
y = 900 (ල. 02)
iii. පිරිමි වැටුප =රු 1200
කාන්තා වැටුප =රු 900
වැටුප අතර ශ්වනස රු1200 - රු 900 = රු 300 (ල. 02)
04. a) 𝑙(𝑘2𝑙 − 𝑚𝑛) − 1(𝑘2𝑙 − 𝑚𝑛) (ල. 02)
(𝑘2𝑙 − 𝑚𝑛)(𝑙 − 1) (ල. 01)
b) 𝑥2 + 4𝑥 = 12 (ල. 01)
𝑥2 + 4𝑥 − 12 = 0 (ල. 01)
(𝑥 + 6)(𝑥 − 2) = 0 (ල. 01)
𝑥 = −6 ශ්හෝ 𝑥 = 2 ශ්ේ (ල. 02)
පළල =2m දිග =6m (ල. 02)
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05.
i. 58 – 62 පන්ිශ්ේය (ල. 01)
ii.
පන්ි
ප්රාන්තරය
ෙධ්ය
අගය සැංඛ්යාතය
(𝑓) අපගෙනය
(𝑑)
(𝑓𝑑)
50 – 54 52 6 -8 -48
54 – 58 56 15 -4 -60
58 – 62 60 16 0
62 – 66 64 13 4 52
66 – 70 68 7 8 56
70 – 74 72 3 12 36
∑(𝑓) = 60 ∑(𝑓𝑑)
= 36
නිවැරදි අගය වගුව (ල.03)
සැබෑ ෙධ්යනය = 𝐴 +∑(𝑓𝑑)
∑(𝑓) (ල. 01)
= 60 +36
60 (ල. 01)
= 60 + 0.6
= 60.6 Kg (ල. 01)
iii. 60.6 ×20 = 1212 Kg (ල. 01)
1212 Kg > 1000 Kg (ල. 02)
∴ යා ශ්නාහැක
06. a) වැහි පීල්ශ්ල් පරිොව =1
2×
22
7× 7 × 7 × 500 (ල. 01)
= 38500cm3 (ල. 01)
ඝනකශ්ේ නගින උස h නම්
500 × h = 38500 (ල. 02)
h = 38500
500
h = 77cm (ල. 01)
log 𝑥 = 1.4346 + 2.1306 − 1.6928 (ල. 02)
log 𝑥 = 3.5652 − 1.6928 (ල. 01)
= 1.8724 (ල. 01)
𝑥 = 𝑎𝑛𝑡𝑖𝑙𝑜𝑔 1.8728
𝑥 = 74.61 (ල. 01)
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07. a)
𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇𝑛 = 15 + 9 × 5
= 15 + 45
𝑇𝑛 = 60 (ල. 02)
𝑆𝑛 =11
2{2𝑎 + (𝑛 − 1)𝑑}
𝑆25 =25
2{2 × 15 + 24 × 5}
𝑆25 =25
2(30 + 120)
= 25× 75
𝑆25 = 1875
1875
08. ඉහත නිර්ෂොණය සදහා ලකුණු ලබා දීශ්ම්දී ප්රශේණ පශ්ත්රේ ශ්කාටසේ සදහා ශ්වන් කල ලකුණු
ප්රොණ නිර්ෂොණශ්ේ නිවැරදි බව හා ුමෙට බව සලකා ලබා ශ්දන්න.
09. i. සේනානය කළ ළමුන් (ල. 02)
(ල. 03)
ii. 45 – 10 = 35 (සම්ූර්ෂණ ප්රකාශයට) (ල. 01)
iii. 𝑥 + 45 + 30 = 80
𝑥 + 75 = 80
𝑥 = 80 − 75
𝑥 = 5 (ල. 02)
b)
(ල. 02)
35 40
30 10
45 5
80
පිරිමි ළමුන්
සේනානය කළ ළමුන්
10. a)
i. 5cm 5000cm ii 1cm 10m
1cm 1000cm 25cm 25×10
1 : 1000 (ල. 02) 250m (ල. 01)
b) i. නිවැරදි පරිොන රූපයට (ල.03)
ii. නිවැරදි අවශ්රෝහණ ශ්කෝණය ලකුණු කර අගය ෙැන ලිවීෙ (ල 01)
iii. නිවැරදි 𝑥 ලක්ෂය ලකුණු කිරීෙ (ල. 01 )
𝑥 සිට 𝑜 ඳි සැබෑ දුර ගණනය (ල.02 )
10
30
40
සේනානය
කළ
පිරිමි
ළෙයි
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11.
a) නිවැරදි ප්රශ්ම්ය සාධ්නය සඳහා (ල. 05)
b)
(ල. 05)
𝑂𝑇 = √102 − 62
𝑂𝑇 = 8𝑐𝑚
𝑂𝑈 = √102 − 82
𝑂𝑈 = 6𝑐𝑚
𝑇𝑢 = 8 − 6
=2cm
x
Q
B
P
A
O
T
U
12.
i. 𝐸�̂�𝐵 =540°
5= 108° ( සවිධි ෂඩාස්රයක් නිසා) ල01
ii. 𝐴�̂�𝐵 =180°−108°
2 (ABE සෙද්විපාද ත්රිශ්කෝණයක් නිසා) ල 02
𝐴�̂�𝐵 = 36°
iii. 𝐵�̂�𝐷 = 108° − 36°
= 72° ල 01
iv. 𝐸�̂�𝐶 = 𝐴�̂�𝐹 + 𝐸�̂�𝐹
= 36° + 72°
= 108° ල02
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v. EFCD චතුරස්රශ්ේ
𝐸�̂�𝐶 = 108°
𝐸�̂�𝐶 = 108°
∴ 𝐸𝐹𝐶𝐷 සොන්තරාස්රයකි (සම්මුඛ් ශ්කෝණ සොන නිසා) ල 02
තවද ED = DC
DC =EF
ED =FC
∴ ශ්ෙය සොන්තරාස්රය ශ්රාම්බසයකි ල 02
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