SCHOOL-BASED ASSESSMENT (SBA) - 2025
END-OF-YEAR ASSESSMENT
SUBJECT: MATHEMATICS FINAL TERM
GRADE-8
[Paper A: 48 Marks, Paper B: 52 Marks, Total: 100 Marks], Time = 3 hours
School Name: GGHS 21/2-L (EMIS: 39330583)
ANSWER KEYS
Q. No.1 : a Q. No.2 : a Q. No.3 : b
Q. No.4 : a Q. No.5 : c Q. No.6 : c
Q. No.7 : d Q. No.8 : b Q. No.9 : c
Q. No.10 : c Q. No.11 : b Q. No.12 : b
Q. No.13 : a Q. No.14 : a Q. No.15 : d
Q. No.16 : c Q. No.17 : a Q. No.18 : d
Q. No.19 : a Q. No.20 : c Q. No.21 : c
Q. No.22 : a Q. No.23 : a Q. No.24 : b
Q. No.25 : a Q. No.26 : d Q. No.27 : a
Q. No.28 : c Q. No.29 : c Q. No.30 : d
Q. No.31 : c Q. No.32 : a
ANSWERS / RUBRICS
Question No: 33
33وسالربمن
a ) If one packet contains 5
1
4
liters of oil, then find the quantity of oil in 5 such packets.(5 marks)
)ربمن5(وٹکیپںںیمانتکلیتوہاگ؟5رٹللیتوہوتاےسی5 ارگاکیٹکیپںیم
1
رٹل=اکیٹکیپںیملیت5 1
Oil in 1 packet = 5 l 1 4
وٹکیپںںیملیت5=5 ×5)ربمن2(
4
1
Oil in 5 packets = 5 ×5 ( 2 marks)
1
4
4
= ×5 (1 mark )
21
4
= ×5)ربمن1(
21
4
= 105
4
(1 mark) = )ربمن1(
105
= 26 l (1mark )
1 4
4
26 ربمن)رٹل1(
1
4
Find Square root of 3364 by prime factorization.(5 marks)
)ربمن5(اکذجرولعممرکںی۔3364رفمدزجتیےکذرےعی
√ 3364 =√2 × 2 × 29 × 29)ربمن1(
√ 3364 =√2 × 2 × 29 × 29 (1 mark)
=√2 × 29 2
(1 mark)
2 =√2 × 29 )ربمن1(
2 2
=√(2 × 29) (1 mark)
2
=√(2 × 29) )ربمن1( 2
=2 × 29 (1 mark) =2 × 29)ربمن1(
= 58 (1 mark)
=58)ربمن1(
Question No: 34
34وسالربمن
a ) If A = {a, b, c, d}, B = {a, e, i}andC = {e, c, g, h}, then verify associative law of intersection. (5 Marks)
) ربمن5( وہوتاقتعطےکاقوننالتزمیکڑپاتلرکںی۔C = {e, c, g, h}اورB = {a, e, i}, A = {a, b, c, d}ارگ
A = {a, b, c, d} , B = {a, e, i} , C = {e, c, g,h}
A∩(B∩C)=(A∩B)∩C)ربمن1(
A = {a, b, c, d} , B = {a, e, i} , C = {e, c, g,h} L.H.S= A∩(B∩C)
A (B C)=(A B) C (1 mark)
∩ ∩ ∩ ∩
L.H.S= A (B C)
∩ ∩ =)}e,c,g,h{∩}a,e,i{(∩}a,b,c,d{
={a,b,c,d} ({a,e,i} {e,c,g,h})
∩ ∩
=}e{∩}a,b,c,d{)ربمن1(
={a,b,c,d} {e} (1 mark)
∩
={ } (1 mark) =} { )ربمن1(
R.H.S= (A B) C
∩ ∩
R.H.S= (A∩B)∩C
=({a,b,c,d} {a,e,i}) {e,c,g,h}
∩ ∩
={a} {e,c,g,h} (1 mark)
∩
=}e,c,g,h{∩)}a,e,i{∩}a,b,c,d{(
={ } (1 mark) =}e,c,g,h{∩}a{)ربمن1(
Hence proved that
A (B C)=(A B) C
∩ ∩ ∩ ∩
=} {)ربمن1(
سپاثتبوہا
A∩(B∩C)=(A∩B)∩C
Feroz bought a car for Rs 75000 and sold it for Rs 100000. Find his profit percentage. (5 Marks)
)ربمن5(روےپںیمرفوتخرکدی۔اساکعفنیفدصولعممرکںی۔100000روےپںیمرخدییاور75000ریفوزےناکیاگڑی
=تمیقرخدی75, 000
Cost price (C.P)= 75000 =تمیقرفوتخ100, 000
Sale Price(S.P)= 100000
Profit=S.P - C.P تمیقرفوتخ=عفن-تمیقرخدی
Profit = 100000 − 75000 = 25000 (2 Marks) =عفن100, 000 − 75, 000= 25, 000 Rs)ربمن2(
Profit %= prof it
× 100
Profit %=
C.P
× 100 = 33. 33% (3 Marks)
25000 عفن%= prof it
× 100
C.P
75000
عفن%= × 100 = 33. 33ربمن) دصیف3(
25000
75000
Question No: 35
35وسالربمن
a ) Find the general term of arithmetic sequence 7,11,15,19,........ . ( 5 marks )
)ربمن5(یکرنجلرمقولعممرکںی۔7,11,15,19,........یعمجےلسلس
7,11,15,19,........,
7,11,15,19,........ a1 = 7, d = 11 − 7 = 4)ربمن1(
(1 mark )
مہاجےتنںیہہک
a1 = 7, d = 11 − 7 = 4
We know that
an = a1 + (n − 1)d (1 mark ) a = a + (n − 1)d)ربمن1(
n 1
an = 7 + (n − 1)(4) (1 mark ) a = 7 + (n − 1)(4))ربمن1(
an = 7 + 4n − 4 ( 1mark ) n
an = 3 + 4n (1 mark ) a = 7 + 4n − 4)ربمن1(
n
a = 3 + 4n)ربمن1(
n
Evaluate (48)
2
by using formula. (5 marks)
)ربمن5((وکافرومالیکدمدےسلحرکںی۔48) 2
2
(48)
2
(48)
= (50 − 2)
2
(1 mark) )ربمن1(
= (50 − 2)
2
(2 marks) − 2(50)(2) + (2) )ربمن2(
2 2 2 2
= (50) − 2(50)(2) + (2) = (50)
= 2500 − 200 + 4 (1 mark) = 2500 − 200 + 4 )ربمن1(
= 2304 (1 mark)
= 2304)ربمن1(
Question No: 36
36وسالربمن
a ) Find solution of 3x + 5y =2 , x+ 2y = 4 with the help of substitution method. (5 marks)
)ربمن5(- اکلحولعممرکںی3x+ 5y =2،x +2y = 4تمیقدرجرکےنےکرطہقییکدمدےس
3x+ 5y = 2 .................. (i)
x+ 2y = 4 .................... (ii)
3x+ 5y = 2 .................. (i)
x+ 2y = 4 .................... (ii) ) ےسii(
from (ii) x = 4 - 2y.......... (A))ربمن1.5(
x = 4 - 2y.......... (A) (1.5 marks)
Put the value of x in (i) )ںیمدرجرکےنےسi( یکتمیقx
3(4- 2y) + 5y = 2 3(4 − 2y) + 5y = 2
12 - 6y + 5y = 2
12 − 6y + 5y = 2
y = 10 (1.5 marks)
put the vlaue of y =10 in (A) y = 10)ربمن1.5(
x = 4 - 2(10) ںیمدرجرکےنےسA یکتمیقy=10
x = 4- 20
x = -16 (2 marks) x = 4 − 2(10)
x = -16 , y = 10 x = 4 − 20
x = −16 )ربمن2(
x = -16 , y = 10
Find the volume of a hemisphere whose radius is 10 cm. (5 marks)
)ربمن5(یٹنیسرٹیمےہ۔10اکیفصنرکہاکمجحولعممرکںیسجاکرداس
Radius of hemisphere = 10 =فصنرکہاکرداس10یٹنیسرٹیم
Volume of hemisphere= πr (1 mark) 2
3
3
=فصنرکہاکمجحπr )ربمن1(
2
3
3
=
2
(
3
22
)(10)
7
(1 mark)3
=
2
( )(10) )ربمن1(
22 3
3 7
× 1000 (1 mark)
44
)ربمن1(
=
21 44
= × 1000
=
44000
(1 mark) 21
)ربمن1(
21
(1 mark)
44000
= 2095. 23cm
3 =
21
= 2095. 23ربمن)بعکمیٹنیسرٹیم1(
Question No: 37
37وسالربمن
¯
LM N mLM = 4cm m∠L = 60
o
a ) Construct a triangle
mLN = 5. 5cm with the help of compass and ruler when , and . (7 marks)
o
¯ )ربمن7(وہ۔mLN
انبںیئہکبجLM N امیپےناوررپاکریکدمدےسثلثم
= 5. 5cm اورm∠L = 60 ,mLM = 4cm
¯ )ربمن1(ےنچنیھکرپmLM = 4cm
On drawing
¯
mLM = 4cm (1 Mark)
On making angle of at point L. (2 Mark) 60
0 )ربمن2(اکزاوہیانبےنرپ60 رپL 0
On drawing an arc of 5.5cm at point L. (2 Mark) )ربمن2(یٹنیسرٹیمیکوقسانبےنرپ5.5 رپLہطقن
On correct completion and labeling. (2 Marks)
)ربمن2(درتسلیمکتاورلبیلرکےنرپ
A fair dice is rolled. Khalid wins the game if the result is less than three. What will be the probability of winning the game for Khalid? (5
Marks)
)ربمن5(اکیڈاسئامھگایایگےہ۔اخدللیھکتیجاجاتےہارگہجیتننیتےسمکوہ۔اخدلےکےیللیھکےنتیجاکایکااکمنوہاگ؟
=لپمیسسیپسS = {1, 2, 3, 4, 5, 6})ربمن1(
Sample Space = S = (1 mark)
n(S) = 6)ربمن1(
{1, 2, 3, 4, 5, 6}
n(S) = 6 (1 mark)
Let, A = {1, 2}less than 3 (1 mark) A = {1, 2}ےسوھچاٹ3)ربمن1(
n(A) = 2 (1 Mark) n(A) = 2)ربمن1(
P(A) = n(A)
n(S) n(A)
P (A) =
= =
2
6
1
3
(1 mark) n(S)
=
2
6
=
1
3
)ربمن1(
