SCHOOL BASED ASSESSMENT 2023-24

First Term Math Grade 6

School Name: GHS FATEH JANG NO.1 (EMIS: 37120001)

ANSWER KEYS
Q. No.1 : b Q. No.2 : c Q. No.3 : b

Q. No.4 : b Q. No.5 : b Q. No.6 : b

Q. No.7 : a Q. No.8 : a Q. No.9 : d

Q. No.10 : b

ANSWERS / RUBRICS
Question No: 11 11‫وسالربمن‬

a ) Find first five multiples of 31. (5 marks)

)‫ربمن‬5(‫ےکےلہپاپچنااعضفولعممرکںی۔‬31
31 × 1 = 31 (1 mark)
31 × 1 = 31 )‫ربمن‬1(
31 × 2 = 62 (1 mark) 31 × 2 = 62 )‫ربمن‬1(
31 × 3 = 93 (1 mark) 31 × 3 = 93 )‫ربمن‬1(
31 × 4 = 124 (1 mark) 31 × 4 = 124 )‫ربمن‬1(
(1 mark)
)‫ربمن‬1(
31 × 5 = 155
31 × 5 = 155

Find L.C.M of 28, 42, 60 by division method. (5 marks)

)‫ربمن‬5(‫اکذوااعضفالقذبرہعیمیسقتولعممرکںی۔‬60‫اور‬42،28

L.C.M = Product of all prime factors

L.C.M = 2 × 2 × 3 × 5 × 7 (1 Mark)
‫امتمرفمدازجاےئرضیباکاحلصرضب=ذوااعضفالق‬-----------
L.C.M = 420 (2 Marks) ‫=ذوااعضفالق‬2 × 2 × 3 × 5 × 7)‫ربمن‬1(
‫=ذوااعضفالق‬420)‫ربمن‬2(
Question No: 12 12‫وسالربمن‬

a ) Arrange the absolute values of the following numbers in ascending order. (5 marks)
−15, 5, 1, 13, −9, −4, 2

)‫ربمن‬5(‫دےیےئگادعادیکقلطموتمیقںوکرتِبیتوعصدیںیمںیھکل۔‬
−15, 5, 1, 13, −9, −4, 2
 ‫روربک‬
Rubric
Given numbers are
‫دےیےئگادعاد‬
−15, 5, 1, 13, −9, −4, 2 −15, 5, 1, 13, −9, −4, 2

Absolute values of given numbers are ‫ادعادیکقلطمںیتمیق‬

15, 5, 1, 13, 9, 4, 2)‫ربمن‬2.5(
15, 5, 1, 13, 9, 4, 2 (2.5 marks)
Arrangment in ascending order
‫رتبیتوعصدی‬
1, 2, 4, 5, 9, 13, 15 (2.5 marks)
1, 2, 4, 5, 9, 13, 15)‫ربمن‬2.5(

Verify that: −20 × (13 + 27) = (−20 × 13) + (−20 × 27) (5 Marks)
)‫ربمن‬5(−20 × (13 + 27) = (−20 × 13) + (−20 × 27) ‫اثتبرکںی۔‬
:‫روربک‬
Rubric:
=
L. H. S −20 × (13 + 27)
L. H. S= −20 × (13 + 27)
= −20 × 40 (1 mark)
= −20 × 40 )‫ربمن‬1(
= −800 (1.5 marks) = −800 )‫ربمن‬1.5(

R. H. S = (−20 × 13) + (−20 × 27)

=
R. H. S (−20 × 13) + (−20 × 27)
= −260 − 540 (1 mark)
= −260 − 540 )‫ربمن‬1(
= −800 (1.5 marks)
= −800 )‫ربمن‬1.5(
L. H . S = R. H . S

L. H . S = R. H . S
Question No: 13 13‫وسالربمن‬

a ) Verify the associative law of multiplication for 12, − 14 and 22. (5 marks)
)‫ربمن‬5(‫ےکےئلاختیصالتزماحلبظرضبدصتقیرکںی۔‬22‫اور‬ ،
− 14 12

Rubric
‫روربک‬
According to associative law L.H.S‫اختیصالتزمےکاطمقب‬
L.H.S (12 × −14) × 19
(12 × −14) × 19
= −168 × 19

)‫ ربمن‬2.5(
= −168 × 19
= −3192
(2.5 marks)
R.H.S
= −3192

R.H.S
= 12 × (−14 × 19) = 12 × (−14 × 19)

= 12 × −266
= 12 × −266
= −3192 (2.5 marks)
= −3192 )‫ ربمن‬2.5(
L. H . S = R. H . S
L. H . S = R. H . S

Umar takes 3 1

2
hours to paint a 14m long wall. What is the rate of painting wall per hour? (5 marks)

)‫ربمن‬5(‫وٹنھگںںیمرگنرکاتےہ۔اسیکیفہٹنھگدویاررگنرکےنیکرشحایکےہ؟‬3 ‫رٹیمیبملدویاروک‬14‫رمع‬
1

‫ےٹنھگںیمرگنیکیئگدویار‬3 12 = 14 ‫رٹیم‬
wall painted in 3 hours= 14m
1

‫ہٹنھگںیمرگنیکیئگدویار‬1= ‫ربمن)رٹیم‬1(
2
1
wall painted in 1 hour= 14 ÷ 3 (1 mark) 1

2
14 ÷ 3
2

wall painted in 1 hour= 14 ÷ (1 mark) ‫ہٹنھگںیمرگنیکیئگدویار‬1= 14 ÷ 2 ‫ربمن)رٹیم‬1(

7 7
2

wall painted in 1 hour= 14 × (1 mark)

‫ہٹنھگںیمرگنیکیئگدویار‬1= 14 × 27 )‫ربمن‬1(
2

wall painted in 1 hour = 4m (2 marks)

‫ہٹنھگںیمرگنیکیئگدویار‬1= 4‫ربمن)رٹیم‬2(
A bus travels 400 km in 5 hours. Find the time taken by the bus to cover the distance of 1600 km. (5 marks)
)‫ربمن‬5(‫ولکرٹیماکافہلصےطرکےنےکےئلدراکروتقولعممرکںی۔‬1600‫ولکرٹیماکرفسرکیتےہ۔‬400‫ےٹنھگںیم‬5‫اکیسب‬
Time taken to travel 400 km = 5 hours ‫ولکرٹیماکافہلصےطرکےنےکےئلدراکروتق‬400=5‫ےٹنھگ‬
Time taken to travel 1 km = hours (1 mark) ‫ولکرٹیماکافہلصےطرکےنےکےئلدراکروتق‬1= 400 ‫ربمن) ےٹنھگ‬1(
5
5
400

Time taken to travel the distance of 1600 km = 5

‫ولکرٹیماکافہلصےطرکےنےکےئلدراکروتق‬1600= 400 ‫ربمن)ےٹنھگ‬1(

× 1600
400 5
× 1600
hours (1 mark)
= 5 × 4 (2 mark) =5 × 4)‫ربمن‬2(
= 20 hours (1 mark) =20‫ربمن)ےٹنھگ‬1(