Mathematics Solved Past Paper 9th Class 2012 Karachi Board

Mathematics Solved Past Paper 9th Class 2012 Karachi Board

Mathematics Solved Past Paper 9th Class 2012 Karachi Board

Mathematics Solved Past Paper 9th Class 2012 Karachi Board

Mathematics Solved Past Paper 9th Class 2012 Karachi Board

SECTION “B” (SHORT-ANSWER QUESTIONS)

NOTE: Answer 10 questions from this section.

2. If A = {2, 4, 6} and B = {x, y} find A x B.

ANSWER:

A = {2, 4, 6} B = {x, y}
A x B = {2, 4, 6} x {x, y}
A x B = {(2, x), (2, y), (4, x), (4, y), (6, x), (6, y)} Ans.

Question no.3

Question no.3

4. Divide 450 rupees among Ahmed, Bashir & Jamil in the ratio of 2, 3 and 4 respectively.

ANSWER:

Rate of the shares 2 : 3 : 4
Sum of the ratios 2 + 3 + 4 = 9
Amount to be distributed = 450
Share of Ahmed = 450 x 2/9 = 100
Share of Bashir = 450 x 3/9 = 150
Share of Jamil = 450 x 4/9 = 200.

Question no.5

Question no.5

6. Factorize the following:
(i) ax + ay – 4x – 4y (ii) X² – 10x + 21

ANSWER:

(i) ax + ay – 4x – 4y
= a (x + y) -4 (x + y)
= (x + y)(a – 4)

(ii) X² – 10x + 21
= X² – 7x – 3x + 21
= x (x – 7) -3 (x – 7)
= (x – 7) (x – 3)

7. The length of the arc of the sector of a circle is 42 meters and diameter of the circle is 60 meters. Find the area of the sector.

ANSWER:

Length of the arc = 42 metre
Diameter of circle = 60 metre
Area of the sector = ?
Formula:
Area of sector = ½ (length x Breadth) = ½ (length of sector x Radius of circle)
Radius = D/2 = 60/2 = 30 metre
:. area = ½ x 42 x 30 = 630 metre Ans.

Question no.8

Question no.8

Question no.9

Question no.9

11. The number of attendance of a student in 6 months is 36, 46, 48, 20, 47, 45 respectively. Find the median of attendance.

ANSWER:

Attendance is
20,36,45, 46,47,48
There is no number in between
Median will be
45+46
2
= 45.5 Ans.

12. Convert 11102 into decimal system.

ANSWER:

1110² = (1 X 2³) + (1 X 2²) + (1 X 2¹) + (0 x 2°)
= (1 x 8) + (1 x 4) + (1 x 2) + (0 x 1)
= 8 + 4 + 2 + 0
= 14
= 14¹° Ans.

Question no.13

Question no.13

Question no.14

Question no.14

Question no.15

Question no.15

16. Find the continued product with the help of formula: (2 – x)(2 + x)(4 + x²)

ANSWER:

(2 – x){2 + x)(4 + x²)
Let 2 = a
X=b
Now (a – b) (a + b) = a² – b²
= (2 – x) (2 + x) (4 + x2)
= (2)² – (x)² (4 + x²)
= (4 – x²) (4 + x²)
= (4)² _ (x²)²
= 16 – x4.

SECTION C (DETAILED-ANSWER QUESTIONS)

NOTE: Answer any 3 questions from this Section. Question No. 19 is compulsory.

Question no.17

Question no.17

Question no.17 (ii)

Question no.17 (ii)

18. A person dies leaving behind him a ‘Tarka’ worth Rs.30,000. Distribute the Tarka among 4 sons and 3 daughters. So that each Daughter gets half of what a son gets. Find the share of each when a debt of Rs.8,000 is also paid from Tarka.

ANSWER:

Amount of Tarka = 30,000
Amount of Debt = 8,000
Number of sons = 4
Number of daughter = 3
Each son gets twice that of a daughter.

SOLUTION:
Amount left after paying back debt = 30,000 – 8,000 = 22,000
Suppose each daughter gets Re.1
Suppose each son gets Rs.2
3 daughters get 3 x 1 = 3 Rs.
4 sons get 4 x 2 = 8 Rs.
Total amount = 3 + 8 = 11
Share of 3 daughters = 22,000 x 3111 = 6,000
So each daughter gets = 6.000/3 = Rs.2,000
Share of 4 sons = 22,000 x 8/11= 16,000
Share of each son = 16,000/4 = Rs.4,000
Hence,
Share of each daughter = Rs.2,000
Share of each son = Rs.4,000.

Question no.19

Question no.19

Question no.20

Question no.20

 

Question no.21

Question no.21

Posted on December 17, 2015 in 2012 Karachi Board Past Papers

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