# Mathematics Solved Past Paper 9th Class 2013 Karachi Board

SECTION “A” (COMPULSORY) (M.C.Q.S)

1. Choose the correct answer for each from the given options:

(1) 32 out of 64 is:
* 16%
* 50%
* 20%
* 80%

(ii) 9x²- 12xy + 4y² = ———
* (3x – 2y)²
* (3x + 2y)²
* (3x + 2y)³
* (x – 2y)²

(iii) √16 x√81 =—–
* 36
* 9
* 4
* 72

(iv) If in a right angled triangle one angle is of 45°, the other angle will be of:
* 30°
* 60°
* 90°
* 45°

(v) If a matrix has all of its elements zero, it is called:
* singular matrix
* equal matrix
* Zero matrix
* square matrix

(vii) Arithmatic mean of -16, -70, 16, 70 is:
* 9
* 0
* -16
* -32

(viii) Union of Two non-collinear rays is called:
* Angle
* Opposite rays
* Straight line
* Rectangle

(ix) If the sum of two angles is 180°, it is called:
* Supplementary angle
* Complementary angle
* Acute angle
* Obtuse angle

(x) The straight line which touches the circle at one point is called:
* Secant
* Tangent
* Chord
* Diameter

(xi) (a + b)³= ———
* a³ + b³ + 3ab (a + b)
* a³ + b³ – 3ab (a – b)
* a²+ 2ab + b²
* a³ + b³ – 3ab

(xii) Area of Sector = π²
* 4
* 2
* 3
* 5

(xiii) The mode of 7, 8, 9, 10, 12, 9, 16 is: ———-
* 9
* 7
* 10
* 16

(xvii) If log2 64 = x, the value of x will be:
* 6
* 16
* 8
* 12

(xix) The decimal system 101²= ——-
* 11
* 7
* 3
* 5

(xx) {2,4,6, —–} is called the set of :
* Even numbers
* Odd numbers
* Natural numbers
* None of them

NOTE: Answer 10 questions from this section.

Question no.2

Question no.3

4. If A = {1 , 2} and B = {4 , 6 , 8}, Find A x B and B x A.

A x B ={1,2} x {4,6,8}
A x B = {(1, 4), (1, 6), (1, 8), (2, 4), (2, 6), (2, 8)}
B x A = {4, 6, 8} x {1, 2}
B x A = {(4, 1), (4, 2), (6, 1), (6, 2), (8, 1), (8, 2)}

Question no.5

6. Find the continued product of (x – y) (x + y) (x2 + y2) (x” + y4) with the help of formula.

(x – y) (x + y) (x² + y²) (x4 + y4)
Formula used
(a + b) (a – b) = a² _ b²
= (x² _ y²) (x² + y²) (x4 + 1)
= (x4 _ y4) (x4 + l)
= (X4)² _ (1)² = x8 -y8.

7. Find the value of x3 + y3 when x + y = 5 and xy = 6

Given x + y = 5
And xy = 6
Formula:
(a + b)³ = a³ + b³ + 3ab (a + b)
Applying this formula in above x + y = 5
(x + y)³= (5)³
x³ + y³ + 3xy (x + y) = 125
Again putting the values of x + y and xy in L.H.S.
x³ + y³ + 3 x 6 (5) = 125
x³ + 1 + 90 = 125
x³ + y³ = 125 – 90
x³ + y³ = 35.

Question no.8

Question no.9

Question no.10

Question no.11

Question no.12

13. Distribute Rs.9000 among Umar, Zaid and Bakr in the ratio of 2:3:4.

Given amount = Rs.9000
Ratio between Umer , Zaid and Bakr
2 : 3 : 4
Sum of terms of Ration = 2 + 3 + 4 = 9
Share of Umer = 9000 x 2/9 = 2000
Share of Zaid = 9000 x 3/9 = 3000
Share of Bakr = 9000 x 4/9 = 4000.

Question no.14

Question no.15

Question no.16

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

Question no.17

18. A man died leaving behind Rs.15,000. He had a loan of Rs.3000. Divide rest of the amount among his wife, 2 sons and 4 daughters in such a way that the wife gets 1/8th of the whole amount and each son gets twice of the each daughter.

Amount of Tarka = 15000
Amount of debt = 3000
No. of sons = 2
No. of daughters = 4
No. of widows = 1
Each son gets twice that of a daughter
SOLUTION:
Amount left after paying the debt = 15000 – 3000 = 12000
Share of widow = 1/8 x 12000 = 1500
Suppose each daughter gets Re.1
Then each son gets Rs.2
4 daughters gets = 4 x 1 = 4 Rs.
2 sons get = 2 x 2 = 4 Rs.
Total amount = 4 + 4 = 8
Share of 4 daughters from Tarka = 418 x 10500 = 5250
(Amount left = 12000 -1500 = 10500)
Each daughter share = 5250 ÷ 4 = 1312.50
Share of 2 sons = 4/8 x 10500 = 5250
Each son gets = 5250 ÷ 2 = 2625.

Question no.19(a)

19.(b) Find the cube of 2a – 3b.

(2a – 3b)³
Formula
(a – b) = a³ – b³ – 3ab (a – b)
(2a – 3b)³
= (2a)³ – (3b)³ – 3(2a) (3b) (2a – 3b)
= 8a³ – 27b³ -18ab (2a – 3b)
= 8a³ _ 27b³ – 36a²b + 54ab².

Question no.20(a)

Question no.20(b)

21.(a) A man saved Rs.20,000 in one year. The cost of the Jewellery of his wife is RS.15,000 .How much Zakat will he pay at the rate of 2.5%?

Saving = Rs.20000
Cost of Jewellery = Rs.15000
Total = 20000 + 15000 = 35000
Zakat Rate 2.5%
Amount of Zakat = 35000 x 2.5%
= 35000 x 2.5/100
= 875.

21.(b) Find the product of: (a + b) (a2 – ab + b2 ) (a – b) (a2 + ab + b2 )

(a + b) (a² – ab + b²) (a – b) (a² + ab + b²)
By formula
a³ + b³ = (a + b) (a²- ab + b²) and
a³ – b³= (a – b) (a² + ab + b²) and
Given sentence becomes
= (a³+ b³) (a³ – b³) = (a³)2 – (b³)2
= a6 – b6.

Posted on December 16, 2015 in 9th Class 2013 Karachi Board Past Papers