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
vi

(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

(xiv)

(xv)

(xvi)

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

(xviii)

(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

SECTION “B” (SHORT-ANSWER QUESTIONS)

NOTE: Answer 10 questions from this section.

Question no.2

Question no.2

Question no.3

Question no.3

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

ANSWER:

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

Question no.5

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

ANSWER:

(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

ANSWER:

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.8

Question no.9

Question no.9

Question no.10

Question no.10

Question no.11

Question no.11

Question no.12

Question no.12

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

ANSWER:

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.14

Question no.15

Question no.15

Question no.16

Question no.16

SECTION C (DETAILED-ANSWER QUESTIONS)

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

Question no.17

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.

ANSWER:

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)

Question no.19(a)

19.(b) Find the cube of 2a – 3b.

ANSWER:

(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(a)

Question no.20(b)

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%?

ANSWER:

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 )

ANSWER:

(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

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