9th Class Maths Paper 2026 fbise Solution complete
Section B (36) marks
9th Class Maths Paper 2026 fbise Solution complete Short Questions
In this post we will see 9th Class Maths Paper 2026 fbise Solution complete. you’ll see written material, notes and video also
Each question carries 4 marks. There are total 9 questions.
Q2
Below are the step-by-step short question answers for this 9th Class Maths Paper 2026 fbise
(i)
$$\sqrt[4]{\frac{a^3}{b^3}\;}.\;\;\sqrt[4]{\frac{b^3}{c^3}\;}.\;\;\sqrt[4]{\frac{c^3}{a^3}}$$
Solution
$$=\;\sqrt[4]{\frac{a^3b^3c^3}{b^3c^3a^3}\;}\\=\;\sqrt[1]1\;=\;1$$
OR
If U= {1,2,3,4,5}
A= {2,5,6}
B={1,2,3}
Use Venn diagram to show that
$$\;(A\cup B)’\;=\;A’\cap B’$$
Solution
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(ii)
A Cyclist travels 10km due south and then 25 km due west. what simple bearing should the cyclist take to return directly.
Solution
$$Tan\theta=\frac{10}{24}\\\theta=22.62$$
So angle from North line = 90-22.6=67.4
OR
In the figure BC is parallel to DE. Find
a) Ratio DE:BC
b) Ratio Area of triangle ADE : Area of triangle ABC
c) Find the area of triangle ADE if area of triangle ABC is 256 centimetre square
d) Area of trapezium DBCE
Solution
Triangle ADE is similar to triangle ABC
In the diagram
AB=AD+DB
=5+3= 8cm
a) Due to similarity the ratio of their corresponding sides is equal
DE:BC=AD:AB=5:8
b) Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides
$$\frac{Area\;of\;\bigtriangleup ADE}{Area\;of\;\bigtriangleup ABC}={(\frac{AD}{AB})}^2$$
Area of triangle ADC : Area of triangle ABC = 25:64
c) Area of triangle ADC : Area of triangle ABC = 25:64
Area of triangle ADC=(25/64)256
Area of triangle ADC= 100 centimetre square
d) Area of trapezium DBCE = Area of triangle ABC – Area of triangle ADE
= 256 – 100 = 156 centimetre square
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(iii)
An earthquake of 1920 was measured about 8.5 on Richter scale. Another earthquake of 1923 was measured 7.8 on that scale. How many times stronger was 1920 earthquake than 1923 earthquake?
Solution
Measurement of earthquake of 1920 =A = 8.5
Measurement of earthquake of 1923 =B = 7.8
Difference=M= 8.5-7.8=0.7
LogM=0.7
M=Antilog 0.7
=5.012
1920’s earthquake was 5 times stronger than 1923’s earthquake.
OR
In a bad with 10 balls, there are 7 black and 3 white balls. If one ball is selected at random from the bad. Calculate
a) Probability of selecting a white ball
b) probability of selecting a black ball
c) Sum of all the probabilities
Solution
P(W) = 3/10=0.3
P(B)= 7/10=0.7
P(W)+P(B)= 0.3+0.7 = 1
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(iv)
Solve the inequality
$$3x+5\leq5-3(x+2)\leq6x-10$$
Solution
First we’ll split it
$$3x+5\leq5-3(x+2)\\3x+5\leq5-3x+6\\3x+5\leq-1-3x\\6x\leq-6\\x\leq-1$$
$$5-3(x+2)\leq6x-10\\5-3x+6\leq6x-10\\-1-3x\leq6x-10\\-9x\leq-9\\x\geq1$$
Hence $$1\leq x\leq-1$$
OR
Factorize
$$(x^2+5x+4)\;(x^2+5x+6)-3$$
Solution
Let
$$x^2+5x=\;y$$
(y+4) (y+6) -3
$$=y^2+6y+4y+24-3\\=y^2+10y+21\\=y^2+7y+3y+21$$
y(y+7) +3(y+7)
(y+7) (y+3)
After putting , we get
$$(x^2+5x+7)\;(x^2+5x+3)$$
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(v)
Verify that
$$\sqrt{\frac{1-Cos\theta}{1+Cos\theta}}=\frac{Sin\theta}{1+Cos\theta}$$
Solution
$$L.H.S\;=\;\sqrt{\frac{1-Cos\theta}{1+Cos\theta}}\\=\;\sqrt{\frac{(1-Cos\theta)\;(1+Cos\theta)}{(1+Cos\theta)\;(1+Cos\theta)}}\\=\sqrt{\frac{1-Cos^2\theta}{{(1+Cos\theta)}^2}}\\=\frac{\sqrt{Sin^2\theta}}{\sqrt{{(1+Cos\theta)}^2}}\\=\;\frac{Sin\;\theta}{1+Cos\theta}\\=R.H.S$$
OR
An Exterior angle of a regular polygon is 12. What is the sum of all the interior angles ? If the length of its side is 1m then find its perimeter.
Solution
Some of exterior angles of convex polygon is always 360
Number of sides = 360/12 = 30
Sum of interior angles =(n-2) 180
= (30-2) 180 =5040
Perimeter = n ( side length )
= 30 ( 1 ) = 30
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(vi)
Lines AB and CD with coordinates A(0,3) , B(2,8) , C(3,10) , D(8,k) are perpendicular. Find k.
Solution
$$slope\;=\frac{y_2-y_1}{x_2-x_1}\\Slope\;of\;AB=\frac{8-3}{2-0}=\frac52\\Solpe\;of\;CD\;=\;\frac{k-10}{8-3}=\frac{k-10}5$$
Lines are perpendicular. Hence slopes will be negatives reciprocals of each other.
$$\frac{k-10}5=-\frac25$$
k-10=-2
k=8
OR
Find the equation of line passing through point of intersection of lines 3x+2y+1=0 , x-2y+3=0 and passing through (-1,0)
Solution
Adding both the equations
3x+2y+1+x-2y+3=0+0
2y will be cancelled with -2y
4x+4=0
x=-1
Both the points have same x coordinate. So line is vertical
Required equation is
x=-1
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(vii)
A company wants to install a new cell phone tower so that it is equidistant from two existing towers located at P(2,5) and Q(8,3). Find equation of locus where the new tower should be placed.
Solution
Let the coordinates of new tower = R(x,y)
Lets apply distance formula
$$PR=\sqrt{{(x-2)}^2+{(y-5)}^2}\\QR=\;\sqrt{{(x-8)}^2+{(y-3)}^2}$$
Since PR=QR. So
$${\small \sqrt{(x – 2)^2 + (y – 5)^2} = \sqrt{(x – 8)^2 + (y – 3)^2}}$$
Taking square on both the side and after simplification we get
3x-y=11
OR
Find the equation of line that crosses the line y=2x-5 at right angle at the point (3,1)
Solution
Comparing y=2x-5 with y=mx+c, we can find the slope
we get m=2
Slope of line perpendicular to y=2x-5 is -1/2
Point slope form
$$y-y_1=m(x-x_1)\\y-1=-\frac12(x-3)\\y=-\frac12x+\frac32+1\\y=-\frac12x+\frac52$$
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(viii)
If A={1,3} and B{2,4}
a) Write the cartesian product AxB
b) Find
$$R_1=\{(x,y)\vert x\in A,y\in B\wedge x>y\}$$
Solution
AxB={(1,2),(1,4),(3,2),(3,4)}
Required relation = {(3,2)}
OR
A fair coin is tossed 150 times and head comes up 70 times. What is the relative frequency getting heads. Also find the relative frequency of getting tails.
Solution
Relative frequency of heads= 70/150=0.4667
Tails= 150-70=80
Relative frequency of tails= 80/150=0.5333
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(ix)
If
$$A=4a^2b\;and\;B=6ab^2\\$$
then show that (A)(B)=(LCM) (HCF)
Solution
$$4a^2b=2.2.a.b\;\\6ab^2=2.3.a.b.b\\\\$$
HCF= 2ab
LCM=(2ab)( 2ab) (6ab)
$$AXB=4a^2b\;X\;6ab^2=24a^3b^3\\LXH=12a^2b^2X2ab=\;24a^3b^3$$
OR
The number of item sold by a shop on different days are 6,8,6,5,7,6,9. Find mode, median and mean
Solution
lets arrange it
5,6,6,6,7,8,9
Mode means most repeated value
Mode=6
Middle means middle observation of the data
Median=6
Mean mean average of the data
$$Mean=\frac{5+6+6+6+7+8+9}7\\Mean\;=\;\frac{47}7$$
You can also watch our video guide explaining this entire 9th Class Maths Paper 2026 fbise.
9th Class Maths Paper 2026 fbise Solution complete Long Questions
Section C (24) marks
Each question carries 8 marks. There are total 3 questions.
Below are the step-by-step short question answers for this 9th Class Maths Paper 2026 fbise
Q3
Find the interior angles of triangles XYZ whose vertices are X(-3,2) , Y(0,-1) , Z(3,3)
Solution
Lets find the slopes first
$$m_1\;is\;the\;slope\;of\;line\;XY\\m_1=\frac{-1-2}{0+3}=\frac{-3}3=-1\\m_2\;is\;the\;slope\;of\;line\;YZ\\m_2=\frac{3+1}{3-0}=\frac43\\m_3\;is\;the\;slope\;of\;line\;XZ\\m_3=\frac{3-2}{3+3}=\frac16$$
Lets find the interior angles now
$$\alpha\;is\;the\;angle\;between\;XY\;and\;XZ\\Tan\alpha=\frac{m_3-m_1}{1+m_3m_1}\\=\frac{{\displaystyle\frac16}-(-1)}{1+(\frac16)(-1)}\\=\frac75\\\alpha=Tan^{-1}(\frac75)\\\alpha=54.46^\circ$$
Similarly we can find the other angles
Other angles will be 81.87 and 43.67
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Q4
From the top of a tower a height of 120 m, angles of depression of two boats on same side of tower at water level are 60 and 45. Find distance between the boats.
Solution
This geometry theorem is a crucial part of the 9th Class Maths Paper 2026 fbise syllabus.
let height of the tower = DC=h=120m
Distance between first boat and the tower=y
Distance between second boat and the tower=z
Distance between the boats =x =?
In triangle BCD
$$Tan60^\circ=\frac{120}y\\\sqrt3=\frac{120}y\\y=\frac{120}{\sqrt3}\\y=69.28$$
In triangle ACD
$$Tan45^\circ=\frac{120}z\\1=\frac{120}z\\z=120$$
x+y=z
x+69.28=120
So
Distance between the boats= x=50.72m
OR
Convert the following equation into different standard forms x-2y+4=0
a) Slope intercept form
b) Two intercept form
c) Symmetric form
d) Normal form
Solution
a) y=mx+c is symmetric form
x-2y+4=0
-2y=-x-4
y=(1/2)x+2
where m=1/2 and c=2
b) x/a + y/b=1 is two intercept form
After dividing our main equation by -4 we get
(x/-4) + (y/2) = 1
a=-4 and b=2
c) Symmetric form is
$$\frac{x-x_1}l=\frac{y-y_1}m$$
First we’ll find the point by putting x=0 in our given equation
0-2y+4=0
y=2
so point is (0,2)
$$\frac{x-0}2=\frac{y-2}1\\\frac x2=y-2$$
d) Normal form is
$$xCos\alpha=ySin\alpha=p$$
where p is perpendicular from origin so it must be positive
x-2y=-4
-x+2y=4
$$\sqrt{{(-1)}^2+{(2)}^2}=\sqrt5\\\frac{-x}{\sqrt5}+\frac{2y}{\sqrt5}=\frac4{\sqrt5}$$
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Q5
For what value of k, the expression becomes a perfect square
$$y^4+4y^2+k+\frac8{y^2}+\frac4{y^4}$$
Solution
OR
The table shows distribution of marks obtained by 50 students in a mathematical test. Calculate median marks
| CI | 01-10 | 11-20 | 21-30 | 31-40 | 41-50 |
|---|---|---|---|---|---|
| f | 5 | 10 | 15 | 12 | 8 |
Solution
Now we will solve the statistics and distribution section of the 9th Class Maths Paper 2026 fbise
| C.I | f | C B | C f |
|---|---|---|---|
| 1-10 | 5 | 0.5-10.5 | 5 |
| 11-20 | 10 | 10.5-20.5 | 15 |
| 21-30 | 15 | 20.5-30.5 | 30 |
| 31-40 | 12 | 30.5-40.5 | 42 |
| 41-50 | 8 | 40.5-50.5 | 50 |
$$ \begin{aligned}
\sum f &= 50 \\
\text{Median} &= l + \frac{h}{f} \left( \frac{n}{2} – c \right) \\
&= 20.5 + \frac{10}{15} (25 – 15) \\
&= 20.5 + \frac{2}{3} (10) \\
&= 20.5 + 0.667(10) \\
&= 20.5 + 6.67 \\
&= 27.17
\end{aligned} $$
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