11th maths public question paper 2022: Questions & Answers

This article provides a comprehensive collection of questions and answers from the 11th Maths public question paper 2022. It aims to help students prepare for their exams by offering an in-depth look at the types of questions asked and detailed solutions.

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Question and Answer Section for 11th Maths Public Question Paper 2022

Question
What is the value of $x$ if $2x+5=15$?

Answer
To solve for $x$, subtract 5 from both sides:
$2x=10$
Now, divide both sides by 2:
$x=5$

Question
Find the slope of the line represented by the equation $3x-4y=12$.

Answer
First, rewrite the equation in slope-intercept form $y=mx+b$.
$3x-4y=12$
Subtract $3x$ from both sides:
$-4y=-3x+12$
Now, divide by -4:
y=34x−3y = \frac{3}{4}x – 3y=43​x−3
Thus, the slope is 34\frac{3}{4}43​.

Question
What is the area of a triangle with base 10 cm and height 6 cm?

Answer
The area of a triangle is given by the formula:
Area=12×base×heightArea = \frac{1}{2} \times base \times heightArea=21​×base×height
Substitute the values:
Area=12×10×6=30 cm2Area = \frac{1}{2} \times 10 \times 6 = 30 \, \text{cm}^2Area=21​×10×6=30cm2

Question
Solve for $x$: $5x-7=18$.

Answer
Add 7 to both sides:
$5x=25$
Now, divide both sides by 5:
$x=5$

Question
Find the value of $\sin \theta$ if θ=30∘\theta = 30^\circθ=30∘.

Answer
Using the trigonometric identity for sin⁡30∘\sin 30^\circsin30∘, we know that:
sin⁡30∘=12\sin 30^\circ = \frac{1}{2}sin30∘=21

​

Question
If the roots of the quadratic equation x2−7x+12=0x^2 – 7x + 12 = 0x2−7x+12=0 are $p$ and $q$, find their sum and product.

Answer
From Vieta’s formula, the sum of the roots is equal to −−71=7-\frac{-7}{1} = 7−1−7​=7 and the product of the roots is 121=12\frac{12}{1} = 12112​=12.

Question
What is the probability of drawing an ace from a deck of cards?

Answer
There are 4 aces in a deck of 52 cards.
So, the probability is:
P=452=113P = \frac{4}{52} = \frac{1}{13}P=524​=131​

Question
Find the length of the hypotenuse of a right triangle with legs 6 cm and 8 cm.

Answer
Use the Pythagorean theorem:
c2=a2+b2c^2 = a^2 + b^2c2=a2+b2
Substitute the values:
c2=62+82=36+64=100c^2 = 6^2 + 8^2 = 36 + 64 = 100c2=62+82=36+64=100
Thus, c=100=10 cmc = \sqrt{100} = 10 \, \text{cm}c=100​=10cm

Question
If $a=3$ and $b=4$, find the value of a2+b2a^2 + b^2a2+b2.

Answer
Substitute the values:
a2+b2=32+42=9+16=25a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25a2+b2=32+42=9+16=25

Question
Solve for $x$ in the equation 2×2−3x−5=02x^2 – 3x – 5 = 02x2−3x−5=0.

Answer
Use the quadratic formula:
x=−(−3)±(−3)2−4(2)(−5)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-5)}}{2(2)}x=2(2)−(−3)±(−3)2−4(2)(−5)​​
Simplify the expression:
x=3±9+404x = \frac{3 \pm \sqrt{9 + 40}}{4}x=43±9+40​​
x=3±494x = \frac{3 \pm \sqrt{49}}{4}x=43±49​​
x=3±74x = \frac{3 \pm 7}{4}x=43±7​
Thus, x=104=2.5x = \frac{10}{4} = 2.5x=410​=2.5 or x=−44=−1x = \frac{-4}{4} = -1x=4−4​=−1.

Question
Find the derivative of f(x)=3×2+2x−1f(x) = 3x^2 + 2x – 1f(x)=3x2+2x−1.

Answer
The derivative of a polynomial is found by applying the power rule:
f′(x)=2(3x)+2=6x+2f'(x) = 2(3x) + 2 = 6x + 2f′(x)=2(3x)+2=6x+2

Question
If a circle has a radius of 5 cm, what is its circumference?

Answer
The formula for the circumference of a circle is:
$C=2\pi r$
Substitute the radius:
$C=2\pi \times 5=10\pi \text{cm}$

Question
Solve for $x$ in the equation log⁡2x=4\log_2 x = 4log2​x=4.

Answer
Convert the logarithmic equation to an exponential form:
x=24=16x = 2^4 = 16x=24=16

Question
Find the value of cos⁡60∘\cos 60^\circcos60∘.

Answer
Using the trigonometric identity for cos⁡60∘\cos 60^\circcos60∘, we know that:
cos⁡60∘=12\cos 60^\circ = \frac{1}{2}cos60∘=21​

Question
Simplify the expression $(x+5)(x-3)$.

Answer
Use the distributive property (FOIL):
(x+5)(x−3)=x2−3x+5x−15(x + 5)(x – 3) = x^2 – 3x + 5x – 15(x+5)(x−3)=x2−3x+5x−15
Simplify:
=x2+2x−15= x^2 + 2x – 15=x2+2x−15

Question
What is the solution to the equation $4x+7=19$?

Answer
Subtract 7 from both sides:
$4x=12$
Now, divide both sides by 4:
$x=3$

Question
Find the equation of the line passing through the points (1, 2) and (3, 6).

Answer
First, find the slope using the formula:
m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​
Substitute the values:
m=6−23−1=42=2m = \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2m=3−16−2​=24​=2
Now, use the point-slope form:
y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)
Substitute $m=2$, x1=1x_1 = 1x1​=1, and y1=2y_1 = 2y1​=2:
$y-2=2(x-1)$
Simplify:
$y=2x$

Question
Find the sum of the angles in a triangle.

Answer
The sum of the angles in any triangle is always 180°.

Question
Find the roots of the quadratic equation x2+6x+9=0x^2 + 6x + 9 = 0x2+6x+9=0.

Answer
This is a perfect square trinomial:
(x+3)2=0(x + 3)^2 = 0(x+3)2=0
So, the root is:
$x=-3$

Question
What is the value of tan⁡45∘\tan 45^\circtan45∘?

Answer
Using the trigonometric identity for tan⁡45∘\tan 45^\circtan45∘, we know that:
tan⁡45∘=1\tan 45^\circ = 1tan45∘=1

Question
Find the length of the diagonal of a rectangle with length 8 cm and width 6 cm.

Answer
Use the Pythagorean theorem:
d2=l2+w2d^2 = l^2 + w^2d2=l2+w2
Substitute the values:
d2=82+62=64+36=100d^2 = 8^2 + 6^2 = 64 + 36 = 100d2=82+62=64+36=100
Thus, d=100=10 cmd = \sqrt{100} = 10 \, \text{cm}d=100​=10cm

Question
Find the value of $x$ in the equation $3x-2=4x+5$.

Answer
Subtract $3x$ from both sides:
$-2=x+5$
Now, subtract 5 from both sides:
$-7=x$

Question
What is the perimeter of a square with side length 5 cm?

Answer
The perimeter of a square is given by:
$P=4\times \text{side length}$
Substitute the side length:
$P=4\times 5=20\text{cm}$

Question
Solve for $x$ in the equation x4=3\frac{x}{4} = 34x​=3.

Answer
Multiply both sides by 4:
$x=12$

By practicing these questions, students can develop a strong understanding of the key mathematical concepts required for the 11th Maths public exam. These questions cover a range of topics, helping students prepare thoroughly for their exams.