Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
Q.6 Area of a Semicircle
The area of a semicircle with diameter \(10 \text{ cm}\) is:
Concept:
1
The area of a full circle is \(A = \pi r^2\). A semicircle is exactly half of a circle, so the formula becomes \(A = \frac{1}{2} \pi r^2\).
www.maheshkumarsir.in
Hint:
1
The question gives you the diameter (\(10 \text{ cm}\)). You must first cut this in half to find the radius before squaring it! Use \(\pi \approx 3.14\).
www.maheshkumarsir.in
Mistakes to Avoid:
!
Diameter Trap: Plugging the number 10 straight into the radius spot (\(\pi \times 10^2\)). Always extract the radius (\(r = d/2\)) first!
!
Forgetting the Half: Calculating the area of the full circle and forgetting to divide by 2 for the semicircle.
www.maheshkumarsir.in
Given: Diameter \( d = 10 \text{ cm} \)
Radius \( r = \frac{d}{2} = \frac{10}{2} = 5 \text{ cm} \)
Area of Semicircle = \( \frac{1}{2} \pi r^2 \)
Using \( \pi \approx 3.14 \):
Area = \( \frac{1}{2} \times 3.14 \times (5)^2 \)
Area = \( \frac{1}{2} \times 3.14 \times 25 \)
Area = \( 1.57 \times 25 \)
Area = \( 39.25 \text{ cm}^2 \) [1 M]
Correct Option: (D) 39.25 cm²
Q.7 Experimental Probability
Two coins are tossed simultaneously 1000 times, and we get:
- Two heads: 200 times
- One head: 600 times
- No head: 200 times
Find the probability of getting two heads.
- Two heads: 200 times
- One head: 600 times
- No head: 200 times
Find the probability of getting two heads.
Concept:
1
Experimental Probability: When data is given from an actual experiment, the probability of an event is calculated strictly as \( \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}} \).
www.maheshkumarsir.in
Hint:
1
Look at the data block. How many times did "Two heads" actually occur? Put that number over the total number of tosses (\(1000\)) and simplify the fraction.
www.maheshkumarsir.in
Mistakes to Avoid:
!
Ignoring the Data: Assuming the answer must be \(1/4\) because theoretically \(P(HH) = 1/4\). In experimental statistics questions, you MUST use the physical counts provided in the problem, not theoretical ideals.
www.maheshkumarsir.in
Total number of tosses = \( 1000 \)
Frequency of getting 'Two heads' = \( 200 \)
\( P(\text{Two heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \)
\( P(\text{Two heads}) = \frac{200}{1000} \)
Simplify by striking out the zeros:
\( P(\text{Two heads}) = \frac{2}{10} \)
\( P(\text{Two heads}) = \frac{1}{5} \) [1 M]
Correct Option: (C) 1/5
Q.8 Coordinate Equivalence
If \((x+2, 4) = (5, y-2)\), then the coordinates \((x, y)\) are:
Concept:
1
Two coordinate points \((x_1, y_1)\) and \((x_2, y_2)\) are strictly equal if and only if their corresponding abscissae (x-values) and ordinates (y-values) are equal. Meaning \(x_1 = x_2\) and \(y_1 = y_2\).
www.maheshkumarsir.in
Hint:
1
Split the problem into two simple linear equations: set the first part equal to the first part (\(x+2 = 5\)), and the second part equal to the second part (\(4 = y-2\)).
www.maheshkumarsir.in
Mistakes to Avoid:
!
Sign errors during shifting: When solving \(y - 2 = 4\), shifting the \(-2\) across the equals sign turns it into a \(+2\). Doing \(4 - 2\) instead of \(4 + 2\) is a common trap.
www.maheshkumarsir.in
Given: \( (x+2, 4) = (5, y-2) \)
Equating the x-coordinates:
\( x + 2 = 5 \)
\( x = 5 - 2 \)
\( x = 3 \)
Equating the y-coordinates:
\( 4 = y - 2 \)
\( 4 + 2 = y \)
\( y = 6 \)
Therefore, the coordinates \((x, y)\) are \((3, 6)\). [1 M]
Correct Option: (C) (3,6)
Q.9 Linear Equations
Which of the following equations represents a line passing through the origin?
Concept:
1
The standard equation of a straight line is \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.
2
For a line to pass perfectly through the origin \((0, 0)\), the y-intercept must be zero (\(c = 0\)). Therefore, the equation will take the simplified form \(y = mx\).
www.maheshkumarsir.in
Hint:
1
The coordinates of the origin are \((0, 0)\). Try substituting \(x = 0\) and \(y = 0\) into each option to see which equation remains mathematically true (\(0 = 0\)).
www.maheshkumarsir.in
Mistakes to Avoid:
!
Assuming all lines cross the origin: Any equation with a standalone constant number (like \(+1\) or \(-3\)) will be shifted away from the origin on a graph. Look for the equation with NO standalone numbers!
www.maheshkumarsir.in
A line passes through the origin if substituting \( x = 0 \) gives \( y = 0 \).
Let's test option (C): \( y = 9x \)
Substitute \( x = 0 \):
\( y = 9(0) \)
\( y = 0 \)
Since it yields the point \((0, 0)\), it passes through the origin.
All other options have a constant term (\(c \neq 0\)), so they will not pass through the origin. [1 M]
Correct Option: (C) \(y=9x\)
Q.10 Decimal Expansions
The decimal expansion of \(\frac{11}{625}\) terminates after:
Concept:
1
A rational number \(\frac{p}{q}\) will terminate if the prime factorization of its denominator \(q\) is of the form \(2^m \times 5^n\).
2
The number of decimal places after which it terminates is determined by the maximum of the powers \(m\) or \(n\).
www.maheshkumarsir.in
Hint:
1
Find the prime factorization of 625. Keep dividing by 5. \(625 = 5 \times 5 \times 5 \times 5\).
www.maheshkumarsir.in
Mistakes to Avoid:
!
Long Division Trap: Trying to actually divide 11 by 625 manually during an exam wastes critical time and is prone to errors. Always use the prime factorization rule for termination questions!
www.maheshkumarsir.in
The fraction is \( \frac{11}{625} \).
Find the prime factorization of the denominator:
\( 625 = 5 \times 5 \times 5 \times 5 \)
\( 625 = 5^4 \)
We can express it in the standard terminating form:
Denominator = \( 2^0 \times 5^4 \)
The highest power of either 2 or 5 is 4.
Therefore, the decimal expansion terminates after 4 places. [1 M]
Correct Option: (C) 4 decimal places
Disclaimer: This content is provided solely for educational purposes to help students with their exam preparations. While we strive for maximum accuracy, this guide does not claim absolute authenticity and has no official affiliation with the school's exam setting authorities. It is a preparatory guide created upon student requests. For any discrepancies or issues regarding the content, please contact Maheshkumar Sir, and it will be modified or removed immediately.
