Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
21 Area of a Rhombus
One diagonal of the rhombus is twice as long as the other diagonal. If the rhombus has area \(128 \text{ cm}^2\), find the length of the shorter diagonal.
Concept:
1
The area of a rhombus is given by the formula: \( \text{Area} = \frac{1}{2} \times d_1 \times d_2 \), where \(d_1\) and \(d_2\) are the lengths of the diagonals.
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Hint:
1
Let the shorter diagonal be \(x\). Then the longer diagonal must be \(2x\). Plug these into the area formula and equate it to 128.
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Mistakes to Avoid:
!
Wrong Formula: A very common error is calculating Area = \( \text{base} \times \text{height} \) or simply multiplying the diagonals without the \(\frac{1}{2}\) factor (\(x \times 2x = 128 \implies x = 8\)). Always remember the \(\frac{1}{2}\) for a rhombus!
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Let the length of the shorter diagonal be \( x \).
Then, the length of the longer diagonal is \( 2x \).
We know, Area of Rhombus = \( \frac{1}{2} \times d_1 \times d_2 \) [½ M]
Substitute the given values:
\( 128 = \frac{1}{2} \times x \times 2x \) [½ M]
The 2s cancel out:
\( 128 = x^2 \) [½ M]
Taking the square root on both sides:
\( x = \sqrt{128} \)
\( x = \sqrt{64 \times 2} \)
\( x = 8\sqrt{2} \text{ cm} \) [½ M]
Shorter diagonal = \(8\sqrt{2}\) cm
22 Fractional Addition
Riya bought a Mathematics notebook for \(26\frac{1}{2}\) and a packet of colour pencils for \(18\frac{3}{4}\) and a geometry box \(14\frac{3}{4}\). How much money did she spend altogether?
Concept:
1
To add mixed fractions, it is usually easiest to convert them into improper fractions first, find a common denominator, add the numerators, and simplify the final result.
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Hint:
1
Convert \(26\frac{1}{2}\) to \(\frac{53}{2}\). To make addition easy, convert it to a denominator of 4 by multiplying top and bottom by 2 (making it \(\frac{106}{4}\)).
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Mistakes to Avoid:
!
Adding denominators: A classic mistake is adding the bottom numbers together. Remember, denominators MUST be the same before adding, and you only add the numerators!
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Total money spent = Sum of all items
Total = \( 26\frac{1}{2} + 18\frac{3}{4} + 14\frac{3}{4} \) [½ M]
Convert to improper fractions:
Total = \( \frac{53}{2} + \frac{75}{4} + \frac{59}{4} \) [½ M]
Make denominators equal (LCM = 4):
Total = \( \frac{106}{4} + \frac{75}{4} + \frac{59}{4} \) [½ M]
Add the numerators:
Total = \( \frac{106 + 75 + 59}{4} \)
Total = \( \frac{240}{4} \)
Total = \( 60 \) [½ M]
Total amount spent = Rs. 60
23 Algebraic Expansion (Internal Choice)
If \((x+k)^2 = x^2+30x+225\), then find the value of k.
OR
Without direct multiplication, find: \(1004 \times 996\)
Concept:
1
Main: Expansion of a binomial square \((a+b)^2 = a^2 + 2ab + b^2\). By expanding the LHS and comparing coefficients with the RHS, we can find the unknown.
2
OR: The difference of squares identity \((a+b)(a-b) = a^2 - b^2\) allows us to calculate large numbers easily by anchoring them to a neat base number like 1000.
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Hint:
1
Main: Expand \((x+k)^2\). The middle term is \(2kx\). Set this equal to the middle term on the right side (\(30x\)).
2
OR: Notice that 1004 is \((1000 + 4)\) and 996 is \((1000 - 4)\).
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Mistakes to Avoid:
!
Direct Math Trap: For the OR question, if you actually multiply \(1004 \times 996\) using traditional long multiplication, you will get zero marks because the question explicitly forbids it!
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MAIN QUESTION SOLUTION:
Given: \( (x+k)^2 = x^2+30x+225 \)
Expand the LHS using \( (a+b)^2 = a^2+2ab+b^2 \):
\( x^2 + 2kx + k^2 = x^2 + 30x + 225 \) [1 M]
Comparing the coefficients of \( x \):
\( 2k = 30 \)
\( k = 15 \) [1 M]
(Checking with the constant term: \(15^2 = 225\), which perfectly matches!)
k = 15
OR
OR QUESTION SOLUTION:
Evaluate: \( 1004 \times 996 \)
Rewrite using a base of 1000:
\(= (1000 + 4) \times (1000 - 4) \) [1 M]
Use the identity: \( (a+b)(a-b) = a^2 - b^2 \)
\(= (1000)^2 - (4)^2 \)
\(= 1000000 - 16 \)
\(= 999984 \) [1 M]
Answer = 999,984
24 Theoretical Probability
What is the probability of getting an even number when rolling a fair 6-sided die?
Concept:
1
Probability: The probability of an event is calculated as \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \).
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Hint:
1
List out the entire sample space (all numbers on a die). Then circle only the even numbers. Count how many you circled, and put that number over the total.
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Mistakes to Avoid:
!
Forgetting to Simplify: Leaving the final answer as \(3/6\) will lose you half a mark. All probability fractions must be reduced to their simplest form!
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Sample space when rolling a die: \( S = \{1, 2, 3, 4, 5, 6\} \)
Total number of outcomes, \( n(S) = 6 \) [½ M]
Event (E) of getting an even number: \( E = \{2, 4, 6\} \)
Number of favorable outcomes, \( n(E) = 3 \) [½ M]
Probability \( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{3}{6} \) [½ M]
Simplify the fraction:
\( P(E) = \frac{1}{2} \) [½ M]
Probability = 1/2
25 Linear Equation Application
A school science exhibition charges 25 per visitor and a fixed registration charge of 120. The total collection from x visitors is represented by: \(C=25x+120\). Find:
(i) The total collection when 32 visitors attend.
(ii) How much additional collection is obtained if the number of visitors increases from 32 to 40?
(i) The total collection when 32 visitors attend.
(ii) How much additional collection is obtained if the number of visitors increases from 32 to 40?
Concept:
1
This is an application of linear equations. Substitute the value of the independent variable (\(x\)) into the equation to find the dependent variable (\(C\)).
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Hint:
1
For part (ii), you don't need to recalculate the entire equation for 40 and subtract. Think about it: the fixed charge (120) is already paid! You only need to calculate the cost of the extra 8 visitors.
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Mistakes to Avoid:
!
Double-counting fixed costs: For part (ii), if you just calculate \(C\) for 8 visitors (\(25(8) + 120\)), you are wrong. The 120 was already charged. The additional revenue comes strictly from the ticket price of the new visitors!
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Given Equation: \( C = 25x + 120 \)
(i) When 32 visitors attend:
Substitute \( x = 32 \) into the equation:
\( C = 25(32) + 120 \)
\( C = 800 + 120 \)
\( C = 920 \) [1 M]
Total Collection = Rs. 920
(ii) Additional collection from 32 to 40 visitors:
Increase in number of visitors = \( 40 - 32 = 8 \)
Since the fixed charge is already collected, the additional revenue comes only from the per-visitor charge.
Additional Collection = \( 8 \times 25 \)
Additional Collection = \( 200 \) [1 M]
Additional Collection = Rs. 200
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