Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
30 Simplification of Rational Numbers
Simplify:
\( \left(\frac{-5}{9} \times \frac{72}{-125}\right) - \left(\frac{11}{17} \times \frac{34}{55}\right) + \left(\frac{28}{-13} \times \frac{-52}{21}\right) \)
\( \left(\frac{-5}{9} \times \frac{72}{-125}\right) - \left(\frac{11}{17} \times \frac{34}{55}\right) + \left(\frac{28}{-13} \times \frac{-52}{21}\right) \)
Concept:
1
BODMAS Rule: Always solve the operations inside the Brackets first. Within each bracket, simplify the multiplication by cross-cancelling common factors between numerators and denominators.
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Hint:
1
In the first bracket, notice that \(-5\) and \(-125\) share a factor of 5, and the negatives cancel. 9 and 72 share a factor of 9. Simplify each bracket individually before adding/subtracting them at the end.
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Mistakes to Avoid:
!
Brute Force Multiplication: Multiplying \( (-5 \times 72) / (9 \times -125) \) to get \( -360 / -1125 \) without simplifying first. This creates unnecessarily huge numbers that are extremely difficult to find an LCM for later! Always cancel early.
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Let's evaluate each bracket individually:
First Term: \( \left(\frac{-5}{9} \times \frac{72}{-125}\right) \)
Cancel negatives. 5 goes into 125 (25 times). 9 goes into 72 (8 times).
\(= \frac{1}{1} \times \frac{8}{25} = \mathbf{\frac{8}{25}} \) [1 M]
Second Term: \( \left(\frac{11}{17} \times \frac{34}{55}\right) \)
11 goes into 55 (5 times). 17 goes into 34 (2 times).
\(= \frac{1}{1} \times \frac{2}{5} = \mathbf{\frac{2}{5}} \) [½ M]
Third Term: \( \left(\frac{28}{-13} \times \frac{-52}{21}\right) \)
Cancel negatives. 13 goes into 52 (4 times). Divide 28 and 21 by 7.
\(= \frac{4}{1} \times \frac{4}{3} = \mathbf{\frac{16}{3}} \) [½ M]
Combine the terms:
Expression = \( \frac{8}{25} - \frac{2}{5} + \frac{16}{3} \)
LCM of denominators 25, 5, and 3 is 75.
\(= \frac{8 \times 3}{75} - \frac{2 \times 15}{75} + \frac{16 \times 25}{75} \)
\(= \frac{24 - 30 + 400}{75} \)
\(= \frac{-6 + 400}{75} \)
\(= \frac{394}{75} \) [1 M]
Answer = \( \frac{394}{75} \)
31 Experimental Statistics
A tyre company records distances before replacement in 1000 cases.
| Distance (km) | Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000 |
|---|---|---|---|---|
| Number of cases | 20 | 210 | 325 | 445 |
Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.
Concept:
1
Experimental Probability: Given empirical data, the probability of an event \(E\) is calculated as \( P(E) = \frac{\text{Frequency of event E occurring}}{\text{Total number of trials}} \).
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Hint:
1
For part (ii) "Between 4000 and 14000 km", notice that this range spans across TWO columns in the table (the 4001-9000 column AND the 9001-14000 column). You must add those two frequencies together!
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Mistakes to Avoid:
!
Incomplete Simplification: Writing the probability as \(\frac{20}{1000}\) and stopping. While technically correct, board exams expect fractions to be fully simplified to their lowest terms (e.g., \(\frac{1}{50}\)) or converted to decimals!
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Total number of cases (trials) \( = 1000 \)
(i) Probability of lasting less than 4000 km:
Number of cases = \( 20 \)
\( P(i) = \frac{20}{1000} = \frac{2}{100} = \frac{1}{50} \) [1 M]
(ii) Probability of lasting between 4000 and 14000 km:
This covers two categories: "4001 to 9000" and "9001 to 14000".
Number of cases = \( 210 + 325 = 535 \)
\( P(ii) = \frac{535}{1000} \)
Divide by 5 to simplify:
\( P(ii) = \frac{107}{200} \) [1 M]
(iii) Probability of lasting more than 14000 km:
Number of cases = \( 445 \)
\( P(iii) = \frac{445}{1000} \)
Divide by 5 to simplify:
\( P(iii) = \frac{89}{200} \) [1 M]
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