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Question 1 of 5
1. Question
Based on the passage, the following assumptions have been made:
- Monitoring frog calls can serve as an early warning system for environmental degradation.
- Human-induced changes in land use directly affect frog behaviour and breeding cycles.
Select the correct answer using the code given below:
Correct
Solution: (c)
Explanation:
Statement 1 is valid — the passage suggests frog calls are ecological indicators, useful in tracking ecosystem health.
Statement 2 is valid — habitat loss and urbanization are cited as factors altering frog breeding behaviour.
Thus, both assumptions logically follow, making (c) correct.Incorrect
Solution: (c)
Explanation:
Statement 1 is valid — the passage suggests frog calls are ecological indicators, useful in tracking ecosystem health.
Statement 2 is valid — habitat loss and urbanization are cited as factors altering frog breeding behaviour.
Thus, both assumptions logically follow, making (c) correct. -
Question 2 of 5
2. Question
In 2018, number of males in a town are 20 percent less than the number of females. The male population decreases by 10 percent every year while the female population increases by 10 percent every year. Find how much percent the population in 2020 is more or less than the population in 2018.
Correct
Answer – A
Solution:Given that,
Males in 2018 are 20 percent less than females.
Male change per year = minus 10 percent.
Female change per year = plus 10 percent.Now,
Let females in 2018 be x.
Then males in 2018 = 0.8x.
Total 2018 = x + 0.8x = 1.8x.In 2020 (two years later):
Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x.
Females = x × 1.1 × 1.1 = 1.21x.
Total 2020 = 0.648x + 1.21x = 1.858x.Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100
= {0.058x ÷ 1.8x} × 100
≈ 3.22 percent increase.Hence option (a) is correct.
Incorrect
Answer – A
Solution:Given that,
Males in 2018 are 20 percent less than females.
Male change per year = minus 10 percent.
Female change per year = plus 10 percent.Now,
Let females in 2018 be x.
Then males in 2018 = 0.8x.
Total 2018 = x + 0.8x = 1.8x.In 2020 (two years later):
Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x.
Females = x × 1.1 × 1.1 = 1.21x.
Total 2020 = 0.648x + 1.21x = 1.858x.Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100
= {0.058x ÷ 1.8x} × 100
≈ 3.22 percent increase.Hence option (a) is correct.
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Question 3 of 5
3. Question
In an 800 metre race, A beats B by 80 metres. In a 1000 metre race, B can give a start of 200 metres to C. By how much distance should A give a start to C so that A beats C by 160 metres in an 800 metre race?
Correct
Answer: (c)
Solution:
From 800 metres: A:B = 800:720 = 10:9.
From 1000 metres: B:C = 1000:800 = 5:4.
So A:B:C = 50:45:36.
When A runs 800, C runs (36/50) × 800 = 576.
To beat by 160, C should be at 640.
Required start = 640 − 576 = 64 metres.Incorrect
Answer: (c)
Solution:
From 800 metres: A:B = 800:720 = 10:9.
From 1000 metres: B:C = 1000:800 = 5:4.
So A:B:C = 50:45:36.
When A runs 800, C runs (36/50) × 800 = 576.
To beat by 160, C should be at 640.
Required start = 640 − 576 = 64 metres. -
Question 4 of 5
4. Question
A box has x red balls, 20 blue balls, and y green balls. The probability of drawing a red ball is 3/10, and the probability of drawing a green ball is 2/5. Find the probability of getting 2 green balls when two balls are picked.
Correct
Answer: (c)
Solution:
Probability of red = x / (x + 20 + y) = 3/10
So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1)Probability of green = y / (x + 20 + y) = 2/5
So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2)Solving (1) and (2): x = 20, y = 40
Total = 20 + 20 + 40 = 80Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190.
Incorrect
Answer: (c)
Solution:
Probability of red = x / (x + 20 + y) = 3/10
So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1)Probability of green = y / (x + 20 + y) = 2/5
So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2)Solving (1) and (2): x = 20, y = 40
Total = 20 + 20 + 40 = 80Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190.
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Question 5 of 5
5. Question
The probability that A, B, and C hit a target is 70%, 50%, and 30% respectively. Find the probability that A and B hit, but C misses.
Correct
Answer: (c)
Solution:
Probability (A hits) = 70/100 = 0.7
Probability (B hits) = 50/100 = 0.5
Probability (C misses) = 1 – 0.3 = 0.7Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5%
Correct answer = 24.5%
Incorrect
Answer: (c)
Solution:
Probability (A hits) = 70/100 = 0.7
Probability (B hits) = 50/100 = 0.5
Probability (C misses) = 1 – 0.3 = 0.7Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5%
Correct answer = 24.5%
