Question : The current population of the town is 68,000. The population increases by 10 percent in the first year and decreases by 30% in the second year. Find the population after 2 years.
Option 1: 52,360
Option 2: 68,500
Option 3: 69,000
Option 4: 70,100
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Correct Answer: 52,360
Solution : Given: The current population of the town is 68,000. Percentage = $\frac{\text{Value}}{\text{Total value}}\times 100$ The population increases by 10% in the first year, then the population becomes = 68000 + $\frac{10}{100}\times 68000$ = 68000 + 6800 = 74,800 The population decreases by 30% in the second year = 74800 – $\frac{30}{100}\times 74800$ = 74800 – 22440 = 52,360 Hence, the correct answer is 52,360.
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Question : If the population of a town is 12,000 and the population increases at the rate of 10% per annum, then find the population after 3 years.
Option 1: 15,972
Option 2: 12,200
Option 3: 11,200
Option 4: 10,200
Question : The population of a town increases at the rate of 20% per annum. If the town's population will be 69,120 after 2 years, then what is the town's present population?
Option 1: 48,000
Option 2: 42,000
Option 3: 40,000
Option 4: 36,000
Question : The value of a machine is 6,250. It decreases by 10% during the first year, 20% during the second year and 30% during the third year. What will be the value of the machine after 3 years?
Option 1: 2,650
Option 2: 3,050
Option 3: 3,150
Option 4: 3,510
Question : Negative feedback in amplifiers
Option 1: increases bandwidth and decreases noise
Option 2: decreases bandwidth and decreases noise
Option 3: increases bandwidth and increases noise
Option 4: decreases bandwidth and increases noise
Question : If the period increases by 5 years, then simple interest increases by Rs. 3600 on a sum of Rs. 6000. What is the annual rate of interest?
Option 1: 15 percent
Option 2: 12 percent
Option 3: 12.5 percent
Option 4: 10 percent
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