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A monkey tries to jump up a 50 m high pole. It jumps up 1.25 m every second, but slides down by 0.5m next second. In how many seconds it will reach the top ?

A133

B131

C134

D132

Answer:

B. 131

Read Explanation:

Understanding the Monkey Jump Problem

  • Problem Type: This is a classic mathematical puzzle that tests logical reasoning and problem-solving skills, often seen in competitive exams under quantitative aptitude sections.

  • Key Concept: The core idea is to calculate the net progress made by the monkey in a two-second cycle and then determine how many such cycles are needed to reach the top, considering the final jump.

Step-by-Step Calculation

  • Net Progress per Cycle:

    • In the first second, the monkey jumps up by 1.25 m.

    • In the second second, it slides down by 0.5 m.

    • Therefore, the net upward movement in a 2-second cycle is 1.25 m - 0.5 m = 0.75 m.

  • Approaching the Top: The pole height is 50 m. The monkey needs to reach a point from where its final jump of 1.25 m will take it to the top. This point is 50 m - 1.25 m = 48.75 m from the bottom.

  • Cycles to Reach 48.75 m:

    • To cover 48.75 m with a net progress of 0.75 m per 2-second cycle, we divide: 48.75 m / 0.75 m/cycle = 65 cycles.

    • The time taken for these 65 cycles is 65 cycles * 2 seconds/cycle = 130 seconds.

  • Final Jump: After 130 seconds (completing 65 cycles), the monkey is at a height of 48.75 m. In the next second (the 131st second), it makes its final jump of 1.25 m.

  • Total Time: 130 seconds (for the cycles) + 1 second (for the final jump) = 131 seconds.


Related Questions:

Find the sum of the numbers lying between 200 and 700 which are multiples of 5.
ഓരോ മുഖത്തിലും 1 മുതൽ 6 വരെയുള്ള എണ്ണൽ സംഖ്യകൾ ഓരോന്നു വീതം എഴുതിയ ഒരു പകിട (dice) എറിഞ്ഞാൽ ഒരു അഭാജ്യസംഖ്യ (prime number) കിട്ടാനുള്ള സാധ്യത എന്ത് ?
എല്ലാ രണ്ട് അക്ക സംഖ്യകളുടെയും ആകെ തുകയെ 7 കൊണ്ട് ഭരിക്കുമ്പോൾ ശേഷിക്കുന്നത് 5 ആണെങ്കിൽ ഏത് സംഖ്യ ഇതിന് തുല്യമായിരിക്കും ?
1 നും 50നും ഇടയ്ക്ക് വരുന്ന ഇരട്ട സംഖ്യകളുടെ തുക :
3 + 6 + 9 + 12 +..........+ 300 എത്ര ?