GATE CS 2016 Set 1 Q.10 (GA Q.10): Quantitative Aptitude

In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.

40.00

46.02

60.01

92.02

Official answer and solution

Official answer: B
From: Official GATE 2016 answer key

Solution

Model exponential decay of failure cycles with load and solve for the unknown load value.

Let $N(L)$ denote the number of cycles to failure at load $L$ . Since $N$ decreases exponentially with $L$ , we write:

N(L) = N_0 e^{-k L}

or equivalently in terms of natural logarithms:

\ln N(L) = \ln N_0 - k L

From the given data points:

At load $L_1 = 80$ : $N(80) = 100$
At halved load $L_2 = 40$ : $N(40) = 10000$

Subtract the two equations:

\ln(10000) - \ln(100) = k(80 - 40)

\ln\left(\frac{10000}{100}\right) = 40 k \implies \ln(100) = 40 k

k = \frac{\ln(100)}{40} = \frac{2 \ln(10)}{40} = \frac{\ln(10)}{20}

We need the load $L$ for which $N(L) = 5000$ :

\ln(10000) - \ln(5000) = k(L - 40)

\ln(2) = k(L - 40)

L - 40 = \frac{\ln(2)}{k} = 40 \times \frac{\ln(2)}{\ln(100)} = 40 \times \frac{\log_{10}(2)}{\log_{10}(100)}

Using $\log_{10}(2) \approx 0.30103$ and $\log_{10}(100) = 2$ :

L - 40 = 40 \times \frac{0.30103}{2} = 40 \times 0.150515 \approx 6.02

Thus:

L = 40 + 6.02 = 46.02\text{ units}

Final answer

Common trap

Assuming linear interpolation between 40 and 80, which would yield $L = 40 + 40 \times \frac{5000}{9900} \approx 60.2$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 10 of 65, printed as GA Q.10
Section: General Aptitude

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