## General Aptitude

 Question 1
What will be the maximum sum of 44, 42, 40, ...... ?
 A 502 B 504 C 506 D 500
GATE CS 2013    General Aptitude
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Question 1 Explanation:
This is a decreasing arithmetic progression with difference as 2. The series is 44, 42, 40 ...... 0, -2, -4...... The sum would be maximum if we consider the series till 0 or 2. So sum would be 506 using the AP Sum formula given here
 Question 2
Find the sum of the expression
 A 7 B 8 C 9 D 10
GATE CS 2013    General Aptitude
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Question 2 Explanation:
The Series can be re-written as (√2-√1)/(√2+√1)(√2-√1) + (√3-√2)/(√2+√2)(√3-√2) + .......... which simplifies to (√2-√1) + (√3-√2) + ..... (√81-√80) which again simplifies to √81 - √1 which is 8
 Question 3
A tourist covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and the rest by cycle at 10 km/h. The average speed of the tourist in km/h during his entire journey is
 A 36 B 30 C 24 D 18
GATE CS 2013    General Aptitude
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Question 3 Explanation:
Let total distance be D Total Time = D(1/2*60 + 1/4*30 + 1/4*10) = D/24 Average Speed = Total distance / Total time = 24
 Question 4
Consider the following logical inferences.
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.
I2: If it rains then the cricket match will not be played.
It did not rain.
Inference: The cricket match was played.
Which of the following is TRUE?
 A Both I1 and I2 are correct inferences B I1 is correct but I2 is not a correct inference C I1 is not correct but I2 is a correct inference D Both I1 and I2 are not correct inferences
GATE CS 2012    General Aptitude
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Question 4 Explanation:
The cricket match may not be played even if doesn't rain.
 Question 5
The cost function for a product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of Rs 50 per unit. The number of units to be produced by the firm such that the profit is maximized is
 A 5 B 10 C 15 D 25
GATE CS 2012    General Aptitude
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Question 5 Explanation:
Profit = Price - Cost
= 50q - 5q2
The value of above expression is maximum at q = 5. 
 Question 6
A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x – 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is
 A 8 meters B 10 meters C 12 meters D 14 meters
GATE CS 2012    General Aptitude
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Question 6 Explanation:
y = 2x – 0.1x2 dy/dx = 2 - 0.2x So the value maximizes at 2 - 0.2x = 0 => x = 10 => y = 20 - 10 = 10 meters
 Question 7
An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X’s shock absorbers, 96% are reliable. Of Y’s shock absorbers, 72% are reliable. The probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is
 A 0.288 B 0.334 C 0.667 D 0.72
GATE CS 2012    General Aptitude
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Question 7 Explanation:
Probability that the absorber is reliable = 0.96*0.6 + 0.72*0.4
= 0.576 + 0.288

Probability that  the absorber is from y and reliable =
(Probability that is made by Y)  X (Probability that it is reliable)
= 0.4 * 0.72
= 0.288

The probability that randomly picked reliable absorber is from y =
(Probability that  the absorber is from y and reliable) / (>Probability that the absorber is reliable )
= (0.288)/ (0.576 + 0.288)
=  0.334 
 Question 8
Which of the following assertions are CORRECT? P: Adding 7 to each entry in a list adds 7 to the mean of the list Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list R: Doubling each entry in a list doubles the mean of the list S: Doubling each entry in a list leaves the standard deviation of the list unchanged
 A P, Q B Q, R C P, R D R, S
GATE CS 2012    General Aptitude
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Question 8 Explanation:
Mean is average.
Let us consider below example
[Tex] 2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9 [/Tex]

These eight data points have the mean (average) of 5:
[Tex] \frac{2 + 4 + 4 + 4 + 5 + 5 + 7 + 9}{8} = 5. [/Tex]

When we add 7 to all numbers, mean becomes 12 so P is TRUE.

If we double all numbers mean becomes double, so R is also
TRUE.

Standard Deviation is square root of variance.
Variance is sum of squares of differences between all numbers
and means.

Deviation for above example
First, calculate the deviations of each data point from the mean,
and square the result of each:

[Tex] \begin{array}{lll} (2-5)^2 = (-3)^2 = 9 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (7-5)^2 = 2^2 = 4 \\ (4-5)^2 = (-1)^2 = 1 && (9-5)^2 = 4^2 = 16. \\ \end{array} [/Tex]

[Tex]variance = \frac{9 + 1 + 1 + 1 + 0 + 0 + 4 + 16}{8} = 4. [/Tex]

[Tex]standard deviation = \sqrt{ 4 } = 2 [/Tex]

If we add 7 to all numbers, standard deviation won't change
as 7 is added to mean also. So Q is FALSE.

If we double all entries, standard deviation also becomes
double.  So S is false. 
References: https://en.wikipedia.org/wiki/Standard_deviation http://staff.argyll.epsb.ca/jreed/math30p/statistics/standardDeviation.htm
 Question 9
Given the sequence of terms, AD CG FK JP, the next term is
 A OV B OW C PV D PW
GATE CS 2012    General Aptitude
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Question 9 Explanation:
Here 1st letter of each term in order are: A C F J Now In alphabets, A and C have a difference of 1 alphabet, C and F have a difference of 2, and F and J have a difference of 3. Hence next term here should contain its 1st letter as an alphabet with a difference of 4 after J, i.e O. Similarly, 2nd letter of each term in order are: D G K P Now In alphabets, D and G have a difference of 2 alphabets, G and K have a difference of 3, and K and P have a difference of 4. Hence next term here should contain its 2nd letter as an alphabet with a difference of 5 after P, i.e V.
 Question 10
If Log(P) = (1/2)Log(Q) = (1/3)Log(R), then which of the following options is TRUE?
 A P2 = Q3R2 B Q2 = PR C Q2 = R3P2 D R = P2Q2
GATE CS 2011    General Aptitude
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Question 10 Explanation:
It is given that Log(P) = (1/2)Log(Q) = (1/3)Log(R)

Let Log(P) = (1/2)Log(Q) = (1/3)Log(R) = C

Let the base of log be B

P = BC
Q = B2C
R = B3C

Which means Q2 = PR
There are 77 questions to complete.

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