• APSEd

GATE 2022 Aptitude | Question Paper with Solution - 5th Feb Session

Updated: Feb 14

APSEd brings you solutions for the GATE 2022 aptitude from the papers conducted on 5th Feb by IIT Kharagpur. The main papers that were conducted on 5th February include CS and EE papers. Aptitude questions that appeared in these papers are discussed further.


In this article, we will share some memory based questions and their solution hints with you.


GATE 2022 Aptitude Solution - 5th February Session
GATE 2022 Aptitude Solution - 5th February Session

Special Announcement!


Before you move on to see the solution, we have something for you.

  • The top 5 students of APSEd, which include all our subscribers will be rewarded hugely on the basis of GATE 2022 results.

  • Other than this, all the GATE top 1000 rankers are eligible to get the APSEd interview course for FREE.

Note: A special surprise reward is waiting for the students who get a top 100 AIR.


For any sort of queries, just hit us an email at support@apsed.in


GATE 2022 Aptitude Questions


As the questions are memory-based, for most of the questions, it is not sure to say whether the question is MCQ or NAT. But, the question is correct to the best of our knowledge.


We will be updating this blog with more memory-based questions. Keep coming back for more. The official key will be out soon on the GATE official website.


Question 1


The ____ is too high to be considered ____.


a) fare/fare

b) fair/fare

c) fair/fair

d) fare/fair


Hint


This is a very simple question. You just need to know about homophones i.e., words that have similar pronunciation but different meanings.


To make it simpler, fare means the price and fair means legit/unbiased etc. Therefore, the correct option is d) fare/fair.


Question 2


Bell-1, bell-2, and bell-3 ring after every 20 min, 30 min and 50 min respectively. If they ring at 12:00 pm, at what time they will ring again together?


a) 5:00 pm b) 6:00 pm

c) 4: 00 pm d) 4:30 pm


Hint


Find the least common multiple (L.C.M.) between the time intervals of bell ringing frequency i.e., 20, 30, and 50. Convert the L.C.M., which will be in minutes, into hours.


LCM (20,30,50) = 300 min. i.e., 5 hours


Add this hour with the initial time when the bells rang together i.e., 12 in this case, to get the time after which the bell will ring together again.


12:00 + 5 = 5:00 pm


Question 3


The ratio of the speeds of P:Q in a 500m race is 3:4. When the race starts P is 140m ahead of Q. Find the distance between P and Q when P wins the race.


a) 40m b) 20m

c) 60m d) 80m


Hint


This question could be easily solved by knowing the concept of the speed-time-distance relationship.


Let the speed of P and Q be 3x and 4x respectively. P covers the remaining distance in time 't'. The distance which P has to cover to win the race is given by,


500 - 140 = 360m


By speed-time relation, time (t) = distance/speed. Therefore, t = 360/3x, which is equal to 120/x or,


x*t = 120


Distance covered by Q will be 4*x*t. As we know, x*t = 120. Therefore, the distance covered by Q is,


4*120 = 480m


The question doesn't end here, we must find the distance between P and Q, which is given by,


500 - 480 = 20m


Question 4


There are two unbiased dices with a single letter on each face. The letters in dice are, Q, R, S, T, U, and V. The dices are thrown independently once. What is the probability that the outcomes were composed only of any of the following combinations of the following outcomes, Q, U, and V?


a) 1/4 b) 3/4

c) 1/6 d) 5/36


Hint


We need to get an outcome of Q, U, and V only in both the dices. The sample space for a single dice is 6. Therefore, the probability of getting a Q, U, and V in dice is 3/6.


But, we have to find the probability of getting this outcome in both the dices (i.e., and condition). Therefore, multiply the probabilities of both the dice,


i.e., (3/6)*(3/6) = 0.25


Question 5


A function of y(x) is defined in the interval [0,1] on the x-axis as,

y(x) = 2 if 0≤x≤1/3

y(x) = 3 if 1/3≤x≤3/4

y(x) = 1 if 3/4≤x≤1


Find the area under the curve for the interval [0,1] on the x-axis.


a) 6/13 b) 6/5

c) 5/6 d) 13/6


Hint


No integration is required in this case, as the curve formed by the function is a simple one.


If you plot the function in a rough graph, you will get three rectangles with dimensions (b * h) as,


(1/3) * 2, (5/12) * 3, and (1/4) * 1.


Find the area of each of the rectangles and add them up to get the final answer. In this case, the answer is 13/6.


Question 6


If, r^2 + 2r + 6 = 0, then find the value of (r+2)*(r+3)*(r+4)*(r+5).


a) 126 b) 52

c) -126 d) -52


Hint


Most of us will think of solving the first equation to find the r value first. But, it will not yield a single value as it is a quadratic equation. And as a result, you will not be able to find the value for the second equation.


Therefore, simplify the second equation first as,


(r+2)(r+3)(r+4)(r+5) = (r^2 + 5r + 6)(r^2 + 9r +20)


From the first equation, r^2 + 2r = -6, substitute this in the simplified second equation as,


=(3r + 6 - 6)(7r + 20 - 6)


Further simplifying we get,


21(r^2 + 2r), which is 21 * -6 = -126.


Question 7


A box contains five balls of the same size and shape. Three of the green coloured balls and two of them are orange-coloured balls. If a green place is drawn is not replaced. If an orange ball is drawn it is replaced with another orange ball. Balls are drawn from the box one at a time. The first ball is drawn. What is the probability of getting an orange ball in the next draw?


a) 19/50 b) 8/50

c) 1/2 d) 23/50


Hint


If a green ball is drawn in the first draw, then the probability for the second draw is,

(3/5)*(2/4) (i.e., 3/5 for the first draw and 2/4 for getting an orange ball in the second draw)


If an orange ball is drawn in the first draw, the ball is replaced with an orange ball. Therefore, the probability will be,


(2/5)*(2/5)


We are not given with outcome of the first draw. But, as only either of these situations actually exists, adding the probabilities of both situations will give the required probability (i.e., OR condition).


(6/20) + (4/25) = 23/50 is the answer


Question 8


As you grow older, an injury to your _____ may take longer to _____


a) heel/heel

b) heal/heel

c) heel/heal

d) heal/heal


Hint


As we saw earlier this is a very simple question. You just need to know about homophones i.e., words that have similar pronunciation but different meanings.


To make it simpler, heal means to recover and heel means foot. Therefore, the correct option is d) heel/heal.


Question 9


The price of an item is 10% cheaper in store S compared to the price of another store M. Store S charges 150 rupees as delivery charge, while store M puts no delivery charge. A person bought the item from store S and saved 100 rupees. What is the price of the item at store M?


a) 1750 b) 2200

c) 1500 d) 2500


Hint


Frame the equation by considering the price of the item at store M as x and by considering delivery charges & 10% cheaper rate at store S. The equation could be framed by cost saved, which is given a