## Infosys Sample Logical Aptitude Questions

Below are three practice logical and/or aptitude questions for you to practice.

Question 1

Saikumar, Varun and Prasanth start from one of the points X and Y and move in their vehicles towards the other point. The distance between X and Y is 550 km. Saikumar and Varun start at 7 am from X and Y respectively whereas Prasanth starts from X at 8 am. Saikumar, Varun and Prasanth maintain constant speed of 60 km, 50 km and 70 km respectively. Which pair Saikumar-Varun or Varun-Prasanth or Prasanth-Saikumar would meet first.

a) Saikumar – Varun b)Varun-Prasanth c) Prasanth-Saikumar d) cannot be determined

Answer : a) Saikumar – Varun

Solution :

Saikumar and Varun start at the same time, the initial distance between them is the distance XY = 550 km
They are moving towards each other i.e. in opposite directions and hence the relative speed = sum of the speeds = 60 + 50 = 110 km.
Hence the time taken for meeting = 550/110 = 5 hours.
i.e. Saikumar and Varun would meet 5 hours from 7 am or at 12 noon.

Saikumar and Prasanth starting from the same point begin their travel at different points of time. Since Prasanth is the late starter, the initial distance between them is the distance between them when Prasanth starts, namely at 8 am.
From 7 to 8 am Saikumar would have travelled 60 km and hence the distance between them at 8 am is 60 km.
Since they are moving in the same direction the relative speed is the difference between their
Speeds --- 70 – 60 = 10 km/hr.
Therefore, time taken for meeting = 60/10 = 6 hours. This implies that Saikumar and Prasanth would meet 6 hours after 8 am. i.e. 2 pm

Varun and Prasnath start from opposite points at different points of time.
At 8 am. The distance between them is 540-50 = 490 km . Since they are moving in opposite directions, the relative speed = 50 + 70 = 120 kmph
The time taken for meeting = 4.08 from 8 am i.e. 12.08 hrs.
Thus, Saikumar and Varun meet first at 12 noon, Varun and Prasanth meet at 12.08 pm and the last pair to meet Sai Kumar and Prasanth at 2 pm.

Question 2

Anna Salai, Madurai has 1100 buildings. A painter was engaged by the Madurai Corporation ( after calling for sealed tenders fixing the last date for receipt of tenders and the tender was opened in the presence of representatives of tenderers). How many zeros will he need to paint.?

a) 212 b) 192 c) 213 d) none of these.

Answer : a) 212.

Solution :

Divide 1100 building numbers into groups of 100 each as follow:
1) [(1..100)],
2) [(101..200), (201..300), ...to (801...900) also (1101-1100)] ...
3) [ (901..1000)]
For the first group, sign-maker will need 11 zeroes.
For group numbers 201 to 900 and 1101 to 1100 , he will require 20 zeroes each.
And for group number 10 (from 901 to 1000), he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + (9 x 20) + 21 = 212

Question 3

St. Xavier Matriculation School, Salem conducted sports meet and awarded prizes. In the sports contest there were m medals awarded on n successive days (n > 1).
1. On the first day 1 medal and 1/6 of the remaining m - 1 medals were awarded.
2. On the second day 2 medals and 1/6 of the now remaining medals was awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?

a) 6 days , 36 medals b) 7 days, 49 medals c) 5 days, 25 medals d) none of these.

Answer : c) 5 days , 25 medals.

Solution :

For this question, we have to go by options. Only for option c) all the conditions in questions will be met as given below. (We recommend you to try with other options, to prove that they are wrong.)
On day 1: Medals awarded = (1 + 24/6) = 5 : Remaining 20 medals
On day 2: Medals awarded = (2 + 18/6) = 5 : Remaining 15 medals
On day 3: Medals awarded = (3 + 12/6) = 5 : Remaining 10 medals
On day 4: Medals awarded = (4 + 6/6) = 5 : Remaining 5 medals
On day 5: Medals awarded = 5

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