# Logical Reasoning capsules by Vikas Saini - Set 9

• Set 1

In a college a class is divided in 4 groups A, B , C and D. Each group comprises of 20% girls

GroupsNumber of students
ABC170
ACD140
BCD130
BDA160

Q1) How many students in group D.
Q2) How many girls in group B.
Q3) How many boys in group C.

Solution :-

A + B + C = 170.
A + C + D = 140.
B + C + D = 130.
B + D + A = 160.
3(A + B + C + D) = 600.
A + B + C + D = 200.
A = 70, B =60, C = 40, D = 30.

(1) 30.
(2) 0.2 x 60 = 12.
(3) 40 x 0.8 = 32.s

Set 2
A coding system performs following operations on the alphabets. P0 = A letter at position odd is substituted with the next letter while the letter at even position is substituted with previous letter. After the substitution, the whole word is written in reverse order. For example coding of the word INDIA will be as follows: I substituted with J, N with M and so on to form the word JMEHB and then the word is written in the reverse order to form BHEMJ.
Pn for n = 1,2,3…… the operation is that every letter is written with n steps ahead. For example ASIA when performed with P1 will be BTJB and then it will also be written in reverse order to form the word BJTB.

I. What will be the code of the word MADAM if the operations P0 and P3 are performed?
II. If operations P0,P1,P3 are performed the word becomes IRFKIL. What is the original word?

Solution :-

I. MADAM after performing P0 it becomes NZEZN. After reversing it is same NZEZN.
After performing P3 it becomes = QCHCQ.

II. After operations P0,P1,P3 any word becomes IRFKIL.
Firstly make it reverse :- LIKFRI
Now before P3 it was :- IFHCOF.
Now again make it reverse :- FOCHFI
Now before P1 it was :- ENBGEH
Now make it reverse again :-HEGBNE
Now before P0 it was :-GFFCMF (Answer)

Set 3

There are two parts in a question paper. Part A has 5 questions, each questions fetches either +2 or 0. Part B has 4 question each question fetches either +10 or 0 and +5 if partly correct.

I. Which of the following score is not possible to achieve in the paper?
(a) 29
(b) 31
(c) 33
(d) 47

II. A student scores 41 marks in all. How many questions has he attempted atleast?

III. A, B, C and D are four students what can be the maximum difference between the scores of (A + B ) and (C + D).

Solution :-

Marks possible in part A = 0,2,4,6,8,10.
Marks possible in part B = 0,5,10,15,20,25,30,35,40.

I. Let’s go through options
(a) 29 = 4+25
(b) 31=6+25
(c) 33 = 8+25
(d) 47 not possible.
47 not possible.

II. 41 = 6 + 35
3 questions from part A and 4 questions from part B.
Total 7 questions atleast.

III. Maxmimum marks can get by A & B = 10 + 40 = 50.
C & D = 0.
Difference = 50 + 50 – 0 = 100.

Set 4

A professional juggler juggles 10 balls of different colors, 1 yellow, 5 black, 2 red and 2 green, but in a specific order, a part of which is – [ B, B, _ , _ , _ , G, _ , _ , _ , R]. The blanks denote that it is not known which ball is juggled in that place. He juggles the first balll at time t = 1 second, the second ball black at time t = 2 second, and so on unitl the tenth ball red at time t = 10 second, after which he again throws the first black ball at t = 11 second. This cycle goes on for a very long time. He throws the balls from his right hand and catches them in his left hand. A ball is air‐borne for exactly 9 seconds.

Some additional facts as given below:

1. He throws the green ball at t = 23 second.
2. He throws the black ball at t = 47 second.
3. He catches the black ball at t = 87 second.
4. No 3 black balls are thrown consecutively.
5. No 2 red balls are thrown consecutively.

Q1) At what time (in seconds) does he catch the green ball for the seventh time?
(a) 32
(b) 33
(c) 42
(d) 43

Q2) Which ball does he throw at t = 5 seconds ?
(a) Black
(b) Red
(c) Green
(d) CBD

Q3) Which ball does he catch at t = 98 seconds ?
(a) Black
(b) Red
(c) Green
(d) Yellow

Solution :-

Order is given as – [ B, B, _ , _ , _ , G, _ , _ , _ , R]
Green ball he throws at t = 23 it means third ball he throws is G.
Black ball he throws at t = 47 seconds, it means 7th ball he throws Black ball.
He catches black balls at t=87 seconds it means he throws this ball at t = 78 seconds.
Hence at 8th order he throws black balls.
Now the order becomes – [B, B, G,_ , _ ,G,B, B,_,R]
It’s given no 3 black balls and no two red balls can be thrown consecutively.
Hence 9th balls shall be Yellow.
Now left 4th and 5th places.
4th – B/R
5th – R/B

Order – [B,B,G,B/R,R/B,G,B,B,Y,R]
He throws green balls at time t = 3,7,13,17,23,27,33.
At t = 33 seconds he will throw the ball for 7th time but he catches it 9 seconds later.
Hence at t = 42 seconds he will catch it.

At t = 5 he throws either Red or Black.
It can’t be determined.

The ball he catches at t = 98 seconds that was thrown at t = 89 seconds.
On 89th seconds he throws 9th no. ball. That ball is yellow.

• awesome article :D

• @Ashu-Singh Thank You :)

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