LOGICAL REASONING – 1
Solutions for 1 to 6: 1.4. (c)
1.5. (b)
Solutions for 1.6 to 1.8:
1. (a)
2. (d)
1.6. (a)
3. (d)
1.7. (a)
4. (d)
1.8. (a)
5. (c)
Solutions for 1.9 to 1.11:
6. (a)
Solutions for 7 to 8:
D/H E/G A
I C B
F D/H E/G
7. (c)
8. (a)
Solutions for 9 to 11: 1.9. (a)
In terms of height, we have : B < F, C < F, C < D, E < D, F < E, A < E, F < 1.10. (c)
A. So, C < F < E < D, B < F, F < A < E.
Thus, the sequence becomes : B < C < F < A < E < D or C < B < F < A < 1.11. (d)
E < D. Type-2
In terms of weight, we have : A < B, A < C, B < D, D < F.
Solutions for 2.1 to 2.3:
So, A < B < D < F, A < C.
From I, we have : X > L.
Thus, the sequence becomes :
From II, we have : M = N.
A < C < B < D < F or A < B < C < F or A < B < D < C < F.
From III, we have : Z is the youngest.
9. (c) From IV, we have : N > Y.
Clearly, D is the tallest. From V, we have : Y > X
10. (d) 2.1. (c)
Clearly, F, B and C are shorter than A. From I and V, we have : Y > X > L or Y > L.
11. (c) 2.2. (a)
Clearly, F is heavier than B and C but shorter than D. From I, II, IV and V, we have : M = N > Y > X > L, i.e. M > L. Thus,
Type-1 L is younger than M. So, III is not required.
ZENITH
Solutions for 1.1 to 1.5: 2.3. (d)
As given, we have :
Prerna > Shaloo > Rakhi; Rakhi = Meghna, Rakhi >
Komal. Thus, the sequence of heights becomes:
Prerna > Shaloo > Rakhi = Meghna > Komal
Clearly, Meghna is shorter than Shaloo.
Solutions for 2.4 to 2.6:
Let C’s weight be x. Then, D’s weight = 2x, E’s weight= 4x, B’s weight
= 4.5x and
1.1. (d) A’s weight = 9x
1.2. (b) So, the order of weights is : A > B > E > D > C.
1.3. (d) 2.4. (c)
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ZENITH
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C is the lightest in weight. 3.8. (c)
2.5. (b) B has got 6th highest marks.
E is heavier than D and C. Solutions for 3.8 to 3.11:
2.6. (a) In terms of height, we have : T > P, T > S, Q > T > P.
Clearly, the descending order of weights is A, B, E, D, C R plays Volleyball, so R is the shortest.
Q plays neither Volleyball nor Basketball.
2.7. (a) So, Q is not the tallest. Thus, U is the tallest.
Let A, B, C and D denote the amounts with these individuals So, the sequence becomes : U > Q > T > P > S > R.
respectively. Then, A + B = C + D, B + D > A + C, A > B. Now, T plays Tennis. U, being tallest, plays Basketball. R plays
Now, B + D > A + C and A > B Volleyball. Q plays Football. Both P and S play either Cricket and
B+D>A+C>B+C ( A > B) Badminton.
B + D > B + C D > C. 3.9. (d)
Also, A + B = C + D, A > B and D > C S is taller than R but shorter than P.
2B < A + B = C + D < 2C
3.10. (a)
2B < 2C C > B. Clearly, P is taller than R.
Now, A > B, C > B, D > C => A > B, D > C > B
Thus, each one of A, D and C has more amount than B. 3.11 .(c)
Hence, B has the least amount. S plays either Cricket or Badminton.
Type-3
Solutions for 3.1 to 3.5:
Days Subjects
Monday Statistics
Tuesday History
Wednesday English
Thursday Economics
Friday Mathematics
3.1. (b)
3.2. (b)
3.3. (c)
3.4. (c)
3.5. (a)
Solutions for 36. to 3.8:
A has Philosophy, B has Biology and D has Physics as his favourite
subject. Now, neither English nor Mathematiccs is C’s favourite
subject. Since C got the second highest marks, so his favourite
subject cannot be Chemistry also. So, C’s favourite subject is Soci-
ology. E’s favourite subject is neither Sociology nor Mathematics.
Also, E got the least marks and so his favourite subject cannot be
ZENITH
Chemistry. So, E’s favourite subject is English.
In terms of marks obtained, it is given that :
D > B, D > G, A > D, B > E, G > B.
Thus, we have : A > D > G > B > E.
Clearly, G did not get the highest marks. So, F’s favourite subject is
Chemistry and G’s favourite subject is Mathematics.
Hence, F got the highest and C got the second highest marks. Thus,
the order of marks obtained becomes: F > C > A > D > G > B > E.
3.6. (c)
Chemistry is the favourite subject of F.
3.7. (a)
E’s favourite subject is English.
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ZENITH
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