Verbal Reasoning - Mental Ability Questions

From I, we find that the weight of one pole is (5 x 4) = 20 kg ,and so the weight of 10 poles is 200 kg.

Thus, I alone is sufficient.

From II. we have : (weight of 3 poles — weight of 2 poles) -= 20 kg or weight of one pole - 20 kg.

So. weight of 10 poles 200 kg.

Thus. II alone is also sufficient 

Let the number of apples in each tree of the 5 farmers be a, b, c, d,e respectively. Therefore total no of apples are 7a, 9b, 11c, 13d and 14e respectively.

Given,
7a+1 = 9b+3 = 11c-1 = 13d+3 = 14e-6 = x

9b+3 = 13d+3
===> 9b = 13d
so take b = 13 and d = 9

9b+3 = 9*13+3 = 120
ie, x=120

substituting
7a+1 = 120
7a = 119
a = 17

11c-1 = 120
11c = 121
c=11

14e-6 = 120
14e = 126
e = 9

Yields per tree in the orchards of d 3rd & 4th farmers= 11,9

Warm and hot are in the same meaning sense , so not so good is same as worse in meaning.

Man's son's sister — Man's daughter. So, the man's daughter is the mother of the woman's husband 

Thus, the woman's husband is the grandson of the man in the photograph. 

Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.

We may consider the two cases as under:

Case I:  ←3C↔8B↔5A→21

Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40

Case II:  ←3C A↔5B

Number of persons between A and C                  

= (8 - 6) = 2   

∵[C↔8B A→21B]

Clearly number of persons in the queue = (3+1+2+1+21) = 28

Now, 28 < 40. So, 28 is the minimum number of persons in the queue.