CAT Questions

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As Tennis is a kind of Sport, similarly, Chemistry is a kind of science.

Hence, Tennis : Sport :: Chemistry : Science.

The given series 404, 415, 402, 413, 400, ? follows a pattern that,

404

404 + 11 = 415

415 - 13 = 402

402 + 11 = 413

413 - 13 = 400

400 + 11 = 411

Hence the next number in the given number series is 411.

The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B.

 Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535

Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 40 km at a speed of 8kmph
time taken indivisually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 40/8 = 5 m

Average speed of man= total distance/ total time
=> 54/13 = 4213 kmph

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

The simplest triangles are AKI, AIL, EKD, LFB, DJC, DKJ, KIJ, ILJ, JLB, BJC, DHC and BCG i.e. 12 in number.
The triangles composed of two components each are AKJ, ALJ, AKL, ADJ, AJB and DBC i.e. 6 in number.
The triangles composed of the three components each are ADC and ABC i.e. 2 in number.
There is only one triangle i.e. ADB composed of four components.
Thus, there are 12 + 6 + 2 + 1 = 21 triangles in the figure.

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

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.