CAT Questions

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The pattern of given series is :

5

14 = (5x3) - 1

45 = (14x3) + 3

130 = (45x3) - 5

(?) = (130x3) + 7

=> (?) = 397

1182 = (397x3) - 9

Thus, the missing number is 397.

We know that,

Odd days --> days more than complete weeks

Number of odd days in 400/800/1200/1600/2000 years are 0.

Hence, the number of odd days in first 1600 years are 0.

Number of odd days in 300 years = 1

Number of odd days in 49 years = (12 x 2 + 37 x 1) = 61 days = 5 odd days

Total number of odd days in 1949 years = 1 + 5 = 6 odd days

Now look at the year 1950

Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days

Total number of odd days = 6 + 5 = 11 => 4 odd days

Odd days :-

0 = sunday ;

1 = monday ;

2 = tuesday ;

3 = wednesday ;

4 = thursday ;

5 = friday ;

6 = saturday

Therefore, Jan 26th 1950 was Thursday.

The given number series is 50, 51, 47, 56, 40, ?

$$\begin{gathered} 50 \\ 50 + 12 = 51 \\ 51 - 22 = 47 \\ 47 + 32 = 56 \\ 56 - 42 = 40 \\ 40 + 52 = 65 \end{gathered}$$

Hence, the next number in the given number series is 65.

Here the given number analogy follows two patterns that,

6 

18 --> 6 x 3

4

12 --> 4 x 3

OR

6

18 ----> 6x6/2 = 18

4

8 ----> 4x4/2 = 8

As 12 is not available in the options, 8 is correct answer.

Time = (100 x 81) / (450 x 4.5) = 4 years.

Since, we have that,

The numerator and denominator is (8+x), which is a polynomial is continuous for all the real numbers i.e. for all x belonging to .

Thus, the limit exists and is given by,

limx→128+x8+x = 8+128+12= 1.

Stars are what that come out at night without being called and are lost in the day without being stolen.

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.