GATE Questions

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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.

How to find the years which have the same Calendars : 

 Leap year calendar repeats every 28 years. 

Here 28 is distributed as 6 + 11 + 11.

Rules:

a) If given year is at 1st position after Leap year then next repeated calendar year is Given Year + 6.

 b) If given year is at 2nd position after Leap year then next repeated calendar year is Given Year + 11.

 c) If given year is at 3rd position after Leap year then next repeated calendar year is Given Year + 11. 

Now, the given year is 2018

We know that 2016 is a Leap year.

2016  2017  2018  2019  2020

 LY    1st     2nd    3rd    LY

Here 2018 is at 2 nd position after the Leap year.

According to rule ( b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.

The given number series is 3, 12, 39, 120, 364

$$\begin{gathered} 3 \\ 3 + 32 = 3 + 9 = 12 \\ 12 + 33 = 12 + 27 = 39 \\ 39 + 34 = 39 + 81 = 120 \\ 120 + 35 = 120 + 243 = 363 (Not 364) \end{gathered}$$

110% of S.P. = Rs. 616

S.P. = (616 x 100)/110 = Rs. 560

C.P = (110 x 560)/112 = Rs. 500

Let sum = S. Then, amount = 7S/6

S.I. = 7S/6 - S = S/6; Time = 3 years.

Rate = (100 x S) / (S x 6 x 3) = 5 5/9 = 50/9 %.

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

                          = 12×47+12×26=1942

The sum of square of two no is 116 i.e no lie 1-11, because 11x11 =121 which is greater than 116
again after cheaking all possible pair of no only 4 and 10 satisfy this condsn , again the progression is of positive integer so 10 cant not be 1st term
so the 1st term is 4,
but from the 2nd we cant able to find the 1st term

so only 1st statement is sufficient to answer the question

Number of persons between Amrita and Mukul = 50 - (10 + 25) = 15. Since Mamta lies in middle of these 15 persons, so Mamta`s position is 8th from Amrita i.e. 18th from the front.