Aptitude and Reasoning Questions

Candidates passed in atleast four subjects

= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)

= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)

= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555    

Here the series follow the rule that

12+22 = 5

32+42 = 25

52+62 = 61

72+82 = 113

SO NEXT IS, 92+102 = 181

Given series follows Cubes on Prime Numbers 

i.e 23, 33, 53,...but  93is 1331

so 1381 is wrong

Arithmetic Series ::

An Arithmetic Series is a series of numbers in which each term increases by a constant amount.

How to find the sum of the Arithmetic Sequence or Series for the given Series ::

When the series contains a large amount of numbers, its impractical to add manually. You can quickly find the sum of any arithmetic sequence by multiplying the average of the first and last term by the number of terms in the sequence.

That is given by Sn = n(a1 + an)2 Where n = number of terms, a1 = first term, an = last term

Here Last term is given by an = a1 + n-1d where d = common difference

Now given Arithmetic Series is 

2, 5, 8, 11,...

Here a1 = 2,  d = 3, n = 36 

Now, $$\begin{gathered} an= a1 + n - 1d \\ a36= 2 + 36 - 13 = 105 + 2 = 107 \end{gathered}$$

Now, Sum to 36 terms is given by

S36 = 36(2 + 107)2 = 36 x 1092 = 39242 = 1962

Therefore, Sum to 36 terms of the series 2, 5, 8, 11,... is 1962.

$$\begin{gathered} ?2 = 2×78-21+8+49-148 \\ = 57 - 21 \\ =36 \\ => ? = ±6 \end{gathered}$$

52000 ÷ 40 ÷ 65 × 30 = ? - √400

(52000 ÷ 40) ÷ 65 × 30 = ? - 20

(1300 ÷ 65) × 30 = ? - 20

20 × 30 = ? - 20

600 = ? - 20

? = 620