Find the odd man out.
2, 5, 10, 17, 26, 37, 50, 64
(1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1
But, 64 is out of pattern.
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41, 43, 47, 53, 61, 71, 73, 81
Each of the numbers except 81 is a prime number.
Simplify
1.53 + 4.73 + 3.83 - 3 x 1.5 x 4.7 x 3.81.52 + 4.72 + 3.82 - 1.5 x 4.7 - 4.7 x 3.8 - 1.5 x 3.8
Apply a3 + b3 + c3 - 3abca2 + b2 + c2 - ab - bc - ca = a + b + c
Hence the answer is 1.5 + 4.7 + 3.8 = 10.
6, 9, 15, 21, 24, 28, 30
Each of the numbers except 28, is a multiple of 3.
92613 x 62742241 = ? + 89
Find the value of '?'
Given 92613 x 62742241 = ? + 89
Using unit digit method,
On LHS,
1 x 1 x 1 = 1
sqrt of last digit 1 => it must end with either 9 or 1
so now
1 x 1 = 1
1 x 9 = 9
On RHS,
? + 89
=> ? = ...1 - 89 = last digit is 11- 9 = 2
or
=> ? = ...9 - 89 = last digit is 9 - 9 = 0
Here in the options last digit 2 is option C.
Today is Monday. After 61 days, it will be :
Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
If the length of the diagonal of a square is 20cm,then its perimeter must be
a. 402cm b. 302cm c. 10cm d. 152cm
We know that d=√2s
Given diagonal = 20 cm
=> s = 20/2 cm
Therefore, perimeter of the square is 4s = 4 x 20/2 = 402 cm.
Find the odd man out?
396, 462, 572, 427, 671, 264.
Here the given series is 396, 462, 572, 427, 671, 264.
In all the terms, the middle digit is the sum of first and third digit except 427.
So the Odd number in the given series is 427.