In following questions, one term in number series is incorrect.

Find out the incorrect number

6, 14, 30, 64, 126

Find out the incorrect number

6, 14, 30, 64, 126

Each term is multiplied by 2 and then increased by 2 to obtain the next term.

So, 64 is wrong and must be replaced by (30 * 2 + 2) i.e. 62..

In following questions, a sequence of groups of alphabets and numbers is provided with one term missing , Choose the missing term from options. FLP, INS, LPV, ?

The first and third letters of each term are moved three steps forward and the second letters is moved two steps forward to obtain the corresponding letters of the next term..

In three coloured boxes - Red, Green and Blue, 108 balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box ?

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then,

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108; 6R = 108 ;R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

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Look at this series: 2, 6, 18, 54, ...

What number should come next?

What number should come next?

This is a simple multiplication series.

Each number is 3 times more than the previous number.

The series is 2, 6, 18, 54, . . .

Rule = (1st number x 3 = 2nd number), (2nd number x 3 = 3rd number)....

Now

(2), (2 x 3 = 6), (6 x 3 = 18), (18 x 3 = 54), (54 x 3 = 162)...

So the answer is 162..

A person travel at the rate of 60 miles per hour and cover 300 miles in 5 hours. If he reducing is paid by 10 miles per hour how long will it take to cover the same distance?

Original speed= 60 miles per hour

If the speed is reduced by 10 miles per hour.

Then the new speed= 50 miles per hour

Distance to be covered= 300 miles

Time= distance/speed

Time=300/50

Time=6 hr

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A man has a daughter and a son. The son is is three years older than the daughter. In one year the man will be six time as old as the daughter is now. In ten years the man will be fourteen years older than the combined ages of his children at that time. what is the man's present age?

Let's the age of daughter=x

age f son = x+3

age of father now=6x-1

condition:

M+0={(S+D)+ 10}+14]

6x-1=[{(x+3)+10}+14]

4x=28

x=7

Father's age=(6x-7)-1=41

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