Quantitative Aptitude - Arithmetic Ability Questions

One way to deal with fractions is to convert them all to decimals. 

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7. 

 Since the month begins with a Sunday, so there will be five Sundays in the month,

Required average = (510 * 5 + 240 * 25) / 30 = 8550/30 = 285.

Given series is 

12        16         32          68         ?

Here it follows a pattern that,

12

12 + (2x2) = 16

16 + (4x4) = 32

32 + (6x6) = 68

68 + (8x8) = 132

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Here the given series 28, 12, 4, 2, -2 follows a pattern that

$$\begin{gathered} 28 \\ 28 - 24 = 28 - 16 = 12 \\ 12 - 23 = 12 - 8 = 4 \\ 4 - 22 = 4 - 4 = 0 (NOT 2) \\ 0 - 21 = 0 - 2 = -2 \end{gathered}$$

Hence, the odd man in the given series is 2.

Let the sum be Rs. x

a. gives, S.I = Rs. 9000 and time = 9 years.

b. gives, Sum + S.I for 6 years = 2 x Sum

--> Sum = S.I for 6 years.

Now, S.I for 9 years = Rs. 9000

S.I for 1 year = Rs. 9000/9 = Rs. 1000.

S.I for 6 years = Rs. (1000 x 6)= Rs. 6000.

--> x = Rs. 6000

Thus, both a and b are necessary to answer the question.