Quantitative Aptitude - Arithmetic Ability Questions

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March. April. May. June.
(31 + 28 + 31 + 30 + 31 + 17) = 168 days
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

This can be easily simplified by Approximation process

Given 39.96% of 649.85 ÷13.07 = 45.14 - ? can be written as 

40% of 650 ÷13 = 45 - ?

$$\begin{gathered} 40 x 65010013 = 45 - ? \\ ? = 45 - 20 \\ ? = 25. \end{gathered}$$

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.

Here the given number series is 3, 14, 36, 67, 113, 168

It follows a pattern that

3

3 + 1 x 11 = 3 + 11 = 14

14 + 2 x 11 = 14 + 22 = 36

36 + 3 x 11 = 36 + 33 = 69

69 + 4 x 11 = 69 + 44 = 113

113 + 5 x 11 = 113 + 55 = 168

Hence, the wrong number is 67.

Given equation can be written as

$$\begin{gathered} 215 x 103 + ? = 24 \\ => 14 + ? = 24 \\ => ? = 24 - 14 \\ => ? = 10 \end{gathered}$$

Total distance travelled in 12 hours    =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance    = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.