Quantitative Aptitude - Arithmetic Ability Questions

56% of 870 = 56x870/100 = 487.20

82% of 180 = 82x180/100 = 147.60

32% of 90 = 32x90/100

487.20 + 147.6 - 28.8  =  ?

? = 634.8  - 28.8

? = 606.

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 x 2000) - (2000 x 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> 14012 - 6002 = 8010.

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.

Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.

So, 9 is wrong.

Let the required number days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

 Spiders1:7Webs7:1⋮⋮ 7:x 

=> 1 * 7 * x = 7 * 1 * 7 

=> x= 7

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.