Aptitude and Reasoning Questions

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.

The sum of square of two no is 116 i.e no lie 1-11, because 11x11 =121 which is greater than 116
again after cheaking all possible pair of no only 4 and 10 satisfy this condsn , again the progression is of positive integer so 10 cant not be 1st term
so the 1st term is 4,
but from the 2nd we cant able to find the 1st term

so only 1st statement is sufficient to answer the question

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.

In the figure, there are

5 columns containing 4 cubes each;

33 columns containing 3 cubes each;

9 columns containing 2 cubes each and 3 columns containing 1 cube each.

∴ Total Number of cubes = ( 5 x 4) + (33 x 3) + (9 x 2) + (3 x 1) = 20 + 99 + 18 + 3 = 140

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

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.