Quantitative Aptitude - Arithmetic Ability Questions

I. 2x2 + 19x + 42 = 0

2x2 + 12x + 7x + 42 = 0

2x (x+6) + 7(x +6) = 0

(2x+7) (x+6) = 0 x = -7/2, -6 I

I. 4y2 + 43y + 30 = 0

4y2 +40y + 3y + 30 = 0

4y(y + 10) + 3(y + 10) (4y + 3)(y + 10) = 0

y = -3/4, -10

So answer is no relation.

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.

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.

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