Quantitative Aptitude - Arithmetic Ability Questions

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

Given Today is 20th January 2017, Sunday

In january, we have 31 days

February - 28 days (Non leap year)

March - 31 days

April - 30 days

=> Remaining days => 31 - 20 = 11 in Jan

+ 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days

More 20 days to complete 90 days => upto 20th April

Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.

Now, the day of the week will be 

90/7 => Remainder '6'

As the day starts with '0' on sunday 

6 => Saturday.

Required day is 20th April, Saturday.

I. gives, S.I for 3 years = Rs. 4500.

II. gives, Rate = 10% p.a.

III. gives, (C.I.) - (S.I.) = Rs. 465.

Clearly, using I and III we get C.I. = Rs. (465 + 4500).

Thus, II is redundant.

Also, from I and II, we get sum =

100 x 4500

= 15000.

10 x 3

Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.

Thus, III is redundant.

Either II or III is redundant.

2l+bb=51      => 2l + 2b = 5b     => 3b = 2l 

b=(2/3)l

Then, Area = 216 cm² 

=> l x b = 216     => l x (2/3)l =216 

l = 18 cm.

Saving  = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864.

(A+B) - (B+C) = 12

A - C = 12.

C is younger than A by 12 years.