Quantitative Aptitude - Arithmetic Ability Questions

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

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

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.

The given number series 11, 13, 30, 96, ? follows a pattern that,

11

11 x 1 + 2 = 13

13 x 2 + 4 = 30

30 x 3 + 6 = 96

96 x 4 + 8 = 392

Hence, the next number in the given series is 392.

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.