Quantitative Aptitude - Arithmetic Ability Questions

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

Next number of the sequence 1, 4, 12, 32,...

1 = 2 x 20

4 = 2 x 21

12 = 3 x 22

32 = 4 x 23

80 = 5 x 24.

The given number series is 90 72 56 42 36 20

9 x 10 = 90

8 x 9 = 72

7 x 8 = 56

6 x 7 = 42

5 x 6 = 30 (not 36) 

4 x 5 = 20

Hence, the odd man in the given number series is 36.

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.

1 x 1 + 1 x 1 = (1 + 1) = 2

2 x 2 + 2 x 2 = (4 + 4) = 8

8 x 3 + 3 x 3 = (24 + 9) = 33

33 x 4 + 4 x 4 = (132 + 16) = 148

148 x 5 + 5 x 5 = (740 + 25) = 765.

Middle numbers  =  [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.

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