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

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.

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.

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 x 2000) - (2000 x 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> 14012 - 6002 = 8010.

Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.

So, 9 is wrong.

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200