Quantitative Aptitude - Arithmetic Ability Questions

Let C.P. be Rs. x.

Then,= >1920-xx*100=x-1280x*100

=>  1920 - x = x - 1280

=>  2x = 3200

=>  x = 1600

 Required S.P. = 125% of Rs. 1600 =Rs(125/100*1600) = Rs2000

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...

Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.

Let the original number be x.

Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x. 

x-(99/100)x=10

=> x =1000

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Total volume of water displaced =(4 x 50) cu.m = 200 cu.m

Rise in water level = 20040×20=0.25m = 25cm

Given numbers are 1.08 , 0.36 and 0.90

 H.C.F of 108, 36 and 90 is 18                  [ ∵ G.C.D is nothing but H.C.F]

 Therefore, H.C.F of given numbers = 0.18                 

Let the required length be x meters

 More men, More length built     (Direct proportion) 

Less days, Less length built      (Direct Proportion)

Men20:35Days6 : 3⋮⋮ 56 :x 

=> (20 x 6 x X)=(35 x 3 x 56) 

=> x = 49 

Hence, the required length is 49 m.

Explanation:

Let rate = R% and time = R years.

Then, 1200*R*R100=432

12R2=432

=>R2=36⇔R=6