Quantitative Aptitude - Arithmetic Ability Questions

 Let   Let x=333= 333

 Then, logx = 33 log3  

= 27 x 0.47712 = 12.88224 

Since the characteristic in the resultant value of log x is 12

∴The number of digits in x is (12 + 1) = 13 

Hence the required number of digits in 333is 13.

fix one person and the brothers B1 P B2 = 2 ways to do so.
other 17 people= 17!

 
Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!

343 = 73,  1331 = 113 , 2197 = 133, 121 = 112

343, 1331, 2197 are perfect cubes, but not 121.

3712 = 752

Required number =  75218 = 75 x 4 = 300

A perfect cube will have prime factors that are in groups of 3; for example 125 has the prime factors 5 x 5 x 5 , and 64 x 125 will also be a cube because its factors will be 4 x 4 x 4 x 5 x 5 x 5

Consider the answer choices in turn.

8 is the cube of 2, and p is a cube, and so the product will also be a cube.

pq will also be a cube as shown above.

pq is a cube and so is 27, but their sum need not be a cube. Consider the case where p =1 and q = 8, the sum of pq and 27 will be 35 which has factors 5 x 7 and is not a cube.

-p will be a cube.

Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).

Let the two numbers be x and y

Given x : y = 3 : 7 .....(1)

Now, x+6 : y+6 = 5 : 9 .....(2)

From (1), x = 3y/7

From (2), 5y - 9x = 24

=> 5y - 9(3y/7) = 24

=> y = 21

=> From(1), x = 9

Hence, the two numbers be 9 and 21

By trial and error method, when we substitute 

x = 0

Both LHS and RHS are equal.

Given that Rs. 1860 will become Rs. 2641.20 at 12%

=> Simple Interest = 2641.20 - 1860 = Rs. 781.20

We know I = PTR/100

=> 781.20 x 100 = 1860 x T x 12

=> T = 78120/1860x12

=> T = 78120/22320

=> T = 3.5 years.