Quantitative Aptitude - Arithmetic Ability Questions

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

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

 L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2 

Therefore, Required  Number = (840 x 2+3)=1683

2ab=a2+b2-a2-b2

29-9=20

=>ab=10

Given a2 + b2 = c2

Now  1logc+ab + 1logc-ab

= logbc+a + logbc-a

= logbc2-a2

= 2logbb = 2

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.

Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.

There is a proposal to increase these seats by 40%, 50% and 75% respectively.

Then Seats for Mathematics, Physics and Biology in a school will be

1.4 X 5x, 1.5 X 7x and 1.75 X 8x and

ratio will be 7 : 10.5 : 14  => 2:3:4.

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million 

i.e. n3=106  (1 million = 106)

  => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

When 3 passengers income was 3x
expense= Rs.30 
profit =20% of 30 = Rs.6 
That means his earning is Rs.36. so that per passenger fare must be Rs.12.

When 4 passengers
earning = 12x4=Rs.48.
expense =Rs.24.
profit = Rs.24 = 100%

 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.