Quantitative Aptitude - Arithmetic Ability Questions

Say total cost price of tea is x.
Then total profit at a rate of 15% is = (15x/100)
According to question,
15x/100 = 60
so x = 400
C.p of the tea is Rs. 400.
so total selling price will be = (400+60) = Rs.460
so the quantity of the tea will be = (460/5.75) = 80kg.

 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.

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

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

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

Let the speed of the boat upstream be p kmph and that of downstream be q kmph

Time for upstream = 8 hrs 48 min = 845hrs

Time for downstream = 4 hrs

Distance in both the cases is same.

=> p x  845= q x 4

=> 44p/5 = 4q

=> q = 11p/5

 Now, the required ratio of Speed of boat : Speed of water current

= q+p2:q-p2

=> (11p/5 + p)/2  : (11p/5 - p)/2

=> 8 : 3

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 C = x.
Then B = x/2
and A = x/4

A:B:C = 1:2:4.
C's share Rs.[(4/7)*700) = 400

We have to arrange 6 books.

The number of permutations of n objects is n! = n. (n – 1) . (n – 2) ... 2.1

Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720