Quantitative Aptitude - Arithmetic Ability Questions

Let the required numbers be 33a and 33b. 

Then 33a +33b= 528   =>   a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways. 

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

 Let the numbers be 37a and 37b. Then , 37a x 37b =4107  => ab = 3.

 Now co-primes with product 3 are (1,3)

 So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111).

 Therefore, Greater number = 111.

Given equation can be written as

$$\begin{gathered} 215 x 103 + ? = 24 \\ => 14 + ? = 24 \\ => ? = 24 - 14 \\ => ? = 10 \end{gathered}$$

Given Today is 20th January 2017, Sunday

In january, we have 31 days

February - 28 days (Non leap year)

March - 31 days

April - 30 days

=> Remaining days => 31 - 20 = 11 in Jan

+ 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days

More 20 days to complete 90 days => upto 20th April

Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.

Now, the day of the week will be 

90/7 => Remainder '6'

As the day starts with '0' on sunday 

6 => Saturday.

Required day is 20th April, Saturday.

48, 54, 63, 81 and 105 are divisible by 3 but not 35.

Let the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have :

Cost of 1 kg of type 1 rice           Cost of 1 kg of type 2 rice  

∴(20-x)/(x-15) = 2/3 

⇒ 60 - 3x = 2x - 30

 ⇒ x = 18.