Quantitative Aptitude - Arithmetic Ability Questions

 Investment ratio of A, B & C = (4000 ×6+3000×9):(5500 ×6+4000×9):4

Let the actual distance be t kms

Now, According to the data,

$$\begin{gathered} 5t + 860 = 6t \\ => 560t + 860 = 6t \\ => 60t + 8 = 72t \\ => 8 = 12t \\ => t = 8/12 = 2/3 hrs \end{gathered}$$

Now, the total distance = 6t = 6 x 2/3 = 2 x 2 = 4 kms.

Clearly, If 60 % is equally divided then remaining 40% of total profit will be divided between 2 persons A and B in ratio of (profit of A) : (profit of B)= 12,500 : 8500 = 125 : 85 = 25:17
now difference in amount = 25-17 = 8 this difference = 240
therefore 40% of total profit = 240*(25+17)/(25-17) = 1260
∴ 100% profit = 1260 / 40 × 100 = 3150

Time taken by both Meghana and Ganesh to work together is given by =

t1 x t2

Where

$$\begin{gathered} t1 = 32 hrs \\ t2 = 12 12 hrs = 252 hrs \end{gathered}$$

Therefore, time took by both to work together = 

32 x 252 = 16 x 25 = 4 x 5 = 20 hrs

Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4

According to the question,

Difference between square of the average of first two odd number and the of the average last

two odd numbers is 396

i.e, x+3 and x-3

$$\begin{gathered} ⇒x + 32 - x - 32 = 396 \\ ⇒ x2 + 9 + 6x - x2 + 6x - 9 = 396 \\ ⇒ 12x = 396 \\ ⇒ x = 33 \end{gathered}$$

Hence, the smallest odd number is 33 - 4 = 29.

Given length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

Then, the diagonal of the rectangle D = 32 + 42 = 25 = 5

Then, it implies side of square = 5 cm

We know that Area of square = S x S = 5 x 5 = 25 sq.cm.

Number of words with 5 letters from given 9 alphabets formed = 95

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 9P5

Number of words can be formed which have at least one letter repeated =  95 - 9P5

= 59049 - 15120

= 43929

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second Or

Case II: 4 students in the first car and 4 in the second

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is

8C3+8C4 = 56 + 70 = 126.