Quantitative Aptitude - Arithmetic Ability Questions

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800x1860x5= 48 sec.

Let the age of the youngest child is 'X' years.
Then the child elder than the yongest is 'X + 3' yrs
Since each have 3yrs difference upto 5 children
And given that
=> X + X+3 + X+6 + X+9 + X+12 = 50
=> 5X + 30 = 50
=> X = 4 yrs.

$$\begin{gathered} 2211 x 75 x 112 \\ => 222 x 75 x 112 \end{gathered}$$

 Total number of prime factors is (22+5+2) = 29

Let the present age's of mother be 'M' and that of her son be 'S'.

According to the given data in the question,

M - 5 = 3(S - 5)

F - 3S = -10  ...............(1)

Then given ,

M + 5 : S + 5 = 11 : 5 

=> 5M - 11S = 30 ..............(2)

From (1) & (2), we get

S = 20

Then by substituting S = 20 in (1) or (2), we get

M = 50.

Hence, the present age of mother is M = 50 years.

The single equivalent payment will be PV + FV.
FV
= Future value of $10,000, 12 months later

$10,000 *(1.0075)/12

$10,938.07
PV=
Present value of $10,000, 24 months earlier

$10,000/(1.0075)24

$8358.31
The equivalent single payment is
$10,938.07 + $8358.31
= $19,296.38

Number of white marbles = 4

Number of Black marbles = 5

Total number of marbles = 9

Number of ways, two marbles picked randomly = 9C2

Now, the required probability of picked marbles are to be of same color = 4C2/9C2  +  5C2/9C2

= 1/6 + 5/18

= 4/9.

The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.

13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.