Quantitative Aptitude - Arithmetic Ability Questions

i=j/m

n
=m(Term)
= 2(15.5)
=31

Fair market value
Present value of the face value

=FV(1+  i)^-n

Number of books that can be placed in the larger box is given by Volume of Box / Volume of book.
Now, n = 25 x 42 x 60 / 7 x 6 x 5
= 63000 / 210
= 300.
Therefore, 300 books can be in that box.

A=P(1+r)^t

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

 Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.

Total repayment amount = $1600 + $368
= $1968
= $82

regular payment= total amount/number of repayments

Given that,

 working days = 5

 working hours = 8

 A man get rupess per hour is Rs.2.40

 So in one day the man get total rupees is 2.40 x 8 = 19.2

 So in 5 days week the man get total rupees is 19.2 x 5 = 96

 So in 4 week the man get total rupees is 96 x 4 = 384

 So the man worked for = 160hours in 4 weeks

 But given that the man earned Rs.432

 Hence remaning money is (432-384 = 48)which is earn by doing overtime work

Overtime hours  = 48/3.20 = 15

So total worked hours is = 15 + 160 = 175.

Here among the given numbers, 3, 19, 67 are prime numbers. And 133 is not a prime number.

Hence, the odd number of the given numbers is 133.

When 16nis divided by 9, we have

 1619, remainder = 7  

1629, remainder = 4 

1639, remainder = 1 

1649, remainder = 7 

1659, remainder = 4 

1669, remainder = 1 

So, we have cyclicity of 3 factors i.e 7,4,1. 

Hence only 3 remainders are possible.