Quantitative Aptitude - Arithmetic Ability Questions

Let total amount = Rs. P

Total expenditure = (23 + 33 + 19 + 16)% of P = 91%ofP

Remaining money = (100 - 91)% of P = 9% of P

But given in the question that,

9% of P = 504

9p/100 = 504

=> P = 50400/9

=> P = Rs. 5600

Required money spent on Food & Insurance = 23 + 33 %

= 56% of 5600

= 56 x 5600/100 = 56 x 56 = Rs. 3136

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.

Sum = Rs.(50*100)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

 C.I = Rs.(551.25-500)= Rs.51.25

Sum of Interior angle is 180 .
and it is given that interior angle is 150 more than exterior,
so let exterior angle is x then interior will be x+150

now , interior + exterior =180
x+150+x=180
x=15

now , number of sides of polygon = 360 / exterior angle
n = 360 / 15
n = 24

The hands of a clock point in opposite directions (in the same straight line) 11 times in every
12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).
So, in a day, the hands point in the opposite directions 22 times

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

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.