Quantitative Aptitude - Arithmetic Ability Questions

S.I. on Rs. 1800 = T.D. on Rs. 1872.

P.W. of Rs. 1872 is Rs. 1800.

Rs. 72 is S.I. on Rs. 1800 at 12%.

Time =[(100 x 72)/ (12x1800)]year

        =1/3year = 4 months.

Let the total number of valid votes be 'V'.

70% of V = 70/100 x V = 7V/10

Number of votes secured by the other candidate = V - 7V/100 = 3V/10

Given, 7V/10 - 3V/10 = 172 => 4V/10 = 172

=> 4V = 1720 => V = 430.

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

Next train comes at 17:00 hrs.
So, last train will be = 17:00hrs - 2:30hrs
= 14:30hrs
Announcement made after 37 min of the last train.
S0, 14:30hrs + 00:37 min
= 15:07 hrs.

We know that

Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.

Total savings = 200 + 200 + 200 + 240 + 280 + ... to n months = 11040

= 120 + 80 + 160 + 40 + 200 + 240 + ....n terms
= 40 + 80 + 120 + 160 + 200 + 240 + ...n terms
= n/2[2a + (n-1)d] + 80 + 40 = 11040
= n/2[240 + (n-1)40] = 10920
= n[6+n-1] = 546
= n[n+5] = 546
n = 21

To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres.

=> the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres

However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity.

i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.

Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

--> 40×518=1009 m/sec

Length of Train= Speed * Time

Length = 1009×6

$$\begin{gathered} ⇒2003=6623 mts \\ \end{gathered}$$