Quantitative Aptitude - Arithmetic Ability Questions

BD = Rs.100

TD = Rs.80

R = 10%

F=BD×TDBD-TD=100×80100-80=Rs.400 

BD = Simple interest on the face value of the bill for unexpired time= FTR/100

$$\begin{gathered} ⇒100=400×T×10100 \\ \end{gathered}$$

=> T = 2.5 years

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

100
65
-------
65 ------- 35
100 ------ ? => 53.84 %

A=S2

given sum =9+34+7+217-9+115

9+7+9+34+217-115

=> 7+765+120-681020=>7+8171020 

Total time taken = k/40 + 2k/20 hours

= 5k/40 = k/8 hours

Average speed = 3k/(k/8) = 24 kmph.

Speed = 78 x 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (900 + x)/60 = 65/3
x = 400 m.

A century means 100 years => 76 normal years and 24 leap years

We know that, a normal year contain 1 odd day (365/7 = 52 weeks + 1 odd day)

a leap year contain 2 odd days (365/7 = 52 weeks + 2 odd days)

100 years => 76 x 1 + 24 x 2 = 124 odd days = 124/7 = (17 weeks + 5 odd days)

200 years => 2 x 5 = 10 = 1 week + 3 odd days

300 years => 3 x 5 = 15 = 2 weeks + 1 odd day

400 years => 4 x 5 = 20 + 1 odd day (leap year) = 3 weeks + 0 odd days

500 years => 5 x 5 = 25+1 = 26/7 = 3 weeks + 5 odd days 

Again repeats...