Quantitative Aptitude - Arithmetic Ability Questions

Let the sum be Rs. x.

I  gives, S.I. = Rs. 7000 and T = 7 years.

II gives, Sum + S.I. for 5 years = 2 x Sum Sum = S.I. for 5 years.

Now, S.I. for 7 years = Rs. 7000.

therefore, S.I. for 1 year =  Rs. 1000.

Thus, I and II both are needed to get the answer.

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

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

As given in the question, Marked price is 25% more than the Cost price.

=> C.P of the article = 34 x 400 = 300

Now, 

Let the original S.P of the article be Rs. P

Now the new S.P = P + 16.666 +13.333300 x P

=> S.P = 7P6

According to the question,

7P6 - 300 = 2P - 300

=> 5P = 1800

=> P = Rs. 360

Hence, the increased S.P = 360 x 7/6 = Rs. 420. 

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.

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days

167 days = (23 weeks + 6 days) =6 odd days.

Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.

Given day is Wednesday.