Quantitative Aptitude - Arithmetic Ability Questions

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800x1860x5= 48 sec.

On 31st December, 2005 it was Saturday.

Number of Odd days from the year 2006 to the year 2009 = (1+1+2+1)=5 days.

 On 31st december 2009, It was Thursday

Thus, on 1st Jan, 2010 it is Friday.

142 x 142 x 11260.25

=> 142 x 142 x 135.5 = 568

what is 70 of 50 means 70% of 50

$$\begin{gathered} ⇒70 x 50100 \\ = 7 x 5 \\ = 35 \end{gathered}$$

Hence, 70 of 50 is 35.

Let N be the required number. 

=> 60100x27x34xN=2664  

=> N = 2664x2x7x5/9  

=> N = 20720.

Let the speed of boat in still water is ‘x’ km/hr & that of stream is ‘y’ km/hr.
Then, ATQ
(x+y)/y = 9/1 9y = x + y
x = 8y
y = 3 km/hr
So, x = 24 km/hr
Upstream speed = 24-3 = 21 km/hr
Hence, distance travelled upstream in 5 hours = 21*5 = 105 km.

We are not given a value of P in this problem, so either pick a value

for P and stick with that throughout the problem, or just let P = P.

We have that t = 1, and r = .055. To find the effective rate of interest,

first find out how much money we have after one year:

A = Pert

A = Pe(.055)(1)

A = 1.056541P.

Therefore, after 1 year, whatever the principal was, we now have 1.056541P.

Next, find out how much interest was earned, I, by subtracting the initial amount of money from the final amount:

I = A − P

  = 1.056541P − P

  = .056541P.

Finally, to find the effective rate of interest, use the simple interest formula, I = Prt. So,

I = Pr(1) = .056541P

.056541 = r.

Therefore, the effective rate of interest is 5.65%

t=I/pr