Bank Exams Questions

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From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of  interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.

Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

As n>0, n will not be -8. Therefore, n=11.

Time taken to meet for the first time anywhere on the track

= length of the track / relative speed

= 300 / (15 + 25)5/18 = 300x 18 / 40 x 5 = 27 seconds.

Since S.P. = 5 (M.P. - S.P.) => 5 M.P. = 6 S.P.
M.P. = (6/5) S.P.
Since the percentage discount = Percentage profit,
65S.P - S.P65S.P×100 = S.P -C.PC.P×100
Therefore, 1/6 = (S.P/C.P) – 1 => S.P = (7/6) C.P
M.P = 6/5 S.P = 6/5 x 7/6 C.P = 7/5 C.P
Therefore, Ratio of discount to C.P
= (M.P – S.P)/C.P = [{(7/5)C.P - (7/6)C.P} / C.P.]
= 7/30

Therefore, the ratio is 7 : 30

1 Joule = 107 ergs = 0.000239 Kilo Calories.

• Joule is the Unit of Energy.

Joule is equal to Energy transferred to an object when a force of One Newton acts on that object in the direction of its motion through a distance of one metre.(1 N-m)