Quantitative Aptitude - Arithmetic Ability Questions

By trial and error method from the given options,

as 54 - 45 =  (1/11) x (45+54)  = 99/11  = 9

Hence this satisfies the given conditions.

Then the woman's husband age is 54.

Marks in English  = 85×7 – 83×6

= 595 – 498

= 97

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.

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

1 < 4n + 7 < 200

n can be 0, or -1

n cannot be -2 or any other negative integer or the expression 4n + 7 will be less than1.

The largest value for n will be an integer < (200 - 7) /4

193/4 = 48.25, hence 48

The number of integers between -1 and 48 inclusive is 50

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...

The Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60.

The Common Multiples of 4 & 6 upto 60 numbers are given as 12, 24, 36, 48, 60.