Quantitative Aptitude - Arithmetic Ability Questions

The given number series is 

71      88       113       150       203          277

   +17    +25       +37      +53        +74

        +8      +12        +16       +21

            +4          +4         +5

The series should be  71  88  112   150    203   276 i.e, the differences of the difference should be +4.

Hence, the odd man in the given series is 277.

Let length of plot = L meters, then breadth = L - 20 meters

and perimeter = 2[L + L - 20] = [4L - 40] meters

[4L - 40]  * 26.50 = 5300

[4L - 40] = 5300 / 26.50 = 200

4L = 240

L = 240/4= 60 meters. 

Given year is divided by 4, and the quotient gives the number of leap years.

Here, 100%4 = 25.

But, as 100 is not a leap year => 25 - 1= 24 leap years.

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4. 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

55 min. spaces are gained in 60 min

40 min. spaces are gained in 6055×40 min  

Let breadth = x m

Then, length = (x+5)m

Area of a rectangle = Length x Breadth

x(x+5) = 750 

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25 or x = -30

Hence, breadth x = 25m

=> Length = x + 5 = 25 + 5 = 30m. 

length of wire = 2πr= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm

area of the room = 544 x 374 sq.cm

size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm

area of 1 tile = 34x34 sq cm

no. of tiles required = (544 x 374) / (34 x 34) = 176