Perimeter of boundary = 2(L + b) = 340

Area of boundary = [(L + 2)(b + 2) - Lb]

                                    = [Lb + 2L + 2b + 4 - Lb]

                                    = 2(L + b) + 4

                                    = 340 + 4 

                                    =344.

Cost of gardening = Rs. (344 X 10)

                                = Rs.3440

12% of 80% of 50

= 12% of `(80/100 ** 50)`

= 12% of 40

= `12/100 ** 40`

= 4.8

Take year 2014 given in choice.

Total odd days in period 2007-2013 = 5 normal years + 2 leap year
= 5 x 1 + 2 x 2 = 9 odd days
= 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Take year 2016 given in choice.

Number of odd days in period 2007-2015 = 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 odd days
= 4 odd days
(Even if odd days were 0, calendar of 2007 will not be same as calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even boring to check odd days)

Take year 2017 given in choice.

Number of odd days in period 2007-2016 = 7 normal years + 3 leap year
= 7 x 1 + 3 x 2 = 13 odd days
= 6 odd days

Take year 2018 given in choice.

Number of odd days in period 2007-2017 = 8 normal years + 3 leap year
= 8 x 1 + 3 x 2 = 14 odd days
= 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Also, both 2007 and 2018 are not leap years.
Since total odd days in period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have same calendar as that of 2007

Answer : 1.

When same distance is covered with different speeds, n average speed = `(2 * x * y) / (x + y)` = 10.88kmph

Leap year Contains 366 days.

Area of floor = (5.5 x 3.75) m² = 20.625 m².

Cost of paving = Rs. (800 x 20.625) = Rs.16500.

Average temperature for Wednesday, Thursday and Friday = 40^oC
Total temperature = 3*40 = 120^oC
Average temperature for Thursday, Friday and Saturday = 41^oC
Total temperature = 41*3 = 123^oC
Temperature on Saturday = 42^oC
Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123-120;
Saturday - Wednesday = 3
Wednesday = 42-3 = 39^oC.

Let speed of bus be x m/s
speed of car = 68km/hr (or) `170/9` m/s
in 10s car covers a distance of 100m (40m + 60m = 100m) at a speed of (`170/9` ? x)m/s
distance = speed*time
`100 = (170/9 -x) **10`
x = `80/9` m/s
Now convert it to km/hr which is 32 km/hr

Cyclicity of 2 = 4
D = d * Q + R
30 = 4 * 7 + 2
30/4 = Remainder 2, 2² = 4

Notes:
Cyclicity of 2, 3, 7, 8 => 4
Cyclicity of 4, 9 => 2
Cyclicity of 1, 5, 6 => 1