On the first day of Ganesh Chaturthi, ladoos and modakas are offered to Lord Ganesha as prasad in the ratio 5:3 respectively which is then mixed and distributed in such a way that each visiting devotee get one count of prasad. If devotees increases daily with a constant number and on last day there are 112 devotees. If prasad last for exactly ten days of festival. Which of the following can be the number of devotees visited on the first day ?
Given, devotees increases constantly on daily basis, which forms an A.P. If number of devotees on 1st day=a, increase per day=d and number of days n=10
Then, number of devotees on last day=a+(n-1)d ⇒ 112=a+9d
Now calculate values of 'a' for different values of 'd'
When (d=1,a=103) (d=2,a=94) (d=3,a=85) (d=4,a=76) (d=5,a=67) (d=6,a=58) (d=7,a=49) (d=8,a=40) (d=9,a=31) (d=10,a=22) (d=11,a=13) (d=12,a=4)
Let total Ladoos=5x and Modakas=3x, then total count of prasad=8x, which last for 10 days.
We know, sum of an A.P. is given by Sn = (n/2)[2a + (n-1)d]
As the total count of prasad is 8x, therefore the sum of an AP should be multiple of 8.
When d=1,a=103, Sum=1075
d=2, a=94 Sum=1030
d=3, a=85 Sum=985
d=4, a=76 Sum=940
d=5, a=67 Sum=895
d=6, a=58 Sum=850
d=7, a=49 Sum=805
d=8, a=40 Sum=760
d=9, a=31 Sum=715
d=10, a=22 Sum=670
d=11, a=13 Sum=625
d=12, a=4 Sum=580
From above, the only sum which is multiple of 8 is 760, hence count of devotees on first day can be 40.
Correct Option 8)