Aptitude and Reasoning Questions

Let the number of students in the class be 'x'.
Given if 1 is absent rest of people can be divided into 6 equal parts i.e (x-1)/6.........(1)
And given if 2 are absent rest of people can be divided into 7 equal parts i.e (x-2)/7.....(2)
From (1)&(2), we get
x-1/6 = ((x-2)/7) + 1
=> x = 37.

P.W = Rs. (1760 - 160) = Rs.1600  

S.I on Rs.1600 at 12% is Rs. 160 

Time = (100 x 160)/(1600 x 12)  = 5/6 years = (5/6) x 12 months = 10 months

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40

The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.

31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)

Odd days in 2000 years = 0

10 years has 2 leap years + 8 ordinary years.

Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.

31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.

Total number of odd days = (0 + 5 + 1) = 6.

Given day is saturday.

CP of 1000gm = Rs. 10
SP of 800gm = Rs. 12
SP of 1000gm =12x1000/800 = Rs. 15

Now take 1000gm as reference to calculate profit.
Profit=SP-CP=15-10=Rs. 5
Profit % = 5x100/10 = 50%

Clearly, we may mark such sets of 3 numbers, in which the middle number is 8 and each of the two numbers on both sides of it is a factor of 8, as shown below :

2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6

So, there are two such 8's. Hence the answer is C.