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.

Given total number of rats = 999919

999919 = 1000000 - 81 = 10002 - 92 = (1000+9)(1000-9) = 1009 x 991.
Since there were more rats than there were cats, 991 cats killed 1009 rats each.

We know that perimeter of rectangle = 2(l+b) = 2(8+7)= 30cm
Given perimeter of square is twice the perimeter of rectangle = 2(30) = 60cm
Therefore, side of the square is = 1/4 x 60 = 15cm
Circumference of the required semicircle = πr + 2r = 227× 152 + 2×152 = 23.57 + 15 = 38.57cm.

If we consider the third term to be ‘x”
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d).
Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
x + 9d will be the 12th term.

Thus, 12th term of the A.P will be zero.

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

 i.e (35,70), (42,63), (45,60)

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

Let the the age of the elder boy = E

Let the the age of the younger boy = Y

Given that Y = cube root of EY

=> Y3 = EY => E = Y2 .....(1)
By the condition of number replacement the age of the father is YE

The Mother's age = EY/2

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

Then now from the given options we can identify which satisfies the all the conditions.

Here Y =2 and E = 4 satisfies all the conditions.

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

If we count letters till the end of the alphabet including that letter in reverse order then
Here the values are nothing but the values corresponding to the alphabets when taken in a reverse order,i.e,A=26 ,B=25...,Z=1
G = 20 and O = 12 totaling 32
S = 8, H = 19 and E = 22 totaling 49

So
SOME = 8 + 12 + 14 + 22 = 56.