Since it is given that xx and yy are integers then |x - 5|×|y - 7| also needs to be an integer.

Thus |x − 5| × |y − 7| = 5 or 6

Case I: |x − 5| × |y − 7| = 5

a) |x - 5| = 5 and |y - 7| = 1 It gives x = 0 or 10 and y = 6 or 8. (10, 6) and (10, 8) are only solution as we are given that x, yx, y are positive

b) |x - 5| = 1 and |y - 7| = 5. It gives x = 6 or 4 and y = 12 or 2. Total of 4 such pairs are possible

Total 6 such pairs possible for case I

Case II: |x − 5| × |y − 7| = 6

a) |x - 5| = 1 and |y - 7| = 6. It gives x = 4 or 6 and y = 1 or 13 Thus 4 pairs are possible

b) |x - 5| = 2 and |y - 7| = 3. It gives x = 3 or 7 and y = 4 or 10. 4 pairs are possible

c) |x - 5| = 3 and |y - 7| = 2. It gives x =  2 or 8 and y = 5 or 9. 4 pairs are possible

d) |x - 5| = 6 and |y - 7| = 1. It gives x = -1 or 11 and y = 6 or 8. Here 2 pairs are possible

Total of 4 + 4 + 4 + 2 = 14 such pairs are possible

Overall 6 + 14 = 20 cases are possible.

Let the cost of pen be xx, a pencil be y and a markers be z

x + 3y + 5z = 100.......(I)

6x + 18y + 2z = 320....(II)

6 × (I) − (II) gives us,

28z = 280 or z = 10

putting z = 10 in (I) we get x + 3y = 50

We have to get the minimum value of x + y + z. z = 10 thus we need to minimize x + y

when y = 16 then x = 2 and x + y + z = 2 + 16 + 10 = 28

When y = 15 then x = 5 and x + y + z = 5 + 15 + 10 = 30

We see that as we keep on decreasing y the sum will increase.

Minimum possible value = 28

According to the question

2000000(1.2)³ = x(1.2)² + x(1.2) + x + 180000 

3456000 − 180000 = x(1.44 + 1.2 + 1)

3276000 = x(1.44 + 1.2 + 1)

3276000 = x × 3.6

3276000/​3.64 = x

x = 900000.

CP of 1 litre = 50

MP of 1 litre = 1.2 x 50 = 60

Also, he mixes water to milk in the ratio of 1:4, so in 1 litre of milk, he mixes 250 mL of water.

Hence, the volume of liquid he is selling = 1.25

MP of 1.25 litres = 60 x 1.25 = 75 Rs.

So, to have a 10% profit, SP = Rs 50 x 1.1 = 55 Rs.

Hence, discount percent = 20/75 x 100 = 80/3%

Let us calculate the actual sum which Ram would have got if he did not have had made the mistake.

We can use the formula to calculate the sum = 30/2 ​ (2(2) + (30 − 1)7)

Thus the actual sum = 15(4 + 29 × 7) = 3105

Excess sum calculated = 3210 - 3105 = 105

Let last n terms be the terms where the common difference was taken as 14 instead of 7

Thus 7 + 14 + ...n terms = 105

n/2(2(7) + (n − 1)7) = 105

n(n + 1)7 = 210

Or 2 + n − 30 = 0 or (n + 6)(n − 5) = 0.

Since n can not be negative, n = 5, Thus last 5 terms had incorrect common differece. x = 30 - n = 30 - 5 = 25

x − 4 = 21 = 3 × 7

Total 4 factors are there.

Both of the series are an Arithmetic Progression

Common difference of the Series are 6 and 7 respectively.

Upon looking, we can see that the first common term is 16.

The series which will have common terms will have first term as 16 and common difference = lcm(6, 7) = 42

Let a_n ​be the n^th and final term of the series

a_{n ​}= 16 + (n − 1)42 ≤ 610

42(n − 1) = 594

n − 1 = 14.14 or n = 15.14. Thus their are total 15 number of common terms.