x² − 10x + 2[x] + 16 = x[x]

We can rewrite it as x² − 10x + 16 = x[x] − 2[x]

(x−2)(x−8) = [x](x−2)

(x−2)(x−8−[x]) =0

There can be two cases: Case 1-> x - 2 =0

When x - 2 = 0, x becomes +2

Other case, Case 2 is x - 2 is not equal to 0, then,

x − 8 − [x] = 0

x − [x] = 8

We know that x − [x] = {x} 

We know that 0 ≤ {x} ≤ 1

Hence, this case becomes invalid.

We get only one value of x, one unique solution.

As AB and CD are two chords that intersect each other, 

AP × PB = CP × PD => 12 × 8 = 6 × CP

CP = 16

From O, we can draw OM, which is perpendicular to AB, and ON, which is perpendicular to CD.

From the centre, a line perpendicular to a chord bisects the chord.

We have, AM = MB = 10cm

MP = 2cm, ON = 2cm, CD = 22cm, CN = 11cm.

ON² + CN² = OC² 

4 + 121 = r² (where r is the radius of the circle)

r² = 125

Let the time taken by a man to build & to break the wall be x days & 10x days respectively.

Let the time taken by a boy to build & to break the wall y days & 10y days respectively.

Let total bricks in wall be = 10xy

This means that a man can build & break 10x bricks & x bricks respectively in a day and a boy can build & break 10y bricks & y bricks respectively in a day.

Since, 1 man & 7 boys can break the wall in 6 days 

i.e. 6(y + 7x) = 10xy   ---- (i)

Also, 1 man & 3 boys can build the wall in 1 day 

i.e. 1(10y + 30x) = 10xy  ---- (ii)

From (i) and (ii), y = 3x --- (iii)

From (i) & (iii), y = 6 & x = 2

Now, time taken by 1 man & 1 boy to break the wall  

There are 20 multiples of 3 from 1 to 60.

If we take only one multiple of 3 out of the 20 multiples, then the product will not be divisible by 36.

Hence, he could have written at most 60 - 19 = 41 numbers.

Let the distance be L meters

When cyclist X finishes the race, cyclist Y has covered (L - 20) meters.

When cyclist Y finishes the race, cyclist Z has covered (L - 15) meters.

When cyclist X finishes the race, cyclist Z has covered (L - 15) meters.

A’s and B’s present ages are 2 and 4 years respectively.

The difference between the ages of any two persons is not more than five years.

Thus 1 cannot be considered.

Let it be 4.

A + B + C + D = 16

C + D = 10

Again it is given that the age difference of any two persons is not more than five years. Thus the ages of C and D could be 7 and 3 years, satisfying all the conditions.3

When it be 9

A + B + C + D = 36

C + D = 30

But the age difference of any two persons now becomes more than five years. Hence this case is discarded.

It is said that the godowns were cleared in 20 days.

He was charged from the 11th day at Rs.2,000 per day.

Total charge = Rs.20,000

Total cost price of the dealer = 15,00,000 + 12 × 2500 × 20 + 20000

= 1500000 + 600000 + 20000 = Rs.21,20,000

Total selling price = Rs.26,50,000