Simplification Questions

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000

We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.

Given teacher drinks 9 ltr
Let number of boys be 'A'.
Let number of girls be 'B'.
Boy drinks 7 ltr and girl drinks 4 ltr
After class total water consumed = 42 ltr
Then,
9 + 7A + 4B = 42
=> 7A + 4B = 33
By trial and error method,
The only integers which satisfy the equation is A = 3 and B = 3
Therefore, number of girls in the class = 3.

Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.

sin 46. cos 44+ cos 46. sin 44

sin 46. sin (90 -44)+ cos 46. cos (90 -44)

= sin^2 46 + cos^2 46

= 1

(4a+7b)(4c-7d) = (4a-7b)(4c+7d)

(4a+7b)/(4a-7b) = (4c+7d)/(4c-7d)

Using componendo and dividendo

(4a+7b)+(4a-7b) / (4a+7b)-(4a-7b)

= (4c+7d)+(4c-7d) / (4c+7d)-(4c-7d)

Or 8a/14b = 8c/14d

Or a/b = c/d

The given equation is in Mixed fraction.

Now to solve this type of fractions,

$$\begin{gathered} 413 + 215 + 414 =? \\ ? = 4 + 2 + 4 + 13 + 15 + 14 \\ ? = 10 + 4760 \\ ? = 104760 \\ \end{gathered}$$