**Q1. If a circle has a square and a hexagon and the vertices of both touch the circle’s circumference then what is the ratio of the area of the two?**

Solution: Ratio of Square and Hexagon = b

^{2}/6 × √3/4 a^{2}
= (4/6√3)b

^{2}/a^{2}
= 4/6√3 (√2r/r

^{2})^{2}
= 4 × 2 × r

^{2}/6 × √3× r^{2}
= 4/3√3

= 4: 3√3

**Q2. Area of which of the following is maximum when perimeter is equal?**

**circle, square, triangle**

Solution: Circle’s area is maximum for a given fixed value
of perimeter.

**Q3. Sum of the squares of three numbers is 323. If the sum of squares of two numbers is twice of the third number, find the product of three numbers.**

Sol: Let the numbers be x, y and z

Given,

x

^{2}+ y^{2}+ z^{2}= 323 — (1)
x

^{2}+ y^{2}= 2z — (2)
Put 2 into 1 and solve it.

**Q4. Find Compound Interest on 4000 rupees for 4 year at 10%rate of interest, compounded annually.**

Solution: CI = P(1 + r/100)

^{n}– P, where P is principal amount; r is the rate of interest; and n is number of years.
Put the values and solve.

**Q5. Find the value of (1+sec20° +cot70°)(1-cosec20° + tan70°)**

Solution: (1 + 1/cos20° + cos70°/sin70°)(1 – 1/sin20° +
sin70°/cos70°)

We know that sin(90 – θ) = cosθ

Hence, the expression will become:

(1 + 1/cos20° + sin20°/cos20°)(1 – 1/sin20° + cos20°/sin20°)

(cos20° + 1 + sin20°)(sin20° – 1 + cos20°)/sin20°cos20°

= [sin220 + cos220 + 2sin20.cos20 – 1]/sin20°cos20°

We know that, sin2θ + cos2θ = 1

⇒ [1 + 2sin20.cos20 –
1]/sin20°cos20°

⇒ 2sin20.cos20/sin20°cos20°

= 2

**Q6. Difference in compound interest is INR 723 at 20% for a yearly or half yearly basis. Find the Sum.**

Sol: Since, the yearly interest rate is 20%, the half yearly
interest rate will be 10%

Given,

P[1 + 10/100]

^{2}– P[1 + 20/100]^{1}= 723
Solve it and get the value of P.

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