Solⁿ:

Let the number be $x$.

According to question,

$x^{2}+x=156$

$\Rightarrow x^{2}+x-156=0$

$\Rightarrow x^{2}+13x-12x-156=0$

$\Rightarrow x(x+13)-12(x+13)=0$

$\Rightarrow (x-12)(x+13)=0$

Either,

$x-12=0 \Rightarrow x=12$

Or,

$x+13=0 \Rightarrow x=-13$

$\therefore x = 12$ or $x = -13$

Hence, the required number is 12 (or $-13$).

[Since both $12$ and $-13$ satisfy the condition ($12^2 + 12 = 156$ and $(-13)^2 + (-13) = 156$)]

2. If the sum of squares of two consecutive odd natural numbers is 290. What are the numbers?

Solⁿ:

Let the two consecutive odd natural numbers be $x$ and $x+2$.

According to question,

$x^{2}+(x+2)^{2}=290$

$\Rightarrow x^{2}+x^{2}+4x+4=290$

$\Rightarrow 2x^{2}+4x-286=0$

$\Rightarrow 2(x^{2}+2x-143)=0$

$\Rightarrow x^{2}+2x-143=0$

$\Rightarrow x^{2}+13x-11x-143=0$

$\Rightarrow x(x+13)-11(x+13)=0$

$\Rightarrow (x-11)(x+13)=0$

Either,

$x-11=0 \Rightarrow x=11 \in \mathbb{N}$

Or,

$x+13=0 \Rightarrow x=-13 \notin \mathbb{N}$

Since $x$ is a natural number, $x \ne -13$.

$\therefore x = 11$

Hence, the required consecutive odd natural numbers are 11 and 13 ($11+2$).

2. If the sum of squares of two consecutive odd natural numbers is 290. What are the numbers?

Solⁿ:

Let the two consecutive odd natural numbers be $x$ and $x+2$.

According to question,

$x^{2}+(x+2)^{2}=290$

$\Rightarrow x^{2}+x^{2}+4x+4=290$

$\Rightarrow 2x^{2}+4x-286=0$

$\Rightarrow 2(x^{2}+2x-143)=0$

$\Rightarrow x^{2}+2x-143=0$

$\Rightarrow x^{2}+13x-11x-143=0$

$\Rightarrow x(x+13)-11(x+13)=0$

$\Rightarrow (x-11)(x+13)=0$

Either,

$x-11=0 \Rightarrow x=11 \in \mathbb{N}$

Or,

$x+13=0 \Rightarrow x=-13 \notin \mathbb{N}$

Since $x$ is a natural number, $x \ne -13$.

$\therefore x = 11$

Hence, the required consecutive odd natural numbers are 11 and 13 ($11+2$).

3. Divide the number 18 into two positive parts so that the sum of their squares is equal to 15 times of the larger part.

Solⁿ:

Let the larger part be $x$.

Then the smaller part is $18-x$.

According to question,

$x^{2}+(18-x)^{2}=15x$

$\Rightarrow x^{2}+324-36x+x^{2}=15x$

$\Rightarrow 2x^{2}-51x+324=0$

$\Rightarrow 2x^{2}-27x-24x+324=0$

$\Rightarrow x(2x-27)-12(2x-27)=0$

$\Rightarrow (x-12)(2x-27)=0$

Either,

$x-12=0 \Rightarrow x=12$

Or,

$2x-27=0 \Rightarrow x=\frac{27}{2}=13.5$

Since x=\frac{27}{2} is not an integer.

$\therefore x = 12$

Larger part $= 12$

Smaller part $= 18 – 12 = 6$

Hence, the required positive parts are 12 and 6.

4. The product of digits of a two digit number is 12. If 36 is added to the number, the digits interchange their positions. What is the number?

Solⁿ:

Let the digit in the units place $= x$

The digit in the tens place $= y$

5. The sum of two numbers is 25 and the sum of their cubes is 4825. What are the numbers?

Solⁿ:

Let one number be $x$. Then the other number is $25-x$.

A/Q

$x^{3}+(25-x)^{3}=4825$

$(x + 25 – x)^{3} – 3 \cdot x \cdot (25-x)(x + 25 – x) = 4825$ [Since, $a^3 + b^3 = (a+b)^3 – 3ab(a+b)$]

$\Rightarrow 25^{3} – 3x(25-x) \times 25 = 4825$

$\Rightarrow 25 \left[ 25^{2} – 3x(25-x) \right] = 4825$

$\Rightarrow 625 – 75x + 3x^{2} = 193$

$\Rightarrow 3x^{2} – 75x + 432 = 0$

$\Rightarrow 3(x^{2} – 25x + 144) = 0$

$\Rightarrow x^{2} – 25x + 144 = 0$

$\Rightarrow x^{2} – 16x – 9x + 144 = 0$

$\Rightarrow x(x-16) – 9(x-16) = 0$

$\Rightarrow (x-16)(x-9) = 0$

Either,

$x-16=0 \Rightarrow x=16$

Or,

$x-9=0 \Rightarrow x=9$

Hence, the required numbers are 16 and 9.

6. The perimeter of a rectangular field is 88 metre and its area is 420 square metre. Find the length and the breadth of the rectangular field.

Solⁿ:

Here, perimeter of the rectangular field $= 88 \text{ m}$.

Half of the perimeter $= \frac{88}{2} = 44 \text{ m}$

Let the length of the field be $x \text{ m}$.

Then the breadth is $(44-x) \text{ m}$.

A/Q

$x(44-x) = 420$

$\Rightarrow 44x – x^{2} = 420$

$\Rightarrow x^{2} – 44x + 420 = 0$

$\Rightarrow x^{2} – 30x – 14x + 420 = 0$

$\Rightarrow x(x-30) – 14(x-30) = 0$

$\Rightarrow (x-30)(x-14) = 0$

Either,

$x-30=0 \Rightarrow x=30$

Or,

$x-14=0 \Rightarrow x=14$

Since length is greater than breadth:

Length $= 30 \text{ m}$

Breadth $= 14 \text{ m}$

Hence, the required length is $30 \text{ m}$ and breadth is $14 \text{ m}$.

7. The sum of areas of two squares is 225 square cm. The length of the side of one is 3cm more than the other. Find the sides of the two squares.

Solⁿ:

Let the side of one square be $x \text{ cm}$.

Then the side of the other square is $(x+3) \text{ cm}$.

A/Q

$x^{2} + (x+3)^{2} = 225$

$\Rightarrow x^{2} + x^{2} + 6x + 9 = 225$

$\Rightarrow 2x^{2} + 6x – 216 = 0$

$\Rightarrow 2(x^{2} + 3x – 108) = 0$

$\Rightarrow x^{2} + 3x – 108 = 0$

$\Rightarrow x^{2} + 12x – 9x – 108 = 0$

$\Rightarrow x(x+12) – 9(x+12) = 0$

$\Rightarrow (x+12)(x-9) = 0$

Either,

$x+12=0 \Rightarrow x=-12$ (Not possible as length cannot be negative)

Or,

$x-9=0 \Rightarrow x=9$

$\therefore x = 9$

Side of other square $= 9 + 3 = 12 \text{ cm}$

Hence, the sides of the two squares are 9 cm and 12 cm respectively.

8. The price of a long piece of cloth is ₹ 440. If the cloth would have been 1 metre longer and the price per metre would have been ₹ 4 less, then cost price of the cloth does not alter. What was the length of the piece of cloth?

Solⁿ:

Let the length of the piece of cloth be $x \text{ m}$.

$\text{Cost per metre} = \frac{440}{x}$

A/Q

$(x+1)\left(\frac{440}{x} – 4\right) = 440$

$\Rightarrow (x+1)\left(\frac{440 – 4x}{x}\right) = 440$

$\Rightarrow (x+1)(440 – 4x) = 440x$

$\Rightarrow 440x – 4x^{2} + 440 – 4x = 440x$

$\Rightarrow -4x^{2} – 4x + 440 = 0$

$\Rightarrow -4(x^{2} + x – 110) = 0$

$\Rightarrow x^{2} + x – 110 = 0$

$\Rightarrow x^{2} + 11x – 10x – 110 = 0$

$\Rightarrow x(x+11) – 10(x+11) = 0$

$\Rightarrow (x+11)(x-10) = 0$

Either,

$x+11=0 \Rightarrow x=-11$ (Not possible, since length cannot be negative)

Or,

$x-10=0 \Rightarrow x=10$

$\therefore x = 10$

Hence, the length of the piece of cloth was 10 m.

9. ₹ 1500 is distributed among some labourers. If the number of labourers would have been 5 more, then each of them would have received ₹ 15 less. Find the number of labourers.

Solⁿ:

Let the number of labourers $= x$.

Payment of each labourer $=  \frac{1500}{x}$

A/Q

$(x+5)\left(\frac{1500}{x} – 15\right) = 1500$

$\Rightarrow (x+5)\left(\frac{1500 – 15x}{x}\right) = 1500$

$\Rightarrow (x+5)(1500 – 15x) = 1500x$

$\Rightarrow 1500x – 15x^{2} + 7500 – 75x = 1500x$

$\Rightarrow -15x^{2} – 75x + 7500 = 0$

$\Rightarrow -15(x^{2} + 5x – 500) = 0$

$\Rightarrow x^{2} + 5x – 500 = 0$

$\Rightarrow x^{2} + 25x – 20x – 500 = 0$

$\Rightarrow x(x+25) – 20(x+25) = 0$

$\Rightarrow (x+25)(x-20) = 0$

Either,

$x+25=0 \Rightarrow x=-25$ (Not possible)

Or,

$x-20=0 \Rightarrow x=20$

$\therefore x = 20$

Hence, the required number of labourers is 20.

10. Find the value:

(i) $\sqrt{20+\sqrt{20+\sqrt{20+……..\infty}}}$

Solⁿ:

Let $x = \sqrt{20+\sqrt{20+\sqrt{20+……..\infty}}}$

Squaring both sides, we get:

Either,

Or,

(ii) $\sqrt{6+\sqrt{6+\sqrt{6+……..\infty}}}$

Solⁿ:

Let $x = \sqrt{6+\sqrt{6+\sqrt{6+……..\infty}}}$

Squaring both sides, we get:

$x^{2} = 6 + \sqrt{6+\sqrt{6+\sqrt{6+……..\infty}}}$

$\Rightarrow x^{2} = 6 + x$

$\Rightarrow x^{2} – x – 6 = 0$

$\Rightarrow x^{2} – 3x + 2x – 6 = 0$$\Rightarrow x(x-3) + 2(x-3) = 0$

$\Rightarrow (x-3)(x+2) = 0$

Either, $x-3=0 \Rightarrow x=3$

Or,

$x+2=0 \Rightarrow x=-2 \quad (\text{Not possible as principal square root is positive})$

$\therefore \sqrt{6+\sqrt{6+\sqrt{6+……..\infty}}} = \mathbf{3}$

(iii) $\sqrt{3+\sqrt{3+\sqrt{3+……..\infty}}}$

Solⁿ:

Let $x = \sqrt{3+\sqrt{3+\sqrt{3+……..\infty}}}$

Squaring both sides, we get:

$x^{2} = 3 + \sqrt{3+\sqrt{3+\sqrt{3+……..\infty}}}$

$\Rightarrow x^{2} = 3 + x$

$\Rightarrow x^{2} – x – 3 = 0$

Here, $a = 1$, $b = -1$, $c = -3$

By using the quadratic formula,

$x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$:

$x = \frac{-(-1) \pm \sqrt{(-1)^{2} – 4(1)(-3)}}{2(1)}$

$\Rightarrow x = \frac{1 \pm \sqrt{1 + 12}}{2}$

$\Rightarrow x = \frac{1 \pm \sqrt{13}}{2}$

Either, $x = \frac{1 + \sqrt{13}}{2}$

Or,

$x = \frac{1 – \sqrt{13}}{2} \quad (\text{Not possible as principal square root is positive})$

$\therefore \sqrt{3+\sqrt{3+\sqrt{3+……..\infty}}} = \mathbf{\frac{1 + \sqrt{13}}{2}}$

(iv) $2+\frac{1}{2+\frac{1}{2+……..\infty}}$

Solⁿ:

Let $x = 2+\frac{1}{2+\frac{1}{2+……..\infty}}$

$\Rightarrow x = 2 + \frac{1}{x}$

$\Rightarrow x = \frac{2x + 1}{x}$

$\Rightarrow x^{2} = 2x + 1$

$\Rightarrow x^{2} – 2x – 1 = 0$

Here, $a = 1$, $b = -2$, $c = -1$

By using the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^{2} – 4(1)(-1)}}{2(1)}$

$\Rightarrow x = \frac{2 \pm \sqrt{4 + 4}}{2}$

$\Rightarrow x = \frac{2 \pm \sqrt{8}}{2}$

$\Rightarrow x = \frac{2 \pm 2\sqrt{2}}{2}$

$\Rightarrow x = 1 \pm \sqrt{2}$

Either, $x = 1 + \sqrt{2}$

Or,

$x = 1 – \sqrt{2} \quad (\text{Not possible as value must be positive})$

$\therefore x = \mathbf{1 + \sqrt{2}}$

(v) $2-\frac{1}{2-\frac{1}{2-……..\infty}}$

Solⁿ:

Let $x = 2-\frac{1}{2-\frac{1}{2-……..\infty}}$

$\Rightarrow x = 2 – \frac{1}{x}$ 

$\Rightarrow x =  \frac{2x – 1}{x}$ 

$\Rightarrow x^{2} = 2x – 1$

$\Rightarrow x^{2} – 2x + 1 = 0$

$\Rightarrow (x – 1)^{2} = 0$

$\Rightarrow x – 1 = 0$

$\Rightarrow x = 1$

$\therefore x = \mathbf{1}$