Solution:
Let the number be \(x\)
According to question,
\(x+x^2=156\)
⇒ \(x^2+x-156=0\)
⇒ \(x^2+13x-12x-156=0\)
⇒ \(x(x+13)-12(x+13)=0\)
⇒ \((x+13)(x-12)=0\)
⇒ \(x=-13\) [Rejected] or \(x=12\)
⸫ The required number is \(12\).
2. If the sum of the squares of two consecutive odd natural numbers is \(290\). What are the numbers?
Solution:
Let the numbers be \(x\) and \(x+2\)
According to question,
\(x^2+(x+2)^2=290\)
⇒ \(x^2+x^2+4+4x-290=0\)
⇒ \(2x^2+4x-286=0\)
⇒ \(x^2+2x-143=0\) [Dividing the equation by \(2\)]
⇒ \(x^2+13x-11x-143=0\)
⇒ \(x(x+13)-11(x+13)=0\)
⇒ \((x+13)(x-11)=0\)
⇒ \(x=-13\) [Rejected] or \(x=11\)
⸫ The required numbers are \(11\) and \(13\).
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.
Solution:
Let the larger of the two parts be \(x\) and the smaller part be \(18-x\)
According to question,
\(x^2+(18-x)^2=15x\)
⇒ \(x^2+324-36x+x^2-15x=0\)
⇒ \(2x^2-51x-324=0\)
⇒ \(2x^2-27x-24x+324=0\)
⇒ \(x(2x-27)-12(2x-27)=0\)
⇒ \((2x-27)(x-12)=0\)
⇒ \(x=\frac{27}{2}\) [Rejected] or \(x=12\)
⸫ The required larger part of the number is \(12\) and the smaller part is \(18-12=6\).
4. The product of the digits of a two digit number is 12. If 36 is added to the number, the digits interchange their position. What is the number?
Solution:
Let the tens digit of the number be \(x\) and the ones digit be \(y\)
⸫ The number = \(10x+y \)
Also, \(xy=12 ⇒ y= \frac{12}{x}\) ...(1)
According to question,
\((10x+y)+36=10y+x\)
⇒ \(10x+y+36=10y+x\)
⇒ \(9x-9y=-36\)
⇒ \(x-y=-4\) [Dividing the equation by \(9\)]
⇒ \(x-\frac{12}{x}=-4\) [using (1)]
⇒ \(\frac{x^2-12}{x}=-4\)
⇒ \(x^2-12=-4x\)
⇒ \(x^2+4x-12=0\)
⇒ \(x^2+6x-2x-12=0\)
⇒ \(x(x+6)-2(x+6)=0\)
⇒ \((x+6)(x-2)=0\)
⇒ \(x=-6\) [Rejected] or \(x=2\)
Now, (1)⇒ \(y=\frac{12}{2}=6\)
⸫ The required number is \((10×2)+6=26\).
5. The sum of the two numbers is \(25\) and the sum of their cubes is \(4825\). What are the numbers?
Solution:
Let the numbers be \(x\) and \(25-x\)
According to question,
\(x^3+(25-x)^3=4825\)
⇒ \((x+25-x)[x^x(25-x)+(25-x)^2]=4825\)
⇒ \(x^2-25x+x^2+625-50x+x^2=4825\)
⇒ \(3x^2-75x=193-625\)
⇒ \(3x^2-75x+432=0\)
⇒ \(x^2-25x+144=0\)
⇒ \(x^2-16x-9x+144=0\)
⇒ \(x(x-16)-9(x-16)=0\)
⇒ \((x-16)(x-9)=0\)
⇒ \(x=16\) or \(x=9\)
⸫ The required numbers are \(16\) and \(9\).
6. The perimeter of a rectangular field is \(88~m\) and its area is \(420~m^2\). Find the length and breadth of the rectangular field.
Solution:
Here, Perimeter of the rectangular field = \(88~m\)
⇒ \(2(length+breadth)=88\)
⇒ \(length+breadth=44\)
Let us now consider the length of the rectangular field be \(x~m\) and so the breadth be \((44-x)~m\)
According to question,
Area of the rectangular field = \(420~m^2\)
⇒ \(x(44-x)=420\)
⇒ \(44x-x^2=420\)
⇒ \(x^2-44x+420=0\)
⇒ \(x^2-30x-14x+420=0\)
⇒ \(x(x-30)-14(x-30)=0\)
⇒ \((x-30)(x-14)=0\)
⇒ \(x=30\) or \(x=14\)
When \(x=30\),
The length of the rectangular field is \(30~m\) and breadth is \(44-30=14~m\).
When \(x=14\),
The length of the rectangular field is \(14~m\) and breadth is \(44-14=30~m\).
7. The sum of the areas of two squares is \(225~cm^2\). The length of the side of one is \(3~cm\) more than the other. Find the sides of two squares.
Solution:
Let the side of one of the squares be \(x~cm\) and that of the other be \((x+3)~cm\)
According to question,
\(x^2+(x+3)^2=225\)
⇒ \(x^2+x^2+9+6x-225=0\)
⇒ \(2x^2+6x-216=0\)
⇒ \(x^2+3x-108=0\) [Dividing the equation by \(2\)]
⇒ \(x^2+12x-9x-108=0\)
⇒ \(x(x+12)-9(x+12)=0\)
⇒ \((x+12)(x-9)=0\)
⇒ \(x=-12\) [Rejected] or \(x=9\)
⸫ The required side of one of the squares is \(9~cm\) and that of the other be \((9+3)=12~cm\)
8. The price of a long piece of cloth is Rs. \(440\). If the cloth would have been \(1\) metre longer and the price per metre would have been Rs. \(4\) less, then cost of the cloth does not alter. What was the length of the piece of the cloth?
Solution:
Let the length of the cloth be \(x~m\)
Given, total price of the cloth= Rs. \(440\)
Cost of each metre = Rs. \(\frac{440}{x}\)
When the length is extended by \(1~m\),
length of the cloth becomes \((x+1)~m\)
Cost of each metre in such case = Rs. \(\frac{440}{x+1}\)
According to question,
\(\frac{440}{x+1}=\frac{440}{x}-4\)
⇒ \(4=\frac{440}{x}-\frac{440}{x+1} \)
⇒ \(4=\frac{440(x+1)-440x}{x(x+1)} \)
⇒ \(4x(x+1)=440x+440-440x \)
⇒ \(4x^2+4x-440=0 \)
⇒ \(x^2+x-110=0 \)
⇒ \(x^2+11x-10x-110=0 \)
⇒ \(x(x+11)-10(x+11)=0 \)
⇒ \((x+11)(x-10)=0 \)
⇒ \(x=-11\) [Rejected] or \(x=10 \)
⸫ The required length of the cloth is \(10~m\)
Alternative Solution:
Let the length of the cloth be \(x~m\)
According to question,
\((x+1)\left(\frac{440}{x}-4\right)=440\)
⇒ \((x+1)\left(\frac{440-4x}{x}\right)=440\)
⇒ \((x+1)(440-4x)=440x\)
⇒ \(440x-4x^2+440-4x=440x\)
⇒ \(-4x^2+440-4x=0\)
⇒ \(4x^2+4x-440=0 \)
⇒ \(x^2+x-110=0 \)
⇒ \(x^2+11x-10x-110=0 \)
⇒ \(x(x+11)-10(x+11)=0 \)
⇒ \((x+11)(x-10)=0 \)
⇒ \(x=-11\) [Rejected] or \(x=10 \)
⸫ The required length of the cloth is \(10~m\)
9. Rs. \(1500\) is distributed among some labours. If the number of labours would have been \(5\) more, then each of them would have been received Rs. \(15\) less. Find the number of labours.
Solution:
Let the number of labour be \(x\)
Given, total amount distributed = Rs. \(1500\)
Amount received by each labour = Rs. \(\frac{1500}{x}\)
When the number of labour is increased by \(5\),
Number of labour becomes \(x+5\)
Amount received by each labour in such case = Rs. \(\frac{1500}{x+5}\)
According to question,
\(\frac{1500}{x+5}=\frac{1500}{x}-15\)
⇒ \(15=\frac{1500}{x}-\frac{1500}{x+5} \)
⇒ \(15=\frac{1500(x+5)-1500x}{x(x+5)} \)
⇒ \(15x(x+5)=1500x+7500-1500x \)
⇒ \(15x^2+75x-7500=0 \)
⇒ \(x^2+5x-500=0 \)
⇒ \(x^2+25x-20x-500=0 \)
⇒ \(x(x+25)-20(x+25)=0 \)
⇒ \((x+25)(x-20)=0 \)
⇒ \(x=-25\) [Rejected] or \(x=20 \)
⸫ The required number of labour is \(20\)
Alternative Solution:
Let the number of labour be \(x\)
According to question,
\((x+5)\left(\frac{1500}{x}-15\right)=1500\)
⇒ \((x+5)\left(\frac{1500-15x}{x}\right)=1500\)
⇒ \((x+5)(1500-15x)=1500x\)
⇒ \(1500x-15x^2+7500-75x=1500x\)
⇒ \(-15x^2+7500-75x=0\)
⇒ \(15x^2+75x-7500=0 \)
⇒ \(x^2+5x-500=0 \)
⇒ \(x^2+25x-20x-500=0 \)
⇒ \(x(x+25)-20(x+25)=0 \)
⇒ \((x+25)(x-20)=0 \)
⇒ \(x=-25\) [Rejected] or \(x=20 \)
⸫ The required number of labour is \(20\)
10. Find the value:
(i) \(\sqrt{20+\sqrt{20+\sqrt{20+....... \infty}}}\)
Solution:
Let \(x=\sqrt{20+\sqrt{20+\sqrt{20+....... \infty}}}\)
⇒ \(x^2=20+\sqrt{20+\sqrt{20+..... \infty}}\)
⇒ \(x^2-20-x=0\)
⇒ \(x^2-x-20=0\)
⇒ \(x^2-5x+4x-20=0\)
⇒ \(x(x-5)+4(x-5)=0\)
⇒ \((x-5)(x+4)=0\)
⇒ \(x=5\) or \(x=-4\)
(ii) \(\sqrt{6+\sqrt{6+\sqrt{6+....... \infty}}}\)
Solution:
Let \(x=\sqrt{6+\sqrt{6+\sqrt{6+....... \infty}}}\)
⇒ \(x^2=6+\sqrt{6+\sqrt{6+..... \infty}}\)
⇒ \(x^2-6-x=0\)
⇒ \(x^2-x-6=0\)
⇒ \(x^2-3x+2x-6=0\)
⇒ \(x(x-3)+2(x-3)=0\)
⇒ \((x-3)(x+2)=0\)
⇒ \(x=3\) or \(x=-2\)
(iii) \(\sqrt{3+\sqrt{3+\sqrt{3+....... \infty}}}\)
Solution:
Let \(x=\sqrt{3+\sqrt{3+\sqrt{3+....... \infty}}}\)
⇒ \(x^2=3+\sqrt{3+\sqrt{3+..... \infty}}\)
⇒ \(x^2-3-x=0\)
⇒ \(x^2-x-3=0\)
Using quadratic formula, we have
\(x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(3)}}{2×1}\)
\(x=\frac{1\pm\sqrt{1+12}}{2}\)
\(x=\frac{1\pm\sqrt{13}}{2}\)
Either \(x=\frac{1+\sqrt{13}}{2}\) or \(x=\frac{1-\sqrt{13}}{2}\)
(iv) \(2+\frac{1}{2+\frac{1}{2+...... \infty}}\)
Solution:
Let \(x=2+\frac{1}{2+\frac{1}{2-...... \infty}}\)
⇒ \(x^2=2+\frac{1}{x}\)
⇒ \(x^2-2x-1=0\)
Using quadratic formula, we have
\(x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2×1}\)
\(x=\frac{2\pm\sqrt{4+4}}{2}\)
\(x=\frac{2\pm\sqrt{8}}{2}\)
\(x=\frac{2\pm2\sqrt{2}}{2}\)
\(x=\frac{2(1\pm\sqrt{2})}{2}\)
⇒ \(x=1\pm\sqrt{2}\)
Either \(x=1+\sqrt{2}\) or \(x=1-\sqrt{2}\)
(v) \(2-\frac{1}{2-\frac{1}{2-...... \infty}}\)
Solution:
Let \(x=2-\frac{1}{2-\frac{1}{2-...... \infty}}\)
⇒ \(x^2=2-\frac{1}{x}\)
⇒ \(x^2-2x+1=0\)
⇒ \((x-1)^2=0\)
⇒ \(x=1\)