Ganit Prakash - Class-X - Let us work out 1.1
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Q1 to Q6(ii) \(7x^5 - x(x + 2)\)
(iii) \(2x(x + 5) + 1\)
(iv) \(2x - 1\)
(i) \(x^2 - 7x + 2\)
The highest power (degree) of the variable \(x\) is \(2\).
Therefore, it is a quadratic polynomial.
(ii) \(7x^5 - x(x + 2)\)
\(= 7x^5 - x^2 - 2x\)
The highest power (degree) of the variable \(x\) is \(5\).
Therefore, it is not a quadratic polynomial.
(iii) \(2x(x + 5) + 1\)
\(= 2x^2 + 10x + 1\)
The highest power (degree) of the variable \(x\) is \(2\).
Therefore, it is a quadratic polynomial.
(iv) \(2x - 1\)
The highest power (degree) of the variable \(x\) is \(1\).
Therefore, it is not a quadratic polynomial (it is a linear polynomial).
(ii) \(x + \frac{3}{x} = x^2, (x \neq 0)\)
(iii) \(x^2 - 6\sqrt{x} + 2 = 0\)
(iv) \((x - 2)^2 = x^2 - 4x + 4\)
(i) \(x - 1 + \frac{1}{x} = 6\)
Multiplying both sides by \(x\):
\(x^2 - x + 1 = 6x\)
or, \(x^2 - 7x + 1 = 0\)
This is in the form of \(ax^2 + bx + c = 0\), where \(a = 1, b = -7, c = 1\).
Therefore, it can be written in the required form.
(ii) \(x + \frac{3}{x} = x^2\)
Multiplying both sides by \(x\):
\(x^2 + 3 = x^3\)
\(x^3 - x^2 - 3 = 0\)
This is a cubic equation, not a quadratic one.
Therefore, it cannot be written in the form of \(ax^2 + bx + c = 0\).
(iii) \(x^2 - 6\sqrt{x} + 2 = 0\)
This equation contains a term with \(\sqrt{x}\) (i.e., \(x^{1/2}\)). A quadratic equation must have variables with non-negative integer powers only.
Therefore, it cannot be written in the form of \(ax^2 + bx + c = 0\).
(iv) \((x - 2)^2 = x^2 - 4x + 4\)
Expanding the left side:
\(x^2 - 4x + 4 = x^2 - 4x + 4\)
\(0 = 0\)
This is an identity, not a quadratic equation (since the \(x^2\) terms cancel out, meaning \(a = 0\)).
Therefore, it cannot be written in the form of \(ax^2 + bx + c = 0\) where \(a \neq 0\).
The given equation is \(x^6 - x^3 - 2 = 0\).
We can rewrite this as:
\((x^3)^2 - (x^3) - 2 = 0\)
Let \(y = x^3\). Then the equation becomes:
\(y^2 - y - 2 = 0\)
This is a quadratic equation in the variable \(y\).
Therefore, the equation becomes a quadratic equation with respect to \(x^3\).
Answer: The power of the variable is \(3\).
(ii) If \(\frac{x}{4 - x} = \frac{1}{3x}\), \((x \neq 0, x \neq 4)\) be expressed in the form of \(ax^2 + bx + c = 0\) (\(a \neq 0\)), then let us determine the co-efficient of \(x\).
(iii) Let us express \(3x^2 + 7x + 23 = (x + 4)(x + 3) + 2\) in the form of the quadratic equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)).
(iv) Let us express the equation \((x + 2)^3 = x(x^2 - 1)\) in the form of \(ax^2 + bx + c = 0\) (\(a \neq 0\)) and write the co-efficients of \(x^2\), \(x\) and \(x^0\).
(i) For the equation \((a - 2)x^2 + 3x + 5 = 0\) to NOT be a quadratic equation, the coefficient of \(x^2\) must be zero.
Therefore, \(a - 2 = 0\)
\(\implies a = 2\)
So, for \(a = 2\), the equation will not be a quadratic equation.
(ii) Given equation: \(\frac{x}{4 - x} = \frac{1}{3x}\)
Cross-multiplying, we get:
\(x \cdot 3x = 1 \cdot (4 - x)\)
\(3x^2 = 4 - x\)
\(3x^2 + x - 4 = 0\)
Comparing this with \(ax^2 + bx + c = 0\), we see the term with \(x\) is \(1x\).
Therefore, the coefficient of \(x\) is \(1\).
(iii) Given equation: \(3x^2 + 7x + 23 = (x + 4)(x + 3) + 2\)
Expanding the right side:
\(3x^2 + 7x + 23 = (x^2 + 3x + 4x + 12) + 2\)
or, \(3x^2 + 7x + 23 = x^2 + 7x + 14\)
Bringing all terms to the left side:
\(3x^2 - x^2 + 7x - 7x + 23 - 14 = 0\)
or, \(2x^2 + 9 = 0\)
This can be written as \(2x^2 + 0x + 9 = 0\), which is in the form \(ax^2 + bx + c = 0\).
(iv) Given equation: \((x + 2)^3 = x(x^2 - 1)\)
Using the formula \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
\(x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = x^3 - x\)
or, \(x^3 + 6x^2 + 12x + 8 = x^3 - x\)
Subtracting \(x^3\) from both sides:
\(6x^2 + 12x + 8 = -x\)
or, \(6x^2 + 13x + 8 = 0\)
This is in the form \(ax^2 + bx + c = 0\).
Comparing with \(ax^2 + bx + cx^0 = 0\) (since \(x^0 = 1\)):
- Coefficient of \(x^2\) is \(6\)
- Coefficient of \(x\) is \(13\)
- Coefficient of \(x^0\) is \(8\)
(ii) The product of two consecutive positive odd numbers is 143.
(iii) The sum of the squares of two consecutive numbers is 313.
(i) Let one part be \(x\).
Then the other part is \(42 - x\).
According to the given condition, one part is equal to the square of the other part.
So, \(x^2 = 42 - x\)
\(\implies x^2 + x - 42 = 0\)
This is the required quadratic equation.
(ii) Let the two consecutive positive odd numbers be \(x\) and \((x + 2)\).
According to the condition, their product is 143.
So, \(x(x + 2) = 143\)
\(\implies x^2 + 2x = 143\)
\(\implies x^2 + 2x - 143 = 0\)
This is the required quadratic equation.
(iii) Let the two consecutive numbers be \(x\) and \((x + 1)\).
According to the condition, the sum of their squares is 313.
So, \(x^2 + (x + 1)^2 = 313\)
\(\implies x^2 + x^2 + 2x + 1 = 313\)
\(\implies 2x^2 + 2x + 1 - 313 = 0\)
\(\implies 2x^2 + 2x - 312 = 0\)
Dividing by 2:
\(\implies x^2 + x - 156 = 0\)
This is the required quadratic equation.
(ii) One person bought some sugar at ₹ \(80\). If he would get \(4\text{ kg}\). more sugar with that money, then the price of \(1\text{ kg}\). of sugar would be less by ₹ \(1\).
(iii) The distance between two stations is \(300\text{ km}\). A train went to second from first station with uniform velocity. If the velocity of the train could be \(5\text{km/hour}\) more, then the time taken by the train to reach the second station would be lesser by \(2\text{ hours}\).
(iv) A clock seller sold a clock by purchasing it at ₹ \(336\). The amount of his profit percentage is as much as the amount with which he bought the clock.
(v) If the velocity of the stream is \(2\text{kms/hr}\), then the total time taken by Ratanmajhi to cover \(21\text{ kms}\) in down stream and the same distance in upstream is \(10\text{ hours}\).
(vi) The time taken to clean out garden by Majid is \(3\text{ hours}\) more than Mahim. Both of them together can complete the work in \(2\text{ hours}\).
(vii) The unit digit of a two digit number exceeds its tens' digit by \(6\) and the product of two digits is less than the number by \(12\).
(viii) There is a road of equal width around a rectangular playground in the outside of it having the length \(45\text{ m}\). and breadth \(40\text{ m}\). and the area of the road is \(450\text{ sqm}\).
(i) Let the breadth of the rectangle be \(x\text{ m}\).
Then its length will be \((x + 3)\text{ m}\).
By Pythagoras theorem, \((\text{Diagonal})^2 = (\text{Length})^2 + (\text{Breadth})^2\)
\(15^2 = (x + 3)^2 + x^2\)
or, \(225 = x^2 + 6x + 9 + x^2\)
or, \(2x^2 + 6x + 9 - 225 = 0\)
or, \(2x^2 + 6x - 216 = 0\)
\(x^2 + 3x - 108 = 0\)
(ii) Let the person buy \(x\text{ kg}\) of sugar for ₹ \(80\).
Price of \(1\text{ kg}\) sugar = ₹ \(\frac{80}{x}\)
If he gets \((x + 4)\text{ kg}\) sugar for ₹ \(80\), the new price of \(1\text{ kg}\) = ₹ \(\frac{80}{x + 4}\)
According to the condition:
\(\frac{80}{x} - \frac{80}{x + 4} = 1\)
or, \(80 \left[ \frac{(x + 4) - x}{x(x + 4)} \right] = 1\)
or, \(\frac{80 \times 4}{x^2 + 4x} = 1\)
or, \(x^2 + 4x = 320\)
\(x^2 + 4x - 320 = 0\)
(iii) Let the uniform velocity of the train be \(x\text{ km/hr}\).
Time taken to travel \(300\text{ km}\) = \(\frac{300}{x}\text{ hours}\).
If velocity is \((x + 5)\text{ km/hr}\), time taken = \(\frac{300}{x + 5}\text{ hours}\).
According to the condition:
\(\frac{300}{x} - \frac{300}{x + 5} = 2\)
or, \(300 \left[ \frac{(x + 5) - x}{x(x + 5)} \right] = 2\)
or, \(\frac{300 \times 5}{x^2 + 5x} = 2\)
or, \(1500 = 2x^2 + 10x\)
or, \(2x^2 + 10x - 1500 = 0\)
\(x^2 + 5x - 750 = 0\)
(iv) Let the purchase price (cost price) of the clock be ₹ \(x\).
Profit percentage = \(x\%\)
Profit amount = \(\frac{x}{100} \times x = \frac{x^2}{100}\)
Selling price = Cost price + Profit
\(336 = x + \frac{x^2}{100}\)
or, \(33600 = 100x + x^2\)
\(x^2 + 100x - 33600 = 0\)
(v) Let the velocity of the boat in still water be \(x\text{ km/hr}\).
Velocity of stream = \(2\text{ km/hr}\).
Velocity downstream = \((x + 2)\text{ km/hr}\).
Velocity upstream = \((x - 2)\text{ km/hr}\).
Total time = Time downstream + Time upstream
\(10 = \frac{21}{x + 2} + \frac{21}{x - 2}\)
or, \(10 = 21 \left[ \frac{(x - 2) + (x + 2)}{(x + 2)(x - 2)} \right]\)
or, \(10 = 21 \left[ \frac{2x}{x^2 - 4} \right]\)
or, \(10(x^2 - 4) = 42x\)
or, \(10x^2 - 40 - 42x = 0\)
or, \(10x^2 - 42x - 40 = 0\)
\(5x^2 - 21x - 20 = 0\)
(vi) Let Mahim take \(x\text{ hours}\) to clean the garden.
Then Majid takes \((x + 3)\text{ hours}\).
In \(1\text{ hour}\), Mahim does \(\frac{1}{x}\) part of the work.
In \(1\text{ hour}\), Majid does \(\frac{1}{x + 3}\) part of the work.
Together, in \(1\text{ hour}\), they do \(\left(\frac{1}{x} + \frac{1}{x + 3}\right)\) part of the work.
Since they complete it in \(2\text{ hours}\), their combined \(1\text{-hour}\) work is \(\frac{1}{2}\).
\(\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2}\)
or, \(\frac{x + 3 + x}{x(x + 3)} = \frac{1}{2}\)
or, \(\frac{2x + 3}{x^2 + 3x} = \frac{1}{2}\)
or, \(x^2 + 3x = 4x + 6\)
\(x^2 - x - 6 = 0\)
(vii) Let the tens' digit be \(x\).
Then the unit digit is \((x + 6)\).
The number is \(10x + (x + 6) = 11x + 6\).
Product of the digits = \(x(x + 6) = x^2 + 6x\).
According to the condition:
\(x^2 + 6x = (11x + 6) - 12\)
or, \(x^2 + 6x = 11x - 6\)
\(x^2 - 5x + 6 = 0\)
(viii) Let the width of the road be \(x\text{ m}\).
Length of the playground with road = \((45 + 2x)\text{ m}\).
Breadth of the playground with road = \((40 + 2x)\text{ m}\).
Area of the road = Area of outer rectangle - Area of inner rectangle
\(450 = (45 + 2x)(40 + 2x) - (45 \times 40)\)
or, \(450 = 1800 + 90x + 80x + 4x^2 - 1800\)
or, \(4x^2 + 170x - 450 = 0\)
\(2x^2 + 85x - 225 = 0\)
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