Ganit Prakash - Class-X - Let us work out 1.3
Let us work out 1.3 Solutions Step-by-Step Approach
Click on “📋 Click for Solution” to reveal the step-by-step answers.
📗 Let us work out 1.3
Q1 to Q9Let the smaller positive whole number be \(x\).
Then, the other positive whole number is \(x + 3\).
According to the given condition, the sum of their squares is \(117\).
\(x^2 + (x + 3)^2 = 117\)
or, \(x^2 + x^2 + 6x + 9 = 117\)
or, \(2x^2 + 6x + 9 - 117 = 0\)
or, \(2x^2 + 6x - 108 = 0\)
Dividing the equation by \(2\):
or, \(x^2 + 3x - 54 = 0\)
or, \(x^2 + 9x - 6x - 54 = 0\)
or, \(x(x + 9) - 6(x + 9) = 0\)
or, \((x - 6)(x + 9) = 0\)
Since the number is positive, \(x \neq -9\). Therefore, \(x = 6\).
The other number is \(6 + 3 = 9\).
Answer: The two numbers are \(6\) and \(9\).
Let the height of the triangle be \(x\text{ m}\).
Then the base of the triangle is \((2x + 18)\text{ m}\).
We know, Area of a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
According to the condition:
\(\frac{1}{2} \times (2x + 18) \times x = 360\)
or, \(\frac{1}{2} \times 2(x + 9) \times x = 360\)
or, \(x(x + 9) = 360\)
or, \(x^2 + 9x - 360 = 0\)
or, \(x^2 + 24x - 15x - 360 = 0\)
or, \(x(x + 24) - 15(x + 24) = 0\)
or, \((x - 15)(x + 24) = 0\)
Since the height of a triangle cannot be negative, \(x \neq -24\). Therefore, \(x = 15\).
Answer: The height of the triangle is \(15\text{ m}\).
Let the positive whole number be \(x\).
According to the given condition:
\(5x = 2x^2 - 3\)
or, \(2x^2 - 5x - 3 = 0\)
or, \(2x^2 - 6x + x - 3 = 0\)
or, \(2x(x - 3) + 1(x - 3) = 0\)
or, \((x - 3)(2x + 1) = 0\)
Either \(x - 3 = 0 \implies x = 3\)
Or \(2x + 1 = 0 \implies x = -\frac{1}{2}\)
Since \(x\) is a positive whole number, \(x \neq -\frac{1}{2}\). Therefore, \(x = 3\).
Answer: The number is \(3\).
Let the speed of the jeep car be \(x\text{ km/hr}\).
Then the speed of the motor car is \((x + 5)\text{ km/hr}\).
Total distance = \(200\text{ km}\).
Time taken by the jeep car = \(\frac{200}{x}\text{ hours}\).
Time taken by the motor car = \(\frac{200}{x + 5}\text{ hours}\).
According to the condition:
\(\frac{200}{x} - \frac{200}{x + 5} = 2\)
or, \(\frac{200(x + 5) - 200x}{x(x + 5)} = 2\)
or, \(\frac{200x + 1000 - 200x}{x^2 + 5x} = 2\)
or, \(1000 = 2(x^2 + 5x)\)
or, \(x^2 + 5x = 500\)
or, \(x^2 + 5x - 500 = 0\)
or, \(x^2 + 25x - 20x - 500 = 0\)
or, \(x(x + 25) - 20(x + 25) = 0\)
\((x - 20)(x + 25) = 0\)
Since speed cannot be negative, \(x \neq -25\). Therefore, \(x = 20\).
Speed of the jeep car is \(20\text{ km/hr}\).
Answer: The speed of the motor car is \(20 + 5 = 25\text{ km/hr}\).
Let the length of the rectangular land be \(x\text{ m}\) and the breadth be \(y\text{ m}\).
Perimeter = \(180\text{ m}\)
\(2(x + y) = 180 \implies x + y = 90 \implies y = 90 - x\)
Area = \(2000\text{ sq.m}\)
\(x \times y = 2000\)
Substituting the value of \(y\):
\(x(90 - x) = 2000\)
or, \(90x - x^2 = 2000\)
or, \(x^2 - 90x + 2000 = 0\)
or, \(x^2 - 50x - 40x + 2000 = 0\)
or, \(x(x - 50) - 40(x - 50) = 0\)
\((x - 50)(x - 40) = 0\)
So, \(x = 50\) or \(x = 40\).
Since length is generally considered greater than breadth, length = \(50\text{ m}\) and breadth = \(40\text{ m}\).
Answer: The length is \(50\text{ m}\) and the breadth is \(40\text{ m}\).
Let the unit digit be \(x\).
Then the tens digit is \((x - 3)\).
The two-digit number is \(10 \times (\text{tens digit}) + (\text{unit digit}) = 10(x - 3) + x = 11x - 30\).
The product of the digits is \(x(x - 3) = x^2 - 3x\).
According to the condition:
\((11x - 30) - (x^2 - 3x) = 15\)
or, \(11x - 30 - x^2 + 3x = 15\)
or, \(-x^2 + 14x - 30 - 15 = 0\)
or, \(-x^2 + 14x - 45 = 0\)
or, \(x^2 - 14x + 45 = 0\)
or, \(x^2 - 9x - 5x + 45 = 0\)
or, \(x(x - 9) - 5(x - 9) = 0\)
\((x - 9)(x - 5) = 0\)
So, either \(x = 9\) or \(x = 5\).
Answer: The unit digit of the number is \(5\) or \(9\).
Let the time taken by the faster pipe to fill the reservoir separately be \(x\text{ minutes}\).
Then the time taken by the other pipe is \((x + 5)\text{ minutes}\).
Together they take \(11\frac{1}{9} = \frac{100}{9}\text{ minutes}\).
In 1 minute, the first pipe fills \(\frac{1}{x}\) part of the reservoir.
In 1 minute, the second pipe fills \(\frac{1}{x + 5}\) part.
Together, in 1 minute, they fill \(\frac{9}{100}\) part.
According to the condition:
\(\frac{1}{x} + \frac{1}{x + 5} = \frac{9}{100}\)
or, \(\frac{x + 5 + x}{x(x + 5)} = \frac{9}{100}\)
or, \(\frac{2x + 5}{x^2 + 5x} = \frac{9}{100}\)
or, \(9(x^2 + 5x) = 100(2x + 5)\)
or, \(9x^2 + 45x = 200x + 500\)
or, \(9x^2 - 155x - 500 = 0\)
or, \(9x^2 - 180x + 25x - 500 = 0\)
or, \(9x(x - 20) + 25(x - 20) = 0\)
\((x - 20)(9x + 25) = 0\)
Since time cannot be negative, \(x \neq -\frac{25}{9}\). Therefore, \(x = 20\).
Answer: The time taken to fill the reservoir separately is \(20\text{ minutes}\) and \((20 + 5) = 25\text{ minutes}\).
Let the time taken by Pijush alone to complete the work be \(x\text{ days}\).
Then the time taken by Porna alone is \((x + 6)\text{ days}\).
In 1 day, Pijush completes \(\frac{1}{x}\) part of the work.
In 1 day, Porna completes \(\frac{1}{x + 6}\) part of the work.
Together they complete the work in 4 days, so their combined 1-day work is \(\frac{1}{4}\).
\(\frac{1}{x} + \frac{1}{x + 6} = \frac{1}{4}\)
or, \(\frac{x + 6 + x}{x(x + 6)} = \frac{1}{4}\)
or, \(\frac{2x + 6}{x^2 + 6x} = \frac{1}{4}\)
or, \(x^2 + 6x = 4(2x + 6)\)
or, \(x^2 + 6x = 8x + 24\)
or, \(x^2 - 2x - 24 = 0\)
or, \(x^2 - 6x + 4x - 24 = 0\)
or, \(x(x - 6) + 4(x - 6) = 0\)
\((x - 6)(x + 4) = 0\)
Since days cannot be negative, \(x \neq -4\). Therefore, \(x = 6\).
Answer: The time taken by Porna alone to complete the work is \(6 + 6 = 12\text{ days}\).
Let the original price of 1 dozen (12) pens be ₹ \(x\).
Original price of 1 pen = ₹ \(\frac{x}{12}\).
Number of pens bought for ₹ 30 originally = \(\frac{30}{\frac{x}{12}} = \frac{360}{x}\).
After reduction, the new price of 1 dozen pens = ₹ \((x - 6)\).
New price of 1 pen = ₹ \(\frac{x - 6}{12}\).
Number of pens bought for ₹ 30 now = \(\frac{30}{\frac{x - 6}{12}} = \frac{360}{x - 6}\).
According to the given condition, we get 3 more pens now:
\(\frac{360}{x - 6} - \frac{360}{x} = 3\)
Dividing the equation by \(3\):
or, \(\frac{120}{x - 6} - \frac{120}{x} = 1\)
or, \(\frac{120x - 120(x - 6)}{x(x - 6)} = 1\)
or, \(\frac{120x - 120x + 720}{x^2 - 6x} = 1\)
or, \(x^2 - 6x = 720\)
or, \(x^2 - 6x - 720 = 0\)
or, \(x^2 - 30x + 24x - 720 = 0\)
or, \(x(x - 30) + 24(x - 30) = 0\)
\((x - 30)(x + 24) = 0\)
Since price cannot be negative, \(x \neq -24\). Therefore, \(x = 30\).
Answer: Before the reduction, the price of 1 dozen pens was ₹ \(30\).
📗 10. V.S.A.
MCQ, True/False & Fill Blanks(a) one
(b) two
(c) three
(d) none of them
A quadratic equation is a polynomial equation of degree 2. Therefore, it always has exactly two roots (real or complex).
Answer: (b) two
(a) \(b \neq 0\)
(b) \(c \neq 0\)
(c) \(a \neq 0\)
(d) none of these
For an equation to be classified as quadratic, the highest power of the variable (which is \(x^2\)) must exist. Therefore, its coefficient cannot be zero.
Answer: (c) \(a \neq 0\)
(a) 1
(b) 2
(c) 3
(d) none of these.
By definition, a quadratic equation has a degree of 2.
Answer: (b) 2
(a) linear
(b) quadratic
(c) 3rd degree
(d) none of these.
Expanding both sides:
\(20x^2 - 28x + 8 = 20x^2 - 30x + 15\)
Subtracting \(20x^2\) from both sides:
or, \(-28x + 8 = -30x + 15\)
or, \(2x - 7 = 0\)
The highest power of \(x\) is 1, so the equation is linear.
Answer: (a) linear
(a) 0
(b) 6
(c) 0 & 6
(d) -6
Given \(\frac{x^2}{x} = 6\).
For the denominator to be valid, \(x \neq 0\).
Simplifying the fraction: \(x = 6\).
Answer: (b) 6
Expanding the left-hand side gives \(x^2 - 6x + 9\).
So, \(x^2 - 6x + 9 = x^2 - 6x + 9\).
Since both sides are exactly identical, it is an algebraic identity, not a conditional equation.
Answer: False
Taking the square root of both sides of \(x^2 = 25\) yields \(x = \pm 5\).
The roots are 5 and -5. So, 5 is not the only root.
Answer: False
If \(a = 0\), the \(x^2\) term disappears and the equation becomes \(bx + c = 0\).
Since the highest power of \(x\) is now 1, it becomes a linear equation.
Answer: linear
If both roots are 1, the factors are \((x - 1)\) and \((x - 1)\).
Equation: \((x - 1)(x - 1) = 0\)
\(x^2 - 2x + 1 = 0\)
Answer: \(x^2 - 2x + 1 = 0\)
\(x^2 - 6x = 0\)
or, \(x(x - 6) = 0\)
So, either \(x = 0\) or \(x - 6 = 0 \implies x = 6\).
Answer: 0 & 6
📗 11. S.A.
Short Answer TypeSince 1 is a root, substituting \(x = 1\) into the equation will satisfy it:
\((1)^2 + a(1) + 3 = 0\)
or, \(1 + a + 3 = 0\)
or, \(a + 4 = 0\)
or, \(a = -4\)
Answer: The value of \(a\) is \(-4\).
Let the roots of the equation be \(\alpha\) and \(\beta\). Let \(\alpha = 2\).
We know that the product of the roots (\(\alpha \times \beta\)) is equal to \(\frac{c}{a}\).
From the equation, \(c = 6\) and \(a = 1\).
\(2 \times \beta = \frac{6}{1}\)
\(2\beta = 6 \implies \beta = 3\)
Answer: The other root is \(3\).
Let the roots of the equation be \(\alpha\) and \(\beta\). Let \(\alpha = 2\).
The product of the roots (\(\alpha \times \beta\)) is equal to \(\frac{c}{a}\).
From the given equation, \(c = 4\) and \(a = 2\).
\(2 \times \beta = \frac{4}{2}\)
\(2\beta = 2 \implies \beta = 1\)
Answer: The other root is \(1\).
Let the proper fraction be \(x\). Its reciprocal is \(\frac{1}{x}\).
Since it is a proper fraction (\(0 < x < 1\)), its reciprocal is always greater than the fraction itself (i.e., \(\frac{1}{x} > x\)).
According to the condition:
\(\frac{1}{x} - x = \frac{9}{20}\)
or, \(\frac{1 - x^2}{x} = \frac{9}{20}\)
or, \(20(1 - x^2) = 9x\)
or, \(20 - 20x^2 = 9x\)
\(20x^2 + 9x - 20 = 0\)
Answer: The required equation is \(20x^2 + 9x - 20 = 0\).
The given roots are \(\alpha = -5\) and \(\beta = -7\).
Product of the roots = \((-5) \times (-7) = 35\).
From the equation, Product of roots = \(\frac{c}{a} = \frac{35}{a}\).
So, \(\frac{35}{a} = 35 \implies a = 1\).
Sum of the roots = \((-5) + (-7) = -12\).
From the equation, Sum of roots = \(-\frac{b}{a}\).
Substituting \(a = 1\): \(-\frac{b}{1} = -b\).
So, \(-b = -12 \implies b = 12\).
Answer: The values are \(a = 1\) and \(b = 12\).
Hi Please, do not Spam in Comments.