10th Std Maths Quarterly Exam 2024 - Salem District | Question Paper with Solutions
PART - A
1. If there are 1024 relations from a set A = {1,2,3,4,5} to a set B, then the number of elements in B is ...
Answer: b) 2
Explanation:
Given, n(A) = 5.
The number of relations from A to B is given by \(2^{n(A) \times n(B)}\).
We have \(2^{n(A) \times n(B)} = 1024\).
We know that \(1024 = 2^{10}\).
So, \(n(A) \times n(B) = 10\).
\(5 \times n(B) = 10 \Rightarrow n(B) = \frac{10}{5} = 2\).
2. If the ordered pairs (a, -1) and (5, b) belong to {(x,y) / y = 2x + 3}, then the values of 'a' and 'b' are ...
- a) (-13, 2)
- b) (2, 13)
- c) (2, -13)
- d) (-2, 13)
Answer: d) (-2, 13)
Explanation:
For (a, -1): Substitute x=a and y=-1 in y = 2x + 3.
\(-1 = 2a + 3 \Rightarrow 2a = -4 \Rightarrow a = -2\).
For (5, b): Substitute x=5 and y=b in y = 2x + 3.
\(b = 2(5) + 3 \Rightarrow b = 10 + 3 = 13\).
So, the values are a = -2 and b = 13.
3. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is ...
- a) 2025
- b) 5220
- c) 5025
- d) 2520
Answer: d) 2520
Explanation:
We need to find the LCM of numbers from 1 to 10.
Prime factors: 2, 3, 5, 7.
Highest powers: \(2^3=8\), \(3^2=9\), \(5^1=5\), \(7^1=7\).
LCM = \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520\).
4. If \(t_n\) is the nth term of an A.P. then \(t_{2n} - t_n\) is ...
- a) nd
- b) 2nd
- c) 2d
- d) 3nd
Answer: a) nd
Explanation:
We know that \(t_k = a + (k-1)d\).
\(t_{2n} = a + (2n-1)d\)
\(t_n = a + (n-1)d\)
\(t_{2n} - t_n = [a + (2n-1)d] - [a + (n-1)d]\)
\(= (2n-1-n+1)d = nd\).
5. The sequence \(\sqrt{11}, \sqrt{55}, 5\sqrt{11}, 5\sqrt{55}, 25\sqrt{11},...\) represents ...
- a) an A.P. only
- b) a G.P. only
- c) neither A.P nor G.P
- d) both A.P and G.P
Answer: b) a G.P. only
Explanation:
Common ratio \(r = \frac{t_2}{t_1} = \frac{\sqrt{55}}{\sqrt{11}} = \sqrt{\frac{55}{11}} = \sqrt{5}\).
\(\frac{t_3}{t_2} = \frac{5\sqrt{11}}{\sqrt{55}} = \frac{5\sqrt{11}}{\sqrt{5}\sqrt{11}} = \frac{5}{\sqrt{5}} = \sqrt{5}\).
Since the common ratio is constant, it is a G.P.
6. If (x - 6) is the HCF of \(x^2 - 2x - 24\) and \(x^2 - kx - 6\) then the value of k is ...
Answer: b) 5
Explanation:
If (x-6) is a factor, then x=6 must be a root of both polynomials.
For \(x^2 - kx - 6\), substitute x = 6:
\((6)^2 - k(6) - 6 = 0\)
\(36 - 6k - 6 = 0\)
\(30 - 6k = 0 \Rightarrow 6k = 30 \Rightarrow k = 5\).
7. Which of the following should be added to make \(x^4 + 64\) a perfect square?
- a) 4x²
- b) 16x²
- c) 8x²
- d) -8x²
Answer: b) 16x²
Explanation:
\(x^4 + 64 = (x^2)^2 + 8^2\).
To make it a perfect square in the form \((a+b)^2 = a^2 + 2ab + b^2\), we need the middle term \(2ab\).
Here \(a = x^2\) and \(b = 8\).
\(2ab = 2(x^2)(8) = 16x^2\).
8. A quadratic equation whose one zero is 5 and the sum of the zeroes is 0 is given by the equation ...
- a) x² - 5x = 0
- b) x² - 5x + 5 = 0
- c) x² - 25 = 0
- d) x² - 5 = 0
Answer: c) x² - 25 = 0
Explanation:
Let the zeroes be \(\alpha\) and \(\beta\).
Given \(\alpha = 5\) and sum of zeroes \(\alpha + \beta = 0\).
\(5 + \beta = 0 \Rightarrow \beta = -5\).
Product of zeroes = \(\alpha\beta = 5 \times (-5) = -25\).
The quadratic equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\).
\(x^2 - (0)x + (-25) = 0 \Rightarrow x^2 - 25 = 0\).
9. The perimeters of two similar triangles \(\triangle ABC\) and \(\triangle PQR\) are 36cm and 24cm respectively. If PQ = 10 cm, the length of AB is ...
- a) \(6\frac{2}{3}\) cm
- b) \(\frac{10\sqrt{6}}{3}\) cm
- c) \(66\frac{2}{3}\) cm
- d) 15 cm
Answer: d) 15 cm
Explanation:
The ratio of the perimeters of similar triangles is equal to the ratio of their corresponding sides.
\(\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)} = \frac{AB}{PQ}\)
\(\frac{36}{24} = \frac{AB}{10} \Rightarrow \frac{3}{2} = \frac{AB}{10}\)
\(AB = \frac{3 \times 10}{2} = 15\) cm.
10. In a \(\triangle ABC\), AD is the bisector of \(\angle BAC\). If AB = 8cm, BD = 6cm and DC = 3cm. The length of the side AC is ...
- a) 6 cm
- b) 4 cm
- c) 3 cm
- d) 8 cm
Answer: b) 4 cm
Explanation:
By Angle Bisector Theorem,
\(\frac{AB}{AC} = \frac{BD}{DC}\)
\(\frac{8}{AC} = \frac{6}{3} \Rightarrow \frac{8}{AC} = 2\)
\(AC = \frac{8}{2} = 4\) cm.
11. The straight line given by the equation x = 11 is ...
- a) Parallel to x axis
- b) parallel to y axis
- c) passing through the origin
- d) passing through the point (0, 11)
Answer: b) parallel to y axis
Explanation:
The equation \(x = c\) (where c is a constant) represents a vertical line that is parallel to the y-axis.
12. If (5,7), (3, p) and (6, 6) are collinear then the value of p is ...
Answer: c) 9
Explanation:
If three points are collinear, their slopes are equal.
Slope of (5,7) and (3,p) = Slope of (3,p) and (6,6)
\(\frac{p-7}{3-5} = \frac{6-p}{6-3}\)
\(\frac{p-7}{-2} = \frac{6-p}{3}\)
\(3(p-7) = -2(6-p) \Rightarrow 3p - 21 = -12 + 2p\)
\(p = 9\).
13. (2,1) is the point of intersection of two lines ...
- a) x - y - 3 = 0; 3x - y - 7 = 0
- b) x + y = 3; 3x + y = 7
- c) 3x + y = 3; x + y = 7
- d) x + 3y - 3 = 0; x - y - 7 = 0
Answer: b) x + y = 3; 3x + y = 7
Explanation:
The point of intersection must satisfy both equations. Substitute (2,1) into the equations in option (b).
Equation 1: \(x + y = 2 + 1 = 3\). (Satisfied)
Equation 2: \(3x + y = 3(2) + 1 = 6 + 1 = 7\). (Satisfied)
Therefore, (2,1) is the point of intersection.
14. If \(5x = \sec\theta\) and \(\frac{5}{y} = \tan\theta\); then \(x^2 - \frac{1}{y^2}\) is equal to ... (Assuming a typo correction from the image)
- a) 25
- b) \(\frac{1}{25}\)
- c) 5
- d) 1
Answer: b) \(\frac{1}{25}\)
Explanation:
Given \(5x = \sec\theta \Rightarrow x = \frac{\sec\theta}{5}\).
Given \(\frac{5}{y} = \tan\theta \Rightarrow \frac{1}{y} = \frac{\tan\theta}{5}\).
We need to find \(x^2 - \frac{1}{y^2}\).
\(x^2 - \frac{1}{y^2} = \left(\frac{\sec\theta}{5}\right)^2 - \left(\frac{\tan\theta}{5}\right)^2\)
\(= \frac{\sec^2\theta}{25} - \frac{\tan^2\theta}{25} = \frac{\sec^2\theta - \tan^2\theta}{25}\)
Since \(\sec^2\theta - \tan^2\theta = 1\), the expression equals \(\frac{1}{25}\).
PART - B
Answer any 10 questions. Question No. 28 is compulsory.
15. A relation R is given by the set \(\{(x,y) / y = x + 3, x \in \{0,1,2,3,4,5\}\}\). Determine its domain and range.
Given relation: \(y = x + 3\) and \(x \in \{0,1,2,3,4,5\}\).
When x=0, y = 0+3 = 3
When x=1, y = 1+3 = 4
When x=2, y = 2+3 = 5
When x=3, y = 3+3 = 6
When x=4, y = 4+3 = 7
When x=5, y = 5+3 = 8
The relation R as a set of ordered pairs is \(\{(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)\}\).
Domain = The set of all first elements = \{0, 1, 2, 3, 4, 5\}.
Range = The set of all second elements = \{3, 4, 5, 6, 7, 8\}.
16. Let f be a function from R to R defined by \(f(x) = 3x - 5\). Find the values of a and b given that (a, 4) and (1, b) belong to f.
Given \(f(x) = 3x - 5\).
Since (a, 4) belongs to f, we have \(f(a) = 4\).
\(3a - 5 = 4 \Rightarrow 3a = 9 \Rightarrow a = 3\).
Since (1, b) belongs to f, we have \(f(1) = b\).
\(b = 3(1) - 5 \Rightarrow b = 3 - 5 \Rightarrow b = -2\).
Therefore, a = 3 and b = -2.
17. If \(f(x) = x^2 - 1\), \(g(x) = x - 2\), find a, if \(g \circ f(a) = 1\).
Given \(f(x) = x^2 - 1\) and \(g(x) = x - 2\).
We need to solve \(g(f(a)) = 1\).
First, find \(f(a) = a^2 - 1\).
Now, \(g(f(a)) = g(a^2 - 1)\).
Substitute \(a^2 - 1\) into g(x):
\(g(a^2 - 1) = (a^2 - 1) - 2 = a^2 - 3\).
Given \(g(f(a)) = 1\), so \(a^2 - 3 = 1\).
\(a^2 = 4 \Rightarrow a = \pm 2\).
18. Solve: \(5x \equiv 4 \pmod{6}\).
The congruence \(5x \equiv 4 \pmod{6}\) can be written as \(5x = 6k + 4\) for some integer k.
We can test values for x:
If x = 1, \(5(1) = 5 \equiv 5 \pmod{6}\).
If x = 2, \(5(2) = 10 \equiv 4 \pmod{6}\). This is a solution.
The solutions are of the form \(x = 2 + 6n\), where n is an integer.
A particular solution is x = 2.
19. In a G.P. 729, 243, 81, ..., find \(t_7\).
The given G.P. is 729, 243, 81, ...
First term, \(a = 729\).
Common ratio, \(r = \frac{243}{729} = \frac{1}{3}\).
The nth term of a G.P. is \(t_n = ar^{n-1}\).
We need to find \(t_7\).
\(t_7 = 729 \times (\frac{1}{3})^{7-1} = 729 \times (\frac{1}{3})^6\).
Since \(729 = 3^6\),
\(t_7 = 3^6 \times \frac{1}{3^6} = 1\).
20. If \(1 + 2 + 3 + \dots + k = 325\), then find \(1^3 + 2^3 + 3^3 + \dots + k^3\).
We are given the sum of the first k natural numbers:
\(1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} = 325\).
We need to find the sum of the cubes of the first k natural numbers:
\(1^3 + 2^3 + 3^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2\).
Substituting the given value:
\(1^3 + 2^3 + \dots + k^3 = (325)^2\).
\(325^2 = 105625\).
21. Find the excluded value of the rational expression \(\frac{t}{t^2-5t+6}\).
The excluded values are the values of 't' for which the denominator is zero.
Set the denominator to zero: \(t^2 - 5t + 6 = 0\).
Factorize the quadratic equation:
\((t - 2)(t - 3) = 0\).
This gives \(t = 2\) or \(t = 3\).
The excluded values are 2 and 3.
22. Simplify: \(\frac{x^3}{x-y} + \frac{y^3}{y-x}\).
\(\frac{x^3}{x-y} + \frac{y^3}{y-x} = \frac{x^3}{x-y} + \frac{y^3}{-(x-y)}\)
\(= \frac{x^3}{x-y} - \frac{y^3}{x-y}\)
\(= \frac{x^3 - y^3}{x-y}\)
Using the formula \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\),
\(= \frac{(x-y)(x^2+xy+y^2)}{x-y}\)
\(= x^2+xy+y^2\).
23. Determine the nature of the roots of the quadratic equation \(15x^2 + 11x + 2 = 0\).
The nature of the roots is determined by the discriminant, \(\Delta = b^2 - 4ac\).
Here, a = 15, b = 11, c = 2.
\(\Delta = (11)^2 - 4(15)(2)\)
\(= 121 - 120 = 1\).
Since \(\Delta = 1 > 0\) and is a perfect square, the roots are real, unequal, and rational.
24. If \(\triangle ABC\) is similar to \(\triangle DEF\) such that BC = 3cm, EF = 4cm and area of \(\triangle ABC = 54cm^2\). Find the area of \(\triangle DEF\).
For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\(\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2\)
\(\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
\(\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96 \, \text{cm}^2\).
25. Find the slope of a line joining the points (-6, 1) and (-3, 2).
The slope \(m\) of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Here, \((x_1, y_1) = (-6, 1)\) and \((x_2, y_2) = (-3, 2)\).
\(m = \frac{2 - 1}{-3 - (-6)} = \frac{1}{-3 + 6} = \frac{1}{3}\).
26. Show that the straight lines \(x - 2y + 3 = 0\) and \(6x + 3y + 8 = 0\) are perpendicular.
For the first line, \(x - 2y + 3 = 0\), the slope \(m_1 = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{1}{-2} = \frac{1}{2}\).
For the second line, \(6x + 3y + 8 = 0\), the slope \(m_2 = -\frac{6}{3} = -2\).
Two lines are perpendicular if the product of their slopes is -1.
\(m_1 \times m_2 = \frac{1}{2} \times (-2) = -1\).
Hence, the lines are perpendicular.
27. Prove that \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta\).
LHS = \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\).
Multiply numerator and denominator inside the square root by the conjugate of the denominator, which is \(1+\sin\theta\).
LHS = \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta}}\)
\(= \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}\)
Using the identity \(1-\sin^2\theta = \cos^2\theta\),
\(= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}\)
\(= \frac{1+\sin\theta}{\cos\theta}\)
\(= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\)
\(= \sec\theta + \tan\theta\) = RHS. (Hence Proved)
28. Find the equation of straight line whose slope is -4 and passing through the point (1,2).
The equation of a straight line with slope \(m\) and passing through a point \((x_1, y_1)\) is given by the point-slope form: \(y - y_1 = m(x - x_1)\).
Given \(m = -4\) and \((x_1, y_1) = (1, 2)\).
Substituting these values:
\(y - 2 = -4(x - 1)\)
\(y - 2 = -4x + 4\)
\(4x + y - 2 - 4 = 0\)
The required equation is \(4x + y - 6 = 0\).
PART - C
Answer any 10 questions. Question No. 42 is compulsory.
29. Let A = {x ∈ W / x < 2}, B = {x ∈ N / 1 < x ≤ 4} and C = {3, 5}. Verify that A × (B ∩ C) = (A × B) ∩ (A × C).
Given sets:
A = {x ∈ W / x < 2} = {0, 1} (W is whole numbers)
B = {x ∈ N / 1 < x ≤ 4} = {2, 3, 4} (N is natural numbers)
C = {3, 5}
LHS: A × (B ∩ C)
First, find B ∩ C = {3}.
A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)}.
RHS: (A × B) ∩ (A × C)
First, find A × B = {0, 1} × {2, 3, 4} = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}.
Next, find A × C = {0, 1} × {3, 5} = {(0, 3), (0, 5), (1, 3), (1, 5)}.
Now, find the intersection: (A × B) ∩ (A × C) = {(0, 3), (1, 3)}.
Since LHS = RHS, the property is verified.
30. Let f : A→B be a function defined by \(f(x) = \frac{x}{2} - 1\), where A = {2,4,6,10,12}, B = {0,1,2,4,5,9}. Represent f by i) a set of ordered pairs, ii) a table, iii) an arrow diagram, iv) a graph.
Given function: \(f(x) = \frac{x}{2} - 1\)
Domain A = {2, 4, 6, 10, 12}
We find the image for each element in A:
- \(f(2) = \frac{2}{2} - 1 = 1 - 1 = 0\)
- \(f(4) = \frac{4}{2} - 1 = 2 - 1 = 1\)
- \(f(6) = \frac{6}{2} - 1 = 3 - 1 = 2\)
- \(f(10) = \frac{10}{2} - 1 = 5 - 1 = 4\)
- \(f(12) = \frac{12}{2} - 1 = 6 - 1 = 5\)
i) A set of ordered pairs:
The function f can be represented as the set of ordered pairs (x, f(x)):
f = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
ii) A table:
The function f can be represented in a tabular form:
| x |
2 |
4 |
6 |
10 |
12 |
| f(x) |
0 |
1 |
2 |
4 |
5 |
iii) An arrow diagram:
(Draw two ovals representing sets A and B. List the elements. Draw arrows from each element in A to its corresponding image in B.)
A = {2, 4, 6, 10, 12}
B = {0, 1, 2, 4, 5, 9}
Arrows should be drawn as follows:
2 → 0
4 → 1
6 → 2
10 → 4
12 → 5
iv) A graph:
(Plot the ordered pairs from (i) on a coordinate plane.)
The points to be plotted are: (2, 0), (4, 1), (6, 2), (10, 4), and (12, 5).
31. If \(f(x) = x^2\), \(g(x) = 3x\) and \(h(x) = x - 2\), prove that \((f \circ g) \circ h = f \circ (g \circ h)\).
Given functions: \(f(x) = x^2\), \(g(x) = 3x\), \(h(x) = x - 2\).
First, let's find the LHS: \((f \circ g) \circ h\)
Step 1: Find \(f \circ g\).
\((f \circ g)(x) = f(g(x)) = f(3x) = (3x)^2 = 9x^2\).
Step 2: Now, find \(((f \circ g) \circ h)(x)\).
\(((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(x-2)\).
Substitute (x-2) into the expression for \((f \circ g)(x)\):
\(= 9(x-2)^2 = 9(x^2 - 4x + 4) = 9x^2 - 36x + 36\). --- (1)
Next, let's find the RHS: \(f \circ (g \circ h)\)
Step 1: Find \(g \circ h\).
\((g \circ h)(x) = g(h(x)) = g(x-2) = 3(x-2) = 3x - 6\).
Step 2: Now, find \((f \circ (g \circ h))(x)\).
\((f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(3x-6)\).
Substitute (3x-6) into \(f(x)\):
\(= (3x-6)^2 = [3(x-2)]^2 = 9(x-2)^2 = 9(x^2 - 4x + 4) = 9x^2 - 36x + 36\). --- (2)
From equations (1) and (2), we see that LHS = RHS.
Therefore, \((f \circ g) \circ h = f \circ (g \circ h)\). (Hence Proved)
32. The 13th term of an A.P is 3 and the sum of first 13 terms is 234. Find the common difference and the sum of first 21 terms.
Given, 13th term \(t_{13} = 3\).
\(a + 12d = 3\) --- (1)
Sum of first 13 terms \(S_{13} = 234\).
\(S_n = \frac{n}{2}(2a + (n-1)d)\)
\(234 = \frac{13}{2}(2a + 12d) = 13(a + 6d)\)
\(a + 6d = \frac{234}{13} = 18\) --- (2)
Subtracting (2) from (1):
\((a + 12d) - (a + 6d) = 3 - 18\)
\(6d = -15 \Rightarrow d = -\frac{15}{6} = -\frac{5}{2}\).
Substitute d in (2): \(a + 6(-\frac{5}{2}) = 18 \Rightarrow a - 15 = 18 \Rightarrow a = 33\).
Common difference \(d = -5/2\).
Sum of first 21 terms \(S_{21}\):
\(S_{21} = \frac{21}{2}(2a + 20d) = 21(a+10d)\)
\(= 21(33 + 10(-\frac{5}{2})) = 21(33 - 25) = 21(8) = 168\).
Sum of first 21 terms is 168.
33. Find the sum of n terms of the series 3 + 33 + 333 + ......... to x terms.
Let \(S_x\) be the sum of the series to x terms.
\(S_x = 3 + 33 + 333 + \dots \text{ to x terms}\)
Step 1: Take 3 as a common factor.
\(S_x = 3(1 + 11 + 111 + \dots \text{ to x terms})\)
Step 2: Multiply and divide by 9.
\(S_x = \frac{3}{9}(9 + 99 + 999 + \dots \text{ to x terms})\)
Step 3: Express each term as a difference involving powers of 10.
\(S_x = \frac{1}{3}[(10-1) + (10^2-1) + (10^3-1) + \dots + (10^x-1)]\)
Step 4: Group the terms.
\(S_x = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^x) - (1 + 1 + 1 + \dots \text{ x times})]\)
Step 5: The first part \((10 + 10^2 + \dots + 10^x)\) is a Geometric Progression (G.P.) with:
- First term, \(a = 10\)
- Common ratio, \(r = 10\)
- Number of terms = x
The sum of this G.P. is given by the formula \(S_{GP} = \frac{a(r^x-1)}{r-1}\).
Sum = \(\frac{10(10^x-1)}{10-1} = \frac{10}{9}(10^x-1)\).
The second part \((1 + 1 + \dots \text{ x times})\) is simply x.
Step 6: Substitute these values back into the equation for \(S_x\).
\(S_x = \frac{1}{3}\left[\frac{10}{9}(10^x-1) - x\right]\)
Step 7: Simplify the expression for the final answer.
\(S_x = \frac{1}{3} \left[ \frac{10(10^x-1) - 9x}{9} \right]\)
\(S_x = \frac{1}{27}[10(10^x-1) - 9x]\)
34. There are 12 pieces of five, ten and twenty rupee currencies whose total value is Rs. 105. When first 2 sorts are interchanged in their numbers its value will be increased by Rs. 20. Find the number of currencies in each sort.
Let x, y, and z be the number of five, ten, and twenty rupee notes, respectively.
From the problem, we get three equations:
1. \(x + y + z = 12\) (Total number of notes)
2. \(5x + 10y + 20z = 105 \Rightarrow x + 2y + 4z = 21\) (Total value)
3. After interchanging x and y, the new value is \(5y + 10x + 20z\). This is Rs. 20 more than the original value.
\(5y + 10x + 20z = 105 + 20 = 125 \Rightarrow 2x + y + 4z = 25\)
Now, solve the system of equations:
Subtract (1) from (2): \((x + 2y + 4z) - (x + y + z) = 21 - 12 \Rightarrow y + 3z = 9\) --- (4)
Multiply (1) by 2 and subtract from (3):
\((2x + y + 4z) - 2(x + y + z) = 25 - 2(12)\)
\((2x + y + 4z) - (2x + 2y + 2z) = 25 - 24 \Rightarrow -y + 2z = 1\) --- (5)
Add (4) and (5):
\((y + 3z) + (-y + 2z) = 9 + 1 \Rightarrow 5z = 10 \Rightarrow z = 2\).
Substitute z=2 into (4): \(y + 3(2) = 9 \Rightarrow y + 6 = 9 \Rightarrow y = 3\).
Substitute y=3 and z=2 into (1): \(x + 3 + 2 = 12 \Rightarrow x = 7\).
Answer: Number of 5 rupee notes = 7, 10 rupee notes = 3, and 20 rupee notes = 2.
35. Find the square root of the polynomial \(37x^2 - 28x^3 + 4x^4 + 42x + 9\) by division method.
First, arrange the polynomial in descending order of its powers (standard form):
\(4x^4 - 28x^3 + 37x^2 + 42x + 9\).
Now, we perform the long division method for square root:
2x² - 7x - 3
_________________________
2x² | 4x⁴ - 28x³ + 37x² + 42x + 9
| 4x⁴
|_________________________
4x²-7x | -28x³ + 37x²
| -28x³ + 49x²
| (-) (-)
|_________________________
4x²-14x-3| -12x² + 42x + 9
| -12x² + 42x + 9
| (+) (-) (-)
|_________________________
| 0
Explanation of Steps:
- The square root of the first term \(4x^4\) is \(2x^2\). Place it as the divisor and the quotient.
- Subtract \((2x^2)^2 = 4x^4\) and bring down the next two terms: \(-28x^3 + 37x^2\).
- Double the quotient \(2x^2\) to get \(4x^2\). Divide \(-28x^3\) by \(4x^2\) to get \(-7x\). This is the next term in the quotient and divisor.
- Multiply the new divisor \(4x^2 - 7x\) by \(-7x\) to get \(-28x^3 + 49x^2\). Subtract this from the dividend.
- Bring down the next two terms \(42x + 9\). The new dividend is \(-12x^2 + 42x + 9\).
- Double the current quotient \(2x^2 - 7x\) to get \(4x^2 - 14x\). Divide \(-12x^2\) by \(4x^2\) to get \(-3\). This is the next term in the quotient and divisor.
- Multiply the new divisor \(4x^2 - 14x - 3\) by \(-3\) to get \(-12x^2 + 42x + 9\). Subtracting this gives a remainder of 0.
Therefore, the square root of the polynomial is \(|2x^2 - 7x - 3|\).
36. The roots of the equation \(x^2 + 6x - 4 = 0\) are \(\alpha, \beta\). Find the quadratic equation whose roots are \(\alpha^2\) and \(\beta^2\).
Given the equation \(x^2 + 6x - 4 = 0\).
Comparing with \(ax^2+bx+c=0\), we have a=1, b=6, c=-4.
For the roots \(\alpha\) and \(\beta\):
Sum of the roots: \(\alpha + \beta = -\frac{b}{a} = -\frac{6}{1} = -6\).
Product of the roots: \(\alpha\beta = \frac{c}{a} = \frac{-4}{1} = -4\).
Now, we need to form a new quadratic equation with roots \(\alpha^2\) and \(\beta^2\).
Sum of new roots:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(= (-6)^2 - 2(-4) = 36 + 8 = 44\).
Product of new roots:
\(\alpha^2 \beta^2 = (\alpha\beta)^2 = (-4)^2 = 16\).
The required quadratic equation is given by:
\(x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0\)
\(x^2 - 44x + 16 = 0\).
37. State and prove Thales theorem.
Statement (Basic Proportionality Theorem or Thales Theorem):
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Given: In \(\triangle ABC\), a line DE is parallel to BC, intersecting AB at D and AC at E. (DE || BC).
To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\).
Construction: Join BE and CD. Draw DM \(\perp\) AC and EN \(\perp\) AB.
Proof:
We know that the area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\).
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AD \times EN\).
Area(\(\triangle BDE\)) = \(\frac{1}{2} \times DB \times EN\).
Dividing these, we get:
\(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}\) --- (1)
Similarly,
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AE \times DM\).
Area(\(\triangle DEC\)) = \(\frac{1}{2} \times EC \times DM\).
Dividing these, we get:
\(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\) --- (2)
Now, \(\triangle BDE\) and \(\triangle DEC\) are on the same base DE and between the same parallel lines DE and BC.
Therefore, Area(\(\triangle BDE\)) = Area(\(\triangle DEC\)).
From (1) and (2), this means:
\(\frac{AD}{DB} = \frac{AE}{EC}\).
Hence, the theorem is proved.
38. Find the area of the quadrilateral formed by the points (8,6), (5, 11), (-5, 12) and (-4, 3).
Let the vertices of the quadrilateral be A(8, 6), B(5, 11), C(-5, 12), and D(-4, 3).
We use the formula for the area of a quadrilateral:
Area = \(\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|\)
Substituting the coordinates in order:
Area = \(\frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((6)(5) + (11)(-5) + (12)(-4) + (3)(8))|\)
Calculate the first part:
\(88 + 60 - 15 - 24 = 148 - 39 = 109\).
Calculate the second part:
\(30 - 55 - 48 + 24 = 54 - 103 = -49\).
Now, substitute these values into the formula:
Area = \(\frac{1}{2} |109 - (-49)| = \frac{1}{2} |109 + 49| = \frac{1}{2} |158|\)
Area = 79 square units.
39. Find the equation of a straight line passing through (1,-4) and has intercepts which are in the ratio 2 : 5.
Let the x-intercept be 'a' and the y-intercept be 'b'.
The equation of the line in intercept form is \(\frac{x}{a} + \frac{y}{b} = 1\).
Given that the ratio of intercepts is 2 : 5, so \(\frac{a}{b} = \frac{2}{5}\).
Let \(a = 2k\) and \(b = 5k\) for some constant k.
The equation becomes \(\frac{x}{2k} + \frac{y}{5k} = 1\).
The line passes through the point (1, -4). Substitute x=1 and y=-4 into the equation:
\(\frac{1}{2k} + \frac{-4}{5k} = 1\)
\(\frac{1}{2k} - \frac{4}{5k} = 1\)
To solve for k, find a common denominator (10k):
\(\frac{5 - 8}{10k} = 1 \Rightarrow \frac{-3}{10k} = 1\)
So, \(10k = -3 \Rightarrow k = -\frac{3}{10}\).
Now find the equation of the line by substituting k back:
\(\frac{x}{2(-\frac{3}{10})} + \frac{y}{5(-\frac{3}{10})} = 1\)
\(\frac{x}{-6/10} + \frac{y}{-15/10} = 1\)
\(\frac{-10x}{6} + \frac{-10y}{15} = 1\)
\(\frac{-5x}{3} - \frac{2y}{3} = 1\)
Multiply the entire equation by 3:
\(-5x - 2y = 3\)
The required equation is \(5x + 2y + 3 = 0\).
40. Find the equation of the perpendicular bisector of the line joining the points A(-4, 2) and B (6, -4).
A perpendicular bisector passes through the midpoint of the line segment and is perpendicular to it.
Step 1: Find the midpoint of AB.
Midpoint M = \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\)
M = \((\frac{-4+6}{2}, \frac{2-4}{2}) = (\frac{2}{2}, \frac{-2}{2}) = (1, -1)\).
Step 2: Find the slope of the line AB.
Slope \(m_{AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{-4-2}{6-(-4)} = \frac{-6}{10} = -\frac{3}{5}\).
Step 3: Find the slope of the perpendicular bisector.
The slope of the perpendicular bisector, \(m_{\perp}\), is the negative reciprocal of \(m_{AB}\).
\(m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3}\).
Step 4: Find the equation of the perpendicular bisector.
Using the point-slope form \(y - y_1 = m(x - x_1)\) with the midpoint M(1, -1) and slope \(m = 5/3\).
\(y - (-1) = \frac{5}{3}(x - 1)\)
\(y + 1 = \frac{5}{3}(x - 1)\)
Multiply by 3 to eliminate the fraction:
\(3(y + 1) = 5(x - 1)\)
\(3y + 3 = 5x - 5\)
Rearranging the terms, the required equation is \(5x - 3y - 8 = 0\).
41. If \(\cot\theta + \tan\theta = x\) and \(\sec\theta - \cos\theta = y\), then prove that \((x^2y)^{2/3} - (xy^2)^{2/3} = 1\).
First, let's simplify the expressions for x and y.
For x:
\(x = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\)
\(x = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}\).
For y:
\(y = \sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta\)
\(y = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}\).
Now, let's evaluate the terms in the expression we need to prove.
Term 1: \(x^2y\)
\(x^2y = (\frac{1}{\sin\theta\cos\theta})^2 \cdot (\frac{\sin^2\theta}{\cos\theta}) = \frac{1}{\sin^2\theta\cos^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \frac{1}{\cos^3\theta}\).
So, \((x^2y)^{2/3} = (\frac{1}{\cos^3\theta})^{2/3} = \frac{1}{(\cos^3\theta)^{2/3}} = \frac{1}{\cos^2\theta} = \sec^2\theta\).
Term 2: \(xy^2\)
\(xy^2 = (\frac{1}{\sin\theta\cos\theta}) \cdot (\frac{\sin^2\theta}{\cos\theta})^2 = \frac{1}{\sin\theta\cos\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta\).
So, \((xy^2)^{2/3} = (\tan^3\theta)^{2/3} = \tan^2\theta\).
Final Proof:
Substitute these results back into the expression:
\((x^2y)^{2/3} - (xy^2)^{2/3} = \sec^2\theta - \tan^2\theta\).
Using the Pythagorean identity \(\sec^2\theta - \tan^2\theta = 1\), we get:
\(\sec^2\theta - \tan^2\theta = 1\).
Hence, \((x^2y)^{2/3} - (xy^2)^{2/3} = 1\). (Proved)
PART - D
Answer all the questions.
42. Swathi has 15 ice cubes of different sizes 9cm, 10cm, 11cm, ......... 23cm. How much volume of ice cubes can be used to prepare some fruit juice with these ice cubes?
The sides of the ice cubes form an arithmetic progression: 9, 10, 11, ..., 23.
The volume of a cube with side 'a' is \(a^3\).
Total volume = \(9^3 + 10^3 + 11^3 + \dots + 23^3\).
We can write this as:
Total Volume = \((1^3 + 2^3 + \dots + 23^3) - (1^3 + 2^3 + \dots + 8^3)\).
Using the formula for the sum of cubes of first n natural numbers: \(S_n = \left(\frac{n(n+1)}{2}\right)^2\).
Sum of cubes up to 23:
\(S_{23} = \left(\frac{23(23+1)}{2}\right)^2 = \left(\frac{23 \times 24}{2}\right)^2 = (23 \times 12)^2 = 276^2 = 76176\).
Sum of cubes up to 8:
\(S_8 = \left(\frac{8(8+1)}{2}\right)^2 = \left(\frac{8 \times 9}{2}\right)^2 = (4 \times 9)^2 = 36^2 = 1296\).
Total Volume = \(S_{23} - S_8 = 76176 - 1296 = 74880\).
Answer: The total volume of the ice cubes is 74880 cm³.
43.
a) Construct a triangle similar to a given triangle ABC with its sides equal to 6/5 of the corresponding sides of the triangle ABC (scale factor 6/5 > 1) (OR)
b) Construct a triangle APQR such that QR = 5cm, ∠P = 30° and the altitude from P to QR is of length 4.2cm.
a) Construction of a similar triangle (Scale factor 6/5)
Steps of Construction:
- Draw any triangle ABC.
- Draw a ray BX starting from B, making an acute angle with BC and on the side opposite to vertex A.
- Since the scale factor is 6/5, locate 6 points (the greater of 6 and 5) B₁, B₂, B₃, B₄, B₅, B₆ on BX such that BB₁ = B₁B₂ = ... = B₅B₆.
- Join B₅ (the 5th point, corresponding to the denominator) to C.
- Draw a line through B₆ parallel to B₅C, to intersect the extended line BC at C'.
- Draw a line through C' parallel to CA to intersect the extended line BA at A'.
- The triangle A'BC' is the required similar triangle.
b) Construction of triangle PQR
Steps of Construction:
- Draw a line segment QR = 5 cm.
- At Q, draw a line QE such that ∠RQE = 30° (equal to the given ∠P).
- At Q, draw a line QF perpendicular to QE (i.e., ∠FQE = 90°).
- Draw the perpendicular bisector of QR, let it intersect QF at O and QR at G.
- With O as the center and OQ as the radius, draw a circle. This circle will pass through Q and R. The major arc of this circle will contain the angle 30°.
- On the perpendicular bisector from G, mark a point H such that GH = 4.2 cm (the length of the altitude).
- Draw a line through H parallel to QR. This line will intersect the circle at two points. Name one of these points as P.
- Join PQ and PR.
- The triangle PQR is the required triangle.
44. (b) The following table shows the data about the number of pipes and the time taken to fill the same tank.
| No. of pipes (X) |
2 |
3 |
6 |
9 |
| Time taken (Y) (in mts) |
45 |
30 |
15 |
10 |
Draw the graph for the above data and hence.
i) Find the time taken to fill the tank when five pipes are used.
ii) Find the number of pipes when the time is 9 minutes.
1. Determine the type of variation:
Calculate the product XY for each pair of values.
\(2 \times 45 = 90\)
\(3 \times 30 = 90\)
\(6 \times 15 = 90\)
\(9 \times 10 = 90\)
Since the product XY is a constant (k=90), this is an Indirect Variation. The relationship is \(XY = 90\).
2. Draw the graph:
- Plot the points (2, 45), (3, 30), (6, 15), (9, 10) on a graph paper.
- Choose an appropriate scale for the X-axis (Number of pipes) and Y-axis (Time in minutes). For example, X-axis: 1 cm = 1 pipe, Y-axis: 1 cm = 5 minutes.
- Join the points with a smooth curve. The resulting graph will be a rectangular hyperbola.
3. Find solutions from the graph/equation:
i) Time taken for 5 pipes:
Using the equation \(XY = 90\), when \(X = 5\):
\(5 \times Y = 90 \Rightarrow Y = \frac{90}{5} = 18\).
On the graph, locate 5 on the X-axis, move vertically up to the curve, and then horizontally to the Y-axis. The reading will be 18.
Answer: 18 minutes.
ii) Number of pipes for 9 minutes:
Using the equation \(XY = 90\), when \(Y = 9\):
\(X \times 9 = 90 \Rightarrow X = \frac{90}{9} = 10\).
On the graph, locate 9 on the Y-axis, move horizontally to the curve, and then vertically down to the X-axis. The reading will be 10.
Answer: 10 pipes.
