10th Maths Quarterly Exam Question Paper 2024 with Answer Key | Samacheer Kalvi
COMMON QUARTERLY EXAMINATION - 2024 | Standard X MATHEMATICS
Part - I
1. If \(A = \{a, b, p\}, B = \{2, 3\}, C = \{p, q, r, s\}\) then \(n[(A \cup C) \times B]\) is
Solution:
Given, \(A = \{a, b, p\}, B = \{2, 3\}, C = \{p, q, r, s\}\).
First, find \(A \cup C\):
\(A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}\)
Now, find the number of elements in \(A \cup C\) and B:
\(n(A \cup C) = 6\)
\(n(B) = 2\)
We need to find \(n[(A \cup C) \times B]\). Using the formula \(n(X \times Y) = n(X) \times n(Y)\):
\(n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12\)
2. \(f(x) = (x + 1)^3 – (x - 1)^3\) represents a function which is
Solution:
Given function: \(f(x) = (x + 1)^3 – (x - 1)^3\)
Using the algebraic identities \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\).
\(f(x) = (x^3 + 3x^2(1) + 3x(1)^2 + 1^3) - (x^3 - 3x^2(1) + 3x(1)^2 - 1^3)\)
\(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)\)
\(f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1\)
\(f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)\)
\(f(x) = 6x^2 + 2\)
The highest power (degree) of x is 2. Therefore, the function is a quadratic function.
3. If \(n(A) = p, n(B) = q\), then the total number of relations that exist from A to B is
Solution:
The total number of relations from a set A to a set B is the number of possible subsets of the Cartesian product \(A \times B\).
Given \(n(A) = p\) and \(n(B) = q\).
The number of elements in the Cartesian product is \(n(A \times B) = n(A) \times n(B) = pq\).
The number of subsets of a set with \(k\) elements is \(2^k\).
Therefore, the total number of relations from A to B is the number of subsets of \(A \times B\), which is \(2^{n(A \times B)} = 2^{pq}\).
4. The sum of the exponents of the prime factors in the prime factorisation of 1729 is
Solution:
First, we find the prime factorization of 1729 (also known as the Hardy-Ramanujan number).
1729 is divisible by 7: \(1729 = 7 \times 247\)
Now, factorize 247. It is divisible by 13: \(247 = 13 \times 19\)
So, the prime factorization of 1729 is \(7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
The sum of the exponents is \(1 + 1 + 1 = 3\).
5. An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P is
Solution:
For an Arithmetic Progression (A.P) with an odd number of terms, the sum is equal to the number of terms multiplied by the middle term.
Here, the number of terms is \(n = 31\), which is odd.
The middle term is the \( \frac{n+1}{2} \)-th term, which is \( \frac{31+1}{2} = \frac{32}{2} = 16\)-th term.
We are given that the 16th term is \(m\).
Sum of all terms, \(S_{31} = (\text{number of terms}) \times (\text{middle term})\)
\(S_{31} = 31 \times m = 31m\)
6. The value of \((1^3 + 2^3 + 3^3 + \dots + 15^3) – (1 + 2 + 3 + \dots + 15)\) is
Solution:
We use the formulas for the sum of the first n natural numbers and the sum of the cubes of the first n natural numbers.
Sum of first n natural numbers: \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
Sum of cubes of first n natural numbers: \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\)
Here, n = 15.
First, calculate \(1 + 2 + \dots + 15\):
\(\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120\)
Next, calculate \(1^3 + 2^3 + \dots + 15^3\):
\(\left(\frac{15(15+1)}{2}\right)^2 = (120)^2 = 14400\)
Now, find the required value:
\((1^3 + \dots + 15^3) - (1 + \dots + 15) = 14400 - 120 = 14280\)
7. \(\frac{3y-3}{y} \div \frac{7y-7}{3y^2}\) is
Solution:
To divide by a fraction, we multiply by its reciprocal.
\(\frac{3y-3}{y} \div \frac{7y-7}{3y^2} = \frac{3y-3}{y} \times \frac{3y^2}{7y-7}\)
Factor out the common terms in the numerators and denominators:
\( = \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)}\)
Cancel the common factor \((y-1)\) and one \(y\):
\( = \frac{3}{1} \times \frac{3y}{7}\)
\( = \frac{9y}{7}\)
8. The square root of \(\frac{256x^8y^4z^{10}}{25x^6y^6z^6}\) is equal to
Solution:
First, simplify the expression inside the square root:
\(\frac{256x^8y^4z^{10}}{25x^6y^6z^6} = \frac{256}{25} \cdot x^{8-6} \cdot y^{4-6} \cdot z^{10-6}\)
\(= \frac{256}{25} x^2 y^{-2} z^4 = \frac{256x^2z^4}{25y^2}\)
Now, take the square root of the simplified expression:
\(\sqrt{\frac{256x^2z^4}{25y^2}} = \frac{\sqrt{256x^2z^4}}{\sqrt{25y^2}}\)
\(= \frac{\sqrt{256}\sqrt{x^2}\sqrt{z^4}}{\sqrt{25}\sqrt{y^2}}\)
\(= \frac{16|x|(z^2)}{5|y|} = \frac{16|xz^2|}{5|y|}\) (Since \(z^2\) is always non-negative)
\(= \frac{16}{5} \left| \frac{xz^2}{y} \right|\)
9. The solution of \((2x - 1)^2 = 9\) is equal to
Solution:
Given the equation: \((2x - 1)^2 = 9\)
Take the square root of both sides:
\(2x - 1 = \pm\sqrt{9}\)
\(2x - 1 = \pm 3\)
This gives two possible cases:
Case 1: \(2x - 1 = 3 \implies 2x = 4 \implies x = 2\)
Case 2: \(2x - 1 = -3 \implies 2x = -2 \implies x = -1\)
The solutions are -1 and 2.
10. If in \(\triangle ABC, DE \parallel BC, AB = 3.6\) cm, \(AC = 2.4\) cm and \(AD = 2.1\) cm then the length of AE is
Solution:
Given that in \(\triangle ABC\), \(DE \parallel BC\).
By the Basic Proportionality Theorem (Thales's Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Therefore, \(\frac{AD}{AB} = \frac{AE}{AC}\)
We are given: \(AD = 2.1\) cm, \(AB = 3.6\) cm, \(AC = 2.4\) cm.
Substitute the values into the formula:
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\)
Solve for AE:
\(AE = \frac{2.1 \times 2.4}{3.6}\)
\(AE = \frac{5.04}{3.6} = 1.4\) cm
11. The point of intersection of \(3x - y = 4\) and \(x + y = 8\) is
Solution:
We have a system of two linear equations:
1) \(3x - y = 4\)
2) \(x + y = 8\)
We can solve this by elimination. Add equation (1) and equation (2):
\((3x - y) + (x + y) = 4 + 8\)
\(4x = 12\)
\(x = 3\)
Substitute the value of x into equation (2):
\(3 + y = 8\)
\(y = 8 - 3 = 5\)
The point of intersection is (3, 5).
12. When proving that a quadrilateral is a parallelogram by using slopes you must find
Solution:
A property of a parallelogram is that its opposite sides are parallel. In coordinate geometry, parallel lines have equal slopes.
Therefore, to prove a quadrilateral is a parallelogram using slopes, one must show that the slopes of both pairs of opposite sides are equal.
13. The area of triangle formed by the points (-2,0), (0,-2) and (2,0) is
Solution:
Let the vertices of the triangle be \(A(-2,0), B(0,-2), C(2,0)\).
The formula for the area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
Substitute the coordinates:
Area \( = \frac{1}{2} |-2(-2 - 0) + 0(0 - 0) + 2(0 - (-2))|\)
Area \( = \frac{1}{2} |-2(-2) + 0 + 2(2)|\)
Area \( = \frac{1}{2} |4 + 0 + 4|\)
Area \( = \frac{1}{2} |8| = 4\) sq.units
14. If \((\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2\alpha + \cot^2\alpha\), then the value of k is equal to
Solution:
Expand the left-hand side (LHS):
LHS \( = (\sin^2 \alpha + \csc^2 \alpha + 2 \sin \alpha \csc \alpha) + (\cos^2 \alpha + \sec^2 \alpha + 2 \cos \alpha \sec \alpha)\)
We know \(\sin \alpha \csc \alpha = 1\) and \(\cos \alpha \sec \alpha = 1\).
LHS \( = (\sin^2 \alpha + \csc^2 \alpha + 2) + (\cos^2 \alpha + \sec^2 \alpha + 2)\)
Rearrange the terms: LHS \( = (\sin^2 \alpha + \cos^2 \alpha) + \csc^2 \alpha + \sec^2 \alpha + 4\)
Using the identities \(\sin^2 \alpha + \cos^2 \alpha = 1\), \(\csc^2 \alpha = 1 + \cot^2 \alpha\), and \(\sec^2 \alpha = 1 + \tan^2 \alpha\).
LHS \( = 1 + (1 + \cot^2 \alpha) + (1 + \tan^2 \alpha) + 4\)
LHS \( = 1 + 1 + 1 + 4 + \tan^2 \alpha + \cot^2 \alpha\)
LHS \( = 7 + \tan^2 \alpha + \cot^2 \alpha\)
Comparing this with the given equation \(k + \tan^2 \alpha + \cot^2 \alpha\), we get \(k = 7\).
Part - II
15. If \(A \times B = \{(3,2), (3,4), (5,2), (5,4)\}\), then find A and B.
Solution:
The set A is the set of all first elements in the ordered pairs of \(A \times B\).
First elements are 3 and 5. So, \(A = \{3, 5\}\).
The set B is the set of all second elements in the ordered pairs of \(A \times B\).
Second elements are 2 and 4. So, \(B = \{2, 4\}\).
16. Let \(X = \{1,2,3,4\}\) and \(Y = \{2,4,6,8,10\}\) and \(R = \{(1,2), (2,4), (3,6), (4,8)\}\), show that R is a function and find its domain, co-domain and range.
Solution:
To show R is a function:
For R to be a function from X to Y, every element in the domain X must have exactly one image in the co-domain Y.
In R = {(1,2), (2,4), (3,6), (4,8)}:
- The image of 1 is 2.
- The image of 2 is 4.
- The image of 3 is 6.
- The image of 4 is 8.
Domain, Co-domain, and Range:
- Domain: The set of all first elements in the ordered pairs of R.
Domain = \{1, 2, 3, 4\} = X. - Co-domain: The entire set Y.
Co-domain = \{2, 4, 6, 8, 10\} = Y. - Range: The set of all second elements (images) in the ordered pairs of R.
Range = \{2, 4, 6, 8\}.
17. Find the 8th term of the G.P. 9, 3, 1, ...
Solution:
The given Geometric Progression (G.P.) is 9, 3, 1, ...
First term, \(a = 9\).
Common ratio, \(r = \frac{3}{9} = \frac{1}{3}\).
The formula for the n-th term of a G.P. is \(t_n = ar^{n-1}\).
We need to find the 8th term (\(n=8\)):
\(t_8 = 9 \times \left(\frac{1}{3}\right)^{8-1} = 9 \times \left(\frac{1}{3}\right)^7\)
\(t_8 = 3^2 \times \frac{1}{3^7} = \frac{3^2}{3^7} = 3^{2-7} = 3^{-5} = \frac{1}{3^5}\)
\(t_8 = \frac{1}{243}\)
18. Find the LCM of \(5x - 10, 5x^2 - 20\)
Solution:
First, factorize each expression.
First expression: \(5x - 10 = 5(x - 2)\)
Second expression: \(5x^2 - 20 = 5(x^2 - 4)\)
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\):
\(5(x^2 - 4) = 5(x-2)(x+2)\)
To find the LCM (Least Common Multiple), we take the highest power of each unique factor.
The factors are 5, \((x-2)\), and \((x+2)\).
LCM = \(5 \times (x-2) \times (x+2) = 5(x-2)(x+2)\)
This can also be written as \(5(x^2-4)\).
19. Simplify: \(\frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4}\)
Solution:
Combine the terms into a single fraction:
\(\frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4} = \frac{4 \times 6 \times x^2 \times x \times y \times z^3}{2 \times 20 \times z^2 \times y^4}\)
Multiply the coefficients and combine the variables:
\( = \frac{24 x^3 y z^3}{40 z^2 y^4}\)
Simplify the numerical coefficient: \(\frac{24}{40} = \frac{3 \times 8}{5 \times 8} = \frac{3}{5}\)
Simplify the variable terms using exponent rules:
\(x^3\)
\(y^{1-4} = y^{-3} = \frac{1}{y^3}\)
\(z^{3-2} = z^1 = z\)
Combine everything:
\( = \frac{3x^3z}{5y^3}\)
20. In the figure, AD is the bisector of \(\angle A\). If BD = 4 cm, DC = 3 cm and AB = 6 cm, find AC.
Solution:
By the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides.
So, we have: \(\frac{AB}{AC} = \frac{BD}{DC}\)
Given values: AB = 6 cm, BD = 4 cm, DC = 3 cm.
Substitute these values into the theorem:
\(\frac{6}{AC} = \frac{4}{3}\)
Now, solve for AC:
\(4 \times AC = 6 \times 3\)
\(4 \times AC = 18\)
\(AC = \frac{18}{4} = 4.5\) cm
21. A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
Let the height of the stick be \(h_1 = 6\) m.
Length of the stick's shadow \(s_1 = 400\) cm = 4 m (since 100 cm = 1 m).
Let the height of the tower be \(h_2\).
Length of the tower's shadow \(s_2 = 28\) m.
At the same time, the angle of elevation of the sun is the same for both the stick and the tower. This forms two similar right-angled triangles.
The ratio of corresponding sides of similar triangles is equal.
\(\frac{\text{Height of stick}}{\text{Shadow of stick}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}\)
\(\frac{h_1}{s_1} = \frac{h_2}{s_2}\)
\(\frac{6}{4} = \frac{h_2}{28}\)
Solve for \(h_2\):
\(h_2 = \frac{6 \times 28}{4} = 6 \times 7 = 42\) m
22. Show that the points P(-1.5, 3), Q(6, -2), R(-3, 4) are collinear.
Solution:
To show that the points are collinear, we can prove that the slope of line segment PQ is equal to the slope of line segment QR.
The formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Slope of PQ:
\(m_{PQ} = \frac{-2 - 3}{6 - (-1.5)} = \frac{-5}{6 + 1.5} = \frac{-5}{7.5} = \frac{-50}{75} = -\frac{2}{3}\)
Slope of QR:
\(m_{QR} = \frac{4 - (-2)}{-3 - 6} = \frac{4 + 2}{-9} = \frac{6}{-9} = -\frac{2}{3}\)
Since the slope of PQ is equal to the slope of QR, and they share a common point Q, the points P, Q, and R are collinear.
23. Find the slope of a line joining the given points (-6, 1) and (14, 10).
Solution:
Let the points be \((x_1, y_1) = (-6, 1)\) and \((x_2, y_2) = (14, 10)\).
The formula for the slope (m) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
\(m = \frac{10 - 1}{14 - (-6)} = \frac{9}{14 + 6} = \frac{9}{20}\)
24. Show that the straight lines \(2x + 3y - 8 = 0\) and \(4x + 6y + 18 = 0\) are parallel.
Solution:
Two lines are parallel if their slopes are equal.
The slope of a line in the form \(Ax + By + C = 0\) is given by \(m = -\frac{A}{B}\).
For the first line, \(2x + 3y - 8 = 0\):
\(m_1 = -\frac{2}{3}\)
For the second line, \(4x + 6y + 18 = 0\):
\(m_2 = -\frac{4}{6} = -\frac{2}{3}\)
Since \(m_1 = m_2\), the two lines have the same slope. Therefore, the lines are parallel.
25. Given the function \(f: x \to x^2 – 5x + 6\), evaluate i) f(-1) and ii) f(2).
Solution:
The function is given by \(f(x) = x^2 - 5x + 6\).
i) f(-1):
Substitute \(x = -1\) into the function:
\(f(-1) = (-1)^2 - 5(-1) + 6\)
\(f(-1) = 1 + 5 + 6 = 12\)
ii) f(2):
Substitute \(x = 2\) into the function:
\(f(2) = (2)^2 - 5(2) + 6\)
\(f(2) = 4 - 10 + 6 = 0\)
26. If \(13824 = 2^a \times 3^b\), then find a and b.
Solution:
To find the values of 'a' and 'b', we need to perform a prime factorization of the number 13824.
We will divide 13824 by the prime number 2 until we get an odd number:
- 13824 ÷ 2 = 6912
- 6912 ÷ 2 = 3456
- 3456 ÷ 2 = 1728
- 1728 ÷ 2 = 864
- 864 ÷ 2 = 432
- 432 ÷ 2 = 216
- 216 ÷ 2 = 108
- 108 ÷ 2 = 54
- 54 ÷ 2 = 27
We divided by 2 a total of 9 times. So, the power of 2 is 9.
Now, we factorize the remaining number, 27:
- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1
We divided by 3 a total of 3 times. So, the power of 3 is 3.
Therefore, the prime factorization of 13824 is \(2^9 \times 3^3\).
Comparing this with \(2^a \times 3^b\), we get:
27. Prove that \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = \csc \theta + \cot \theta\)
Solution:
We will start with the Left-Hand Side (LHS) and simplify it to get the Right-Hand Side (RHS).
LHS = \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\)
To remove the square root, we multiply the numerator and the denominator inside the root by the conjugate of the denominator, which is \((1 + \cos \theta)\).
LHS = \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}\)
LHS = \(\sqrt{\frac{(1+\cos \theta)^2}{1^2 - \cos^2 \theta}}\)
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we know that \(1 - \cos^2 \theta = \sin^2 \theta\).
LHS = \(\sqrt{\frac{(1+\cos \theta)^2}{\sin^2 \theta}}\)
Now, we can take the square root of the numerator and the denominator:
LHS = \(\frac{1+\cos \theta}{\sin \theta}\)
Split the fraction into two parts:
LHS = \(\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}\)
Using the reciprocal and quotient identities, we know that \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).
LHS = \(\csc \theta + \cot \theta\)
This is equal to the RHS.
28. Find the sum of 1 + 3 + 5 + ... + 51.
Solution:
The given series 1, 3, 5, ..., 51 is an Arithmetic Progression (A.P.) of odd numbers.
Here, the first term \(a = 1\).
The common difference \(d = 3 - 1 = 2\).
The last term \(l = 51\).
First, we need to find the number of terms (n) in this series. The formula for the last term is \(l = a + (n-1)d\).
\(51 = 1 + (n-1)2\)
\(51 - 1 = (n-1)2\)
\(50 = (n-1)2\)
\(\frac{50}{2} = n-1\)
\(25 = n-1 \implies n = 26\)
So, there are 26 terms in the series.
Now, we can find the sum of the series using the formula \(S_n = \frac{n}{2}(a+l)\).
\(S_{26} = \frac{26}{2}(1+51)\)
\(S_{26} = 13(52)\)
\(S_{26} = 676\)
Alternatively, the sum of the first 'n' odd natural numbers is given by the formula \(n^2\). Since we found n=26, the sum is \(26^2 = 676\).
Part - III
29. Let \(f : A \to B\) 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) Set of ordered pairs, ii) A table, iii) An arrow diagram, iv) A graph.
Solution:
First, we find the image of each element of A under f.
- \(f(2) = \frac{2}{2} - 1 = 1 - 1 = 0\)
- \(f(4) = \frac{4}{2} - 1 = 2 - 1 = 1\)
- \(f(6) = \frac{6}{2} - 1 = 3 - 1 = 2\)
- \(f(10) = \frac{10}{2} - 1 = 5 - 1 = 4\)
- \(f(12) = \frac{12}{2} - 1 = 6 - 1 = 5\)
i) Set of ordered pairs:
The function f can be represented as a set of ordered pairs (x, f(x)).
f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\}
ii) A table:
iii) An arrow diagram:
(An arrow diagram would show two ovals for sets A and B, with arrows connecting each element in A to its corresponding image in B.)
Set A: {2, 4, 6, 10, 12} -> Set B: {0, 1, 2, 4, 5, 9}
iv) A graph:
(A graph would plot the ordered pairs as points on a Cartesian plane.)
The points to be plotted are (2, 0), (4, 1), (6, 2), (10, 4), and (12, 5).
30. Find x if gff(x) = fgg(x), given f(x) = 3x + 1 and g(x) = x + 3.
Solution:
Given functions are \(f(x) = 3x + 1\) and \(g(x) = x + 3\).
We are given the condition \(g(f(f(x))) = f(g(g(x)))\).
Calculate the Left-Hand Side (LHS): gff(x)
First, find \(f(f(x))\):
\(f(f(x)) = f(3x + 1) = 3(3x + 1) + 1 = 9x + 3 + 1 = 9x + 4\)
Now, find \(g(f(f(x)))\):
\(g(9x + 4) = (9x + 4) + 3 = 9x + 7\)
Calculate the Right-Hand Side (RHS): fgg(x)
First, find \(g(g(x))\):
\(g(g(x)) = g(x + 3) = (x + 3) + 3 = x + 6\)
Now, find \(f(g(g(x)))\):
\(f(x + 6) = 3(x + 6) + 1 = 3x + 18 + 1 = 3x + 19\)
Equate LHS and RHS and solve for x:
\(9x + 7 = 3x + 19\)
\(9x - 3x = 19 - 7\)
\(6x = 12\)
\(x = \frac{12}{6} = 2\)
31. If \(p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400\) where \(p_1, p_2, p_3, p_4\) are prime numbers in ascending order. Find the values of \(p_1, p_2, p_3, p_4\) and \(x_1, x_2, x_3, x_4\).
Solution:
We need to find the prime factorization of 113400.
\(113400 = 1134 \times 100\)
\( = 1134 \times 10^2 = 1134 \times (2 \times 5)^2 = 1134 \times 2^2 \times 5^2\)
Now, factorize 1134:
\(1134 = 2 \times 567\)
\(567 = 9 \times 63 = 3^2 \times (9 \times 7) = 3^2 \times 3^2 \times 7 = 3^4 \times 7\)
Combine all the factors:
\(113400 = (2 \times 3^4 \times 7) \times 2^2 \times 5^2\)
\(113400 = 2^{1+2} \times 3^4 \times 5^2 \times 7^1 = 2^3 \times 3^4 \times 5^2 \times 7^1\)
The prime factors \(p_1, p_2, p_3, p_4\) in ascending order are 2, 3, 5, 7.
The corresponding exponents are \(x_1, x_2, x_3, x_4\).
By comparing \(2^3 \times 3^4 \times 5^2 \times 7^1\) with \(p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4}\):
\(x_1 = 3, x_2 = 4, x_3 = 2, x_4 = 1\)
32. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
Solution:
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The area of a square with side 's' is \(s^2\).
The total area is the sum of the areas of all the squares:
Total Area \( = 10^2 + 11^2 + 12^2 + \dots + 24^2\)
We use the formula for the sum of the squares of the first n natural numbers: \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).
We can write the required sum as \((\sum_{k=1}^{24} k^2) - (\sum_{k=1}^{9} k^2)\).
Step 1: Calculate \(\sum_{k=1}^{24} k^2\)
\(\frac{24(24+1)(2 \times 24 + 1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900\)
Step 2: Calculate \(\sum_{k=1}^{9} k^2\)
\(\frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = 3 \times 5 \times 19 = 285\)
Step 3: Find the total area
Total Area = 4900 - 285 = 4615 cm².
33. In an A.P, sum of 4 consecutive terms is 28 and the sum of their squares is 276. Find the four numbers.
Solution:
Let the four consecutive terms of the A.P. be \(a-3d, a-d, a+d, a+3d\).
Condition 1: Sum of the terms is 28
\((a-3d) + (a-d) + (a+d) + (a+3d) = 28\)
\(4a = 28 \implies a = 7\)
Condition 2: Sum of their squares is 276
\((a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 276\)
Expanding this: \((a^2 - 6ad + 9d^2) + (a^2 - 2ad + d^2) + (a^2 + 2ad + d^2) + (a^2 + 6ad + 9d^2) = 276\)
\(4a^2 + 20d^2 = 276\)
Substitute \(a = 7\):
\(4(7^2) + 20d^2 = 276\)
\(4(49) + 20d^2 = 276\)
\(196 + 20d^2 = 276\)
\(20d^2 = 276 - 196 = 80\)
\(d^2 = 4 \implies d = \pm 2\)
Case 1: d = 2
The terms are: \(7-3(2), 7-2, 7+2, 7+3(2) \implies 1, 5, 9, 13\)
Case 2: d = -2
The terms are: \(7-3(-2), 7-(-2), 7+(-2), 7+3(-2) \implies 13, 9, 5, 1\)
In both cases, the set of numbers is the same.
34. If \(9x^4 + 12x^3 + 28x^2 + ax + b\) is a perfect square, find the values of a and b.
Solution:
We use the long division method to find the square root.
The square root of the polynomial will be of the form \(px^2 + qx + r\).
\((px^2 + qx + r)^2 = p^2x^4 + 2pqx^3 + (q^2+2pr)x^2 + 2qrx + r^2\)
Comparing the coefficients with \(9x^4 + 12x^3 + 28x^2 + ax + b\):
- Coefficient of \(x^4\): \(p^2 = 9 \implies p=3\)
- Coefficient of \(x^3\): \(2pq = 12 \implies 2(3)q = 12 \implies 6q = 12 \implies q=2\)
- Coefficient of \(x^2\): \(q^2 + 2pr = 28 \implies (2)^2 + 2(3)r = 28 \implies 4 + 6r = 28 \implies 6r = 24 \implies r=4\)
- Coefficient of \(x\): \(a = 2qr \implies a = 2(2)(4) \implies a = 16\)
- Constant term: \(b = r^2 \implies b = 4^2 \implies b = 16\)
35. There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Solution:
Let's define the variables for the number of currencies of each type:
- Let x be the number of five-rupee notes.
- Let y be the number of ten-rupee notes.
- Let z be the number of twenty-rupee notes.
From the problem statement, we can form three linear equations:
1. Total number of currencies is 12:
\(x + y + z = 12\) --- (1)
2. Total value of the currencies is ₹105:
\(5x + 10y + 20z = 105\)
Dividing the equation by 5 to simplify:
\(x + 2y + 4z = 21\) --- (2)
3. Value increases by ₹20 when the number of five and ten-rupee notes are interchanged:
The new value is \(5y + 10x + 20z\). This is equal to the original value plus 20.
\(5y + 10x + 20z = (5x + 10y + 20z) + 20\)
Simplifying the equation:
\(10x - 5x + 5y - 10y = 20\)
\(5x - 5y = 20\)
Dividing by 5:
\(x - y = 4 \implies x = y + 4\) --- (3)
Now, we solve the system of equations.
Substitute equation (3) into equation (1):
\((y + 4) + y + z = 12\)
\(2y + z = 8 \implies z = 8 - 2y\) --- (4)
Substitute equation (3) into equation (2):
\((y + 4) + 2y + 4z = 21\)
\(3y + 4z = 17\) --- (5)
Now, substitute equation (4) into equation (5):
\(3y + 4(8 - 2y) = 17\)
\(3y + 32 - 8y = 17\)
\(-5y = 17 - 32\)
\(-5y = -15 \implies y = 3\)
Find the values of x and z:
Using \(y=3\) in equation (3): \(x = 3 + 4 \implies x = 7\)
Using \(y=3\) in equation (4): \(z = 8 - 2(3) = 8 - 6 \implies z = 2\)
Number of ten-rupee notes (y) = 3
Number of twenty-rupee notes (z) = 2
36. State and prove Basic Proportionality Theorem.
Solution:
In ∆ABC, D is a point on AB and E is a point on AC.
To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Construction: Draw a line DE || BC
| No. | Statement | Reason |
|---|---|---|
| 1. | ∠ABC = ∠ADE = ∠1 | Corresponding angles are equal because DE || BC |
| 2. | ∠ACB = ∠AED = ∠2 | Corresponding angles are equal because DE || BC |
| 3. | ∠DAE = ∠BAC = ∠3 | Both triangles have a common angle |
| ∆ABC ~ ∆ADE | By AAA similarity | |
| \(\frac{AB}{AD} = \frac{AC}{AE}\) | Corresponding sides are proportional | |
| \(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\) | Split AB and AC using the points D and E. | |
| \(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\) | On simplification | |
| \(\frac{DB}{AD} = \frac{EC}{AE}\) | Cancelling 1 on both sides | |
| \(\frac{AD}{DB} = \frac{AE}{EC}\) | Taking reciprocals. Hence proved. |
37. Find the area of the quadrilateral formed by the points (8,6), (5,11), (-5,12) and (-4,3).
Solution:
Let the vertices be A(8,6), B(5,11), C(-5,12), and D(-4,3). The points are in counter-clockwise order.
The formula for the area of a quadrilateral with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)\) is:
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:
Area \( = \frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((6)(5) + (11)(-5) + (12)(-4) + (3)(8))|\)
Area \( = \frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|\)
Area \( = \frac{1}{2} |(148 - 39) - (54 - 103)|\)
Area \( = \frac{1}{2} |109 - (-49)| = \frac{1}{2} |109 + 49| = \frac{1}{2} |158|\)
Area = 79 sq. units.
38. A cat is located at the point (-6,-4) in xy plane. A bottle of milk is kept at (5,11). The cat wishes to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take.
Solution:
The shortest path between two points is a straight line. We need to find the equation of the line passing through points A(-6, -4) and B(5, 11).
Using the two-point form of a linear equation: \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
Here, \((x_1, y_1) = (-6, -4)\) and \((x_2, y_2) = (5, 11)\).
\(\frac{y - (-4)}{11 - (-4)} = \frac{x - (-6)}{5 - (-6)}\)
\(\frac{y + 4}{15} = \frac{x + 6}{11}\)
Cross-multiply:
\(11(y + 4) = 15(x + 6)\)
\(11y + 44 = 15x + 90\)
Rearrange into the standard form \(Ax + By + C = 0\):
\(15x - 11y + 90 - 44 = 0\)
\(15x - 11y + 46 = 0\)
39. Let A(3,-4), B(9,-4), C(5,−7) and D(7,-7). Show that ABCD is a trapezium.
Solution:
A trapezium is a quadrilateral with exactly one pair of parallel sides. We can check this by calculating the slopes of the four sides.
Slope formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
- Slope of AB: \(m_{AB} = \frac{-4 - (-4)}{9 - 3} = \frac{0}{6} = 0\).
- Slope of BC: \(m_{BC} = \frac{-7 - (-4)}{5 - 9} = \frac{-3}{-4} = \frac{3}{4}\).
- Slope of CD: \(m_{CD} = \frac{-7 - (-7)}{7 - 5} = \frac{0}{2} = 0\).
- Slope of DA: \(m_{DA} = \frac{-4 - (-7)}{3 - 7} = \frac{3}{-4} = -\frac{3}{4}\).
Comparing the slopes:
\(m_{AB} = m_{CD} = 0\), which means side AB is parallel to side CD.
\(m_{BC} \neq m_{DA}\), which means side BC is not parallel to side DA.
Since one pair of opposite sides is parallel, the quadrilateral ABCD is a trapezium.
40. Simplify: \(\frac{b^2+3b-28}{b^2+4b+4} + \frac{b^2-49}{b^2-5b-14}\)
Solution:
First, factorize all the quadratic expressions.
- \(b^2+3b-28 = (b+7)(b-4)\)
- \(b^2+4b+4 = (b+2)^2\)
- \(b^2-49 = (b-7)(b+7)\)
- \(b^2-5b-14 = (b-7)(b+2)\)
Substitute the factored forms back into the expression:
\(\frac{(b+7)(b-4)}{(b+2)^2} + \frac{(b-7)(b+7)}{(b-7)(b+2)}\)
Simplify the second term by canceling \((b-7)\):
\(\frac{(b+7)(b-4)}{(b+2)^2} + \frac{b+7}{b+2}\)
To add the fractions, find a common denominator, which is \((b+2)^2\).
\(\frac{(b+7)(b-4)}{(b+2)^2} + \frac{(b+7)(b+2)}{(b+2)^2}\)
Combine the numerators over the common denominator:
\(\frac{(b+7)(b-4) + (b+7)(b+2)}{(b+2)^2}\)
Factor out the common term \((b+7)\) in the numerator:
\(\frac{(b+7)[(b-4) + (b+2)]}{(b+2)^2} = \frac{(b+7)(2b-2)}{(b+2)^2} = \frac{2(b+7)(b-1)}{(b+2)^2}\)
41. Prove the following identity. \(\frac{\sin^3 A + \cos^3 A}{\sin A + \cos A} + \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 2\)
Solution:
We use the algebraic identities for the sum and difference of cubes:
\(a^3+b^3 = (a+b)(a^2-ab+b^2)\)
\(a^3-b^3 = (a-b)(a^2+ab+b^2)\)
Applying these to the Left-Hand Side (LHS):
LHS = \(\frac{(\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A)}{\sin A + \cos A} + \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A - \cos A}\)
Cancel the common factors in each term:
LHS = \((\sin^2 A - \sin A \cos A + \cos^2 A) + (\sin^2 A + \sin A \cos A + \cos^2 A)\)
Group the terms and use the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\):
LHS = \((\sin^2 A + \cos^2 A) - \sin A \cos A + (\sin^2 A + \cos^2 A) + \sin A \cos A\)
LHS = \((1) - \sin A \cos A + (1) + \sin A \cos A\)
LHS = \(1 + 1 = 2\)
Since LHS = RHS, the identity is proved.
Part - IV
42. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime numbers. Verify that \(A \times (B - C) = (A \times B) - (A \times C)\).
Solution:
First, let's list the elements of each set:
- A = \{1, 2, 3, 4, 5, 6, 7\}
- B = \{2, 3, 5, 7\}
- C = \{2\} (since 2 is the only even prime number)
Calculate the Left-Hand Side (LHS): \(A \times (B - C)\)
First, find \(B - C\):
\(B - C = \{2, 3, 5, 7\} - \{2\} = \{3, 5, 7\}\)
Now, find \(A \times (B - C)\):
\(A \times \{3, 5, 7\} = \{(1,3), (1,5), (1,7), (2,3), (2,5), (2,7), (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3), (6,5), (6,7), (7,3), (7,5), (7,7)\}\)
Calculate the Right-Hand Side (RHS): \((A \times B) - (A \times C)\)
First, find \(A \times B\):
\(A \times B = \{(1,2), (1,3), (1,5), (1,7), (2,2), ..., (7,7)\}\)
Next, find \(A \times C\):
\(A \times C = A \times \{2\} = \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)\}\)
Now, compute \((A \times B) - (A \times C)\). This means we take all elements from \(A \times B\) and remove the elements that are also in \(A \times C\). The elements to be removed are all ordered pairs where the second element is 2.
The remaining elements will be the ordered pairs from \(A \times B\) where the second element is 3, 5, or 7. This is exactly the set \(A \times \{3, 5, 7\}\).
Thus, \((A \times B) - (A \times C) = A \times \{3, 5, 7\}\).
Verification:
The set obtained for the LHS is identical to the set obtained for the RHS.
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{7}{3}\) of the corresponding sides of the triangle PQR (Scale factor \(\frac{7}{3} > 1\)).
Solution:
Since the scale factor is \(\frac{7}{3}\), which is greater than 1, the new triangle will be larger than the original triangle PQR.
Steps of Construction:
- Draw a triangle PQR with any suitable measurements.
- Draw a ray QX making an acute angle with QR, on the side opposite to vertex P.
- Locate 7 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\) on the ray QX such that the distances between them are equal (\(QQ_1 = Q_1Q_2 = \dots = Q_6Q_7\)).
- Join \(Q_3\) (the 3rd point, as 3 is the denominator) to R.
- Draw a line through \(Q_7\) parallel to \(Q_3R\). This line will intersect the extended line segment QR at a point R'.
- Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
- \(\triangle P'QR'\) is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of \(\triangle PQR\).
43. b) Construct a triangle \(\triangle PQR\) such that QR = 5 cm, \(\angle P = 30^\circ\) and the altitude from P to QR is of length 4.2 cm.
Solution:
This construction involves using the concept that angles in the same segment of a circle are equal.
Steps of Construction:
- Draw a line segment QR of length 5 cm.
- At point Q, draw a line QE such that \(\angle RQE = 30^\circ\).
- Draw a line QF perpendicular to QE at Q.
- Draw the perpendicular bisector of the line segment QR. Let it intersect QR at G and the line QF at O.
- With O as the center and OQ as the radius, draw a circle. This circle will pass through Q and R. The major segment QR contains the angle \(30^\circ\).
- On the perpendicular bisector, mark a point M at a distance of 4.2 cm from QR.
- Draw a line through M parallel to QR. This line will intersect the circle at two points, P and S.
- Join PQ and PR (or SQ and SR). \(\triangle PQR\) is the required triangle.
44. a) A bus is travelling at a uniform speed of 50 km/hr. Draw the distance-time graph and hence find i) The constant of variation, ii) How far will it travel in 90 minutes?, iii) The time required to cover a distance of 300 km from the graph.
Solution:
Let x be the time in hours and y be the distance in km. The relationship is given by Distance = Speed \(\times\) Time, so \(y = 50x\). This is a case of direct variation.
Table of Values:
| Time (x) in hours | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Distance (y) in km | 50 | 100 | 150 | 200 | 250 |
The graph is a straight line passing through the origin (0,0) and the points from the table.
From the graph and calculation:
i) The constant of variation (k): Since \(y = kx\), we have \(k = \frac{y}{x} = \frac{50}{1} = 50\). The constant of variation is 50 km/hr.
ii) How far will it travel in 90 minutes? 90 minutes = 1.5 hours. From the graph, at x = 1.5 hours, the corresponding y value is 75 km. (Calculation: \(y = 50 \times 1.5 = 75\) km).
iii) The time required to cover a distance of 300 km. From the graph, for a distance y = 300 km, the corresponding time x is 6 hours. (Calculation: \(300 = 50x \implies x = 6\) hours).
44. b) A school announces that for a certain competition, the cash price will be distributed for all the participants equally as shown below. i) Find the constant of variation, ii) Graph the above data and hence, find how much will each participant get if the number of participants are 12.
Solution:
Let X be the number of participants and y be the amount for each participant in Rs.
Data Table:
| No. of Participants (X) | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|
| Amount for each (y) | 180 | 90 | 60 | 45 | 36 |
Let's check the product \(X \times y\):
\(2 \times 180 = 360\); \(4 \times 90 = 360\); \(6 \times 60 = 360\); etc.
Since the product \(Xy\) is constant, this is an inverse variation, where \(Xy = k\).
i) Find the constant of variation (k):
The constant of variation is \(k = Xy = 360\).
ii) Graph and find the amount for 12 participants:
The graph will be a hyperbola representing the equation \(y = \frac{360}{X}\).
To find the amount for 12 participants (X=12), we can use the graph or the equation.
From the graph, when we locate X=12 on the x-axis, the corresponding point on the curve gives a y-value of 30.
Calculation: \(y = \frac{360}{12} = 30\).