10th Std Maths Quarterly Exam Question Paper with Answers 2024
PART I: Choose the correct answer (14 x 1 = 14)
1. If \( n(A \times B) = 6 \) and \( A = \{1, 3\} \), then \( n(B) \) is
Given, \( n(A \times B) = 6 \) and \( A = \{1, 3\} \).
From set A, we can see that the number of elements in A is \( n(A) = 2 \).
We know the formula \( n(A \times B) = n(A) \times n(B) \).
Substituting the given values: \( 6 = 2 \times n(B) \).
\( n(B) = \frac{6}{2} = 3 \).
Answer: c) 3
2. 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
Given, \( A = \{1, 2, 3, 4, 5\} \), so \( n(A) = 5 \).
Let the number of elements in B be \( n(B) = n \).
The total number of relations from set A to set B is given by \( 2^{n(A) \times n(B)} \).
Given number of relations is 1024.
So, \( 2^{5 \times n} = 1024 \).
We know that \( 1024 = 2^{10} \).
Therefore, \( 2^{5n} = 2^{10} \).
Equating the powers: \( 5n = 10 \), which gives \( n = 2 \).
Answer: b) 2
3. Let \( n(A) = m \) and \( n(B) = n \). Then the total number of non-empty relations that can be defined from A to B is
The total number of relations from A to B is \( 2^{n(A) \times n(B)} = 2^{mn} \).
This includes the empty relation (a relation with no ordered pairs).
The number of non-empty relations is the total number of relations minus the one empty relation.
Number of non-empty relations = \( 2^{mn} - 1 \).
Answer: c) \( 2^{mn} - 1 \)
4. If \( \{(a,8), (6,b)\} \) represents an identity function, then the value of a and b are respectively
An identity function is a function where \( f(x) = x \) for all x in the domain. This means the input and output values are the same for each ordered pair.
For the ordered pair \( (a, 8) \), we must have \( a = 8 \).
For the ordered pair \( (6, b) \), we must have \( b = 6 \).
So, the values are \( a = 8 \) and \( b = 6 \).
Answer: a) (8,6)
5. If \( f: A \to B \) is a bijective function and if \( n(B) = 7 \), then \( n(A) \) is equal to
A bijective function is both one-to-one (injective) and onto (surjective).
For a function to be bijective, there must be a one-to-one correspondence between the elements of the domain (A) and the codomain (B).
This implies that the number of elements in the domain must be equal to the number of elements in the codomain.
Therefore, \( n(A) = n(B) \).
Given \( n(B) = 7 \), so \( n(A) = 7 \).
Answer: a) 7
6. The sum of the exponents of the prime factors in the prime factorization of 1729 is
First, find the prime factorization of 1729.
\( 1729 = 7 \times 247 \)
\( 1729 = 7 \times 13 \times 19 \)
The prime factorization 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 \( 1 + 1 + 1 = 3 \).
Answer: c) 3
7. Given \( F_1 = 1, F_2 = 3 \) and \( F_n = F_{n-1} + F_{n-2} \), then \( F_5 \) is
We are given the recurrence relation and initial terms:
\( F_1 = 1 \)
\( F_2 = 3 \)
\( F_3 = F_2 + F_1 = 3 + 1 = 4 \)
\( F_4 = F_3 + F_2 = 4 + 3 = 7 \)
\( F_5 = F_4 + F_3 = 7 + 4 = 11 \)
Answer: d) 11
8. If 6 times of 6th term of an A.P is equal to 7 times 7th term, then the 13th term of the A.P is
Let the A.P have first term 'a' and common difference 'd'.
The nth term is \( T_n = a + (n-1)d \).
Given, \( 6 \times T_6 = 7 \times T_7 \).
\( 6(a + (6-1)d) = 7(a + (7-1)d) \)
\( 6(a + 5d) = 7(a + 6d) \)
\( 6a + 30d = 7a + 42d \)
Rearranging the terms: \( 7a - 6a + 42d - 30d = 0 \)
\( a + 12d = 0 \)
We need to find the 13th term, \( T_{13} \).
\( T_{13} = a + (13-1)d = a + 12d \)
Since we found that \( a + 12d = 0 \), the 13th term is 0.
Answer: a) 0
9. The value of \( (1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15) \) is
We use the formulas for the sum of first n natural numbers and the sum of cubes of first n natural numbers.
Sum of first n numbers: \( \sum n = \frac{n(n+1)}{2} \)
Sum of cubes of first n numbers: \( \sum n^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
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 \( (1^3 + 2^3 + \dots + 15^3) \):
\( \left(\frac{15(15+1)}{2}\right)^2 = (120)^2 = 14400 \)
The required value is \( 14400 - 120 = 14280 \).
Answer: c) 14280
10. The solution of the system \( x + y - 3z = -6, -7y + 7z = 7, 3z = 9 \) is
We have a system of three linear equations:
1) \( x + y - 3z = -6 \)
2) \( -7y + 7z = 7 \)
3) \( 3z = 9 \)
From equation (3): \( z = \frac{9}{3} = 3 \).
Substitute \( z = 3 \) into equation (2):
\( -7y + 7(3) = 7 \)
\( -7y + 21 = 7 \)
\( -7y = 7 - 21 = -14 \)
\( y = \frac{-14}{-7} = 2 \)
Substitute \( y = 2 \) and \( z = 3 \) into equation (1):
\( x + (2) - 3(3) = -6 \)
\( x + 2 - 9 = -6 \)
\( x - 7 = -6 \)
\( x = -6 + 7 = 1 \)
The solution is \( x=1, y=2, z=3 \).
Answer: a) x=1, y=2, z=3
11. Simplify: \( \frac{3y-3}{y} \div \frac{7y-7}{3y^2} \)
To divide by a fraction, we multiply by its reciprocal.
\( \frac{3y-3}{y} \times \frac{3y^2}{7y-7} \)
Factor out the common terms in the numerators and denominators:
\( \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)} \)
Cancel out the common term \( (y-1) \):
\( \frac{3}{y} \times \frac{3y^2}{7} \)
Cancel out one 'y' from the denominator and numerator:
\( 3 \times \frac{3y}{7} = \frac{9y}{7} \)
Answer: a) \( \frac{9y}{7} \)
12. In \( \Delta LMN \), \( \angle L = 60^\circ \), \( \angle M = 50^\circ \). If \( \Delta LMN \sim \Delta PQR \), then the value of \( \angle R \) is
First, find the third angle in \( \Delta LMN \). The sum of angles in a triangle is \( 180^\circ \).
\( \angle N = 180^\circ - (\angle L + \angle M) \)
\( \angle N = 180^\circ - (60^\circ + 50^\circ) = 180^\circ - 110^\circ = 70^\circ \)
When two triangles are similar (\( \Delta LMN \sim \Delta PQR \)), their corresponding angles are equal.
\( \angle P = \angle L \), \( \angle Q = \angle M \), and \( \angle R = \angle N \).
Therefore, \( \angle R = \angle N = 70^\circ \).
Answer: b) \( 70^\circ \)
13. The straight line given by the equation \( x = 11 \) is
The equation \( x = 11 \) represents a set of all points where the x-coordinate is always 11, regardless of the y-coordinate. Examples of points on this line are (11, 0), (11, 2), (11, -5), etc.
When these points are plotted, they form a vertical line.
A vertical line is parallel to the y-axis.
Answer: b) Parallel to y-axis
14. If the slope of the line PQ is \( \frac{1}{\sqrt{3}} \), then the slope of the perpendicular bisector of PQ is
The perpendicular bisector of a line segment PQ is a line that is perpendicular to PQ.
If two lines are perpendicular, the product of their slopes is -1. Let the slope of PQ be \( m_1 \) and the slope of the perpendicular line be \( m_2 \).
Then \( m_1 \times m_2 = -1 \).
Given, \( m_1 = \frac{1}{\sqrt{3}} \).
\( \frac{1}{\sqrt{3}} \times m_2 = -1 \)
\( m_2 = -1 \times \sqrt{3} = -\sqrt{3} \)
Answer: b) \( -\sqrt{3} \)
PART II: Answer any 10 questions (10 x 2 = 20)
Question 28 is compulsory.
15. If \( B \times A = \{(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)\} \), find A and B.
In the Cartesian product \( B \times A \), the first element of each ordered pair belongs to set B, and the second element belongs to set A.
Set B is the set of all first elements: \( \{-2, 0, 3\} \).
Set A is the set of all second elements: \( \{3, 4\} \).
Answer: \( A = \{3, 4\} \), \( B = \{-2, 0, 3\} \)
16. \( A = \{1, 2, 3, 4, \dots, 45\} \) and R be the relation defined as "is square of a number" on A. Write R as a subset of \( A \times A \). Also find the domain and range of R.
The relation R consists of ordered pairs \( (x, y) \) where \( x \in A, y \in A \) and \( y = x^2 \).
We find the pairs:
- If \( x=1, y=1^2=1 \). \( (1,1) \in R \)
- If \( x=2, y=2^2=4 \). \( (2,4) \in R \)
- If \( x=3, y=3^2=9 \). \( (3,9) \in R \)
- If \( x=4, y=4^2=16 \). \( (4,16) \in R \)
- If \( x=5, y=5^2=25 \). \( (5,25) \in R \)
- If \( x=6, y=6^2=36 \). \( (6,36) \in R \)
- If \( x=7, y=7^2=49 \). \( 49 \notin A \), so we stop.
R as a subset of \( A \times A \): \( R = \{(1,1), (2,4), (3,9), (4,16), (5,25), (6,36)\} \)
Domain of R: The set of all first elements of the pairs in R. Domain = \( \{1, 2, 3, 4, 5, 6\} \).
Range of R: The set of all second elements of the pairs in R. Range = \( \{1, 4, 9, 16, 25, 36\} \).
17. Let \( A = \{1, 2, 3\}, B = \{4, 5, 6, 7\} \) and \( f = \{(1,4), (2,5), (3,6)\} \) be a function from A to B. Show that f is one-one but not onto function.
One-one function: A function is one-one if every distinct element in the domain A has a distinct image in the codomain B.
Here, f(1) = 4, f(2) = 5, f(3) = 6. Since the images (4, 5, 6) are all different for different inputs (1, 2, 3), the function f is one-one.
Onto function: A function is onto if every element in the codomain B has at least one pre-image in the domain A. In other words, the range of the function must be equal to the codomain.
The codomain is \( B = \{4, 5, 6, 7\} \).
The range of f is the set of all images, which is \( \{4, 5, 6\} \).
Since Range \( \neq \) Codomain (the element 7 in B has no pre-image in A), the function f is not onto.
18. Find k if \( f \circ f(k) = 5 \), where \( f(k) = 2k - 1 \).
We are given \( f(k) = 2k - 1 \).
First, find the expression for \( f \circ f(k) \), which is \( f(f(k)) \).
\( f(f(k)) = f(2k-1) \)
Substitute \( (2k-1) \) into the function \( f(k) \):
\( f(2k-1) = 2(2k-1) - 1 = 4k - 2 - 1 = 4k - 3 \)
We are given that \( f \circ f(k) = 5 \).
So, \( 4k - 3 = 5 \).
\( 4k = 5 + 3 = 8 \)
\( k = \frac{8}{4} = 2 \)
Answer: \( k = 2 \)
19. 'a' and 'b' are two positive integers such that \( a^b \times b^a = 800 \). Find 'a' and 'b'.
We need to find the prime factorization of 800.
\( 800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2 \).
Now, we compare this with the given expression \( a^b \times b^a \).
\( a^b \times b^a = 2^5 \times 5^2 \)
By direct comparison, we can see that if we let \( a = 2 \) and \( b = 5 \), the expression matches.
Let's check: \( 2^5 \times 5^2 = 32 \times 25 = 800 \).
Answer: \( a = 2 \) and \( b = 5 \) (or a=5, b=2).
20. Complete x such that \( 10^4 \equiv x \pmod{19} \).
We need to find the remainder when \( 10^4 \) is divided by 19.
Let's calculate powers of 10 modulo 19:
\( 10^1 \equiv 10 \pmod{19} \)
\( 10^2 = 100 \). When 100 is divided by 19, \( 100 = 5 \times 19 + 5 \). So, \( 10^2 \equiv 5 \pmod{19} \).
Now we can find \( 10^4 \):
\( 10^4 = (10^2)^2 \equiv 5^2 \pmod{19} \)
\( 10^4 \equiv 25 \pmod{19} \)
Since 25 is greater than 19, we find its remainder when divided by 19. \( 25 = 1 \times 19 + 6 \).
So, \( 25 \equiv 6 \pmod{19} \).
Therefore, \( 10^4 \equiv 6 \pmod{19} \).
Answer: \( x = 6 \)
21. Find \( a_8 \) and \( a_{15} \) whose nth term is \( a_n = \begin{cases} \frac{n^2-1}{n+3} & \text{if n is even, } n \in \mathbb{N} \\ \frac{n^2}{2n+1} & \text{if n is odd, } n \in \mathbb{N} \end{cases} \)
To find \( a_8 \), we use the formula for even n, since 8 is even.
\( a_8 = \frac{8^2 - 1}{8 + 3} = \frac{64 - 1}{11} = \frac{63}{11} \)
To find \( a_{15} \), we use the formula for odd n, since 15 is odd.
\( a_{15} = \frac{15^2}{2(15) + 1} = \frac{225}{30 + 1} = \frac{225}{31} \)
Answer: \( a_8 = \frac{63}{11} \), \( a_{15} = \frac{225}{31} \)
22. If \( 3+k, 18-k, 5k+1 \) are in A.P then find k.
If three terms \( t_1, t_2, t_3 \) are in an Arithmetic Progression (A.P), then the common difference is constant, which means \( t_2 - t_1 = t_3 - t_2 \), or \( 2t_2 = t_1 + t_3 \).
Here, \( t_1 = 3+k, t_2 = 18-k, t_3 = 5k+1 \).
\( 2(18-k) = (3+k) + (5k+1) \)
\( 36 - 2k = 6k + 4 \)
Rearranging the terms to solve for k:
\( 36 - 4 = 6k + 2k \)
\( 32 = 8k \)
\( k = \frac{32}{8} = 4 \)
Answer: \( k = 4 \)
23. Find the excluded values of \( \frac{y}{y^2-25} \).
An expression is undefined when its denominator is equal to zero. These values of the variable are called excluded values.
Set the denominator to zero: \( y^2 - 25 = 0 \).
This is a difference of squares: \( (y-5)(y+5) = 0 \).
This gives two possible solutions:
\( y - 5 = 0 \implies y = 5 \)
\( y + 5 = 0 \implies y = -5 \)
Answer: The excluded values are 5 and -5.
24. Simplify \( \frac{5t^3}{4t-8} \times \frac{6t-12}{10t} \).
First, factor the expressions in the denominators:
\( \frac{5t^3}{4(t-2)} \times \frac{6(t-2)}{10t} \)
Now, we can cancel the common terms. The term \( (t-2) \) appears in the numerator and denominator, so it cancels out.
\( \frac{5t^3}{4} \times \frac{6}{10t} \)
Combine the fractions:
\( \frac{5 \times 6 \times t^3}{4 \times 10 \times t} = \frac{30 t^3}{40 t} \)
Simplify the numerical coefficient and the variable part:
\( \frac{30}{40} = \frac{3}{4} \)
\( \frac{t^3}{t} = t^{3-1} = t^2 \)
The simplified expression is \( \frac{3}{4} t^2 \).
Answer: \( \frac{3t^2}{4} \)
25. In \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If \( \frac{AD}{DB} = \frac{3}{4} \) and \( AC = 15 \) cm. Find AE.
By the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides the two sides in the same ratio. An important corollary is that \( \frac{AD}{AB} = \frac{AE}{AC} \).
We are given \( \frac{AD}{DB} = \frac{3}{4} \). This means we can write \( AD = 3x \) and \( DB = 4x \) for some constant x.
Then, the full length of side AB is \( AB = AD + DB = 3x + 4x = 7x \).
Now we find the ratio \( \frac{AD}{AB} = \frac{3x}{7x} = \frac{3}{7} \).
Using the theorem: \( \frac{AD}{AB} = \frac{AE}{AC} \).
\( \frac{3}{7} = \frac{AE}{15} \)
Solving for AE:
\( AE = \frac{3 \times 15}{7} = \frac{45}{7} \)
Answer: \( AE = \frac{45}{7} \) cm.
26. Show that the points (-3,-4), (7,2) and (12, 5) are collinear.
Three points are collinear if they lie on the same straight line. This can be shown if the area of the triangle formed by these points is zero.
The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Let \( (-3, -4) = (x_1, y_1), (7, 2) = (x_2, y_2), (12, 5) = (x_3, y_3) \).
Area \( = \frac{1}{2} |-3(2 - 5) + 7(5 - (-4)) + 12(-4 - 2)| \)
Area \( = \frac{1}{2} |-3(-3) + 7(5 + 4) + 12(-6)| \)
Area \( = \frac{1}{2} |9 + 7(9) - 72| \)
Area \( = \frac{1}{2} |9 + 63 - 72| \)
Area \( = \frac{1}{2} |72 - 72| = \frac{1}{2} |0| = 0 \)
Since the area of the triangle is 0, the points are collinear.
27. Find the slope of a line joining the points \( (5, \sqrt{5}) \) with the origin.
The two points are \( P_1 = (5, \sqrt{5}) \) and the origin \( P_2 = (0, 0) \).
The formula for the slope (m) of a line joining points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
\( m = \frac{\sqrt{5} - 0}{5 - 0} = \frac{\sqrt{5}}{5} \)
This can also be written as \( \frac{\sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{1}{\sqrt{5}} \).
Answer: The slope is \( \frac{\sqrt{5}}{5} \) or \( \frac{1}{\sqrt{5}} \).
28. (Compulsory) The father's age is six times his son's age. Six years hence the age of father will be four times his son's age. Find the present age (in years) of the son and father.
Let the present age of the father be F and the present age of the son be S.
From the first statement: "The father's age is six times his son's age."
\( F = 6S \) --- (1)
From the second statement: "Six years hence..."
Father's age in 6 years = \( F + 6 \)
Son's age in 6 years = \( S + 6 \)
"...the age of father will be four times his son's age."
\( F + 6 = 4(S + 6) \) --- (2)
Now, substitute the value of F from equation (1) into equation (2):
\( (6S) + 6 = 4(S + 6) \)
\( 6S + 6 = 4S + 24 \)
Bring the S terms to one side and constants to the other:
\( 6S - 4S = 24 - 6 \)
\( 2S = 18 \)
\( S = 9 \)
So, the son's present age is 9 years.
Now find the father's age using equation (1):
\( F = 6S = 6 \times 9 = 54 \)
The father's present age is 54 years.
Answer: The present age of the son is 9 years and the father is 54 years.
PART III: Answer any 10 of the following questions (10 x 5 = 50)
Question 42 is compulsory.
29. Let \( A = \{x \in \mathbb{N} | 1 < x < 4\}, B = \{x \in W | 0 \le x < 2\} \) and \( C = \{x \in \mathbb{N} | x < 3\} \). Verify that \( A \times (B \cap C) = (A \times B) \cap (A \times C) \).
First, let's list the elements of each set:
- \( A = \{x \in \mathbb{N} | 1 < x < 4\} \implies A = \{2, 3\} \)
- \( B = \{x \in W | 0 \le x < 2\} \implies B = \{0, 1\} \)
- \( C = \{x \in \mathbb{N} | x < 3\} \implies C = \{1, 2\} \)
L.H.S (Left Hand Side): \( A \times (B \cap C) \)
First, find \( B \cap C \):
\( B \cap C = \{0, 1\} \cap \{1, 2\} = \{1\} \)
Now, find \( A \times (B \cap C) \):
\( A \times (B \cap C) = \{2, 3\} \times \{1\} = \{(2, 1), (3, 1)\} \) --- (1)
R.H.S (Right Hand Side): \( (A \times B) \cap (A \times C) \)
First, find \( A \times B \):
\( A \times B = \{2, 3\} \times \{0, 1\} = \{(2, 0), (2, 1), (3, 0), (3, 1)\} \)
Next, find \( A \times C \):
\( A \times C = \{2, 3\} \times \{1, 2\} = \{(2, 1), (2, 2), (3, 1), (3, 2)\} \)
Now, find the intersection of these two sets:
\( (A \times B) \cap (A \times C) = \{(2, 1), (3, 1)\} \) --- (2)
Conclusion:
From (1) and (2), we see that L.H.S = R.H.S.
Hence, \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is verified.
30. Let \( A = \{2, 4, 6, 10, 12\}, B = \{0, 1, 2, 4, 5, 9\} \) be two sets. Let \( f: A \to B \) be a function given by \( f(x) = \frac{x}{2} - 1 \). Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.
Given \( f(x) = \frac{x}{2} - 1 \). We calculate 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) By arrow diagram:
[An image placeholder for an arrow diagram showing mappings from A to B: 2→0, 4→1, 6→2, 10→4, 12→5]
(ii) In a table form:
| x (Input) | f(x) (Output) |
|---|---|
| 2 | 0 |
| 4 | 1 |
| 6 | 2 |
| 10 | 4 |
| 12 | 5 |
(iii) As a set of ordered pairs:
\( f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\} \)
(iv) In a graphical form:
[An image placeholder for a graph plotting the points (2,0), (4,1), (6,2), (10,4), (12,5)]
31. If \( f(x) = 2x + 3, g(x) = 1 - 2x \) and \( h(x) = 3x \), prove that \( f \circ (g \circ h) = (f \circ g) \circ h \).
L.H.S: \( f \circ (g \circ h) \)
First, find \( g \circ h(x) = g(h(x)) \):
\( g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x \)
Now, find \( f \circ (g \circ h)(x) = f(g \circ h(x)) \):
\( f(1 - 6x) = 2(1 - 6x) + 3 = 2 - 12x + 3 = 5 - 12x \) --- (1)
R.H.S: \( (f \circ g) \circ h \)
First, find \( f \circ g(x) = f(g(x)) \):
\( f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 2 - 4x + 3 = 5 - 4x \)
Now, find \( (f \circ g) \circ h(x) = (f \circ g)(h(x)) \):
\( (f \circ g)(3x) = 5 - 4(3x) = 5 - 12x \) --- (2)
Conclusion:
From (1) and (2), L.H.S = R.H.S. Hence, the associative property of composition of functions is proved.
32. If \( P_1^{x_1} \times P_2^{x_2} \times P_3^{x_3} \times P_4^{x_4} = 113400 \), where P₁, P₂, P₃, P₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers, find the value of P₁, P₂, P₃, P₄ and x₁, x₂, x₃, x₄.
The problem requires us to find the prime factorization of the number 113400.
Step 1: Break down the number into simpler factors.
\( 113400 = 1134 \times 100 \)
Step 2: Find the prime factors of each part.
Factorizing 100:
\( 100 = 10 \times 10 = (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2 \)
Factorizing 1134:
\( 1134 = 2 \times 567 \)
To factor 567, we check for divisibility. Sum of digits is 5+6+7=18, which is divisible by 9.
\( 567 = 9 \times 63 = 3^2 \times (9 \times 7) = 3^2 \times 3^2 \times 7 = 3^4 \times 7 \)
So, \( 1134 = 2 \times 3^4 \times 7 \)
Step 3: Combine all the prime factors.
\( 113400 = (2 \times 3^4 \times 7) \times (2^2 \times 5^2) \)
Combine the powers of the same bases:
\( 113400 = 2^{1+2} \times 3^4 \times 5^2 \times 7^1 = 2^3 \times 3^4 \times 5^2 \times 7^1 \)
Step 4: Compare the result with the given expression \( P_1^{x_1} \times P_2^{x_2} \times P_3^{x_3} \times P_4^{x_4} \).
Since P₁, P₂, P₃, P₄ are prime numbers in ascending order, we have:
- \( P_1 = 2 \)
- \( P_2 = 3 \)
- \( P_3 = 5 \)
- \( P_4 = 7 \)
The corresponding exponents (integers x₁, x₂, x₃, x₄) are:
- \( x_1 = 3 \)
- \( x_2 = 4 \)
- \( x_3 = 2 \)
- \( x_4 = 1 \)
Answer: The values are P₁=2, P₂=3, P₃=5, P₄=7 and x₁=3, x₂=4, x₃=2, x₄=1.
33. Find the sum of all natural numbers between 300 and 600 which are divisible by 7.
The numbers form an Arithmetic Progression (A.P.).
First number greater than 300 divisible by 7: \( 301 \) (since \( 300 \div 7 = 42 \) with remainder 6). So, \( a = 301 \).
Last number less than 600 divisible by 7: \( 595 \) (since \( 600 \div 7 = 85 \) with remainder 5). So, \( l = 595 \).
The common difference is \( d = 7 \).
First, find the number of terms (n) using the formula \( l = a + (n-1)d \):
\( 595 = 301 + (n-1)7 \)
\( 595 - 301 = (n-1)7 \)
\( 294 = (n-1)7 \)
\( n-1 = \frac{294}{7} = 42 \)
\( n = 43 \)
Now, find the sum of these 43 terms using the formula \( S_n = \frac{n}{2}(a+l) \):
\( S_{43} = \frac{43}{2}(301 + 595) = \frac{43}{2}(896) = 43 \times 448 = 19264 \)
Answer: The sum is 19264.
34. Find the sum of to n terms of the series 3 + 33 + 333 + ... + n terms.
This is a special type of series. Let's denote the sum by \( S_n \).
\( S_n = 3 + 33 + 333 + \dots \) to n terms.
Step 1: Factor out the common term.
We can take '3' common from each term:
\( S_n = 3(1 + 11 + 111 + \dots \) to n terms)
Step 2: Multiply and divide by 9.
This is a standard technique to convert the terms into a form involving powers of 10.
\( S_n = \frac{3}{9}(9 + 99 + 999 + \dots \) to n terms)
Step 3: Express each term as a difference involving powers of 10.
We can write \( 9 = 10 - 1 \), \( 99 = 100 - 1 = 10^2 - 1 \), \( 999 = 1000 - 1 = 10^3 - 1 \), and so on.
\( S_n = \frac{1}{3}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)] \)
Step 4: Separate the series into two parts.
Group the powers of 10 together and the '-1' terms together.
\( S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{n times})] \)
Step 5: Calculate the sum of each part.
The first part, \( (10 + 10^2 + \dots + 10^n) \), is a Geometric Progression (G.P.) with:
- First term, \( a = 10 \)
- Common ratio, \( r = 10 \)
- Number of terms = \( n \)
The sum of this G.P is given by the formula \( S_{GP} = a\frac{(r^n - 1)}{r-1} \):
Sum = \( 10 \frac{(10^n - 1)}{10 - 1} = \frac{10}{9}(10^n - 1) \)
The second part, \( (1 + 1 + \dots \text{n times}) \), is simply \( n \).
Step 6: Substitute back and simplify.
Now, substitute the sums back into the expression for \( S_n \):
\( S_n = \frac{1}{3} \left[ \frac{10}{9}(10^n - 1) - n \right] \)
This can be further simplified if needed, but it is often left in this form.
\( S_n = \frac{10}{27}(10^n - 1) - \frac{n}{3} \)
Answer: The sum to n terms is \( \frac{1}{3} \left[ \frac{10}{9}(10^n - 1) - n \right] \).
35. Rekha has 15 square colour papers of sizes 10cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The area of each square is the side squared. So the total area is the sum of the squares of the sides:
Total Area = \( 10^2 + 11^2 + 12^2 + \dots + 24^2 \)
We can write this sum as the difference of two sums:
Total Area = \( (1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2) \)
Using the formula for the sum of the first n squares, \( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \):
Sum up to 24: \( S_{24} = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900 \)
Sum up to 9: \( S_9 = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} = 3 \times 5 \times 19 = 285 \)
Total Area = \( S_{24} - S_9 = 4900 - 285 = 4615 \)
Answer: 4615 cm² of area can be decorated.
36. Find the GCD of the polynomials \( x^3 + x^2 - x + 2 \) and \( 2x^3 - 5x^2 + 5x - 3 \).
Let \( P(x) = 2x^3 - 5x^2 + 5x - 3 \) and \( Q(x) = x^3 + x^2 - x + 2 \).
We use the division algorithm to find the GCD.
Divide P(x) by Q(x):
\( 2x^3 - 5x^2 + 5x - 3 = 2(x^3 + x^2 - x + 2) - 7x^2 + 7x - 7 \)
The remainder is \( R_1(x) = -7x^2 + 7x - 7 = -7(x^2 - x + 1) \).
Now, divide Q(x) by \( (x^2 - x + 1) \) (we can ignore the constant factor -7).
Dividing \( x^3 + x^2 - x + 2 \) by \( x^2 - x + 1 \):
x + 2
x²-x+1 | x³ + x² - x + 2
-(x³ - x² + x)
----------------
2x² - 2x + 2
-(2x² - 2x + 2)
----------------
0
The remainder is 0. The last non-zero divisor is the GCD.
Answer: The GCD is \( x^2 - x + 1 \).
37. If \( A = \frac{2x+1}{2x-1} \) and \( B = \frac{2x-1}{2x+1} \), find \( \frac{1}{A-B} - \frac{2B}{A^2-B^2} \).
We need to simplify the expression \( \frac{1}{A-B} - \frac{2B}{A^2-B^2} \).
Step 1: Simplify the algebraic expression in terms of A and B.
First, notice that the denominator \( A^2 - B^2 \) is a difference of squares, which can be factored as \( (A-B)(A+B) \).
The expression becomes:
\( \frac{1}{A-B} - \frac{2B}{(A-B)(A+B)} \)
To combine these fractions, we find a common denominator, which is \( (A-B)(A+B) \).
\( = \frac{1(A+B)}{(A-B)(A+B)} - \frac{2B}{(A-B)(A+B)} \)
\( = \frac{(A+B) - 2B}{(A-B)(A+B)} \)
\( = \frac{A - B}{(A-B)(A+B)} \)
Now, we can cancel the common factor \( (A-B) \) from the numerator and the denominator, assuming \( A \neq B \).
\( = \frac{1}{A+B} \)
Step 2: Calculate the value of A + B.
Now we substitute the given expressions for A and B into \( A+B \):
\( A+B = \frac{2x+1}{2x-1} + \frac{2x-1}{2x+1} \)
The common denominator is \( (2x-1)(2x+1) \).
\( A+B = \frac{(2x+1)(2x+1)}{(2x-1)(2x+1)} + \frac{(2x-1)(2x-1)}{(2x-1)(2x+1)} \)
\( A+B = \frac{(2x+1)^2 + (2x-1)^2}{(2x-1)(2x+1)} \)
Expand the squares in the numerator using \( (a+b)^2=a^2+2ab+b^2 \) and \( (a-b)^2=a^2-2ab+b^2 \):
Numerator = \( (4x^2 + 4x + 1) + (4x^2 - 4x + 1) = 4x^2 + 4x + 1 + 4x^2 - 4x + 1 = 8x^2 + 2 \)
The denominator is a difference of squares: \( (2x)^2 - 1^2 = 4x^2 - 1 \).
So, \( A+B = \frac{8x^2 + 2}{4x^2 - 1} \)
Step 3: Find the final answer.
The original expression simplified to \( \frac{1}{A+B} \). We just need to find the reciprocal of the result from Step 2.
\( \frac{1}{A+B} = \frac{1}{\frac{8x^2 + 2}{4x^2 - 1}} = \frac{4x^2 - 1}{8x^2 + 2} \)
We can factor out a 2 from the denominator:
\( = \frac{4x^2 - 1}{2(4x^2 + 1)} \)
Answer: \( \frac{4x^2 - 1}{8x^2 + 2} \) or \( \frac{4x^2 - 1}{2(4x^2 + 1)} \).
38. Find the square root of \( 37x^2 - 28x^3 + 4x^4 + 42x + 9 \) by division method.
First, arrange the polynomial in descending order of powers:
\( 4x^4 - 28x^3 + 37x^2 + 42x + 9 \)
Now, perform the long division:
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
Answer: The square root is \( |2x^2 - 7x - 3| \).
39. State and prove Basic Proportionality Theorem.
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. |
40. Find the area of the quadrilateral formed by the points (8, 6), (5, 11), (-5, 12) and (-4, 3).
Let the vertices be A(8, 6), B(5, 11), C(-5, 12), and D(-4, 3). The points are already in counter-clockwise order.
Area of quadrilateral = \( \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \)
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 = \( \frac{1}{2} |(109) - (-49)| \)
Area = \( \frac{1}{2} |109 + 49| = \frac{1}{2} |158| = 79 \)
Answer: The area is 79 square units.
41. Find the equation of the perpendicular bisector of the line joining the points A(-4, 2) and B(6, -4).
1. Find the midpoint of AB:
Midpoint M = \( \left(\frac{-4+6}{2}, \frac{2+(-4)}{2}\right) = \left(\frac{2}{2}, \frac{-2}{2}\right) = (1, -1) \)
2. Find the slope of the line AB:
Slope \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 2}{6 - (-4)} = \frac{-6}{10} = -\frac{3}{5} \)
3. Find the slope of the perpendicular bisector:
The slope of the perpendicular line, \( m_{\perp} \), is the negative reciprocal of \( m_{AB} \).
\( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3} \)
4. Find the equation of the line:
The perpendicular bisector passes through the midpoint M(1, -1) and has a slope of \( \frac{5}{3} \). Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - (-1) = \frac{5}{3}(x - 1) \)
\( y + 1 = \frac{5}{3}(x - 1) \)
Multiply by 3 to eliminate the fraction:
\( 3(y + 1) = 5(x - 1) \)
\( 3y + 3 = 5x - 5 \)
\( 5x - 3y - 8 = 0 \)
Answer: The equation of the perpendicular bisector is \( 5x - 3y - 8 = 0 \).
42. (Compulsory) A straight line cuts the co-ordinate axes at A and B. If the mid point of AB is (2,3), find the equation of AB.
Let the line intersect the x-axis at A(a, 0) and the y-axis at B(0, b).
The midpoint of the line segment AB is given by the formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
Midpoint of AB = \( \left(\frac{a+0}{2}, \frac{0+b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \)
We are given that the midpoint is (2, 3).
Comparing the coordinates:
\( \frac{a}{2} = 2 \implies a = 4 \)
\( \frac{b}{2} = 3 \implies b = 6 \)
So, the x-intercept is a = 4 and the y-intercept is b = 6.
The equation of a line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substituting the values of a and b:
\( \frac{x}{4} + \frac{y}{6} = 1 \)
To write this in the standard form \( Ax + By + C = 0 \), we multiply by the LCM of 4 and 6, which is 12.
\( 12 \left(\frac{x}{4}\right) + 12 \left(\frac{y}{6}\right) = 12(1) \)
\( 3x + 2y = 12 \)
\( 3x + 2y - 12 = 0 \)
Answer: The equation of the line AB is \( 3x + 2y - 12 = 0 \).
PART IV: Answer the following questions (2 x 8 = 16)
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 7/3 of the corresponding sides of the triangle PQR (Scale factor 7/3 > 1).
Steps of Construction:
- Draw any triangle PQR.
- Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
- Locate 7 points (the greater of 7 and 3 in 7/3) Q₁, Q₂, ..., Q₇ on QX so that all segments QQ₁, Q₁Q₂, etc., are equal.
- Join Q₃ (the smaller of 7 and 3) to R.
- Draw a line through Q₇ parallel to Q₃R to intersect the extended line QR at R'. (This is done by copying the angle at Q₃ to Q₇).
- Draw a line through R' parallel to RP to intersect the extended line QP at P'.
- The triangle P'QR' is the required similar triangle, with each side being 7/3 times the corresponding side of \( \Delta PQR \).
[An image placeholder for the geometric construction would be here.]
(OR)
b) Construct a \( \Delta PQR \) in which the base PQ = 4.5 cm, \( \angle R = 35^\circ \) and the median from R to RG is 6 cm.
Steps of Construction:
- Draw a line segment PQ = 4.5 cm.
- At P, draw a line PE such that \( \angle QPE = 35^\circ \).
- Draw a line PF perpendicular to PE at P.
- Draw the perpendicular bisector of PQ. Let it intersect PQ at G and the line PF at O.
- With O as center and OP as radius, draw a circle. This circle passes through P and Q. The points on the major arc PQ will form an angle of 35° with the segment PQ.
- The median from R is RG = 6 cm. G is the midpoint of PQ.
- With G as the center and a radius of 6 cm, draw an arc. This arc will cut the circle at two points, R and R'.
- Join PR and QR.
- \( \Delta PQR \) is the required triangle.
[An image placeholder for the geometric construction would be here.]
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 1.5 hour (iii) the time required to cover a distance of 300 km from the graph.
Let x be the time in hours and y be the distance in km.
The relationship is y = 50x, which is a direct variation.
Table of Values:
| Time (x) (hr) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance (y) (km) | 50 | 100 | 150 | 200 |
Graph: Plot the points (1, 50), (2, 100), etc., on a graph with an appropriate scale (e.g., x-axis: 1 cm = 1 hr, y-axis: 1 cm = 50 km) and draw a straight line through them passing through the origin (0,0).
[An image placeholder for the distance-time graph would be here.]
Solutions from the Graph:
(i) The constant of variation (k): Since y = kx, \( k = \frac{y}{x} = \frac{50}{1} = 50 \). The constant of variation is 50 km/hr.
(ii) Distance travelled in 1.5 hours: On the graph, locate x = 1.5 on the x-axis. Move vertically up to the line and then horizontally to the y-axis. The corresponding y-value will be 75. Answer: 75 km.
(iii) Time required to cover 300 km: On the graph, locate y = 300 on the y-axis. Move horizontally to the line and then vertically down to the x-axis. The corresponding x-value will be 6. Answer: 6 hours.
(OR)
b) Draw the graph of \( xy = 24 \), x, y > 0. Using the graph find (i) y when x = 3, (ii) x when y = 12.
The equation is \( y = \frac{24}{x} \). This is an indirect variation.
Table of Values:
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
Graph: Plot these points on a graph and join them with a smooth curve (a rectangular hyperbola in the first quadrant).
[An image placeholder for the graph of xy = 24 would be here.]
Solutions from the Graph:
(i) Find y when x = 3: On the graph, find x = 3 on the x-axis. Move vertically to the curve and then horizontally to the y-axis. The y-value is 8. Answer: y = 8.
(ii) Find x when y = 12: On the graph, find y = 12 on the y-axis. Move horizontally to the curve and then vertically down to the x-axis. The x-value is 2. Answer: x = 2.