Tenkasi District
Common First Mid Term Test - 2024
STANDARD 10 - MATHS
Part - A
7 × 1 = 7Choose the best answer:
- 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:
- If $\{(a, 8), (6, b)\}$ represents an identity function, then the value of a and b are respectively:
- If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then $f \circ g$ is:
- Using Euclid's division lemma, if the cube of any positive integer is divided by 9, then the possible remainders are:
- If 6 times the 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P is:
- The value of $(1^3+2^3+3^3+\dots+15^3) – (1+2+3+\dots+15)$ is:
- The equation $xy - 7 = 3$ is a:
Part - B
5 × 2 = 10Answer any 5 of the following questions. [Q.No. 14 is compulsory]
- 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.
- Find $f \circ g$, if $f(x) = x-6$ and $g(x) = x^2$.
- If $f(x) = 2x-x^2$, find (i) $f(1)$ (ii) $f(2)$.
- If d is the highest common factor of 32 and 60, find x and y satisfying $d = 32x+60y$.
- Which term of the A.P $16, 11, 6, 1, \dots$ is -54?
- Find the sum to infinity of $9+3+1+\dots$
- Solve: $x+y = 5$; $x-y = 1$.
Part - C
5 × 5 = 25Answer any 5 of the following questions. [Q.No. 21 is compulsory]
- Let $A = \{x \in W / x < 2\}$, $B = \{x \in N / 1 < x \le 4\}$ and $C = \{3, 5\}$. Verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
- If $f(x) = 2x+3$, $g(x) = 1-2x$ and $h(x) = 3x$, prove that $f \circ (g \circ h) = (f \circ g) \circ h$.
- The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
- The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
- Find the sum to n terms of the series $5+55+555+\dots$
- Solve the following system of Linear equations: $3x-2y+z = 2$, $2x+3y-z = 5$, $x+y+z = 6$.
- Find the sum of the series $6^2+7^2+8^2+\dots+21^2$.
Part - D
1 × 8 = 8Answer the following question:
-
a) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{3}{5}$ of the corresponding sides of the triangle PQR. (Scale factor $\frac{3}{5} < 1$)
(OR)b) Construct a triangle similar to a given triangle ABC with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle ABC. (Scale factor $\frac{7}{3} > 1$)
Complete Step-by-Step Solutions
Part - A Solutions
7 × 1 = 7If 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:
Show Solution
Step 1: Find the number of elements in set A.
Given $A = \{1, 2, 3, 4, 5\}$, the number of elements in A is $n(A) = 5$.
Step 2: Understand the formula for the number of relations.
The total number of possible relations from a set A to a set B is given by the formula $2^{n(A) \times n(B)}$.
Step 3: Set up the equation.
We are given that the number of relations is 1024.
So, $2^{n(A) \times n(B)} = 1024$.
Step 4: Solve for the exponent.
We need to express 1024 as a power of 2. We know that $1024 = 2^{10}$.
Therefore, $2^{n(A) \times n(B)} = 2^{10}$.
Step 5: Equate the exponents and solve for n(B).
Since the bases are equal, the exponents must be equal:
$n(A) \times n(B) = 10$.
Substitute the value of $n(A)$:
$5 \times n(B) = 10$.
$n(B) = \frac{10}{5} = 2$.
If $\{(a, 8), (6, b)\}$ represents an identity function, then the value of a and b are respectively:
Show Solution
Step 1: Understand the definition of an identity function.
An identity function is a function that maps every element of a set to itself. It is denoted by $f(x) = x$. This means the output of the function is always equal to the input.
Step 2: Apply the definition to the first ordered pair.
The ordered pair $(a, 8)$ means that the input is 'a' and the output is '8'. For an identity function, input must equal output.
Therefore, $a = 8$.
Step 3: Apply the definition to the second ordered pair.
The ordered pair $(6, b)$ means that the input is '6' and the output is 'b'. For an identity function, input must equal output.
Therefore, $b = 6$.
Step 4: State the final values.
The values are $a=8$ and $b=6$. The question asks for the values of a and b respectively, which is the pair $(8, 6)$.
If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then $f \circ g$ is:
Show Solution
Step 1: Understand the composition of functions $f \circ g$.
The notation $f \circ g$ means $f(g(x))$. We first apply the function $g$ to $x$, and then apply the function $f$ to the result.
Step 2: Substitute $g(x)$ into $f(x)$.
We have $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$.
$f(g(x))$ means we replace every 'x' in the function $f$ with the expression for $g(x)$.
$$ f(g(x)) = f\left(\frac{1}{3x}\right) $$
Step 3: Evaluate the expression.
Now, substitute $\frac{1}{3x}$ into $f(x) = 2x^2$:
$$ f\left(\frac{1}{3x}\right) = 2\left(\frac{1}{3x}\right)^2 $$
$$ = 2\left(\frac{1^2}{(3x)^2}\right) = 2\left(\frac{1}{9x^2}\right) $$
$$ = \frac{2}{9x^2} $$
Using Euclid's division lemma, if the cube of any positive integer is divided by 9, then the possible remainders are:
Show Solution
Step 1: Understand the principle.
By Euclid's division lemma, any positive integer 'n' can be written in the form $n = bq + r$, where $0 \le r < b$. When dividing by 3, any integer can be expressed as $3q$, $3q+1$, or $3q+2$. We will cube each of these forms and check the remainder when divided by 9.
Step 2: Case 1: The integer is of the form $3q$.
Let $n = 3q$. Then its cube is:
$$ n^3 = (3q)^3 = 27q^3 = 9(3q^3) $$
This is of the form $9k$. The remainder is 0.
Step 3: Case 2: The integer is of the form $3q+1$.
Let $n = 3q+1$. Then its cube is:
$$ n^3 = (3q+1)^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3 $$
$$ = 27q^3 + 27q^2 + 9q + 1 $$
$$ = 9(3q^3 + 3q^2 + q) + 1 $$
This is of the form $9k+1$. The remainder is 1.
Step 4: Case 3: The integer is of the form $3q+2$.
Let $n = 3q+2$. Then its cube is:
$$ n^3 = (3q+2)^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3 $$
$$ = 27q^3 + 54q^2 + 36q + 8 $$
$$ = 9(3q^3 + 6q^2 + 4q) + 8 $$
This is of the form $9k+8$. The remainder is 8.
Step 5: Conclude the possible remainders.
From the three cases, the possible remainders when the cube of a positive integer is divided by 9 are 0, 1, and 8.
If 6 times the 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P is:
Show Solution
Step 1: Write the formula for the nth term of an A.P.
The nth term, $t_n$, of an Arithmetic Progression (A.P) is given by $t_n = a + (n-1)d$, where 'a' is the first term and 'd' is the common difference.
Step 2: Express the 6th and 7th terms.
$t_6 = a + (6-1)d = a + 5d$
$t_7 = a + (7-1)d = a + 6d$
Step 3: Set up the equation based on the given condition.
The problem states that 6 times the 6th term equals 7 times the 7th term.
$$ 6 \times t_6 = 7 \times t_7 $$
$$ 6(a + 5d) = 7(a + 6d) $$
Step 4: Simplify the equation.
$$ 6a + 30d = 7a + 42d $$
Move all terms to one side to find a relationship between 'a' and 'd'.
$$ 0 = (7a - 6a) + (42d - 30d) $$
$$ 0 = a + 12d $$
Step 5: Find the 13th term.
The 13th term, $t_{13}$, is given by:
$$ t_{13} = a + (13-1)d = a + 12d $$
From Step 4, we found that $a + 12d = 0$.
Therefore, $t_{13} = 0$.
The value of $(1^3+2^3+3^3+\dots+15^3) – (1+2+3+\dots+15)$ is:
Show Solution
Step 1: Identify the required formulas.
Sum of the first n natural numbers: $S_1 = 1+2+...+n = \frac{n(n+1)}{2}$.
Sum of the cubes of the first n natural numbers: $S_3 = 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$.
Step 2: Calculate the sum of the first 15 natural numbers.
Here, $n=15$.
$$ 1+2+3+\dots+15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120 $$
Step 3: Calculate the sum of the cubes of the first 15 natural numbers.
Using the formula from Step 1 and the result from Step 2:
$$ 1^3+2^3+3^3+\dots+15^3 = \left(\frac{15(15+1)}{2}\right)^2 = (120)^2 = 14400 $$
Step 4: Calculate the final value.
The problem asks for the difference between the two sums:
$$ (1^3+2^3+\dots+15^3) – (1+2+\dots+15) = 14400 - 120 = 14280 $$
The equation $xy - 7 = 3$ is a:
Show Solution
Step 1: Simplify the given equation.
$$ xy - 7 = 3 $$
$$ xy = 3 + 7 $$
$$ xy = 10 $$
Step 2: Understand the definition of a linear equation.
A linear equation in two variables (say, x and y) is an equation of the form $Ax + By = C$, where A, B, and C are constants, and A and B are not both zero. The key characteristic is that the highest power (degree) of any term is 1.
Step 3: Determine the degree of the term in the simplified equation.
The equation is $xy = 10$. The term on the left is $xy$. The degree of a term is the sum of the exponents of its variables. Here, the degree of $x$ is 1 and the degree of $y$ is 1.
The degree of the term $xy$ is $1 + 1 = 2$.
Step 4: Conclude based on the degree.
Since the degree of the equation is 2, it is not a linear equation. A linear equation must have a degree of 1.
Part - B Solutions
5 × 2 = 10A 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.
Show Solution
Step 1: Find the ordered pairs in the relation R.
We are given the rule $y = x+3$ and the set of possible x-values, $x \in \{0, 1, 2, 3, 4, 5\}$. We calculate the corresponding y-value for each x.
- If $x=0$, $y = 0+3 = 3$. The pair is $(0, 3)$.
- If $x=1$, $y = 1+3 = 4$. The pair is $(1, 4)$.
- If $x=2$, $y = 2+3 = 5$. The pair is $(2, 5)$.
- If $x=3$, $y = 3+3 = 6$. The pair is $(3, 6)$.
- If $x=4$, $y = 4+3 = 7$. The pair is $(4, 7)$.
- If $x=5$, $y = 5+3 = 8$. The pair is $(5, 8)$.
So, the relation R is the set of these ordered pairs: $R = \{(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)\}$.
Step 2: Determine the domain.
The domain of a relation is the set of all first elements (x-coordinates) of the ordered pairs.
From the set R, the first elements are $\{0, 1, 2, 3, 4, 5\}$.
Step 3: Determine the range.
The range of a relation is the set of all second elements (y-coordinates) of the ordered pairs.
From the set R, the second elements are $\{3, 4, 5, 6, 7, 8\}$.
Range = $\{3, 4, 5, 6, 7, 8\}$
Find $f \circ g$, if $f(x) = x-6$ and $g(x) = x^2$.
Show Solution
Step 1: Understand the composition of functions $f \circ g$.
The notation $f \circ g$ means $f(g(x))$. This involves substituting the entire function $g(x)$ into the variable $x$ of the function $f(x)$.
Step 2: Perform the substitution.
We are given $f(x) = x-6$ and $g(x) = x^2$.
$$ f(g(x)) = f(x^2) $$
To find $f(x^2)$, we replace $x$ in the expression for $f(x)$ with $x^2$.
$$ f(x^2) = (x^2) - 6 $$
Step 3: State the final result.
The composed function is $f \circ g(x) = x^2 - 6$.
If $f(x) = 2x-x^2$, find (i) $f(1)$ (ii) $f(2)$.
Show Solution
The given function is $f(x) = 2x-x^2$.
(i) Find f(1)
Step 1: Substitute $x=1$ into the function.
$$ f(1) = 2(1) - (1)^2 $$
Step 2: Simplify the expression.
$$ f(1) = 2 - 1 = 1 $$
(ii) Find f(2)
Step 1: Substitute $x=2$ into the function.
$$ f(2) = 2(2) - (2)^2 $$
Step 2: Simplify the expression.
$$ f(2) = 4 - 4 = 0 $$
(ii) $f(2) = 0$
If d is the highest common factor of 32 and 60, find x and y satisfying $d = 32x+60y$.
Show Solution
Step 1: Find the HCF (d) of 32 and 60 using Euclid's Division Algorithm.
We apply the division lemma $a = bq + r$ repeatedly.
Let $a = 60$ and $b = 32$.
$60 = 1 \times 32 + 28$ (Equation 1)
Now, let $a=32$ and $b=28$.
$32 = 1 \times 28 + 4$ (Equation 2)
Now, let $a=28$ and $b=4$.
$28 = 7 \times 4 + 0$
The last non-zero remainder is the HCF. So, $d = 4$.
Step 2: Use the Extended Euclidean Algorithm to express d in the form $32x+60y$.
We work backwards from the second-to-last step of the algorithm (Equation 2).
From Equation 2, express the remainder 4: $$ 4 = 32 - 1 \times 28 $$
From Equation 1, express the remainder 28: $28 = 60 - 1 \times 32$.
Now substitute this expression for 28 into the equation for 4:
$$ 4 = 32 - 1 \times (60 - 1 \times 32) $$
Step 3: Simplify the expression to the required form.
Distribute the $-1$:
$$ 4 = 32 - 1 \times 60 + 1 \times 32 $$
Group the terms with 32 and 60:
$$ 4 = (1+1) \times 32 + (-1) \times 60 $$
$$ 4 = 2 \times 32 + (-1) \times 60 $$
Step 4: Identify x and y.
Comparing this with $d = 32x + 60y$, we get $x=2$ and $y=-1$.
(Note: Other solutions exist, but this is one valid pair.)
Which term of the A.P $16, 11, 6, 1, \dots$ is -54?
Show Solution
Step 1: Identify the first term (a) and common difference (d) of the A.P.
The given A.P is $16, 11, 6, 1, \dots$
First term, $a = 16$.
Common difference, $d = (\text{second term}) - (\text{first term}) = 11 - 16 = -5$.
Step 2: Set up the formula for the nth term.
The formula for the nth term of an A.P is $t_n = a + (n-1)d$.
We want to find the term number 'n' for which the value $t_n$ is -54.
So, we set $t_n = -54$.
Step 3: Substitute the known values and solve for n.
$$ -54 = 16 + (n-1)(-5) $$
Subtract 16 from both sides:
$$ -54 - 16 = (n-1)(-5) $$
$$ -70 = (n-1)(-5) $$
Divide both sides by -5:
$$ \frac{-70}{-5} = n-1 $$
$$ 14 = n-1 $$
Add 1 to both sides:
$$ n = 14 + 1 = 15 $$
Find the sum to infinity of $9+3+1+\dots$
Show Solution
Step 1: Identify the type of series.
Let's check the ratio between consecutive terms:
$\frac{3}{9} = \frac{1}{3}$
$\frac{1}{3} = \frac{1}{3}$
Since there is a common ratio, this is a Geometric Progression (G.P).
Step 2: Identify the first term (a) and common ratio (r).
First term, $a = 9$.
Common ratio, $r = \frac{1}{3}$.
Step 3: Check the condition for the sum to infinity.
The sum to infinity of a G.P exists only if the absolute value of the common ratio is less than 1, i.e., $|r| < 1$.
Here, $|r| = |\frac{1}{3}| = \frac{1}{3}$, which is less than 1. So, the sum to infinity exists.
Step 4: Apply the formula for the sum to infinity.
The formula is $S_\infty = \frac{a}{1-r}$.
Substitute the values of a and r:
$$ S_\infty = \frac{9}{1 - \frac{1}{3}} $$
$$ S_\infty = \frac{9}{\frac{3}{3} - \frac{1}{3}} = \frac{9}{\frac{2}{3}} $$
$$ S_\infty = 9 \times \frac{3}{2} = \frac{27}{2} $$
Solve: $x+y = 5$; $x-y = 1$.
Show Solution
Step 1: Write down the given equations.
Equation (1): $x+y = 5$
Equation (2): $x-y = 1$
Step 2: Use the elimination method to solve for x.
Notice that the coefficients of 'y' are +1 and -1. By adding the two equations, the 'y' terms will cancel out.
Add Equation (1) and Equation (2): $$(x+y) + (x-y) = 5 + 1$$ $$2x = 6$$ $$x = \frac{6}{2} = 3$$
Step 3: Substitute the value of x back into one of the original equations to solve for y.
Let's use Equation (1): $x+y = 5$.
Substitute $x=3$:
$$ 3 + y = 5 $$
Subtract 3 from both sides:
$$ y = 5 - 3 = 2 $$
Step 4: State the solution.
The solution is $x=3, y=2$. We can verify this using Equation (2): $3-2=1$, which is correct.
Part - C Solutions
5 × 5 = 25Let $A = \{x \in W / x < 2\}$, $B = \{x \in N / 1 < x \le 4\}$ and $C = \{3, 5\}$. Verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Show Solution
Step 1: List the elements of sets A and B.
Set A: $A = \{x \in W / x < 2\}$. W represents Whole Numbers (0, 1, 2, ...). So, $A = \{0, 1\}$.
Set B: $B = \{x \in N / 1 < x \le 4\}$. N represents Natural Numbers (1, 2, 3, ...). So, $B = \{2, 3, 4\}$.
Set C is given as $C = \{3, 5\}$.
Step 2: Calculate the Left Hand Side (LHS): $A \times (B \cup C)$
First, find the union of B and C:
$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$.
Next, find the Cartesian product of A and $(B \cup C)$:
$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (1)
Step 3: Calculate the Right Hand Side (RHS): $(A \times B) \cup (A \times C)$
First, find the Cartesian product $A \times B$:
$A \times B = \{0, 1\} \times \{2, 3, 4\}$
$= \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\}$.
Next, find the Cartesian product $A \times C$:
$A \times C = \{0, 1\} \times \{3, 5\}$
$= \{(0,3), (0,5), (1,3), (1,5)\}$.
Finally, find the union of these two sets:
$(A \times B) \cup (A \times C) = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\} \cup \{(0,3), (0,5), (1,3), (1,5)\}$.
Combining all unique elements:
$= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (2)
Step 4: Compare LHS and RHS.
Comparing the results from (1) and (2), we see that the sets are identical.
$A \times (B \cup C) = (A \times B) \cup (A \times C)$.
If $f(x) = 2x+3$, $g(x) = 1-2x$ and $h(x) = 3x$, prove that $f \circ (g \circ h) = (f \circ g) \circ h$.
Show Solution
This question asks to prove the associative property of function composition.
Step 1: Calculate the Left Hand Side (LHS): $f \circ (g \circ h)$
First, we need to find the composition $g \circ h$.
$g \circ h(x) = g(h(x)) = g(3x)$.
Substitute $3x$ into $g(x) = 1-2x$:
$g(3x) = 1 - 2(3x) = 1 - 6x$.
So, $g \circ h(x) = 1 - 6x$.
Now, we find $f \circ (g \circ h)(x) = f(g \circ h(x)) = f(1 - 6x)$.
Substitute $1 - 6x$ into $f(x) = 2x+3$:
$f(1 - 6x) = 2(1 - 6x) + 3 = 2 - 12x + 3 = 5 - 12x$.
So, $LHS = 5 - 12x$. --- (1)
Step 2: Calculate the Right Hand Side (RHS): $(f \circ g) \circ h$
First, we need to find the composition $f \circ g$.
$f \circ g(x) = f(g(x)) = f(1 - 2x)$.
Substitute $1 - 2x$ into $f(x) = 2x+3$:
$f(1 - 2x) = 2(1 - 2x) + 3 = 2 - 4x + 3 = 5 - 4x$.
So, $f \circ g(x) = 5 - 4x$.
Now, we find $(f \circ g) \circ h(x) = (f \circ g)(h(x)) = (f \circ g)(3x)$.
Substitute $3x$ into our result for $f \circ g(x)$:
$(f \circ g)(3x) = 5 - 4(3x) = 5 - 12x$.
So, $RHS = 5 - 12x$. --- (2)
Step 3: Compare LHS and RHS.
From (1) and (2), we have $LHS = 5 - 12x$ and $RHS = 5 - 12x$.
Since LHS = RHS, the associative property is proven for these functions.
The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
Show Solution
Step 1: Represent the three consecutive terms in an A.P.
Let the three terms be $a-d, a, a+d$, where 'a' is the middle term and 'd' is the common difference.
Step 2: Use the given sum to find the value of 'a'.
Sum of the terms = 27
$$ (a-d) + a + (a+d) = 27 $$
$$ 3a = 27 $$
$$ a = \frac{27}{3} = 9 $$
So, the middle term is 9.
Step 3: Use the given product to find the value of 'd'.
Product of the terms = 288
$$ (a-d)(a)(a+d) = 288 $$
Substitute $a=9$:
$$ (9-d)(9)(9+d) = 288 $$
Divide both sides by 9:
$$ (9-d)(9+d) = \frac{288}{9} $$
$$ (9-d)(9+d) = 32 $$
Using the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:
$$ 9^2 - d^2 = 32 $$
$$ 81 - d^2 = 32 $$
$$ d^2 = 81 - 32 $$
$$ d^2 = 49 $$
$$ d = \pm\sqrt{49} \implies d = \pm 7 $$
Step 4: Find the three terms for each value of 'd'.
Case 1: $d = 7$
The terms are $a-d, a, a+d$, which are $9-7, 9, 9+7$.
The terms are $2, 9, 16$.
Case 2: $d = -7$
The terms are $a-d, a, a+d$, which are $9-(-7), 9, 9+(-7)$.
The terms are $16, 9, 2$.
In both cases, the set of three terms is the same.
The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
Show Solution
Step 1: Recall the property of similar triangles.
If two triangles are similar, the ratio of their perimeters is equal to the ratio of their corresponding sides.
Step 2: Set up the ratio equation.
Given that $\triangle ABC \sim \triangle PQR$.
Therefore,
$$ \frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle PQR} = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} $$
Step 3: Substitute the given values into the equation.
Perimeter of $\triangle ABC = 36$ cm.
Perimeter of $\triangle PQR = 24$ cm.
Side $PQ = 10$ cm.
We need to find side AB.
$$ \frac{36}{24} = \frac{AB}{10} $$
Step 4: Solve for AB.
First, simplify the ratio of the perimeters:
$$ \frac{36}{24} = \frac{12 \times 3}{12 \times 2} = \frac{3}{2} $$
Now, the equation becomes:
$$ \frac{3}{2} = \frac{AB}{10} $$
To find AB, multiply both sides by 10:
$$ AB = \frac{3}{2} \times 10 = 3 \times 5 = 15 $$
Find the sum to n terms of the series $5+55+555+\dots$
Show Solution
Step 1: Write the sum and factor out the common term.
Let $S_n = 5+55+555+\dots$ to n terms.
$$ S_n = 5(1+11+111+\dots \text{ to n terms}) $$
Step 2: Multiply and divide by 9 to create a standard form.
This trick allows us to express the terms as powers of 10.
$$ S_n = \frac{5}{9}(9+99+999+\dots \text{ to n terms}) $$
Step 3: Rewrite each term inside the bracket.
$9 = 10 - 1$
$99 = 100 - 1 = 10^2 - 1$
$999 = 1000 - 1 = 10^3 - 1$
And so on.
$$ S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \dots \text{ to n terms}) $$
Step 4: Separate the terms into two groups.
Group the powers of 10 and the '-1's separately.
$$ S_n = \frac{5}{9} [ (10 + 10^2 + 10^3 + \dots + 10^n) - (1+1+1+\dots \text{n times}) ] $$
Step 5: Calculate the sum of each group.
The first group is a Geometric Progression (G.P) with first term $a=10$, common ratio $r=10$, and $n$ terms. The sum is $S_{GP} = \frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.
The second group is simply the sum of 'n' ones, which is $n$.
Step 6: Substitute back and write the final formula.
$$ S_n = \frac{5}{9} \left[ \frac{10}{9}(10^n-1) - n \right] $$
We can also distribute the outer fraction for an alternative form:
$$ S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9} $$
Solve the following system of Linear equations: $3x-2y+z = 2$, $2x+3y-z = 5$, $x+y+z = 6$.
Show Solution
Step 1: Label the equations.
(1) $3x-2y+z = 2$
(2) $2x+3y-z = 5$
(3) $x+y+z = 6$
Step 2: Eliminate one variable (e.g., z) using two pairs of equations.
Pair 1: Add Equation (1) and Equation (2). The 'z' terms will cancel.
$$(3x-2y+z) + (2x+3y-z) = 2 + 5$$
$$5x + y = 7 \quad \text{--- (4)}$$
Pair 2: Add Equation (2) and Equation (3). The 'z' terms will cancel again.
$$(2x+3y-z) + (x+y+z) = 5 + 6$$
$$3x + 4y = 11 \quad \text{--- (5)}$$
Step 3: Solve the new system of two linear equations (4) and (5).
From equation (4), we can express y in terms of x:
$y = 7 - 5x$.
Substitute this expression for y into equation (5):
$$3x + 4(7 - 5x) = 11$$
$$3x + 28 - 20x = 11$$
$$-17x = 11 - 28$$
$$-17x = -17$$
$$x = 1$$
Step 4: Substitute the value of x back to find y.
Using $y = 7 - 5x$:
$$y = 7 - 5(1) = 7 - 5 = 2$$
Step 5: Substitute the values of x and y back into one of the original equations to find z.
Using Equation (3) is simplest: $x+y+z=6$.
$$1 + 2 + z = 6$$
$$3 + z = 6$$
$$z = 6 - 3 = 3$$
Step 6: State the final solution.
Check the solution in another equation, e.g., (1): $3(1)-2(2)+3 = 3-4+3 = 2$. It is correct.
Find the sum of the series $6^2+7^2+8^2+\dots+21^2$.
Show Solution
Step 1: Express the series as a difference of two sums.
The given series does not start from 1. We can find its sum by taking the sum from 1 to 21 and subtracting the sum from 1 to 5.
$$ 6^2+7^2+\dots+21^2 = (1^2+2^2+\dots+21^2) - (1^2+2^2+\dots+5^2) $$
Step 2: Use the formula for the sum of the squares of the first n natural numbers.
The formula is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Step 3: Calculate the first sum (from 1 to 21).
Here, $n=21$.
$$ \sum_{k=1}^{21} k^2 = \frac{21(21+1)(2(21)+1)}{6} = \frac{21(22)(43)}{6} $$
$$ = \frac{7 \times 3 \times 11 \times 2 \times 43}{3 \times 2} = 7 \times 11 \times 43 $$
$$ = 77 \times 43 = 3311 $$
Step 4: Calculate the second sum (from 1 to 5).
Here, $n=5$.
$$ \sum_{k=1}^{5} k^2 = \frac{5(5+1)(2(5)+1)}{6} = \frac{5(6)(11)}{6} $$
$$ = 5 \times 11 = 55 $$
Step 5: Find the difference to get the final answer.
Required sum = (Sum from 1 to 21) - (Sum from 1 to 5)
$$ = 3311 - 55 = 3256 $$
Part - D Solutions
1 × 8 = 8a) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{3}{5}$ of the corresponding sides of the triangle PQR. (Scale factor $\frac{3}{5} < 1$)
Show Solution Steps
Since the scale factor $\frac{3}{5}$ is less than 1, the new triangle (let's call it $\triangle P'QR'$) will be smaller than the original $\triangle PQR$.
Steps of Construction:
- Draw the given triangle $\triangle PQR$ with any measurements.
- From vertex Q, draw a ray QX making an acute angle with the side QR (on the side opposite to vertex P).
- On the ray QX, mark 5 equidistant points (since 5 is the denominator of the scale factor). Label them $Q_1, Q_2, Q_3, Q_4, Q_5$ such that $QQ_1 = Q_1Q_2 = \dots = Q_4Q_5$.
- Join the last point, $Q_5$, to the vertex R. This forms the line segment $Q_5R$.
- From the point $Q_3$ (since 3 is the numerator of the scale factor), draw a line parallel to $Q_5R$. This parallel line must intersect the side QR. Use a compass and ruler to construct the parallel line (by copying the angle at $Q_5$). Let the point of intersection on QR be R'.
- From the point R', draw a line parallel to the side PR. This line must intersect the side PQ. Let the point of intersection on PQ be P'.
- The triangle $\triangle P'QR'$ is the required triangle, whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle PQR$.
b) Construct a triangle similar to a given triangle ABC with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle ABC. (Scale factor $\frac{7}{3} > 1$)
Show Solution Steps
Since the scale factor $\frac{7}{3}$ is greater than 1, the new triangle (let's call it $\triangle AB'C'$) will be larger than the original $\triangle ABC$.
Steps of Construction:
- Draw the given triangle $\triangle ABC$ with any measurements.
- Extend the sides AB and AC. Draw rays from B and C away from the triangle along the lines AB and AC.
- From vertex A, draw a ray AX making an acute angle with the side AB (on the side opposite to vertex C).
- On the ray AX, mark 7 equidistant points (since 7 is the numerator of the scale factor). Label them $A_1, A_2, \dots, A_7$ such that $AA_1 = A_1A_2 = \dots = A_6A_7$.
- Join the point $A_3$ (since 3 is the denominator of the scale factor) to the vertex B. This forms the line segment $A_3B$.
- From the point $A_7$, draw a line parallel to $A_3B$. This parallel line will intersect the extended side AB at a point. Label this point B'.
- From the point B', draw a line parallel to the side BC. This parallel line will intersect the extended side AC at a point. Label this point C'.
- The triangle $\triangle AB'C'$ is the required triangle, whose sides are $\frac{7}{3}$ of the corresponding sides of $\triangle ABC$.