THOOTHUKUDI DISTRICT
FIRST MID TERM TEST - 2024
Standard X - MATHEMATICS
Time: 1.30 hrs.
Reg. No:
Marks: 50
Part - I
4 x 1 = 4I. Choose the correct answer:
1.
The range of the relation $R = \{(x, x^2) \mid x \text{ is a prime number less than 13}\}$ is
2.
The value of $(1^3 + 2^3 + 3^3 + \dots + 15^3) – (1 + 2 + 3 + \dots + 15)$ is
3.
The solution of the system $3z = 9, -7y + 7z = 7, x + y - 3z = -6$ is
4.
In $\triangle LMN$, $\angle L = 60^\circ, \angle M = 50^\circ$. If $\triangle LMN \sim \triangle PQR$, then the value of $\angle R$ is
Part - II
5 x 2 = 10II. Answer any 5 questions. (Q.No.11 is compulsory)
5.
A function $f: [-5, 9] \rightarrow R$ is defined as follows:
$$ f(x) =
\begin{cases}
6x+1 & ;-5 \le x < 2 \\
5x^2-1 & ; 2 \le x < 6 \\
3x-4 & ; 6 \le x \le 9
\end{cases}
$$
Find $2f(4) + f(8)$.
6.
If $13824 = 2^a \times 3^b$, then find a and b.
7.
Use Euclid's Division Algorithm to find the Highest Common Factor (HCF) of 396, 504, 636.
8.
A boy of height 90 cm is walking away from the base of a lamppost at a speed of 1.2 m/sec. If the lamppost is 3.6 m above the ground, find the length of his shadow cast after 4 seconds.
9.
Simplify: $\displaystyle \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4}$
10.
Find the LCM of the polynomials $a^2 + 4a - 12$ and $a^2 - 5a + 6$ whose GCD is $a - 2$.
11.
Represent the function $f = \{(1,2), (2,2), (3,2), (4,3), (5,4)\}$ through:
i) an arrow diagram
ii) a table form
iii) a graph
Part - III
4 x 5 = 20III. Answer any 4 questions. (Q.No.17 is compulsory)
12.
Given $A = \{x \in W \mid x < 2\}$, $B = \{x \in N \mid 1 < x \le 4\}$ and $C = \{3,5\}$, verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
13.
Find the sum to n terms of the series $6 + 66 + 666 + \dots$
14.
The ratio of the 6th and 8th term of an A.P is 7:9. Find the ratio of the 9th term to the 13th term.
15.
Simplify: $\displaystyle \frac{1}{x^2 - 5x + 6} + \frac{1}{x^2 - 3x + 2} - \frac{1}{x^2 - 8x + 15}$
16.
Find the GCD of $6x^3 - 30x^2 + 60x - 48$ and $3x^3 - 12x^2 + 21x - 18$.
17.
The data in the adjacent table depicts the length of a person's forehand and their corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as $y = ax + b$, where a, b are constants.
| Length 'x' of forehand (in cm) | Height 'y' (in inches) |
|---|---|
| 35 | 56 |
| 45 | 65 |
| 50 | 69.5 |
| 55 | 74 |
i) Check if this relation is a function.
ii) Find a and b.
iii) Find the height of a person whose forehand length is 40 cm.
iv) Find the length of the forehand of a person if the height is 53.3 inches.
18.
i) Find the least positive value of x such that $67 + x \equiv 1 \pmod 4$.
ii) Solve: $5x \equiv 4 \pmod 6$.
Part - IV
2 x 8 = 16IV. Answer the following questions.
19.
a) Construct a triangle similar to a given triangle PQR with its sides equal to $\displaystyle\frac{7}{3}$ of the corresponding sides of the triangle PQR. (scale factor $\displaystyle\frac{7}{3} > 1$)
(OR)
b) Construct a triangle similar to a given triangle ABC with its sides equal to $\displaystyle\frac{3}{5}$ of the corresponding sides of the triangle ABC. (scale factor $\displaystyle\frac{3}{5} < 1$)
20.
a) A two-wheeler parking zone near a bus stand charges as below:
Check if the amount charged is in direct variation or in inverse variation to the parking time. Graph the data. Also,
| Time (in hours) (x) | Amount (₹) (y) |
|---|---|
| 4 | 60 |
| 8 | 120 |
| 12 | 180 |
| 24 | 360 |
i) Find the amount to be paid when parking time is 6 hr.
ii) Find the parking duration when the amount paid is ₹150.
(OR)
b) Graph the following linear function $\displaystyle y = \frac{1}{2}x$. Identify the constant of variation and verify it with the graph. Also,
(i) find y when x = 9
(ii) find x when y = 7.5
*** END OF QUESTIONS ***
Solutions
Part - I Solutions
4 x 1 = 4Solution 1
Step 1: Identify the domain.
The relation is defined for $x$ where $x$ is a prime number less than 13. The prime numbers less than 13 are 2, 3, 5, 7, and 11.
The relation is defined for $x$ where $x$ is a prime number less than 13. The prime numbers less than 13 are 2, 3, 5, 7, and 11.
Step 2: Calculate the corresponding values.
The relation is $R = \{(x, x^2)\}$. We need to find $x^2$ for each prime number.
The relation is $R = \{(x, x^2)\}$. We need to find $x^2$ for each prime number.
- If $x=2$, $x^2 = 4$
- If $x=3$, $x^2 = 9$
- If $x=5$, $x^2 = 25$
- If $x=7$, $x^2 = 49$
- If $x=11$, $x^2 = 121$
Step 3: Determine the range.
The range is the set of all second elements (the $x^2$ values).
Range = $\{4, 9, 25, 49, 121\}$.
The range is the set of all second elements (the $x^2$ values).
Range = $\{4, 9, 25, 49, 121\}$.
c) $\{4,9,25,49,121\}$
Solution 2
Step 1: Identify the formulas.
We need the formula for the sum of the first n natural numbers and the sum of the cubes of the first n natural numbers.
We need the formula 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 numbers: $1 + 2 + \dots + n = \sum_{k=1}^n k = \frac{n(n+1)}{2}$
- Sum of cubes: $1^3 + 2^3 + \dots + n^3 = \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
Step 2: Calculate for n=15.
Here, $n=15$.
Sum of numbers: $1+2+\dots+15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120$.
Sum of cubes: $1^3+2^3+\dots+15^3 = \left(\frac{15(15+1)}{2}\right)^2 = (120)^2 = 14400$.
Here, $n=15$.
Sum of numbers: $1+2+\dots+15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120$.
Sum of cubes: $1^3+2^3+\dots+15^3 = \left(\frac{15(15+1)}{2}\right)^2 = (120)^2 = 14400$.
Step 3: Find the difference.
The required value is $(1^3 + \dots + 15^3) - (1 + \dots + 15) = 14400 - 120 = 14280$.
The required value is $(1^3 + \dots + 15^3) - (1 + \dots + 15) = 14400 - 120 = 14280$.
d) 14280
Solution 3
Step 1: Solve the first equation.
Given $3z = 9$.
$z = \frac{9}{3} \implies z=3$.
Given $3z = 9$.
$z = \frac{9}{3} \implies z=3$.
Step 2: Substitute z into the second equation.
Given $-7y + 7z = 7$. Substitute $z=3$.
$-7y + 7(3) = 7$
$-7y + 21 = 7$
$-7y = 7 - 21$
$-7y = -14 \implies y=2$.
Given $-7y + 7z = 7$. Substitute $z=3$.
$-7y + 7(3) = 7$
$-7y + 21 = 7$
$-7y = 7 - 21$
$-7y = -14 \implies y=2$.
Step 3: Substitute y and z into the third equation.
Given $x + y - 3z = -6$. Substitute $y=2$ and $z=3$.
$x + (2) - 3(3) = -6$
$x + 2 - 9 = -6$
$x - 7 = -6$
$x = -6 + 7 \implies x=1$.
Given $x + y - 3z = -6$. Substitute $y=2$ and $z=3$.
$x + (2) - 3(3) = -6$
$x + 2 - 9 = -6$
$x - 7 = -6$
$x = -6 + 7 \implies x=1$.
b) $x = 1, y = 2, z = 3$
Solution 4
Step 1: Find the third angle of $\triangle LMN$.
The sum of angles in a triangle is $180^\circ$.
In $\triangle LMN$, $\angle L + \angle M + \angle N = 180^\circ$.
$60^\circ + 50^\circ + \angle N = 180^\circ$
$110^\circ + \angle N = 180^\circ$
$\angle N = 180^\circ - 110^\circ = 70^\circ$.
The sum of angles in a triangle is $180^\circ$.
In $\triangle LMN$, $\angle L + \angle M + \angle N = 180^\circ$.
$60^\circ + 50^\circ + \angle N = 180^\circ$
$110^\circ + \angle N = 180^\circ$
$\angle N = 180^\circ - 110^\circ = 70^\circ$.
Step 2: Use the property of similar triangles.
Given that $\triangle LMN \sim \triangle PQR$. This means their corresponding angles are equal.
$\angle L = \angle P = 60^\circ$
$\angle M = \angle Q = 50^\circ$
$\angle N = \angle R = 70^\circ$
Given that $\triangle LMN \sim \triangle PQR$. This means their corresponding angles are equal.
$\angle L = \angle P = 60^\circ$
$\angle M = \angle Q = 50^\circ$
$\angle N = \angle R = 70^\circ$
b) $70^\circ$
Part - II Solutions
5 x 2 = 10Solution 5
Step 1: Find f(4).
For $x=4$, the condition $2 \le x < 6$ applies. So we use the rule $f(x) = 5x^2 - 1$.
$f(4) = 5(4^2) - 1 = 5(16) - 1 = 80 - 1 = 79$.
For $x=4$, the condition $2 \le x < 6$ applies. So we use the rule $f(x) = 5x^2 - 1$.
$f(4) = 5(4^2) - 1 = 5(16) - 1 = 80 - 1 = 79$.
Step 2: Find f(8).
For $x=8$, the condition $6 \le x \le 9$ applies. So we use the rule $f(x) = 3x - 4$.
$f(8) = 3(8) - 4 = 24 - 4 = 20$.
For $x=8$, the condition $6 \le x \le 9$ applies. So we use the rule $f(x) = 3x - 4$.
$f(8) = 3(8) - 4 = 24 - 4 = 20$.
Step 3: Calculate the final expression.
We need to find $2f(4) + f(8)$.
$2(79) + 20 = 158 + 20 = 178$.
We need to find $2f(4) + f(8)$.
$2(79) + 20 = 158 + 20 = 178$.
178
Solution 6
Step 1: Perform prime factorization of 13824.
We divide 13824 by the smallest prime factors repeatedly.
We divide 13824 by the smallest prime factors repeatedly.
2 | 13824
---
2 | 6912
---
2 | 3456
---
2 | 1728
---
2 | 864
---
2 | 432
---
2 | 216
---
2 | 108
---
2 | 54
---
3 | 27
---
3 | 9
---
3 | 3
---
1
Counting the factors, we have nine 2s and three 3s.
Step 2: Write in the form $2^a \times 3^b$.
$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^9 \times 3^3$.
$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^9 \times 3^3$.
Step 3: Compare and find a and b.
Comparing $13824 = 2^9 \times 3^3$ with the given $13824 = 2^a \times 3^b$, we get $a=9$ and $b=3$.
Comparing $13824 = 2^9 \times 3^3$ with the given $13824 = 2^a \times 3^b$, we get $a=9$ and $b=3$.
a = 9, b = 3
Solution 7
Step 1: Find HCF of the first two numbers (504, 396).
Using Euclid's Division Algorithm ($a = bq + r$):
$504 = 1 \times 396 + 108$
$396 = 3 \times 108 + 72$
$108 = 1 \times 72 + 36$
$72 = 2 \times 36 + 0$
The last non-zero remainder is 36. So, HCF(504, 396) = 36.
Using Euclid's Division Algorithm ($a = bq + r$):
$504 = 1 \times 396 + 108$
$396 = 3 \times 108 + 72$
$108 = 1 \times 72 + 36$
$72 = 2 \times 36 + 0$
The last non-zero remainder is 36. So, HCF(504, 396) = 36.
Step 2: Find HCF of the result and the third number (636, 36).
Now we find the HCF of 636 and 36.
$636 = 17 \times 36 + 24$
$36 = 1 \times 24 + 12$
$24 = 2 \times 12 + 0$
The last non-zero remainder is 12.
Now we find the HCF of 636 and 36.
$636 = 17 \times 36 + 24$
$36 = 1 \times 24 + 12$
$24 = 2 \times 12 + 0$
The last non-zero remainder is 12.
The HCF of 396, 504, and 636 is 12.
Solution 8
Step 1: Set up the problem and draw a diagram.
Let AB be the lamppost and DE be the boy. Let C be the tip of the shadow. Height of lamppost, AB = 3.6 m.
Height of boy, DE = 90 cm = 0.9 m.
Speed of walking = 1.2 m/s.
Time = 4 seconds.
Let AB be the lamppost and DE be the boy. Let C be the tip of the shadow. Height of lamppost, AB = 3.6 m.
Height of boy, DE = 90 cm = 0.9 m.
Speed of walking = 1.2 m/s.
Time = 4 seconds.
Step 2: Calculate the distance walked.
Distance BD = Speed × Time = $1.2 \text{ m/s} \times 4 \text{ s} = 4.8 \text{ m}$.
Distance BD = Speed × Time = $1.2 \text{ m/s} \times 4 \text{ s} = 4.8 \text{ m}$.
Step 3: Use similar triangles.
In the diagram, $\triangle CDE$ and $\triangle CBA$ are similar because $\angle C$ is common and $\angle CED = \angle CBA = 90^\circ$. Therefore, the ratio of corresponding sides is equal: $$ \frac{DE}{AB} = \frac{CE}{CB} $$ Let the length of the shadow CE be $x$. Then $CB = CE + EB = x + 4.8$.
$$ \frac{0.9}{3.6} = \frac{x}{x + 4.8} $$
In the diagram, $\triangle CDE$ and $\triangle CBA$ are similar because $\angle C$ is common and $\angle CED = \angle CBA = 90^\circ$. Therefore, the ratio of corresponding sides is equal: $$ \frac{DE}{AB} = \frac{CE}{CB} $$ Let the length of the shadow CE be $x$. Then $CB = CE + EB = x + 4.8$.
$$ \frac{0.9}{3.6} = \frac{x}{x + 4.8} $$
Step 4: Solve for x.
$$ \frac{1}{4} = \frac{x}{x + 4.8} $$ $1 \times (x + 4.8) = 4 \times x$
$x + 4.8 = 4x$
$4.8 = 3x$
$x = \frac{4.8}{3} = 1.6 \text{ m}$.
$$ \frac{1}{4} = \frac{x}{x + 4.8} $$ $1 \times (x + 4.8) = 4 \times x$
$x + 4.8 = 4x$
$4.8 = 3x$
$x = \frac{4.8}{3} = 1.6 \text{ m}$.
The length of the shadow is 1.6 m.
Solution 9
Step 1: Rearrange the terms for simplification.
$$ \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4} = \left(\frac{4 \times 6}{2 \times 20}\right) \times \left(\frac{x^2 \cdot x}{1}\right) \times \left(\frac{y}{y^4}\right) \times \left(\frac{z^3}{z^2}\right) $$
$$ \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4} = \left(\frac{4 \times 6}{2 \times 20}\right) \times \left(\frac{x^2 \cdot x}{1}\right) \times \left(\frac{y}{y^4}\right) \times \left(\frac{z^3}{z^2}\right) $$
Step 2: Simplify each part.
Coefficients: $\frac{24}{40} = \frac{3 \times 8}{5 \times 8} = \frac{3}{5}$
x-terms: $x^2 \cdot x = x^{2+1} = x^3$
y-terms: $\frac{y}{y^4} = y^{1-4} = y^{-3} = \frac{1}{y^3}$
z-terms: $\frac{z^3}{z^2} = z^{3-2} = z^1 = z$
Coefficients: $\frac{24}{40} = \frac{3 \times 8}{5 \times 8} = \frac{3}{5}$
x-terms: $x^2 \cdot x = x^{2+1} = x^3$
y-terms: $\frac{y}{y^4} = y^{1-4} = y^{-3} = \frac{1}{y^3}$
z-terms: $\frac{z^3}{z^2} = z^{3-2} = z^1 = z$
Step 3: Combine the simplified parts.
$$ \frac{3}{5} \times x^3 \times \frac{1}{y^3} \times z = \frac{3x^3z}{5y^3} $$
$$ \frac{3}{5} \times x^3 \times \frac{1}{y^3} \times z = \frac{3x^3z}{5y^3} $$
$\displaystyle \frac{3x^3z}{5y^3}$
Solution 10
Step 1: Recall the relationship between LCM, GCD, and polynomials.
For two polynomials $p(x)$ and $q(x)$, the relationship is: $$ \text{LCM}[p(x), q(x)] \times \text{GCD}[p(x), q(x)] = p(x) \times q(x) $$
For two polynomials $p(x)$ and $q(x)$, the relationship is: $$ \text{LCM}[p(x), q(x)] \times \text{GCD}[p(x), q(x)] = p(x) \times q(x) $$
Step 2: Factorize the given polynomials.
Let $p(a) = a^2 + 4a - 12$. We need two numbers that multiply to -12 and add to 4. These are 6 and -2. So, $p(a) = (a+6)(a-2)$.
Let $q(a) = a^2 - 5a + 6$. We need two numbers that multiply to 6 and add to -5. These are -2 and -3. So, $q(a) = (a-2)(a-3)$.
Let $p(a) = a^2 + 4a - 12$. We need two numbers that multiply to -12 and add to 4. These are 6 and -2. So, $p(a) = (a+6)(a-2)$.
Let $q(a) = a^2 - 5a + 6$. We need two numbers that multiply to 6 and add to -5. These are -2 and -3. So, $q(a) = (a-2)(a-3)$.
Step 3: Calculate the LCM.
Given GCD = $a-2$. $$ \text{LCM} = \frac{p(a) \times q(a)}{\text{GCD}} = \frac{(a+6)(a-2) \times (a-2)(a-3)}{a-2} $$ Canceling one $(a-2)$ term: $$ \text{LCM} = (a+6)(a-2)(a-3) $$
Given GCD = $a-2$. $$ \text{LCM} = \frac{p(a) \times q(a)}{\text{GCD}} = \frac{(a+6)(a-2) \times (a-2)(a-3)}{a-2} $$ Canceling one $(a-2)$ term: $$ \text{LCM} = (a+6)(a-2)(a-3) $$
LCM = $(a+6)(a-2)(a-3)$
Solution 11
i) An arrow diagram:
Draw two ovals. The first (Domain) contains {1, 2, 3, 4, 5}. The second (Codomain) contains {2, 3, 4}. Draw arrows from the Domain to the Codomain:
Draw two ovals. The first (Domain) contains {1, 2, 3, 4, 5}. The second (Codomain) contains {2, 3, 4}. Draw arrows from the Domain to the Codomain:
- 1 → 2
- 2 → 2
- 3 → 2
- 4 → 3
- 5 → 4
ii) A table form:
| x | f(x) |
|---|---|
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
iii) A graph:
Plot the following points on a Cartesian coordinate plane: (1, 2), (2, 2), (3, 2), (4, 3), and (5, 4).
Plot the following points on a Cartesian coordinate plane: (1, 2), (2, 2), (3, 2), (4, 3), and (5, 4).
Part - III Solutions
4 x 5 = 20Solution 12
Step 1: List the elements of the sets A, B, and C.
$A = \{x \in W \mid x < 2\} = \{0, 1\}$ (W = Whole Numbers starting from 0)
$B = \{x \in N \mid 1 < x \le 4\} = \{2, 3, 4\}$ (N = Natural Numbers starting from 1)
$C = \{3, 5\}$
$A = \{x \in W \mid x < 2\} = \{0, 1\}$ (W = Whole Numbers starting from 0)
$B = \{x \in N \mid 1 < x \le 4\} = \{2, 3, 4\}$ (N = Natural Numbers starting from 1)
$C = \{3, 5\}$
Step 2: Calculate the Left Hand Side (LHS): $A \times (B \cup C)$
First, find $B \cup C$:
$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$
Next, find the Cartesian product $A \times (B \cup C)$:
$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (1)
First, find $B \cup C$:
$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$
Next, find the Cartesian product $A \times (B \cup C)$:
$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (1)
Step 3: Calculate the Right Hand Side (RHS): $(A \times B) \cup (A \times C)$
First, find $A \times B$:
$A \times B = \{0, 1\} \times \{2, 3, 4\} = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\}$
Next, find $A \times C$:
$A \times C = \{0, 1\} \times \{3, 5\} = \{(0,3), (0,5), (1,3), (1,5)\}$
Finally, find the union $(A \times B) \cup (A \times C)$:
$= \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4), (0,5), (1,5)\}$
Arranging in order: $= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (2)
First, find $A \times B$:
$A \times B = \{0, 1\} \times \{2, 3, 4\} = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\}$
Next, find $A \times C$:
$A \times C = \{0, 1\} \times \{3, 5\} = \{(0,3), (0,5), (1,3), (1,5)\}$
Finally, find the union $(A \times B) \cup (A \times C)$:
$= \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4), (0,5), (1,5)\}$
Arranging in order: $= \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$ --- (2)
Step 4: Verify LHS = RHS.
Comparing equations (1) and (2), we see that LHS = RHS.
Comparing equations (1) and (2), we see that LHS = RHS.
Hence, $A \times (B \cup C) = (A \times B) \cup (A \times C)$ is verified.
Solution 13
Step 1: Write the sum and factor out the common term.
$S_n = 6 + 66 + 666 + \dots$ to n terms.
$S_n = 6(1 + 11 + 111 + \dots)$
$S_n = 6 + 66 + 666 + \dots$ to n terms.
$S_n = 6(1 + 11 + 111 + \dots)$
Step 2: Multiply and divide by 9.
$S_n = \frac{6}{9}(9 + 99 + 999 + \dots) = \frac{2}{3}(9 + 99 + 999 + \dots)$
$S_n = \frac{6}{9}(9 + 99 + 999 + \dots) = \frac{2}{3}(9 + 99 + 999 + \dots)$
Step 3: Rewrite the terms in the bracket.
$S_n = \frac{2}{3}((10-1) + (10^2-1) + (10^3-1) + \dots \text{ to n terms})$
$S_n = \frac{2}{3}((10-1) + (10^2-1) + (10^3-1) + \dots \text{ to n terms})$
Step 4: Separate the terms into two series.
$S_n = \frac{2}{3} \left[ (10 + 10^2 + 10^3 + \dots) - (1 + 1 + 1 + \dots) \right]$
The first part is a Geometric Progression (GP) with first term $a=10$ and common ratio $r=10$.
The second part is the sum of 'n' ones, which is just $n$.
$S_n = \frac{2}{3} \left[ (10 + 10^2 + 10^3 + \dots) - (1 + 1 + 1 + \dots) \right]$
The first part is a Geometric Progression (GP) with first term $a=10$ and common ratio $r=10$.
The second part is the sum of 'n' ones, which is just $n$.
Step 5: Apply the sum formula for a GP.
The sum of the first n terms of a GP is $S_{GP} = \frac{a(r^n-1)}{r-1}$.
$S_{GP} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.
Substituting back: $S_n = \frac{2}{3} \left[ \frac{10}{9}(10^n-1) - n \right]$.
The sum of the first n terms of a GP is $S_{GP} = \frac{a(r^n-1)}{r-1}$.
$S_{GP} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.
Substituting back: $S_n = \frac{2}{3} \left[ \frac{10}{9}(10^n-1) - n \right]$.
$S_n = \frac{20}{27}(10^n - 1) - \frac{2n}{3}$
Solution 14
Step 1: Set up the equation from the given ratio.
The nth term of an A.P. is $t_n = a + (n-1)d$. Given $\frac{t_6}{t_8} = \frac{7}{9}$.
$$ \frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9} $$
The nth term of an A.P. is $t_n = a + (n-1)d$. Given $\frac{t_6}{t_8} = \frac{7}{9}$.
$$ \frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9} $$
Step 2: Solve for 'a' in terms of 'd'.
Cross-multiply: $9(a+5d) = 7(a+7d)$
$9a + 45d = 7a + 49d$
$9a - 7a = 49d - 45d$
$2a = 4d \implies a = 2d$.
Cross-multiply: $9(a+5d) = 7(a+7d)$
$9a + 45d = 7a + 49d$
$9a - 7a = 49d - 45d$
$2a = 4d \implies a = 2d$.
Step 3: Find the required ratio.
We need to find the ratio of the 9th term to the 13th term, which is $\frac{t_9}{t_{13}}$. $$ \frac{t_9}{t_{13}} = \frac{a+8d}{a+12d} $$
We need to find the ratio of the 9th term to the 13th term, which is $\frac{t_9}{t_{13}}$. $$ \frac{t_9}{t_{13}} = \frac{a+8d}{a+12d} $$
Step 4: Substitute $a=2d$ into the ratio.
$$ \frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{10}{14} = \frac{5}{7} $$
$$ \frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{10}{14} = \frac{5}{7} $$
The ratio of the 9th term to the 13th term is 5:7.
Solution 15
Step 1: Factorize the denominators.
$x^2 - 5x + 6 = (x-2)(x-3)$
$x^2 - 3x + 2 = (x-1)(x-2)$
$x^2 - 8x + 15 = (x-3)(x-5)$
$x^2 - 5x + 6 = (x-2)(x-3)$
$x^2 - 3x + 2 = (x-1)(x-2)$
$x^2 - 8x + 15 = (x-3)(x-5)$
Step 2: Rewrite the expression with factored denominators.
$$ \frac{1}{(x-2)(x-3)} + \frac{1}{(x-1)(x-2)} - \frac{1}{(x-3)(x-5)} $$
$$ \frac{1}{(x-2)(x-3)} + \frac{1}{(x-1)(x-2)} - \frac{1}{(x-3)(x-5)} $$
Step 3: Find the Least Common Multiple (LCM) and combine the fractions.
The LCM of the denominators is $(x-1)(x-2)(x-3)(x-5)$.
$$ \frac{(x-1)(x-5) + (x-3)(x-5) - (x-1)(x-2)}{(x-1)(x-2)(x-3)(x-5)} $$
The LCM of the denominators is $(x-1)(x-2)(x-3)(x-5)$.
$$ \frac{(x-1)(x-5) + (x-3)(x-5) - (x-1)(x-2)}{(x-1)(x-2)(x-3)(x-5)} $$
Step 4: Expand and simplify the numerator.
Numerator $= (x^2-6x+5) + (x^2-8x+15) - (x^2-3x+2)$
$= x^2 - 6x + 5 + x^2 - 8x + 15 - x^2 + 3x - 2$
$= (x^2+x^2-x^2) + (-6x-8x+3x) + (5+15-2)$
$= x^2 - 11x + 18$
Numerator $= (x^2-6x+5) + (x^2-8x+15) - (x^2-3x+2)$
$= x^2 - 6x + 5 + x^2 - 8x + 15 - x^2 + 3x - 2$
$= (x^2+x^2-x^2) + (-6x-8x+3x) + (5+15-2)$
$= x^2 - 11x + 18$
Step 5: Factor the simplified numerator and simplify the final expression.
$x^2 - 11x + 18 = (x-2)(x-9)$.
The expression becomes: $$ \frac{(x-2)(x-9)}{(x-1)(x-2)(x-3)(x-5)} $$ Cancel the common factor $(x-2)$.
$x^2 - 11x + 18 = (x-2)(x-9)$.
The expression becomes: $$ \frac{(x-2)(x-9)}{(x-1)(x-2)(x-3)(x-5)} $$ Cancel the common factor $(x-2)$.
$\displaystyle \frac{x-9}{(x-1)(x-3)(x-5)}$
Solution 16
Step 1: Factor out the common numerical coefficients.
Let $p(x) = 6x^3 - 30x^2 + 60x - 48 = 6(x^3 - 5x^2 + 10x - 8)$.
Let $q(x) = 3x^3 - 12x^2 + 21x - 18 = 3(x^3 - 4x^2 + 7x - 6)$.
The GCD of the coefficients 6 and 3 is 3.
Let $p(x) = 6x^3 - 30x^2 + 60x - 48 = 6(x^3 - 5x^2 + 10x - 8)$.
Let $q(x) = 3x^3 - 12x^2 + 21x - 18 = 3(x^3 - 4x^2 + 7x - 6)$.
The GCD of the coefficients 6 and 3 is 3.
Step 2: Apply the division algorithm to the polynomial parts.
Let $f(x) = x^3 - 5x^2 + 10x - 8$ and $g(x) = x^3 - 4x^2 + 7x - 6$. Divide $f(x)$ by $g(x)$.
Let $f(x) = x^3 - 5x^2 + 10x - 8$ and $g(x) = x^3 - 4x^2 + 7x - 6$. Divide $f(x)$ by $g(x)$.
1
x³-4x²+7x-6 | x³-5x²+10x-8
-(x³-4x²+7x-6)
----------------
-x² + 3x - 2
The remainder is $-x^2+3x-2$. We can factor out -1 to get $x^2-3x+2$.
Step 3: Divide the previous divisor by the new remainder.
Now, divide $g(x) = x^3 - 4x^2 + 7x - 6$ by $x^2-3x+2$.
Now, divide $g(x) = x^3 - 4x^2 + 7x - 6$ by $x^2-3x+2$.
x - 1
x²-3x+2 | x³-4x²+7x-6
-(x³-3x²+2x)
-------------
-x²+5x-6
-(-x²+3x-2)
----------
2x-4
The remainder is $2x-4 = 2(x-2)$.
Step 4: Divide the previous divisor by the new remainder.
Now, divide $x^2-3x+2$ by $x-2$. Since $x^2-3x+2 = (x-1)(x-2)$, the remainder is 0.
Now, divide $x^2-3x+2$ by $x-2$. Since $x^2-3x+2 = (x-1)(x-2)$, the remainder is 0.
Step 5: Combine the results.
The last non-zero remainder (ignoring constant factors) is $x-2$. This is the GCD of the polynomial parts. The final GCD is the product of the numerical GCD and the polynomial GCD. GCD = $3 \times (x-2) = 3x-6$.
The last non-zero remainder (ignoring constant factors) is $x-2$. This is the GCD of the polynomial parts. The final GCD is the product of the numerical GCD and the polynomial GCD. GCD = $3 \times (x-2) = 3x-6$.
The GCD is $3x-6$.
Solution 17
i) Check if this relation is a function.
Yes, the relation is a function. For each unique input value 'x' (length of forehand), there is exactly one unique output value 'y' (height).
Yes, the relation is a function. For each unique input value 'x' (length of forehand), there is exactly one unique output value 'y' (height).
ii) Find a and b.
The relationship is given by the linear equation $y = ax + b$. Let's use the first two data points: (35, 56) and (45, 65). For (35, 56): $56 = 35a + b$ --- (1)
For (45, 65): $65 = 45a + b$ --- (2)
Subtract equation (1) from (2):
$(65 - 56) = (45a - 35a) + (b - b)$
$9 = 10a \implies a = 0.9$.
Substitute $a=0.9$ into equation (1):
$56 = 35(0.9) + b \implies 56 = 31.5 + b \implies b = 56 - 31.5 = 24.5$.
So, the relation is $y = 0.9x + 24.5$.
The relationship is given by the linear equation $y = ax + b$. Let's use the first two data points: (35, 56) and (45, 65). For (35, 56): $56 = 35a + b$ --- (1)
For (45, 65): $65 = 45a + b$ --- (2)
Subtract equation (1) from (2):
$(65 - 56) = (45a - 35a) + (b - b)$
$9 = 10a \implies a = 0.9$.
Substitute $a=0.9$ into equation (1):
$56 = 35(0.9) + b \implies 56 = 31.5 + b \implies b = 56 - 31.5 = 24.5$.
So, the relation is $y = 0.9x + 24.5$.
iii) Find the height of a person whose forehand length is 40 cm.
Here, $x=40$. We need to find $y$.
$y = 0.9(40) + 24.5 = 36 + 24.5 = 60.5$.
The height is 60.5 inches.
Here, $x=40$. We need to find $y$.
$y = 0.9(40) + 24.5 = 36 + 24.5 = 60.5$.
The height is 60.5 inches.
iv) Find the length of the forehand of a person if the height is 53.3 inches.
Here, $y=53.3$. We need to find $x$.
$53.3 = 0.9x + 24.5$
$53.3 - 24.5 = 0.9x$
$28.8 = 0.9x$
$x = \frac{28.8}{0.9} = \frac{288}{9} = 32$.
The forehand length is 32 cm.
Here, $y=53.3$. We need to find $x$.
$53.3 = 0.9x + 24.5$
$53.3 - 24.5 = 0.9x$
$28.8 = 0.9x$
$x = \frac{28.8}{0.9} = \frac{288}{9} = 32$.
The forehand length is 32 cm.
i) Yes, it's a function.
ii) a = 0.9, b = 24.5
iii) 60.5 inches
iv) 32 cm
ii) a = 0.9, b = 24.5
iii) 60.5 inches
iv) 32 cm
Solution 18
i) Find the least positive value of x such that $67 + x \equiv 1 \pmod 4$.
The congruence means that $67+x-1$ is a multiple of 4.
$66 + x$ is a multiple of 4.
We find the remainder of 66 when divided by 4: $66 = 4 \times 16 + 2$. So, $66 \equiv 2 \pmod 4$.
The congruence becomes $2 + x \equiv 0 \pmod 4$.
This means $2+x$ must be a multiple of 4. The smallest positive multiple of 4 is 4 itself.
$2 + x = 4 \implies x = 2$.
The least positive value of x is 2.
The congruence means that $67+x-1$ is a multiple of 4.
$66 + x$ is a multiple of 4.
We find the remainder of 66 when divided by 4: $66 = 4 \times 16 + 2$. So, $66 \equiv 2 \pmod 4$.
The congruence becomes $2 + x \equiv 0 \pmod 4$.
This means $2+x$ must be a multiple of 4. The smallest positive multiple of 4 is 4 itself.
$2 + x = 4 \implies x = 2$.
The least positive value of x is 2.
ii) Solve: $5x \equiv 4 \pmod 6$.
This means $5x-4$ is a multiple of 6, or $5x-4 = 6k$ for some integer k. This is equivalent to $5x = 6k+4$. We can find a solution by testing integer values for x.
This means $5x-4$ is a multiple of 6, or $5x-4 = 6k$ for some integer k. This is equivalent to $5x = 6k+4$. We can find a solution by testing integer values for x.
- If $x=1$, $5(1) = 5$. $5 \equiv 5 \pmod 6$. Not a solution.
- If $x=2$, $5(2) = 10$. $10 = 1 \times 6 + 4$, so $10 \equiv 4 \pmod 6$. This is a solution.
i) x = 2
ii) x = 2 (or any integer of the form $2+6k$)
ii) x = 2 (or any integer of the form $2+6k$)
Part - IV Solutions
2 x 8 = 16Solution 19
a) Construction for scale factor $\frac{7}{3} > 1$
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 $\frac{7}{3}$) $Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7$ on QX so that $QQ_1 = Q_1Q_2 = \dots = Q_6Q_7$.
- Join $Q_3$ (the 3rd point, corresponding to the denominator) to R.
- Draw a line through $Q_7$ (the 7th point, corresponding to the numerator) parallel to $Q_3R$. This line will intersect the extended line segment QR at a point R'. (To draw the parallel line, construct an angle at $Q_7$ equal to $\angle QQ_3R$).
- Draw a line through R' parallel to RP. This line will intersect the extended line segment QP at a point P'.
- The triangle $\triangle P'QR'$ is the required triangle, which is similar to $\triangle PQR$ and whose sides are $\frac{7}{3}$ times the corresponding sides of $\triangle PQR$.
b) Construction for scale factor $\frac{3}{5} < 1$
Steps of Construction:
- Draw any triangle ABC.
- Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
- Locate 5 points (the greater of 3 and 5 in $\frac{3}{5}$) $B_1, B_2, B_3, B_4, B_5$ on BX so that $BB_1 = B_1B_2 = \dots = B_4B_5$.
- Join $B_5$ (the 5th point, corresponding to the denominator) to C.
- Draw a line through $B_3$ (the 3rd point, corresponding to the numerator) parallel to $B_5C$. This line will intersect BC at a point C'. (To draw the parallel line, construct an angle at $B_3$ equal to $\angle BB_5C$).
- Draw a line through C' parallel to CA. This line will intersect BA at a point A'.
- The triangle $\triangle A'BC'$ is the required triangle, which is similar to $\triangle ABC$ and whose sides are $\frac{3}{5}$ times the corresponding sides of $\triangle ABC$.
Solution 20
a) Two-wheeler parking zone
Check for Variation: We check if the ratio $\frac{y}{x}$ (for direct variation) or the product $xy$ (for inverse variation) is constant.
- $\frac{60}{4} = 15$
- $\frac{120}{8} = 15$
- $\frac{180}{12} = 15$
- $\frac{360}{24} = 15$
Since the ratio $\frac{y}{x} = 15$ is constant, the amount charged is in direct variation to the parking time. The constant of variation is $k=15$, and the equation is $y=15x$.
Graph: Plot the points (4, 60), (8, 120), (12, 180), and (24, 360). Draw a straight line passing through these points and the origin (0,0). The x-axis represents Time (in hours) and the y-axis represents Amount (in ₹).
i) Amount for 6 hours:
Using the equation $y=15x$:
$y = 15(6) = 90$.
The amount to be paid is ₹90.
ii) Parking duration for ₹150:
Using the equation $y=15x$:
$150 = 15x \implies x = \frac{150}{15} = 10$.
The parking duration is 10 hours.
b) Graph $y = \frac{1}{2}x$
The function $y = \frac{1}{2}x$ is in the form $y=kx$. This is a direct variation.
Constant of variation: $k = \frac{1}{2}$.
Table of Values:
| x | 0 | 2 | 4 | 6 | 9 |
|---|---|---|---|---|---|
| y = (1/2)x | 0 | 1 | 2 | 3 | 4.5 |
Graph: Plot the points (0,0), (2,1), (4,2), (6,3), etc., on a graph. Draw a line through them. The graph is a straight line passing through the origin, which verifies it is a direct variation.
(i) Find y when x = 9:
$y = \frac{1}{2}(9) = 4.5$.
(ii) Find x when y = 7.5:
$7.5 = \frac{1}{2}x \implies x = 7.5 \times 2 = 15$.