FIRST MID TERM TEST - 2024
Standard X
MATHEMATICS
Question Paper
Part - I
(4 x 1 = 4)
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I.Choose the correct answer:
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1.The range of the relation $R = \{(x, x^2) \mid x \text{ is a prime number less than 13}\}$ is
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2.The value of $(1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15)$ is
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3.The solution of the system $3z = 9, -7y + 7z = 7, x + y - 3z = -6$ is
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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
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Part - II
(5 x 2 = 10)
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II.Answer any 5 questions. (Q.No.11 is compulsory)
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5.A function $f: [-5, 9] \to R$ 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 $2f(4) + f(8)$.
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6.If $13824 = 2^a \times 3^b$, then find a and b.
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7.Use Euclid's Division Algorithm to find the Highest Common Factor (HCF) of 396, 504, 636.
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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.
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9.Simplify: $\displaystyle \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4}$
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10.Find the LCM of the polynomials $a^2 + 4a - 12$ and $a^2 - 5a + 6$ whose GCD is $a - 2$.
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11.Represent the function $f = \{(1,2), (2,2), (3,2), (4,3), (5,4)\}$ through:
- an arrow diagram
- a table form
- a graph
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Part - III
(4 x 5 = 20)
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III.Answer any 4 questions. (Q.No.17 is compulsory)
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12.Given $A = \{x \in W \mid x < 2\}$, $B = \{x \in N \mid 1 < x \le 4\}$ and $C = \{3,5\}$, verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
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13.Find the sum to n terms of the series $6 + 66 + 666 + \dots$
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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.
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15.Simplify: $\displaystyle \frac{1}{x^2-5x+6} + \frac{1}{x^2-3x+2} - \frac{1}{x^2-8x+15}$
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16.Find the GCD of $6x^3 - 30x^2 + 60x – 48$ and $3x^3 - 12x^2 + 21x-18$.
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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 and b are constants.
Length 'x' of forehand (in cm) Height 'y' (in inches) 35 56 45 65 50 69.5 55 74 - Check if this relation is a function.
- Find a and b.
- Find the height of a person whose forehand length is 40 cm.
- Find the length of the forehand of a person if the height is 53.3 inches.
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18.
- Find the least positive value of x such that $67 + x \equiv 1 \pmod{4}$.
- Solve: $5x \equiv 4 \pmod{6}$.
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Part - IV
(2 x 8 = 16)
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IV.Answer the following questions.
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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 $\displaystyle \frac{7}{3} > 1$)
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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 $\displaystyle \frac{3}{5} < 1$)
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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 with the parking time. Graph the data. Also,Time (in hours) (x) Amount ₹ (y) 4 60 8 120 12 180 24 360 - Find the amount to be paid when parking time is 6 hr.
- Find the parking duration when the amount paid is ₹150.
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b)Graph the following linear function $\displaystyle y = \frac{1}{2}x$. Identify the constant of variation and verify it with the graph. Also
- find y when x = 9
- find x when y = 7.5
(OR)(OR) -
Solutions
Part - I Solutions
Solution for Question 1
1. The range of the relation $R = \{(x, x^2) \mid x \text{ is a prime number less than 13}\}$ is
The domain consists of all prime numbers less than 13. The prime numbers are numbers greater than 1 that have only two factors: 1 and themselves. The prime numbers less than 13 are: 2, 3, 5, 7, 11.
The relation is defined by $R = \{(x, x^2)\}$. We need to find the square of each prime number in the domain.
- For $x = 2$, $y = 2^2 = 4$
- For $x = 3$, $y = 3^2 = 9$
- For $x = 5$, $y = 5^2 = 25$
- For $x = 7$, $y = 7^2 = 49$
- For $x = 11$, $y = 11^2 = 121$
The range is the set of all the second elements (y-values) of the ordered pairs. Range = $\{4, 9, 25, 49, 121\}$.
Correct answer is c) $\{4,9,25,49,121\}$
Solution for Question 2
2. The value of $(1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15)$ is
The formula for the sum of the cubes of the first n natural numbers is $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$. Here, $n=15$. $$ \sum_{k=1}^{15} k^3 = \left[\frac{15(15+1)}{2}\right]^2 = \left[\frac{15 \times 16}{2}\right]^2 = (15 \times 8)^2 = 120^2 = 14400 $$
The formula for the sum of the first n natural numbers is $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. Here, $n=15$. $$ \sum_{k=1}^{15} k = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120 $$
Subtract the sum of the numbers from the sum of their cubes. $$ 14400 - 120 = 14280 $$
Correct answer is d) 14280
Solution for Question 3
3. The solution of the system $3z = 9, -7y + 7z = 7, x + y - 3z = -6$ is
Given the equation $3z = 9$. $$ z = \frac{9}{3} = 3 $$
The second equation is $-7y + 7z = 7$. Substitute $z=3$. $$ -7y + 7(3) = 7 $$ $$ -7y + 21 = 7 $$ $$ -7y = 7 - 21 $$ $$ -7y = -14 $$ $$ y = \frac{-14}{-7} = 2 $$
The third equation is $x + y - 3z = -6$. Substitute $y=2$ and $z=3$. $$ x + (2) - 3(3) = -6 $$ $$ x + 2 - 9 = -6 $$ $$ x - 7 = -6 $$ $$ x = -6 + 7 = 1 $$
The solution is $(x, y, z) = (1, 2, 3)$.
Correct answer is b) $x = 1, y = 2, z = 3$
Solution for Question 4
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
The sum of angles in any triangle is $180^\circ$. $$ \angle L + \angle M + \angle N = 180^\circ $$ $$ 60^\circ + 50^\circ + \angle N = 180^\circ $$ $$ 110^\circ + \angle N = 180^\circ $$ $$ \angle N = 180^\circ - 110^\circ = 70^\circ $$
When two triangles are similar, their corresponding angles are equal. The similarity is given as $\triangle LMN \sim \triangle PQR$. This means: $$ \angle P = \angle L = 60^\circ $$ $$ \angle Q = \angle M = 50^\circ $$ $$ \angle R = \angle N $$
From Step 1 and 2, we have $\angle R = \angle N = 70^\circ$.
Correct answer is b) $70^\circ$
Part - II Solutions
Solution for Question 5
5. Find $2f(4) + f(8)$ for the given piecewise function.
The function is defined as: $$ f(x) = \begin{cases} 6x+1; & -5 \le x < 2 \\ 5x^2-1; & 2 \le x < 6 \\ 3x-4; & 6 \le x \le 9 \end{cases} $$
The value $x=4$ falls in the interval $2 \le x < 6$. So, we use the rule $f(x) = 5x^2 - 1$. $$ f(4) = 5(4)^2 - 1 = 5(16) - 1 = 80 - 1 = 79 $$
The value $x=8$ falls in the interval $6 \le x \le 9$. So, we use the rule $f(x) = 3x - 4$. $$ f(8) = 3(8) - 4 = 24 - 4 = 20 $$
Substitute the values found in the previous steps. $$ 2f(4) + f(8) = 2(79) + 20 = 158 + 20 = 178 $$
The value of $2f(4) + f(8)$ is 178.
Solution for Question 6
6. If $13824 = 2^a \times 3^b$, then find a and b.
We need to find the prime factorization of 13824.
We can do this by repeatedly dividing by the smallest prime factors (2 and 3).
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
From the factorization, we count the number of 2s and 3s. There are nine 2s and three 3s.
So, $13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^9 \times 3^3$.
We are given $13824 = 2^a \times 3^b$. Comparing this with our result $13824 = 2^9 \times 3^3$, we can see that:
$a = 9$ and $b = 3$.
$a = 9, b = 3$.
Solution for Question 7
7. Use Euclid's Division Algorithm to find the HCF of 396, 504, 636.
According to Euclid's division lemma, $a = bq + r$, where $0 \le r < b$. We apply this repeatedly.
- $636 = 1 \times 504 + 132$
- $504 = 3 \times 132 + 108$
- $132 = 1 \times 108 + 24$
- $108 = 4 \times 24 + 12$
- $24 = 2 \times 12 + 0$
The last non-zero remainder is the HCF. So, HCF(636, 504) = 12.
Now we find HCF(12, 396).
- $396 = 33 \times 12 + 0$
The remainder is 0 in the first step. Therefore, the HCF is the divisor, 12.
The HCF of 396, 504, and 636 is the HCF of HCF(636, 504) and 396.
The HCF of 396, 504, and 636 is 12.
Solution for Question 8
8. Shadow problem with a lamppost and a walking boy.
- Lamppost height = 3.6 m
- Boy's height = 90 cm = 0.9 m
- Boy's speed = 1.2 m/sec
- Time = 4 seconds
Distance walked by the boy from the base of the lamppost = Speed × Time $$ \text{Distance} = 1.2 \, \text{m/s} \times 4 \, \text{s} = 4.8 \, \text{m} $$
Let AB be the lamppost, and DE be the boy. Let C be the tip of the shadow. The ground is represented by the line BC.
We have two right-angled triangles: $\triangle ABC$ and $\triangle DEC$.
$$ AB = 3.6 \, \text{m} $$
$$ DE = 0.9 \, \text{m} $$
$$ BE = 4.8 \, \text{m} $$
Let the length of the shadow, EC, be $x$. Then $BC = BE + EC = 4.8 + x$.
Since the lamppost and the boy are both vertical to the ground, $AB \parallel DE$.
Therefore, $\triangle ABC \sim \triangle DEC$ by AA similarity ($\angle C$ is common and $\angle B = \angle E = 90^\circ$).
The ratio of corresponding sides is equal. $$ \frac{AB}{DE} = \frac{BC}{EC} $$ $$ \frac{3.6}{0.9} = \frac{4.8 + x}{x} $$ $$ 4 = \frac{4.8 + x}{x} $$ $$ 4x = 4.8 + x $$ $$ 4x - x = 4.8 $$ $$ 3x = 4.8 $$ $$ x = \frac{4.8}{3} = 1.6 \, \text{m} $$
The length of the shadow is 1.6 meters.
Solution for Question 9
9. Simplify: $\displaystyle \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4}$
- Constants: $ \displaystyle \frac{24}{40} = \frac{3 \times 8}{5 \times 8} = \frac{3}{5} $
- x terms: $ \displaystyle x^2 \times x = x^{2+1} = x^3 $
- y terms: $ \displaystyle \frac{y}{y^4} = y^{1-4} = y^{-3} = \frac{1}{y^3} $
- z terms: $ \displaystyle \frac{z^3}{z^2} = z^{3-2} = z^1 = z $
The simplified expression is $\displaystyle \frac{3x^3z}{5y^3}$.
Solution for Question 10
10. Find the LCM of $p(a) = a^2 + 4a - 12$ and $q(a) = a^2 - 5a + 6$ whose GCD is $a - 2$.
For any two polynomials $p(x)$ and $q(x)$, the relationship is: $$ \text{LCM}[p(x), q(x)] \times \text{GCD}[p(x), q(x)] = p(x) \times q(x) $$ Therefore, $$ \text{LCM} = \frac{p(a) \times q(a)}{\text{GCD}} $$
This is a good way to verify the given GCD and easily find the LCM.
$p(a) = a^2 + 4a - 12$. We need two numbers that multiply to -12 and add to 4. These are +6 and -2. $$ p(a) = (a+6)(a-2) $$
$q(a) = a^2 - 5a + 6$. We need two numbers that multiply to 6 and add to -5. These are -2 and -3. $$ q(a) = (a-2)(a-3) $$
The GCD is the common factor, which is indeed $(a-2)$. The LCM is the product of the GCD and all other remaining factors. $$ \text{LCM} = \text{GCD} \times (\text{non-common factors}) $$ $$ \text{LCM} = (a-2) \times (a+6) \times (a-3) $$
Alternatively, using the formula from Step 1: $$ \text{LCM} = \frac{(a+6)(a-2) \times (a-2)(a-3)}{a-2} = (a+6)(a-2)(a-3) $$
The LCM is $(a-2)(a-3)(a+6)$.
Solution for Question 11
11. Represent the function $f = \{(1,2), (2,2), (3,2), (4,3), (5,4)\}$ in different forms.
An arrow diagram shows the mapping from the domain to the co-domain. The domain is $A = \{1, 2, 3, 4, 5\}$ and the range is $B = \{2, 3, 4\}$.
The function can be represented as a table with inputs (x) in the first column and outputs (f(x)) in the second column.
| x | f(x) |
|---|---|
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
The function is represented by plotting the ordered pairs as points on a Cartesian coordinate plane.
Part - III Solutions
Solution for Question 12
12. Verify $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
$A = \{x \in W \mid x < 2\}$. W represents whole numbers, so $A = \{0, 1\}$.
$B = \{x \in N \mid 1 < x \le 4\}$. N represents natural numbers, so $B = \{2, 3, 4\}$.
$C = \{3, 5\}$.
First, find $B \cup C$: $$ B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\} $$ Now, find the Cartesian product $A \times (B \cup C)$: $$ A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\} $$ $$ = \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\} $$
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 $A \times C$: $$ A \times C = \{0, 1\} \times \{3, 5\} = \{(0,3), (0,5), (1,3), (1,5)\} $$ Now, find the union of these two sets: $$ (A \times B) \cup (A \times C) = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\} \cup \{(0,3), (0,5), (1,3), (1,5)\} $$ $$ = \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\} $$
The set of ordered pairs calculated for the LHS is identical to the set calculated for the RHS.
Since LHS = RHS, the property $A \times (B \cup C) = (A \times B) \cup (A \times C)$ is verified.
Solution for Question 13
13. Find the sum to n terms of the series $6 + 66 + 666 + \dots$
The first part, $(10 + 10^2 + \dots + 10^n)$, is a Geometric Progression (GP) with first term $a=10$, common ratio $r=10$. The sum of this GP is $S_{GP} = a\frac{r^n-1}{r-1} = 10 \frac{10^n-1}{10-1} = \frac{10}{9}(10^n-1)$.
The second part is the sum of '1' repeated n times, which is simply $n$.
The sum to n terms is $S_n = \frac{2}{3} \left[ \frac{10}{9}(10^n - 1) - n \right]$.
Solution for Question 14
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.
Let the first term of the A.P. be 'a' and the common difference be 'd'. The nth term is given by $t_n = a + (n-1)d$. We are given $\frac{t_6}{t_8} = \frac{7}{9}$. $$ \frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9} $$
Cross-multiply the equation: $$ 9(a+5d) = 7(a+7d) $$ $$ 9a + 45d = 7a + 49d $$ $$ 9a - 7a = 49d - 45d $$ $$ 2a = 4d \implies a = 2d $$
The ratio we need to find is: $$ \frac{t_9}{t_{13}} = \frac{a+(9-1)d}{a+(13-1)d} = \frac{a+8d}{a+12d} $$
Substitute $a=2d$ into the ratio: $$ \frac{t_9}{t_{13}} = \frac{(2d)+8d}{(2d)+12d} = \frac{10d}{14d} $$ Since $d \neq 0$ (for a non-constant A.P.), we can cancel 'd'. $$ \frac{10}{14} = \frac{5}{7} $$
The ratio of the 9th term to the 13th term is 5:7.
Solution for Question 15
15. Simplify: $\displaystyle \frac{1}{x^2-5x+6} + \frac{1}{x^2-3x+2} - \frac{1}{x^2-8x+15}$
- $x^2-5x+6 = (x-2)(x-3)$
- $x^2-3x+2 = (x-1)(x-2)$
- $x^2-8x+15 = (x-3)(x-5)$
The LCM is the product of each unique factor raised to its highest power. $$ \text{LCM} = (x-1)(x-2)(x-3)(x-5) $$
Numerator = $(x^2-6x+5) + (x^2-8x+15) - (x^2-3x+2)$
= $x^2-6x+5 + x^2-8x+15 - x^2+3x-2$
= $(x^2+x^2-x^2) + (-6x-8x+3x) + (5+15-2)$
= $x^2 - 11x + 18$
The numerator $x^2 - 11x + 18$ factors into $(x-2)(x-9)$. The full expression is now: $$ \frac{(x-2)(x-9)}{(x-1)(x-2)(x-3)(x-5)} $$ Cancel the common factor $(x-2)$: $$ \frac{x-9}{(x-1)(x-3)(x-5)} $$
The simplified expression is $\displaystyle \frac{x-9}{(x-1)(x-3)(x-5)}$.
Solution for Question 16
16. Find the GCD of $6x^3 - 30x^2 + 60x – 48$ and $3x^3 - 12x^2 + 21x-18$.
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 numerical coefficients 6 and 3 is 3.
Let $f(x) = x^3 - 5x^2 + 10x - 8$ and $g(x) = x^3 - 4x^2 + 7x - 6$. We will divide $f(x)$ by $g(x)$. $$ (x^3 - 5x^2 + 10x - 8) = 1 \cdot (x^3 - 4x^2 + 7x - 6) + (-x^2 + 3x - 2) $$ The remainder is $r_1(x) = -x^2 + 3x - 2 = -(x^2 - 3x + 2)$. We can ignore the negative sign for the next division step.
Now, we divide $g(x) = x^3 - 4x^2 + 7x - 6$ by $x^2 - 3x + 2$. Using polynomial long division:
x - 1
____________
x²-3x+2 | x³ - 4x² + 7x - 6
-(x³ - 3x² + 2x)
-----------------
-x² + 5x - 6
-(-x² + 3x - 2)
----------------
2x - 4
The remainder is $r_2(x) = 2x - 4 = 2(x-2)$.
Now, we divide the previous divisor $x^2 - 3x + 2$ by the polynomial part of the new remainder, which is $x-2$. $$ (x^2 - 3x + 2) \div (x-2) $$ We can factor $x^2-3x+2$ as $(x-1)(x-2)$. $$ \frac{(x-1)(x-2)}{x-2} = x-1 $$ The remainder is 0. The last non-zero remainder was $2(x-2)$, so the polynomial GCD is $(x-2)$.
The final GCD is the product of the GCD of the numerical coefficients and the GCD of the polynomial parts. $$ \text{GCD} = 3 \times (x-2) = 3(x-2) $$
The GCD is $3(x-2)$.
Solution for Question 17
17. Analysis of the relationship between forehand length (x) and height (y).
The given relationship is $y = ax + b$. The data points are (35, 56), (45, 65), (50, 69.5), (55, 74).
A relation is a function if each input (x-value) corresponds to exactly one output (y-value). In the given table, each forehand length (35, 45, 50, 55) is unique and is paired with only one height. Therefore, yes, this relation is a function.
We can create a system of two linear equations using two data points. Let's use (35, 56) and (45, 65).
Eq 1: $56 = a(35) + b \implies 35a + b = 56$
Eq 2: $65 = a(45) + b \implies 45a + b = 65$
Subtract Eq 1 from Eq 2:
$(45a + b) - (35a + b) = 65 - 56$
$10a = 9 \implies a = 0.9$
Substitute $a = 0.9$ into Eq 1:
$35(0.9) + b = 56 \implies 31.5 + b = 56 \implies b = 56 - 31.5 = 24.5$
So, the relationship is $y = 0.9x + 24.5$.
a = 0.9 and b = 24.5.
Here, $x=40$. We use the formula $y = 0.9x + 24.5$. $$ y = 0.9(40) + 24.5 = 36 + 24.5 = 60.5 $$
The height of the person is 60.5 inches.
Here, $y=53.3$. We solve for $x$ in the formula $y = 0.9x + 24.5$. $$ 53.3 = 0.9x + 24.5 $$ $$ 53.3 - 24.5 = 0.9x $$ $$ 28.8 = 0.9x $$ $$ x = \frac{28.8}{0.9} = \frac{288}{9} = 32 $$
The length of the forehand is 32 cm.
Solution for Question 18
18. Modular Arithmetic problems.
First, simplify 67 (mod 4).
$$ 67 \div 4 = 16 \text{ with a remainder of } 3. $$
So, $67 \equiv 3 \pmod{4}$.
The congruence becomes:
$$ 3 + x \equiv 1 \pmod{4} $$
Subtract 3 from both sides:
$$ x \equiv 1 - 3 \pmod{4} $$
$$ x \equiv -2 \pmod{4} $$
To find the least positive value, we can add the modulus (4) to the negative result.
$$ x \equiv -2 + 4 \pmod{4} $$
$$ x \equiv 2 \pmod{4} $$
The least positive value of x is 2.
This congruence means that $5x - 4$ is a multiple of 6. $$ 5x - 4 = 6k, \text{ for some integer } k $$ We can test integer values for x to find a solution.
- If $x=0$, $5(0) = 0 \not\equiv 4 \pmod{6}$.
- If $x=1$, $5(1) = 5 \not\equiv 4 \pmod{6}$.
- If $x=2$, $5(2) = 10$. Since $10 = 1 \times 6 + 4$, we have $10 \equiv 4 \pmod{6}$. This is a solution.
The solution is $x \equiv 2 \pmod{6}$.
Part - IV Solutions
Solution for Question 19
19. a) Construct a triangle similar to $\triangle PQR$ with sides $\frac{7}{3}$ of the corresponding sides.
Since the scale factor $\frac{7}{3} > 1$, the new triangle $\triangle P'QR'$ will be larger than $\triangle PQR$.
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 7 (the larger number in the fraction) equally spaced points, namely $Q_1, Q_2, \dots, Q_7$, such that $QQ_1 = Q_1Q_2 = \dots = Q_6Q_7$.
- Join the 3rd point (from the denominator), $Q_3$, to the vertex R.
- Extend the side QR beyond R.
- From the 7th point (from the numerator), $Q_7$, draw a line parallel to $Q_3R$. This line will intersect the extended side QR at a new point, R'. (To draw the parallel line, you can construct an equal angle at $Q_7$ as the one at $Q_3$).
- Extend the side QP beyond P.
- From the new point R', draw a line parallel to PR. This line will intersect the extended side QP at a new point, P'.
The triangle $\triangle P'QR'$ is the required similar triangle whose sides are $\frac{7}{3}$ times the corresponding sides of $\triangle PQR$.
19. b) Construct a triangle similar to $\triangle ABC$ with sides $\frac{3}{5}$ of the corresponding sides.
Since the scale factor $\frac{3}{5} < 1$, the new triangle $\triangle A'BC'$ will be smaller than $\triangle ABC$.
Steps of Construction:
- Draw the given triangle $\triangle ABC$ with any measurements.
- From vertex B, draw a ray BX making an acute angle with the side BC (on the side opposite to vertex A).
- On the ray BX, mark 5 (the larger number in the fraction) equally spaced points, namely $B_1, B_2, \dots, B_5$, such that $BB_1 = B_1B_2 = \dots = B_4B_5$.
- Join the 5th point (from the denominator), $B_5$, to the vertex C.
- From the 3rd point (from the numerator), $B_3$, draw a line parallel to $B_5C$. This line will intersect the side BC at a new point, C'.
- From the new point C', draw a line parallel to AC. This line will intersect the side AB at a new point, A'.
The triangle $\triangle A'BC'$ is the required similar triangle whose sides are $\frac{3}{5}$ times the corresponding sides of $\triangle ABC$.
Solution for Question 20
20. a) Parking charges analysis and graph.
Let Time = x and Amount = y.
For direct variation, the ratio $k = \frac{y}{x}$ should be constant.
For inverse variation, the product $k = xy$ should be constant.
- $\frac{60}{4} = 15$
- $\frac{120}{8} = 15$
- $\frac{180}{12} = 15$
- $\frac{360}{24} = 15$
Since the ratio $\frac{y}{x}$ is a constant (15), the amount charged is in direct variation with the parking time. The equation of variation is $y = 15x$.
Plot the points (4, 60), (8, 120), (12, 180), (24, 360) and draw a straight line through them and the origin (0,0).
Using the equation $y = 15x$:
$y = 15(6) = 90$.
On the graph, find x=6, move up to the line, and read the corresponding y-value.
The amount to be paid for 6 hours is ₹90.
Using the equation $y = 15x$:
$150 = 15x \implies x = \frac{150}{15} = 10$.
On the graph, find y=150, move across to the line, and read the corresponding x-value.
The parking duration for ₹150 is 10 hours.
20. b) Graph the linear function $\displaystyle y = \frac{1}{2}x$.
The equation is in the form $y=kx$. By comparing, the constant of variation is $k = \frac{1}{2}$.
Let's create a table of values to plot the graph.
| x | 0 | 2 | 4 | 6 | 9 |
|---|---|---|---|---|---|
| y | 0 | 1 | 2 | 3 | 4.5 |
Plot these points and draw a straight line through them and the origin.
Verification: Pick any point on the line, for example (4, 2). Calculate the ratio $\frac{y}{x} = \frac{2}{4} = \frac{1}{2}$. This matches our constant of variation, k. This verifies the graph.
$y = \frac{1}{2}(9) = 4.5$.
When x=9, y=4.5.
$7.5 = \frac{1}{2}x \implies x = 7.5 \times 2 = 15$.
When y=7.5, x=15.