10th Maths Quarterly Exam 2025 - Complete Solutions
PART I - Answer all the questions (14 x 1 = 14)
1. If the ordered pairs $(a + 2, 4)$ and $(5, 2a + b)$ are equal then $(a, b)$ is
Solution:
Given that the ordered pairs are equal: $(a + 2, 4) = (5, 2a + b)$.
This means their corresponding elements are equal.
Equating the first elements:
$a + 2 = 5$
$a = 5 - 2$
$a = 3$
Equating the second elements:
$4 = 2a + b$
Substitute the value of $a = 3$ into the second equation:
$4 = 2(3) + b$
$4 = 6 + b$
$b = 4 - 6$
$b = -2$
So, the pair $(a, b)$ is $(3, -2)$.
Correct Answer: (D) (3, -2)
2. $A = \{a, b, p\}$, $B = \{2, 3\}$, $C = \{p, q, r, s\}$ then $n[(A \cup C) \times B]$ is
Solution:
Given sets: $A = \{a, b, p\}$, $B = \{2, 3\}$, $C = \{p, q, r, s\}$.
First, find the union of A and C:
$A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}$
Now, find the number of elements in $(A \cup C)$ and B:
$n(A \cup C) = 6$
$n(B) = 2$
The number of elements in the Cartesian product is given by:
$n[(A \cup C) \times B] = n(A \cup C) \times n(B)$
$n[(A \cup C) \times B] = 6 \times 2 = 12$
Correct Answer: (C) 12
3. If $g = \{(1, 1), (2, 3), (3, 5), (4, 7)\}$ is a function given by $g(x) = \alpha x + \beta$ then the values of $\alpha$ and $\beta$ are
Solution:
The function is $g(x) = \alpha x + \beta$.
From the set of ordered pairs, we can take any two points. Let's use $(1, 1)$ and $(2, 3)$.
For $(1, 1)$: $g(1) = 1$. So, $\alpha(1) + \beta = 1 \implies \alpha + \beta = 1$ --- (1)
For $(2, 3)$: $g(2) = 3$. So, $\alpha(2) + \beta = 3 \implies 2\alpha + \beta = 3$ --- (2)
Subtracting equation (1) from equation (2):
$(2\alpha + \beta) - (\alpha + \beta) = 3 - 1$
$\alpha = 2$
Substitute $\alpha = 2$ into equation (1):
$2 + \beta = 1$
$\beta = 1 - 2$
$\beta = -1$
So, the values are $\alpha = 2$ and $\beta = -1$.
Correct Answer: (B) (2, -1)
4. The domain of the function $f(x) = \frac{1}{3x-1}$ is
Solution:
The function $f(x) = \frac{1}{3x-1}$ is a rational function. The domain of a rational function is all real numbers except for the values of $x$ that make the denominator zero.
Set the denominator to zero:
$3x - 1 = 0$
$3x = 1$
$x = 1/3$
The function is undefined at $x = 1/3$. Therefore, the domain is the set of all real numbers (R) except for $1/3$.
Correct Answer: (C) R - {1/3}
5. If the HCF of 65 and 117 is expressible in the form of $65m - 117$, then the value of m is
Solution:
First, find the HCF of 65 and 117 using Euclid's division algorithm.
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
The HCF is the last non-zero remainder, which is 13.
Now, we are given that the HCF is expressible as $65m - 117$.
So, $65m - 117 = 13$
$65m = 13 + 117$
$65m = 130$
$m = \frac{130}{65}$
$m = 2$
Correct Answer: (B) 2
6. If 6 times of 6th term of an A.P. is equal to 7 times the 7th term, then the 13th term of the A.P. is
Solution:
Let the A.P. have first term 'a' and common difference 'd'. The nth term is given by $t_n = a + (n-1)d$.
Given: $6 \times t_6 = 7 \times t_7$
$6 \times (a + (6-1)d) = 7 \times (a + (7-1)d)$
$6(a + 5d) = 7(a + 6d)$
$6a + 30d = 7a + 42d$
$6a - 7a = 42d - 30d$
$-a = 12d \implies a = -12d$
We need to find the 13th term, $t_{13}$.
$t_{13} = a + (13-1)d = a + 12d$
Substitute $a = -12d$ into the expression for $t_{13}$:
$t_{13} = (-12d) + 12d = 0$
Correct Answer: (A) 0
7. $7^{4k} \equiv$ _____ (mod 100)
Solution:
We need to find the remainder when $7^{4k}$ is divided by 100.
Let's calculate powers of 7 modulo 100:
$7^1 \equiv 7$ (mod 100)
$7^2 = 49 \equiv 49$ (mod 100)
$7^3 = 7^2 \times 7 = 49 \times 7 = 343 \equiv 43$ (mod 100)
$7^4 = 7^2 \times 7^2 = 49 \times 49 = 2401$
$2401 = 24 \times 100 + 1$. So, $7^4 \equiv 1$ (mod 100).
Now, we can find $7^{4k}$:
$7^{4k} = (7^4)^k$
Since $7^4 \equiv 1$ (mod 100),
$(7^4)^k \equiv 1^k$ (mod 100)
$7^{4k} \equiv 1$ (mod 100)
Correct Answer: (A) 1
8. A system of three linear equations in three variables is inconsistent if their planes
Solution:
An inconsistent system of linear equations is one that has no solution. Geometrically, for three equations in three variables, this means the three planes represented by the equations do not have any point in common.
The planes might be parallel, or two might intersect in a line that is parallel to the third plane, but in all inconsistent cases, there is no common intersection point for all three.
Correct Answer: (D) do not intersect
9. The solution of $(2x – 1)^2 = 9$ is equal to
Solution:
Given the equation: $(2x - 1)^2 = 9$.
Take the square root of both sides:
$2x - 1 = \pm\sqrt{9}$
$2x - 1 = \pm3$
This gives two possible cases:
Case 1: $2x - 1 = 3$
$2x = 3 + 1$
$2x = 4$
$x = 2$
Case 2: $2x - 1 = -3$
$2x = -3 + 1$
$2x = -2$
$x = -1$
The solutions are $x = -1$ and $x = 2$.
Correct Answer: (C) -1, 2
10. In $\triangle LMN$, $\angle L = 60^\circ$, $\angle M = 50^\circ$. If $\triangle LMN \sim \triangle PQR$ then the value of $\angle R$ is
Solution:
In $\triangle LMN$, the sum of angles is $180^\circ$.
$\angle L + \angle M + \angle N = 180^\circ$
$60^\circ + 50^\circ + \angle N = 180^\circ$
$110^\circ + \angle N = 180^\circ$
$\angle N = 180^\circ - 110^\circ = 70^\circ$
It is given that $\triangle LMN \sim \triangle PQR$. When two triangles are similar, their corresponding angles are equal.
$\angle L = \angle P = 60^\circ$
$\angle M = \angle Q = 50^\circ$
$\angle N = \angle R = 70^\circ$
Therefore, the value of $\angle R$ is $70^\circ$.
Correct Answer: (B) 70°
11. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
Solution:
Let the two poles be AB and CD, with heights 6 m and 11 m respectively. The distance between their feet, BD, is 12 m.
Draw a line AE parallel to BD from the top of the shorter pole to the taller pole. This forms a rectangle ABDE and a right-angled triangle AEC.
In rectangle ABDE, $AE = BD = 12$ m and $ED = AB = 6$ m.
The length of CE is the difference in the heights of the poles: $CE = CD - ED = 11 - 6 = 5$ m.
Now, in the right-angled triangle AEC, the distance between the tops is the hypotenuse AC.
By Pythagoras theorem: $AC^2 = AE^2 + CE^2$
$AC^2 = 12^2 + 5^2$
$AC^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13$ m.
Correct Answer: (A) 13 m
12. The slope of the line which is perpendicular to a line joining the points (0, 0) and (-8, 8) is
Solution:
First, find the slope of the line joining the points $(0, 0)$ and $(-8, 8)$. Let this slope be $m_1$.
$m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1$
Let the slope of the perpendicular line be $m_2$.
The condition for two lines to be perpendicular is $m_1 \times m_2 = -1$.
$(-1) \times m_2 = -1$
$m_2 = \frac{-1}{-1} = 1$
The slope of the perpendicular line is 1.
Correct Answer: (B) 1
13. The area of triangle formed by the points (-5, 0), (0, -5) and (5, 0) is
Solution:
Let the vertices be $A(-5, 0)$, $B(0, -5)$, and $C(5, 0)$.
The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area = $\frac{1}{2} |-5(-5 - 0) + 0(0 - 0) + 5(0 - (-5))|$
Area = $\frac{1}{2} |-5(-5) + 0 + 5(5)|$
Area = $\frac{1}{2} |25 + 0 + 25|$
Area = $\frac{1}{2} |50| = 25$ sq.units.
Correct Answer: (B) 25 sq.units
14. The y-intercept of the equation of the straight line $y = 12x$ is
Solution:
The slope-intercept form of a straight line is $y = mx + c$, where 'm' is the slope and 'c' is the y-intercept.
The given equation is $y = 12x$.
We can write this as $y = 12x + 0$.
Comparing this with $y = mx + c$, we get $m = 12$ and $c = 0$.
The y-intercept is $c = 0$.
Correct Answer: (C) 0
PART II - Answer any 10 of the following. (Q.No 28 is Compulsory) (10 x 2 = 20)
15. 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.
Solution:
The relation is defined by the rule $y = x + 3$, where the values of $x$ are from the set $\{0, 1, 2, 3, 4, 5\}$.
The domain is the set of all possible input values ($x$), which is given as $\{0, 1, 2, 3, 4, 5\}$.
To find the range, we calculate the corresponding $y$ value for each $x$ in the domain:
- If $x = 0$, $y = 0 + 3 = 3$
- If $x = 1$, $y = 1 + 3 = 4$
- If $x = 2$, $y = 2 + 3 = 5$
- If $x = 3$, $y = 3 + 3 = 6$
- If $x = 4$, $y = 4 + 3 = 7$
- If $x = 5$, $y = 5 + 3 = 8$
The range is the set of all resulting output values ($y$).
Domain = {0, 1, 2, 3, 4, 5}
Range = {3, 4, 5, 6, 7, 8}
16. A function f is defined by $f(x) = 3 - 2x$. Find x such that $f(x^2) = [f(x)]^2$.
Solution:
Given the function $f(x) = 3 - 2x$.
We are given the condition $f(x^2) = [f(x)]^2$.
First, find the expressions for both sides.
LHS: $f(x^2) = 3 - 2(x^2) = 3 - 2x^2$
RHS: $[f(x)]^2 = (3 - 2x)^2 = 3^2 - 2(3)(2x) + (2x)^2 = 9 - 12x + 4x^2$
Now, equate the LHS and RHS:
$3 - 2x^2 = 9 - 12x + 4x^2$
Rearrange the terms to form a quadratic equation:
$4x^2 + 2x^2 - 12x + 9 - 3 = 0$
$6x^2 - 12x + 6 = 0$
Divide the entire equation by 6:
$x^2 - 2x + 1 = 0$
This is a perfect square trinomial: $(x - 1)^2 = 0$
Taking the square root, we get $x - 1 = 0$, which gives $x = 1$.
The value of x is 1.
17. If $f(x) = 2x - 1$, $g(x) = \frac{x+1}{2}$, show that $f \circ g = g \circ f = x$.
Solution:
We need to show that the composition of the functions $f$ and $g$ in both orders results in the identity function, $x$.
Part 1: Find $f \circ g$
$(f \circ g)(x) = f(g(x))$
Substitute the expression for $g(x)$ into $f(x)$:
$f(g(x)) = 2 \left( \frac{x+1}{2} \right) - 1$
$= (x+1) - 1$
$= x$
So, $f \circ g = x$. --- (1)
Part 2: Find $g \circ f$
$(g \circ f)(x) = g(f(x))$
Substitute the expression for $f(x)$ into $g(x)$:
$g(f(x)) = \frac{(2x - 1) + 1}{2}$
$= \frac{2x}{2}$
$= x$
So, $g \circ f = x$. --- (2)
From (1) and (2), we have shown that $f \circ g = g \circ f = x$. This means $f$ and $g$ are inverse functions of each other.
18. 'a' and 'b' are two positive integers such that $a^b \times b^a = 800$. Find 'a' and 'b'.
Solution:
Given the equation $a^b \times b^a = 800$.
First, find the prime factorization of 800.
$800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2$
So, we have $a^b \times b^a = 2^5 \times 5^2$.
By comparing the bases and exponents, we can see a direct match.
Let's try setting $a=2$ and $b=5$.
$a^b \times b^a = 2^5 \times 5^2$
This perfectly matches the prime factorization of 800.
We can also try $a=5$ and $b=2$.
$a^b \times b^a = 5^2 \times 2^5 = 2^5 \times 5^2$, which is the same.
Since 'a' and 'b' are positive integers, the solution is the pair {2, 5}.
The values are a = 2, b = 5 (or a = 5, b = 2).
19. Find the first four terms of the sequences whose nth term is given by $a_n = 2n^2 - 6$.
Solution:
The nth term is given by the formula $a_n = 2n^2 - 6$.
To find the first four terms, we substitute $n = 1, 2, 3, 4$.
- First term (n=1):
$a_1 = 2(1)^2 - 6 = 2(1) - 6 = 2 - 6 = -4$ - Second term (n=2):
$a_2 = 2(2)^2 - 6 = 2(4) - 6 = 8 - 6 = 2$ - Third term (n=3):
$a_3 = 2(3)^2 - 6 = 2(9) - 6 = 18 - 6 = 12$ - Fourth term (n=4):
$a_4 = 2(4)^2 - 6 = 2(16) - 6 = 32 - 6 = 26$
The first four terms are -4, 2, 12, 26.
20. Find the sum to infinity of $9 + 3 + 1 + \dots$
Solution:
The given series is a Geometric Progression (G.P.).
The first term is $a = 9$.
The common ratio is $r = \frac{\text{second term}}{\text{first term}} = \frac{3}{9} = \frac{1}{3}$.
The sum to infinity of a G.P. exists if $|r| < 1$. Here, $|1/3| < 1$, so the sum exists.
The formula for the sum to infinity is $S_\infty = \frac{a}{1 - r}$.
$S_\infty = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{3-1}{3}} = \frac{9}{\frac{2}{3}}$
$S_\infty = 9 \times \frac{3}{2} = \frac{27}{2}$ or $13.5$.
The sum to infinity is 27/2.
21. Find L.C.M of $16m$, $12m^2n^2$, $8m$.
Solution:
First, find the L.C.M of the numerical coefficients: 16, 12, and 8.
Prime factorization:
$16 = 2^4$
$12 = 2^2 \times 3^1$
$8 = 2^3$
L.C.M of coefficients = (highest power of 2) $\times$ (highest power of 3) = $2^4 \times 3^1 = 16 \times 3 = 48$.
Next, find the L.C.M of the variables: $m$, $m^2n^2$, $m$.
L.C.M of variables = (highest power of m) $\times$ (highest power of n) = $m^2 \times n^2 = m^2n^2$.
Combining the results:
L.C.M = $48m^2n^2$.
22. Simplify: $\frac{x^3}{x-y} + \frac{y^3}{y-x}$
Solution:
The expression is $\frac{x^3}{x-y} + \frac{y^3}{y-x}$.
Notice that the denominators are negatives of each other: $y-x = -(x-y)$.
We can rewrite the second term:
$\frac{y^3}{y-x} = \frac{y^3}{-(x-y)} = -\frac{y^3}{x-y}$
Now substitute this back into the original expression:
$\frac{x^3}{x-y} - \frac{y^3}{x-y}$
Since the denominators are the same, we can combine the numerators:
$\frac{x^3 - y^3}{x-y}$
Use the formula for the difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.
$\frac{(x-y)(x^2 + xy + y^2)}{x-y}$
Cancel out the $(x-y)$ term from the numerator and denominator.
$= x^2 + xy + y^2$
The simplified expression is $x^2 + xy + y^2$.
23. If the difference between a number and its reciprocal is $\frac{24}{5}$, find the number.
Solution:
Let the number be $x$. Its reciprocal is $\frac{1}{x}$.
The difference is given as $\frac{24}{5}$. So, $x - \frac{1}{x} = \frac{24}{5}$.
To solve for $x$, first create a common denominator on the left side:
$\frac{x^2 - 1}{x} = \frac{24}{5}$
Cross-multiply:
$5(x^2 - 1) = 24x$
$5x^2 - 5 = 24x$
Rearrange into a standard quadratic equation:
$5x^2 - 24x - 5 = 0$
We can solve this by factoring. We need two numbers that multiply to $5 \times (-5) = -25$ and add to $-24$. These numbers are $-25$ and $1$.
$5x^2 - 25x + 1x - 5 = 0$
$5x(x - 5) + 1(x - 5) = 0$
$(5x + 1)(x - 5) = 0$
This gives two possible solutions for $x$:
$5x + 1 = 0 \implies x = -1/5$
$x - 5 = 0 \implies x = 5$
The number is 5 or -1/5.
24. In triangle ABC, AD is the bisector of $\angle A$. If BD = 4 cm, DC = 3 cm and AB = 6 cm, find AC.
Solution:
This problem uses the Angle Bisector Theorem.
The theorem states that if a line bisects an angle of a triangle, it divides the opposite side into two segments that are proportional to the other two sides of the triangle.
In $\triangle ABC$, since AD bisects $\angle A$, we have:
$\frac{AB}{AC} = \frac{BD}{DC}$
Substitute the given values:
$\frac{6}{AC} = \frac{4}{3}$
To find AC, cross-multiply:
$6 \times 3 = 4 \times AC$
$18 = 4 \times AC$
$AC = \frac{18}{4} = \frac{9}{2} = 4.5$ cm
The length of AC is 4.5 cm.
25. Find the area of the triangle whose vertices are (1, -1), (-4, 6) and (-3, -5).
Solution:
Let the vertices be $A(x_1, y_1) = (1, -1)$, $B(x_2, y_2) = (-4, 6)$, and $C(x_3, y_3) = (-3, -5)$.
The formula for the area of a triangle is:
Area $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)|$
Area $= \frac{1}{2} |(1 \times 6 + (-4) \times (-5) + (-3) \times (-1)) - ((-1) \times (-4) + 6 \times (-3) + (-5) \times 1)|$
Area $= \frac{1}{2} |(6 + 20 + 3) - (4 - 18 - 5)|$
Area $= \frac{1}{2} |(29) - (-19)|$
Area $= \frac{1}{2} |29 + 19|$
Area $= \frac{1}{2} |48| = 24$ sq. units.
The area of the triangle is 24 sq. units.
26. What is the slope of a line whose inclination with positive direction of x-axis is (i) 90° (ii) 0°
Solution:
The slope ($m$) of a line is related to its angle of inclination ($\theta$) by the formula $m = \tan(\theta)$.
(i) Inclination is 90°
$m = \tan(90°)$
The value of $\tan(90°)$ is undefined. This corresponds to a vertical line.
(ii) Inclination is 0°
$m = \tan(0°)$
The value of $\tan(0°)$ is 0. This corresponds to a horizontal line.
(i) The slope is undefined. (ii) The slope is 0.
27. Find the equation of a straight line which has slope $\frac{-5}{4}$ and passing through the point (-1, 2).
Solution:
We are given the slope $m = -\frac{5}{4}$ and a point $(x_1, y_1) = (-1, 2)$.
We can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$.
Substitute the given values:
$y - 2 = -\frac{5}{4}(x - (-1))$
$y - 2 = -\frac{5}{4}(x + 1)$
To eliminate the fraction, multiply both sides by 4:
$4(y - 2) = -5(x + 1)$
$4y - 8 = -5x - 5$
Rearrange the equation into the general form $Ax + By + C = 0$:
$5x + 4y - 8 + 5 = 0$
$5x + 4y - 3 = 0$
The equation of the straight line is $5x + 4y - 3 = 0$.
28. A boy named Mani saves Rs. 5 on the first day. If he saves Rs. 3 more every day. What is the total amount he saved in 15 days.
Solution:
This is an arithmetic progression (A.P.) problem.
The amount saved on the first day (first term) is $a = 5$.
The amount he saves increases by Rs. 3 each day, so the common difference is $d = 3$.
We need to find the total amount saved in 15 days, which is the sum of the first 15 terms ($S_{15}$).
The number of days (terms) is $n = 15$.
The formula for the sum of an A.P. is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substitute the values:
$S_{15} = \frac{15}{2}[2(5) + (15-1)3]$
$S_{15} = \frac{15}{2}[10 + (14)3]$
$S_{15} = \frac{15}{2}[10 + 42]$
$S_{15} = \frac{15}{2}[52]$
$S_{15} = 15 \times 26$
$S_{15} = 390$
The total amount Mani saved in 15 days is Rs. 390.
PART III - Answer any 10 of the following. (Q.No 42 is Compulsory) (10 x 5 = 50)
29. Let $A = \{x \in W / x < 2\}$, $B = \{x \in N / 1 < x \le 4\}$, $C = \{3, 5\}$. Verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Solution:
First, let's list the elements of each set:
- $A = \{x \in W / x < 2\}$: Whole numbers less than 2 are 0 and 1. So, $A = \{0, 1\}$.
- $B = \{x \in N / 1 < x \le 4\}$: Natural numbers greater than 1 and less than or equal to 4 are 2, 3, 4. So, $B = \{2, 3, 4\}$.
- $C = \{3, 5\}$.
LHS: $A \times (B \cup C)$
First, find $B \cup C$:
$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$.
Now, find $A \times (B \cup C)$:
$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$= \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$ --- (1)
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 $A \times C$:
$A \times C = \{0, 1\} \times \{3, 5\}$
$= \{(0, 3), (0, 5), (1, 3), (1, 5)\}$.
Now, find the union of these two sets:
$(A \times B) \cup (A \times C)$
$= \{(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)\} \cup \{(0, 3), (0, 5), (1, 3), (1, 5)\}$
$= \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$ --- (2)
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.
30. Forensic scientists can determine the height (in cm) of a person based on the length of the thigh bone. They usually do so using the function $h(b) = 2.47b + 54.10$ where b is the length of the thigh bone.
(i) Verify the function h is one-one or not.
(ii) Also find the height of a person if the length of his thigh bone is 50 cm.
(iii) Find the length of the thigh bone if the height of a person is 147.96 cm.
Solution:
The function is $h(b) = 2.47b + 54.10$.
(i) Verify if the function is one-one.
To check if $h$ is one-one, we assume $h(b_1) = h(b_2)$ for two different thigh bone lengths $b_1$ and $b_2$. If this implies $b_1 = b_2$, then the function is one-one.
$h(b_1) = 2.47b_1 + 54.10$
$h(b_2) = 2.47b_2 + 54.10$
Set $h(b_1) = h(b_2)$:
$2.47b_1 + 54.10 = 2.47b_2 + 54.10$
$2.47b_1 = 2.47b_2$
$b_1 = b_2$
Since $h(b_1) = h(b_2)$ implies $b_1 = b_2$, the function is one-one. This means that for every distinct thigh bone length, there is a distinct height.
(ii) Find the height for a thigh bone of 50 cm.
Here, $b = 50$ cm. We need to calculate $h(50)$.
$h(50) = 2.47(50) + 54.10$
$h(50) = 123.5 + 54.10$
$h(50) = 177.6$ cm.
(iii) Find the thigh bone length for a height of 147.96 cm.
Here, $h(b) = 147.96$ cm. We need to solve for $b$.
$147.96 = 2.47b + 54.10$
$147.96 - 54.10 = 2.47b$
$93.86 = 2.47b$
$b = \frac{93.86}{2.47}$
$b = 38$ cm.
(i) The function h is one-one.
(ii) The height of the person is 177.6 cm.
(iii) The length of the thigh bone is 38 cm.
31. If $f(x) = x^2$, $g(x) = 2x$, $h(x) = x + 4$ then prove that $f \circ (g \circ h) = (f \circ g) \circ h$.
Solution:
We need to prove the associative property of function composition for the given functions.
Given functions: $f(x) = x^2$, $g(x) = 2x$, $h(x) = x + 4$.
LHS: $f \circ (g \circ h)$
First, we find the composition of $g$ and $h$, which is $(g \circ h)(x)$.
$(g \circ h)(x) = g(h(x))$
Substitute $h(x) = x + 4$ into $g(x)$:
$(g \circ h)(x) = g(x + 4) = 2(x + 4) = 2x + 8$.
Now, we find $f \circ (g \circ h)$:
$(f \circ (g \circ h))(x) = f((g \circ h)(x))$
Substitute $(g \circ h)(x) = 2x + 8$ into $f(x)$:
$(f \circ (g \circ h))(x) = f(2x + 8) = (2x + 8)^2$
Using the identity $(a+b)^2 = a^2 + 2ab + b^2$:
$(2x + 8)^2 = (2x)^2 + 2(2x)(8) + 8^2 = 4x^2 + 32x + 64$. --- (1)
RHS: $(f \circ g) \circ h$
First, we find the composition of $f$ and $g$, which is $(f \circ g)(x)$.
$(f \circ g)(x) = f(g(x))$
Substitute $g(x) = 2x$ into $f(x)$:
$(f \circ g)(x) = f(2x) = (2x)^2 = 4x^2$.
Now, we find $(f \circ g) \circ h$:
$((f \circ g) \circ h)(x) = (f \circ g)(h(x))$
Substitute $h(x) = x + 4$ into the expression for $(f \circ g)(x)$:
$((f \circ g) \circ h)(x) = (f \circ g)(x + 4) = 4(x + 4)^2$
Using the identity $(a+b)^2 = a^2 + 2ab + b^2$:
$4(x^2 + 2(x)(4) + 4^2) = 4(x^2 + 8x + 16) = 4x^2 + 32x + 64$. --- (2)
Conclusion:
From equations (1) and (2), we see that the expressions for both LHS and RHS are identical.
$4x^2 + 32x + 64 = 4x^2 + 32x + 64$
LHS = RHS. Hence, $f \circ (g \circ h) = (f \circ g) \circ h$ is proved.
32. Find the HCF of 396, 504, 636
Solution:
We will use Euclid's division algorithm to find the HCF.
Step 1: Find HCF of 504 and 396.
$504 = 396 \times 1 + 108$
$396 = 108 \times 3 + 72$
$108 = 72 \times 1 + 36$
$72 = 36 \times 2 + 0$
The HCF of 504 and 396 is 36.
Step 2: Find HCF of the result (36) and the third number (636).
$636 = 36 \times 17 + 24$
$36 = 24 \times 1 + 12$
$24 = 12 \times 2 + 0$
The HCF of 36 and 636 is 12.
Therefore, the HCF of 396, 504, and 636 is 12.
The HCF is 12.
33. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution:
Let the three consecutive terms in an Arithmetic Progression (A.P.) be $(a - d)$, $a$, and $(a + d)$, where 'a' is the middle term and 'd' is the common difference.
Condition 1: The sum is 27.
$(a - d) + a + (a + d) = 27$
$3a = 27$
$a = \frac{27}{3} = 9$
So, the middle term is 9.
Condition 2: The product is 288.
$(a - d) \times a \times (a + d) = 288$
Substitute the value of $a = 9$:
$(9 - d) \times 9 \times (9 + d) = 288$
$(9 - d)(9 + d) = \frac{288}{9}$
$9^2 - d^2 = 32$
$81 - d^2 = 32$
$d^2 = 81 - 32$
$d^2 = 49$
$d = \pm\sqrt{49} = \pm7$
Case 1: If d = 7
The terms are $(a - d), a, (a + d) \implies (9 - 7), 9, (9 + 7) \implies 2, 9, 16$.
Case 2: If d = -7
The terms are $(a - d), a, (a + d) \implies (9 - (-7)), 9, (9 + (-7)) \implies 16, 9, 2$.
In both cases, the set of three terms is the same.
The three terms are 2, 9, and 16.
34. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
Solution:
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The area of a square with side 's' is $s^2$.
The areas of the papers are $10^2, 11^2, 12^2, \dots, 24^2$.
To find the total area, we need to calculate the sum: $S = 10^2 + 11^2 + 12^2 + \dots + 24^2$.
We use the formula for the sum of the squares of the first n natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
We can write our sum as: $S = (1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2)$.
$S = \sum_{k=1}^{24} k^2 - \sum_{k=1}^{9} k^2$
Calculate $\sum_{k=1}^{24} k^2$:
$= \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900$.
Calculate $\sum_{k=1}^{9} k^2$:
$= \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} = 3 \times 5 \times 19 = 285$.
Calculate the total area:
$S = 4900 - 285 = 4615$.
The total area that can be decorated is 4615 cm².
35. Solve the system of linear equations:
$x + 2y - z = 5$ --- (1)
$x - y + z = -2$ --- (2)
$-5x - 4y + z = -11$ --- (3)
Solution:
We use the elimination method to solve the system.
Step 1: Eliminate 'z' using equations (1) and (2).
Add equation (1) and (2):
$(x + 2y - z) + (x - y + z) = 5 + (-2)$
$2x + y = 3$ --- (4)
Step 2: Eliminate 'z' using equations (2) and (3).
Subtract equation (3) from (2):
$(x - y + z) - (-5x - 4y + z) = -2 - (-11)$
$x - y + z + 5x + 4y - z = -2 + 11$
$6x + 3y = 9$
Divide by 3: $2x + y = 3$ --- (5)
Step 3: Analyze equations (4) and (5).
We observe that equations (4) and (5) are identical: $2x + y = 3$.
When solving a system of three equations and we end up with two identical equations in two variables, it means the system has infinitely many solutions. The planes intersect in a line.
To express the solution, let $y = k$, where $k$ is any real number.
From $2x + y = 3$, we have $2x + k = 3 \implies 2x = 3 - k \implies x = \frac{3-k}{2}$.
Now substitute $x$ and $y$ into one of the original equations to find $z$. Let's use equation (2):
$x - y + z = -2$
$\frac{3-k}{2} - k + z = -2$
$z = -2 - \frac{3-k}{2} + k$
$z = \frac{-4 - (3-k) + 2k}{2} = \frac{-4 - 3 + k + 2k}{2} = \frac{3k - 7}{2}$
The system has infinitely many solutions given by $(x, y, z) = (\frac{3-k}{2}, k, \frac{3k-7}{2})$ for any real number $k$.
36. Find the square root of $64x^4 - 16x^3 + 17x^2 - 2x + 1$.
Solution:
We use the long division method to find the square root of the polynomial.
8x² -x +1
_____________________
8x² | 64x⁴ - 16x³ + 17x² - 2x + 1
|-(64x⁴)
|_____________________
16x²-x | -16x³ + 17x²
| -(-16x³ + x²)
|_____________________
16x²-2x+1| 16x² - 2x + 1
| -(16x² - 2x + 1)
|_____________________
| 0
Step-by-step explanation:
- The square root of the first term, $64x^4$, is $8x^2$. Write $8x^2$ as the first term of the root and as the divisor.
- Multiply $8x^2$ by $8x^2$ to get $64x^4$. Subtract it from the polynomial.
- Bring down the next two terms: $-16x^3 + 17x^2$.
- Double the current root ($8x^2$) to get the new divisor's first term: $16x^2$.
- Divide the first term of the new dividend ($-16x^3$) by the first term of the new divisor ($16x^2$) to get $-x$. This is the second term of the root.
- Write $-x$ in the root and in the divisor. The new divisor is $16x^2 - x$.
- Multiply the new divisor ($16x^2 - x$) by the new term of the root ($-x$) to get $-16x^3 + x^2$. Subtract this.
- The remainder is $(17x^2 - 2x + 1) - (x^2) = 16x^2 - 2x + 1$. Bring down the remaining term $+1$.
- Double the current root ($8x^2 - x$) to get $16x^2 - 2x$.
- Divide the first term of the new dividend ($16x^2$) by the first term of the new divisor ($16x^2$) to get $+1$. This is the third term of the root.
- Write $+1$ in the root and in the divisor. The new divisor is $16x^2 - 2x + 1$.
- Multiply the new divisor by $+1$ to get $16x^2 - 2x + 1$. Subtract this. The remainder is 0.
The square root is the expression on top.
The square root is $|8x^2 - x + 1|$.
37. If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ are real and equal, prove that either $a=0$ or $a^3 + b^3 + c^3 = 3abc$.
Solution:
For a quadratic equation $Ax^2 + Bx + C = 0$ to have real and equal roots, its discriminant must be zero: $B^2 - 4AC = 0$.
In the given equation:
- $A = c^2 - ab$
- $B = -2(a^2 - bc)$
- $C = b^2 - ac$
Set the discriminant to zero:
$[-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0$
$4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0$
Divide by 4:
$(a^2 - bc)^2 - (c^2 - ab)(b^2 - ac) = 0$
Expand the terms:
$(a^4 - 2a^2bc + b^2c^2) - (c^2b^2 - ac^3 - ab^3 + a^2bc) = 0$
$a^4 - 2a^2bc + b^2c^2 - c^2b^2 + ac^3 + ab^3 - a^2bc = 0$
Combine like terms:
$a^4 + ab^3 + ac^3 - 3a^2bc = 0$
Factor out 'a':
$a(a^3 + b^3 + c^3 - 3abc) = 0$
This equation holds true if either of the factors is zero.
Case 1: $a = 0$
Case 2: $a^3 + b^3 + c^3 - 3abc = 0 \implies a^3 + b^3 + c^3 = 3abc$
Hence, it is proved that if the roots are real and equal, then either $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.
38. State and Prove Basic Proportionality theorem.
Solution:
Statement (Basic Proportionality Theorem or Thales' Theorem):
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Proof:
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. |
39. Find the area of the quadrilateral whose vertices are at (-9, -2), (-8, -4), (2, 2) and (1, -3).
Solution:
Let the vertices of the quadrilateral be $A(x_1, y_1) = (-9, -2)$, $B(x_2, y_2) = (-8, -4)$, $C(x_3, y_3) = (2, 2)$, and $D(x_4, y_4) = (1, -3)$.
The formula for the area of a quadrilateral is:
Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$
Step 1: Calculate the first part of the formula.
$x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1$
$= (-9)(-4) + (-8)(2) + (2)(-3) + (1)(-2)$
$= 36 - 16 - 6 - 2$
$= 12$
Step 2: Calculate the second part of the formula.
$y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1$
$= (-2)(-8) + (-4)(2) + (2)(1) + (-3)(-9)$
$= 16 - 8 + 2 + 27$
$= 37$
Step 3: Substitute the values back into the area formula.
Area = $\frac{1}{2} |12 - 37|$
Area = $\frac{1}{2} |-25|$
Area = $\frac{1}{2} \times 25 = 12.5$
The area of the quadrilateral is 12.5 sq. units.
40. Prove analytically that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to half of its length.
Solution:
The solution is provided above. For a more in-depth understanding, please see the detailed explanation that follows.
This is the Midpoint Theorem, which we will prove using coordinate geometry.
Step 1: Set up the triangle on a coordinate plane.
To simplify calculations, let the vertices of the triangle ABC be $A(0, 0)$, $B(2a, 0)$, and $C(2b, 2c)$.
Step 2: Find the midpoints of two sides.
Let D be the midpoint of side AC.
$D = \left(\frac{0+2b}{2}, \frac{0+2c}{2}\right) = (b, c)$.
Let E be the midpoint of side BC.
$E = \left(\frac{2a+2b}{2}, \frac{0+2c}{2}\right) = (a+b, c)$.
Step 3: Prove that DE is parallel to the third side AB.
To prove parallelism, we show that their slopes are equal.
Slope of DE ($m_{DE}$) = $\frac{c - c}{(a+b) - b} = \frac{0}{a} = 0$.
Slope of AB ($m_{AB}$) = $\frac{0 - 0}{2a - 0} = \frac{0}{2a} = 0$.
Since $m_{DE} = m_{AB} = 0$, the line segment DE is parallel to the side AB.
Step 4: Prove that the length of DE is half the length of AB.
We use the distance formula, $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Length of DE = $\sqrt{((a+b) - b)^2 + (c-c)^2} = \sqrt{a^2 + 0^2} = \sqrt{a^2} = a$.
Length of AB = $\sqrt{(2a-0)^2 + (0-0)^2} = \sqrt{(2a)^2} = 2a$.
Comparing the lengths, we see that $DE = a$ and $AB = 2a$. Therefore, $DE = \frac{1}{2} AB$.
We have proved analytically that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half its length. Hence proved.
41. A (-3, 0), B (10, -2) and C (12, 3) are the vertices of $\triangle ABC$. Find the equation of the altitude through A.
Solution:
The altitude through vertex A is a line segment that starts from A and is perpendicular to the opposite side, BC.
Step 1: Find the slope of the side BC.
The vertices are $B(10, -2)$ and $C(12, 3)$.
Slope of BC ($m_{BC}$) = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2}$.
Step 2: Find the slope of the altitude from A.
Since the altitude is perpendicular to BC, its slope ($m_{alt}$) is the negative reciprocal of the slope of BC.
$m_{alt} = -\frac{1}{m_{BC}} = -\frac{1}{5/2} = -\frac{2}{5}$.
Step 3: Find the equation of the altitude.
We have the slope of the altitude ($m_{alt} = -2/5$) and the point it passes through, which is vertex $A(-3, 0)$.
Using the point-slope form of a line, $y - y_1 = m(x - x_1)$:
$y - 0 = -\frac{2}{5}(x - (-3))$
$y = -\frac{2}{5}(x + 3)$
Step 4: Convert the equation to the general form $Ax + By + C = 0$.
Multiply both sides by 5 to eliminate the fraction:
$5y = -2(x + 3)$
$5y = -2x - 6$
Move all terms to one side:
$2x + 5y + 6 = 0$
The equation of the altitude through A is $2x + 5y + 6 = 0$.
42. If $\alpha$ and $\beta$ are the roots of the equation $x^2 + x - 6 = 0$ then find the values of the following.
i) $\alpha - \beta$
ii) $\alpha^2 + \beta^2$
iii) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$
iv) $\alpha^2\beta + \beta^2\alpha$
Solution:
For the quadratic equation $x^2 + x - 6 = 0$, we have $a=1, b=1, c=-6$.
Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{1}{1} = -1$.
Product of roots: $\alpha\beta = \frac{c}{a} = \frac{-6}{1} = -6$.
i) $\alpha - \beta$
We use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
$(\alpha - \beta)^2 = (-1)^2 - 4(-6) = 1 + 24 = 25$.
$\alpha - \beta = \pm\sqrt{25} = \pm 5$.
ii) $\alpha^2 + \beta^2$
We use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
$\alpha^2 + \beta^2 = (-1)^2 - 2(-6) = 1 + 12 = 13$.
iii) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$
First, find a common denominator: $\frac{\alpha^2 + \beta^2}{\alpha\beta}$.
From part (ii), we know $\alpha^2 + \beta^2 = 13$. We know $\alpha\beta = -6$.
So, $\frac{13}{-6} = -\frac{13}{6}$.
iv) $\alpha^2\beta + \beta^2\alpha$
Factor out the common term $\alpha\beta$: $\alpha\beta(\alpha + \beta)$.
Substitute the known values:
$(-6)(-1) = 6$.
i) $\alpha - \beta = \pm 5$
ii) $\alpha^2 + \beta^2 = 13$
iii) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{13}{6}$
iv) $\alpha^2\beta + \beta^2\alpha = 6$
PART IV - Answer all the questions (2 x 8 = 16)
43.
a) Construct a triangle similar to a given triangle LMN with its sides equal to 4/5 of the corresponding sides of the triangle LMN (scale factor 4/5 < 1).
(OR)
b) Construct a $\triangle PQR$ such that QR = 6.5 cm, $\angle P = 60^\circ$ and the altitude from P to QR is of length 4.5 cm.
Solution for 43 (a):
Objective: To construct a triangle $\triangle LM'N'$ similar to $\triangle LMN$ such that each of its sides is 4/5 of the corresponding sides of $\triangle LMN$.
The solution is provided above. For a more in-depth understanding, please see the detailed explanation that follows.
Steps of Construction:
- Draw any triangle LMN of any measurement (since it is a "given" triangle).
- Draw a ray LX starting from L, making an acute angle with the side LM, on the side opposite to vertex N.
- Since the scale factor is 4/5, locate 5 points (the greater of 4 and 5) on the ray LX, namely L₁, L₂, L₃, L₄, and L₅, such that $LL_1 = L_1L_2 = L_2L_3 = L_3L_4 = L_4L_5$.
- Join the 5th point (denominator of the ratio), L₅, to the vertex M.
- Draw a line from the 4th point (numerator of the ratio), L₄, parallel to L₅M. Let this line intersect the side LM at point M'. (This can be done by constructing an angle at L₄ equal to $\angle LL_5M$).
- From point M', draw a line parallel to the side MN. Let this line intersect the side LN at point N'.
- The triangle $\triangle LM'N'$ is the required similar triangle whose sides are 4/5 of the corresponding sides of $\triangle LMN$.
Solution for 43 (b):
Objective: To construct a $\triangle PQR$ with base QR = 6.5 cm, vertex angle $\angle P = 60^\circ$, and altitude from P to QR of 4.5 cm.
Steps of Construction:
- Draw a line segment QR of length 6.5 cm.
- At point Q, draw a ray QX making an angle of 60° with QR (i.e., $\angle RQX = 60^\circ$).
- Draw a ray QY perpendicular to QX at Q (i.e., $\angle XQY = 90^\circ$).
- Draw the perpendicular bisector of the line segment QR. Let it intersect QR at point M.
- Let the perpendicular bisector and the ray QY intersect at point O.
- With O as the center and OQ (or OR) as the radius, draw a circle. This circle will pass through points Q and R. The major arc QR of this circle contains the point P such that $\angle QPR = 60^\circ$.
- From point M on the perpendicular bisector, mark a point G at a distance of 4.5 cm (the length of the altitude).
- Draw a line through G parallel to QR. This line represents all points that are 4.5 cm away from the base QR.
- This parallel line will intersect the circle at two points. Mark these intersection points as P and P'.
- Join PQ and PR (or P'Q and P'R).
- The triangle $\triangle PQR$ (or $\triangle P'QR$) is the required triangle.
44. a) Draw the graph of $xy = 24$, $x, y > 0$. Using the graph find,
(i) y when x = 3
(ii) x when y = 6
Solution:
The solution is provided above. For a more in-depth understanding, please see the detailed explanation that follows.
The equation is $xy = 24$, or $y = \frac{24}{x}$. This is the equation of a rectangular hyperbola. Since $x, y > 0$, we will only draw the curve in the first quadrant.
Step 1: Create a table of values.
Choose positive values for $x$ and calculate the corresponding $y$ values.
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 | 24 |
|---|---|---|---|---|---|---|---|---|
| y = 24/x | 24 | 12 | 8 | 6 | 4 | 3 | 2 | 1 |
Step 2: Draw the graph.
Plot the points from the table on a graph paper and join them with a smooth curve. The x-axis and y-axis will be asymptotes to the curve.
Step 3: Use the graph to find the required values.
(i) Find y when x = 3:
On the graph, locate $x = 3$ on the x-axis. Draw a vertical line up to the curve. From that point on the curve, draw a horizontal line to the y-axis. The point where it meets the y-axis is the value of y. From the graph, we find that when $x=3$, $y=8$.
Verification: $y = 24/3 = 8$.
(ii) Find x when y = 6:
On the graph, locate $y = 6$ on the y-axis. Draw a horizontal line to the curve. From that point on the curve, draw a vertical line down to the x-axis. The point where it meets the x-axis is the value of x. From the graph, we find that when $y=6$, $x=4$.
Verification: $x = 24/6 = 4$.
(i) From the graph, when x = 3, y = 8.
(ii) From the graph, when y = 6, x = 4.
44. b) A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the Marked Price and the Discount. Hence find
(i) the marked price when a customer gets a discount of ₹3250 (from graph)
(ii) the discount when the marked price is ₹2500
Solution:
The solution is provided above. For a more in-depth understanding, please see the detailed explanation that follows.
Let the Marked Price be $x$ and the Discount be $y$.
The discount is 50% of the marked price.
So, the relation is $y = 50\% \text{ of } x = \frac{50}{100}x = \frac{1}{2}x$.
This is a linear equation of the form $y=mx$, representing a straight line passing through the origin with a slope of $1/2$.
Step 1: Create a table of values.
| Marked Price (x) | 0 | 1000 | 2000 | 3000 | 4000 | 5000 | 6000 |
|---|---|---|---|---|---|---|---|
| Discount (y = x/2) | 0 | 500 | 1000 | 1500 | 2000 | 2500 | 3000 |
Step 2: Draw the graph.
Plot the points from the table on a graph paper and join them with a straight line. (Scale: X-axis 1 cm = ₹1000, Y-axis 1 cm = ₹500)
Step 3: Use the graph to find the required values.
(i) Find the marked price when the discount is ₹3250.
On the graph, locate $y = 3250$ on the y-axis. Draw a horizontal line to the right to meet the graphed line. From this point of intersection, draw a vertical line down to the x-axis. The value on the x-axis is the marked price.
From the graph, we find the marked price to be ₹6500.
Verification: $x = 2y = 2 \times 3250 = 6500$.
(ii) Find the discount when the marked price is ₹2500.
On the graph, locate $x = 2500$ on the x-axis. Draw a vertical line up to meet the graphed line. From this point of intersection, draw a horizontal line to the left to meet the y-axis. The value on the y-axis is the discount.
From the graph, we find the discount to be ₹1250.
Verification: $y = x/2 = 2500 / 2 = 1250$.
(i) The marked price is ₹6500 when the discount is ₹3250.
(ii) The discount is ₹1250 when the marked price is ₹2500.