Standard 10 Maths - Mid Term Test 2024 Maths Important Questions Samacheer Kalvi

Tenkasi District

Common First Mid Term Test - 2024

Standard: 10 Subject: MATHS Time: 1.30 Hrs. Marks: 50

Part - A

7 x 1 = 7

Choose the best answer.

1. 1. If there are 1024 relations from a set $A = \{1, 2, 3, 4, 5\}$ to a set B, then the number of elements in B is
   • a) 3
   • b) 2
   • c) 4
   • d) 8
   Solution:

   Given, $n(A) = 5$. Let $n(B) = m$. The number of relations is $2^{n(A) \times n(B)} = 2^{5m}$. We are given $2^{5m} = 1024 = 2^{10}$. Equating powers, $5m = 10 \implies m = 2$.
2. 2. If $\{(a, 8), (6, b)\}$ represents an identity function, then the value of a and b are respectively.
   • a) (8, 6)
   • b) (8, 8)
   • c) (6, 8)
   • d) (6, 6)
   Solution:

   In an identity function, $f(x)=x$. So, for $(a, 8)$, $f(a)=8 \implies a=8$. For $(6, b)$, $f(6)=b \implies b=6$. The values are $(8, 6)$.
3. 3. If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then $f \circ g$ is
   • a) $\frac{3}{2x^2}$
   • b) $\frac{2}{3x^2}$
   • c) $\frac{2}{9x^2}$
   • d) $\frac{1}{6x^2}$
   Solution:

   $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{3x}\right) = 2 \left(\frac{1}{3x}\right)^2 = 2 \left(\frac{1}{9x^2}\right) = \frac{2}{9x^2}$.
4. 4. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
   • a) 0, 1, 8
   • b) 1, 4, 8
   • c) 0, 1, 3
   • d) 1, 3, 5
   Solution:

   Any integer is of the form $3q, 3q+1,$ or $3q+2$. Case 1: $(3q)^3 = 27q^3 = 9(3q^3)$. Remainder is 0. Case 2: $(3q+1)^3 = 9(\dots)+1$. Remainder is 1. Case 3: $(3q+2)^3 = 9(\dots)+8$. Remainder is 8. The remainders are 0, 1, 8.
5. 5. If 6 times of 6^th term of an A.P is equal to 7 times of 7^th term, then the 13^th term of the A.P is
   • a) 0
   • b) 6
   • c) 7
   • d) 13
   Solution:

   Given $6 \times t_6 = 7 \times t_7 \implies 6(a+5d) = 7(a+6d) \implies 6a+30d = 7a+42d \implies -a = 12d \implies a+12d=0$. The 13th term is $t_{13} = a+12d$, which is 0.
6. 6. The value of $(1^3+2^3+3^3+...+15^3) - (1+2+3+...+15)$ is
   • a) 14400
   • b) 14200
   • c) 14280
   • d) 14520
   Solution:

   Sum of cubes: $(\frac{n(n+1)}{2})^2 = (\frac{15(16)}{2})^2 = 120^2 = 14400$. Sum of numbers: $\frac{n(n+1)}{2} = 120$. Difference = $14400 - 120 = 14280$.
7. 7. The equation $xy - 7 = 3$ is a
   • a) linear equation
   • b) equation of circle
   • c) cubic equation
   • d) not a linear equation
   Solution:

   The equation is $xy=10$. The term $xy$ has a degree of 2 (degree of x is 1, degree of y is 1, total is 1+1=2). A linear equation has a maximum degree of 1. Therefore, it is not a linear equation.

Part - B

5 x 2 = 10

Answer any 5 questions. [Q.No. 14 is compulsory]

1. 8. A Relation R is given by the set $\{(x, y) \mid y = x+3, x \in \{0, 1, 2, 3, 4, 5\}\}$. Determine its domain and range.
   Solution:

   Domain is the set of input values $x$. Domain = $\{0, 1, 2, 3, 4, 5\}$.

   Range is the set of output values $y=x+3$. Range = $\{0+3, 1+3, 2+3, 3+3, 4+3, 5+3\} = \{3, 4, 5, 6, 7, 8\}$.
2. 9. Find $f \circ g$, if $f(x) = x-6$ and $g(x) = x^2$.
   Solution:

   $(f \circ g)(x) = f(g(x)) = f(x^2)$. Apply the rule for $f(x)$: $f(x^2) = (x^2) - 6 = x^2 - 6$.
3. 10. If $f(x) = 2x-x^2$, find (i) $f(1)$ (ii) $f(2)$.
   Solution:

   (i) $f(1) = 2(1) - (1)^2 = 2 - 1 = 1$.

   (ii) $f(2) = 2(2) - (2)^2 = 4 - 4 = 0$.
4. 11. If d is the highest common factor of 32 and 60. Find $x$ and $y$ satisfying $d = 32x+60y$.
   Solution:

   Using Euclid's Algorithm: $60 = 1 \times 32 + 28$; $32 = 1 \times 28 + 4$; $28 = 7 \times 4 + 0$. HCF, $d=4$.

   Working backwards: $4 = 32 - 1 \times 28 = 32 - 1 \times (60 - 1 \times 32) = 32 - 60 + 32 = 2 \times 32 - 1 \times 60$.

   So, $4 = 32(2) + 60(-1)$. Comparing this to $d = 32x+60y$, we get $x=2, y=-1$.
5. 12. Which term of an A.P $16, 11, 6, 1, \dots$ is $-54$?
   Solution:

   Here $a=16, d=11-16=-5$. Let $t_n = -54$.

   $t_n = a + (n-1)d \implies -54 = 16 + (n-1)(-5) \implies -70 = -5(n-1) \implies 14 = n-1 \implies n=15$. The 15th term is -54.
6. 13. Find the sum to infinity of $9+3+1+\dots$
   Solution:

   This is a G.P with $a=9$ and $r = 3/9 = 1/3$. Since $|r|<1$, the sum to infinity is $S_\infty = \frac{a}{1-r} = \frac{9}{1 - 1/3} = \frac{9}{2/3} = \frac{27}{2}$.
7. 14. [Compulsory] Solve: $x+y = 5$; $x-y = 1$
   Solution:

   Adding the two equations: $(x+y) + (x-y) = 5+1 \implies 2x = 6 \implies x=3$.

   Substitute $x=3$ into the first equation: $3+y=5 \implies y=2$. The solution is $x=3, y=2$.

Part - C

5 x 5 = 25

Answer any 5 questions. [Q.No. 21 is compulsory]

1. 15. Let $A = \{x \in W \mid x < 2\}$, $B = \{x \in N \mid 1 < x \le 4\}$ and $C = \{3, 5\}$. Verify $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
   Solution:

   Sets are: $A = \{0, 1\}$, $B = \{2, 3, 4\}$, $C = \{3, 5\}$.

   LHS: $B \cup C = \{2, 3, 4, 5\}$. So, $A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\} = \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\}$.

   RHS: $A \times B = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\}$. $A \times C = \{(0,3), (0,5), (1,3), (1,5)\}$.

   $(A \times B) \cup (A \times C) = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4), (0,5), (1,5)\}$. LHS=RHS, hence verified.
2. 16. If $f(x) = 2x+3$, $g(x) = 1-2x$ and $h(x) = 3x$, prove that $f \circ (g \circ h) = (f \circ g) \circ h$.
   Solution:

   LHS: $(g \circ h)(x) = g(h(x)) = g(3x) = 1-2(3x) = 1-6x$. Then, $f(g \circ h)(x) = f(1-6x) = 2(1-6x)+3 = 2-12x+3 = 5-12x$.

   RHS: $(f \circ g)(x) = f(g(x)) = f(1-2x) = 2(1-2x)+3 = 2-4x+3 = 5-4x$. Then, $(f \circ g)(h(x)) = (f \circ g)(3x) = 5-4(3x) = 5-12x$. LHS=RHS, hence proved.
3. 17. 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 terms be $a-d, a, a+d$. Sum: $(a-d)+a+(a+d) = 3a = 27 \implies a=9$.

   Product: $(a-d)(a)(a+d) = 288 \implies (9-d)(9)(9+d)=288 \implies 81-d^2=32 \implies d^2=49 \implies d=\pm 7$.

   If d=7, terms are 2, 9, 16. If d=-7, terms are 16, 9, 2. The terms are 2, 9, 16.
4. 18. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
   Solution:

   For similar triangles, the ratio of perimeters equals the ratio of corresponding sides. $\frac{\text{Perimeter(ABC)}}{\text{Perimeter(PQR)}} = \frac{AB}{PQ}$.

   $\frac{36}{24} = \frac{AB}{10} \implies \frac{3}{2} = \frac{AB}{10} \implies AB = \frac{3 \times 10}{2} = 15$. So, AB = 15 cm.
5. 19. Find the sum to n terms of the series $5+55+555+\dots$
   Solution:

   $S_n = 5(1+11+111+\dots) = \frac{5}{9}(9+99+999+\dots)$.

   $S_n = \frac{5}{9}((10-1)+(10^2-1)+(10^3-1)+\dots)$.

   $S_n = \frac{5}{9}[(10+10^2+\dots+10^n) - n]$. The part in brackets is a GP.

   $S_n = \frac{5}{9}\left[ \frac{10(10^n-1)}{10-1} - n \right] = \frac{5}{9}\left[ \frac{10(10^n-1)}{9} - n \right]$.
6. 20. [Compulsory] Solve: $3x-2y+z = 2$, $2x+3y-z = 5$, $x+y+z = 6$.
   Solution:

   (1)+(2): $5x+y=7 \implies y=7-5x$.

   (2)+(3): $3x+4y=11$.

   Substitute y: $3x+4(7-5x)=11 \implies 3x+28-20x=11 \implies -17x=-17 \implies x=1$.

   Then $y=7-5(1)=2$. From (3), $1+2+z=6 \implies z=3$. Solution: $(1, 2, 3)$.
7. 21. Find the sum of the series $6^2+7^2+8^2+\dots+21^2$.
   Solution:

   Sum = $(1^2+2^2+\dots+21^2) - (1^2+2^2+\dots+5^2)$.

   Using formula $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.

   $\sum_{1}^{21} k^2 = \frac{21(22)(43)}{6} = 3311$.

   $\sum_{1}^{5} k^2 = \frac{5(6)(11)}{6} = 55$.

   Sum = $3311 - 55 = 3256$.

Part - D

1 x 8 = 8

Answer the following question.

1. 22.

   a) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{3}{5}$ of the corresponding sides of the triangle PQR. (Scale factor $\frac{3}{5} < 1$)

   (OR)

   b) Construct a triangle similar to a given triangle ABC with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle ABC. (Scale factor $\frac{7}{3} > 1$)

   Solution (Construction Steps):

   a) Construction for Scale Factor $\frac{3}{5} < 1$

   1. Draw any triangle PQR.
   2. Draw a ray QX from Q, making an acute angle with QR.
   3. Mark 5 (the larger number) equally spaced points on QX. Label them $Q_1, \dots, Q_5$.
   4. Join the last point, $Q_5$, to R.
   5. From $Q_3$ (the numerator), draw a line parallel to $Q_5R$ that intersects QR at R'.
   6. From R', draw a line parallel to RP that intersects QP at P'.
   7. $\triangle P'QR'$ is the required smaller triangle.

   ---

   b) Construction for Scale Factor $\frac{7}{3} > 1$

   1. Draw any triangle ABC.
   2. Extend the sides BA and BC.
   3. Draw a ray BX from B, making an acute angle with BC.
   4. Mark 7 (the larger number) equally spaced points on BX. Label them $B_1, \dots, B_7$.
   5. Join $B_3$ (the denominator) to C.
   6. From $B_7$ (the numerator), draw a line parallel to $B_3C$ that intersects the extended side BC at C'.
   7. From C', draw a line parallel to CA that intersects the extended side BA at A'.
   8. $\triangle A'BC'$ is the required larger triangle.