Virudhunagar District Common First Mid Term Test - 2024
Standard 10 - MATHS
Time: 1.30 Hrs. | Marks: 50
Question Paper
- 1
- 2
- 3
- 6
- linear
- cubic
- reciprocal
- quadratic
- 1
- 2
- 3
- 4
- 0
- 6
- 7
- 13
- 7
- 49
- 1
- 14
- 2025
- 5220
- 5025
- 2520
- \( \frac{1}{24} \)
- \( \frac{1}{27} \)
- \( \frac{2}{3} \)
- \( \frac{1}{81} \)
8) If \( B \times A = \{(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)\} \) find A and B.
9) If \( A = \{-2, -1, 0, 1, 2\} \) and \( f: A \to B \) is an onto function defined by \( f(x) = x^2 + x + 1 \) then find B.
10) Compute x, such that \( 10^4 \equiv x \pmod{19} \).
11) Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
12) If \( 1+2+3+\dots+k = 325 \), then find \( 1^3+2^3+3^3+\dots+k^3 \).
13) 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.
14) Find the sum \( 3+1+\frac{1}{3}+\dots \).
15) Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that \( (A \cap B) \times C = (A \times C) \cap (B \times C) \).
16) Let \( A = \{1, 2, 3, 4\} \) and \( B = \{2, 5, 8, 11, 14\} \) be two sets. Let \( f: A \to B \) be a function given by \( f(x) = 3x-1 \). Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.
17) Find the sum of all natural numbers between 600 and 800 which are divisible by 11.
18) Find the sum to n terms of the series \( 5+55+555+\dots \).
19) 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?
20) The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
21) If \( f(x) = x-4 \), \( g(x) = x^2 \) and \( h(x) = 3x-5 \), prove that \( (f \circ g) \circ h = f \circ (g \circ h) \).
22) Construct a triangle similar to a given triangle PQR with its sides equal to \( \frac{7}{4} \) of the corresponding sides of the triangle PQR. (scale factor \( \frac{7}{4} > 1 \)).
Construct a triangle similar to a given triangle LMN with its sides equal to \( \frac{4}{5} \) of the corresponding sides of the triangle LMN. (scale factor \( \frac{4}{5} < 1 \)).
Solutions
1. If \( n(A \times B) = 6 \) and \( A = \{1, 3\} \) then \( n(B) \) is
Given, \( n(A \times B) = 6 \) and \( A = \{1, 3\} \), so \( n(A) = 2 \).
Substituting the values: \( 6 = 2 \times n(B) \).
Therefore, \( n(B) = \frac{6}{2} = 3 \).
Correct answer: c) 3
2. \( f(x) = (x+1)^3 - (x-1)^3 \) represents a function which is
Using the formulas \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \) and \( (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \).
\( f(x) = (x^3 + 3x^2(1) + 3x(1)^2 + 1^3) - (x^3 - 3x^2(1) + 3x(1)^2 - 1^3) \)
\( f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \)
\( f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \)
\( f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1) \)
\( f(x) = 6x^2 + 2 \)
Since the highest power of x is 2, the function is a quadratic function.
Correct answer: d) quadratic
3. The sum of the exponents of the prime factors in the prime factorization of 1729 is
1729 is divisible by 7: \( 1729 \div 7 = 247 \).
247 is divisible by 13: \( 247 \div 13 = 19 \).
19 is a prime number.
So, the prime factorization is \( 1729 = 7^1 \times 13^1 \times 19^1 \).
The exponents of the prime factors are 1, 1, and 1.
The sum of the exponents is \( 1 + 1 + 1 = 3 \).
Correct answer: c) 3
4. 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
The nth term of an A.P is given by \( t_n = a + (n-1)d \).
Given: \( 6 \times t_6 = 7 \times t_7 \)
\( 6(a + (6-1)d) = 7(a + (7-1)d) \)
\( 6(a + 5d) = 7(a + 6d) \)
\( 6a + 30d = 7a + 42d \)
\( 6a - 7a = 42d - 30d \)
\( -a = 12d \) or \( a + 12d = 0 \).
We need to find the 13th term, \( t_{13} \).
\( t_{13} = a + (13-1)d = a + 12d \).
Since we found \( a + 12d = 0 \), the 13th term is 0.
Correct answer: a) 0
5. If \( f: A \to B \) is a bijective function and if \( n(B) = 7 \), then \( n(A) \) is equal to
For a function to be bijective between two finite sets, the number of elements in the domain must be equal to the number of elements in the codomain.
Therefore, \( n(A) = n(B) \).
Given \( n(B) = 7 \), so \( n(A) = 7 \).
Correct answer: a) 7
6. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
First, write the prime factorization of each number:
1 = 1
2 = 2
3 = 3
4 = \(2^2\)
5 = 5
6 = 2 x 3
7 = 7
8 = \(2^3\)
9 = \(3^2\)
10 = 2 x 5
To find the LCM, take the highest power of each prime factor present: \(2^3, 3^2, 5^1, 7^1\).
LCM = \( 2^3 \times 3^2 \times 5 \times 7 \)
LCM = \( 8 \times 9 \times 5 \times 7 \)
LCM = \( 72 \times 35 \)
LCM = 2520.
Correct answer: d) 2520
7. The next term of the sequence \( \frac{1}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, \dots \) is
Ratio 1: \( \frac{t_2}{t_1} = \frac{1/8}{1/16} = \frac{16}{8} = 2 \).
Ratio 2: \( \frac{t_3}{t_2} = \frac{1/12}{1/8} = \frac{8}{12} = \frac{2}{3} \).
Ratio 3: \( \frac{t_4}{t_3} = \frac{1/18}{1/12} = \frac{12}{18} = \frac{2}{3} \).
The pattern is that after the first term, each subsequent term is obtained by multiplying the previous term by \( \frac{2}{3} \).
So, the next term \( t_5 \) will be \( t_4 \times \frac{2}{3} \).
\( t_5 = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27} \).
Correct answer: b) \( \frac{1}{27} \)
8. If \( B \times A = \{(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)\} \) find A and B.
Set B is the set of all first coordinates: \( B = \{-2, 0, 3\} \).
Set A is the set of all second coordinates: \( A = \{3, 4\} \).
A = {3, 4}, B = {-2, 0, 3}
9. If \( A = \{-2, -1, 0, 1, 2\} \) and \( f: A \to B \) is an onto function defined by \( f(x) = x^2 + x + 1 \) then find B.
We need to find the value of \( f(x) \) for each element in A.
\( f(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3 \)
\( f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1 \)
\( f(0) = (0)^2 + 0 + 1 = 1 \)
\( f(1) = (1)^2 + 1 + 1 = 1 + 1 + 1 = 3 \)
\( f(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7 \)
The range of the function is the set of unique values: \( \{1, 3, 7\} \).
Since the function is onto, \( B = \text{Range} \).
B = {1, 3, 7}
10. Compute x, such that \( 10^4 \equiv x \pmod{19} \).
Let's calculate powers of 10 modulo 19.
\( 10^1 \equiv 10 \pmod{19} \)
\( 10^2 = 100 \). When 100 is divided by 19, \( 100 = 5 \times 19 + 5 \). So, \( 10^2 \equiv 5 \pmod{19} \).
Now, we can find \( 10^4 \):
\( 10^4 = (10^2)^2 \equiv 5^2 \pmod{19} \)
\( 10^4 \equiv 25 \pmod{19} \)
When 25 is divided by 19, \( 25 = 1 \times 19 + 6 \). So, \( 25 \equiv 6 \pmod{19} \).
Therefore, \( 10^4 \equiv 6 \pmod{19} \).
x = 6
11. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
In an A.P., any term is the arithmetic mean of its neighbours. So, y is the arithmetic mean of 10 and 24.
\( y = \frac{10 + 24}{2} = \frac{34}{2} = 17 \).
Now we can find the common difference (d):
\( d = t_3 - t_2 = 17 - 10 = 7 \). (Check: \( t_4 - t_3 = 24 - 17 = 7 \)).
Now we can find x and z:
\( x = t_1 = t_2 - d = 10 - 7 = 3 \).
\( z = t_5 = t_4 + d = 24 + 7 = 31 \).
x = 3, y = 17, z = 31
12. If \( 1+2+3+\dots+k = 325 \), then find \( 1^3+2^3+3^3+\dots+k^3 \).
\( \sum_{i=1}^{k} i = 1+2+3+\dots+k = 325 \).
We need to find the sum of the cubes of the first k natural numbers:
\( \sum_{i=1}^{k} i^3 = 1^3+2^3+3^3+\dots+k^3 \).
The formula for the sum of cubes is related to the sum of numbers:
$$ \sum_{i=1}^{k} i^3 = \left( \sum_{i=1}^{k} i \right)^2 $$ Substituting the given value:
$$ 1^3+2^3+3^3+\dots+k^3 = (325)^2 $$ $$ (325)^2 = 105625 $$ The sum is 105625.
13. 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.
Domain = {0, 1, 2, 3, 4, 5}
The range is the set of all resulting output values for y. We calculate y for each x in the domain using \( y = x+3 \).
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 set of y values is the range.
Range = {3, 4, 5, 6, 7, 8}
14. Find the sum \( 3+1+\frac{1}{3}+\dots \).
The first term is \( a = 3 \).
The common ratio is \( r = \frac{t_2}{t_1} = \frac{1}{3} \).
Since the absolute value of the common ratio \( |r| = |\frac{1}{3}| < 1 \), the sum to infinity exists.
The formula for the sum to infinity of a G.P. is \( S_\infty = \frac{a}{1-r} \).
$$ S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} $$ The sum is \( \frac{9}{2} \) or 4.5.
15. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that \( (A \cap B) \times C = (A \times C) \cap (B \times C) \).
A = {Natural numbers < 8} = \( \{1, 2, 3, 4, 5, 6, 7\} \)
B = {Prime numbers < 8} = \( \{2, 3, 5, 7\} \)
C = {Even prime number} = \( \{2\} \)
LHS: \( (A \cap B) \times C \)
First, find \( A \cap B \):
\( A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 3, 5, 7\} = \{2, 3, 5, 7\} \)
Now, find the Cartesian product with C:
\( (A \cap B) \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\} \)
RHS: \( (A \times C) \cap (B \times C) \)
First, find \( A \times C \):
\( A \times C = \{1, 2, 3, 4, 5, 6, 7\} \times \{2\} = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)\} \)
Next, find \( B \times C \):
\( B \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\} \)
Now, find the intersection of these two sets:
\( (A \times C) \cap (B \times C) = \{(2, 2), (3, 2), (5, 2), (7, 2)\} \)
Verification:
LHS = \( \{(2, 2), (3, 2), (5, 2), (7, 2)\} \)
RHS = \( \{(2, 2), (3, 2), (5, 2), (7, 2)\} \)
Since LHS = RHS, the property is verified.
16. Let \( A = \{1, 2, 3, 4\} \) and \( B = \{2, 5, 8, 11, 14\} \) be two sets. Let \( f: A \to B \) be a function given by \( f(x) = 3x-1 \). Represent this function.
\( f(1) = 3(1) - 1 = 2 \)
\( f(2) = 3(2) - 1 = 5 \)
\( f(3) = 3(3) - 1 = 8 \)
\( f(4) = 3(4) - 1 = 11 \)
(i) By arrow diagram:
| x | f(x) = 3x - 1 |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 11 |
\( f = \{(1, 2), (2, 5), (3, 8), (4, 11)\} \) (iv) In a graphical form:
Plot the points (1, 2), (2, 5), (3, 8), and (4, 11) on a Cartesian plane.
17. Find the sum of all natural numbers between 600 and 800 which are divisible by 11.
First term (a): The first number greater than 600 divisible by 11. \( 600 \div 11 \approx 54.54 \). So, the next integer is 55. \( 11 \times 55 = 605 \). Thus, \( a = 605 \).
Last term (l): The last number less than 800 divisible by 11. \( 800 \div 11 \approx 72.72 \). So, the previous integer is 72. \( 11 \times 72 = 792 \). Thus, \( l = 792 \).
The common difference \( d = 11 \).
Number of terms (n): Using the formula \( l = a + (n-1)d \).
\( 792 = 605 + (n-1)11 \)
\( 792 - 605 = (n-1)11 \)
\( 187 = (n-1)11 \)
\( n-1 = \frac{187}{11} = 17 \)
\( n = 17 + 1 = 18 \).
Sum of the terms (S_n): Using the formula \( S_n = \frac{n}{2}(a+l) \).
\( S_{18} = \frac{18}{2}(605 + 792) \)
\( S_{18} = 9(1397) \)
\( S_{18} = 12573 \).
The sum is 12573.
18. Find the sum to n terms of the series \( 5+55+555+\dots \).
Take 5 as a common factor: $$ S_n = 5(1+11+111+\dots+ n \text{ terms}) $$ Multiply and divide by 9: $$ S_n = \frac{5}{9}(9+99+999+\dots+ n \text{ terms}) $$ Rewrite the terms inside the bracket: $$ S_n = \frac{5}{9}((10-1)+(100-1)+(1000-1)+\dots+ n \text{ terms}) $$ Group the powers of 10 and the -1s: $$ S_n = \frac{5}{9}((10+10^2+10^3+\dots+ n \text{ terms}) - (1+1+1+\dots+ n \text{ times})) $$ The first part in the bracket is a G.P. with first term \( a=10 \) and common ratio \( r=10 \). The sum is \( \frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10(10^n-1)}{9} \). The second part is simply \( n \).
$$ S_n = \frac{5}{9}\left[\frac{10(10^n-1)}{9} - n\right] $$ \( S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9} \)
19. 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?
The areas of these squares are \( 10^2, 11^2, 12^2, \dots, 24^2 \).
The total area is the sum of these squares: $$ \text{Total Area} = 10^2 + 11^2 + \dots + 24^2 $$ We use the formula for the sum of squares of first n natural numbers: \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \).
We can write the required sum as: $$ (\sum_{k=1}^{24} k^2) - (\sum_{k=1}^{9} k^2) $$ Sum up to 24: $$ \sum_{k=1}^{24} k^2 = \frac{24(24+1)(2 \cdot 24+1)}{6} = \frac{24 \cdot 25 \cdot 49}{6} = 4 \cdot 25 \cdot 49 = 100 \cdot 49 = 4900 $$ Sum up to 9: $$ \sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \cdot 9+1)}{6} = \frac{9 \cdot 10 \cdot 19}{6} = 3 \cdot 5 \cdot 19 = 285 $$ Total Area = 4900 - 285 = 4615.
The total area is 4615 cm².
20. The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
Sum of the terms: \( (a-d) + a + (a+d) = 27 \)
\( 3a = 27 \)
\( a = 9 \)
Product of the terms: \( (a-d) \cdot a \cdot (a+d) = 288 \)
Substitute \( a=9 \): \( (9-d) \cdot 9 \cdot (9+d) = 288 \)
\( (9-d)(9+d) = \frac{288}{9} \)
\( 81 - d^2 = 32 \)
\( d^2 = 81 - 32 = 49 \)
\( d = \pm\sqrt{49} = \pm 7 \)
If \( d = 7 \), the terms are \( 9-7, 9, 9+7 \), which are \( 2, 9, 16 \).
If \( d = -7 \), the terms are \( 9-(-7), 9, 9-7 \), which are \( 16, 9, 2 \).
In both cases, the set of terms is the same.
The three terms are 2, 9, and 16.
21. If \( f(x) = x-4 \), \( g(x) = x^2 \) and \( h(x) = 3x-5 \), prove that \( (f \circ g) \circ h = f \circ (g \circ h) \).
LHS: \( (f \circ g) \circ h \)
First, find \( f \circ g \): \( (f \circ g)(x) = f(g(x)) = f(x^2) = x^2 - 4 \).
Now, apply h to this function: \( ((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(3x-5) \)
Substitute \( 3x-5 \) into the expression for \( (f \circ g)(x) \): \( = (3x-5)^2 - 4 \)
\( = (9x^2 - 30x + 25) - 4 \)
\( = 9x^2 - 30x + 21 \)
RHS: \( f \circ (g \circ h) \)
First, find \( g \circ h \): \( (g \circ h)(x) = g(h(x)) = g(3x-5) = (3x-5)^2 \).
Now, apply this function to f: \( (f \circ (g \circ h))(x) = f((g \circ h)(x)) = f((3x-5)^2) \)
Substitute \( (3x-5)^2 \) into the expression for \( f(x) \): \( = (3x-5)^2 - 4 \)
\( = (9x^2 - 30x + 25) - 4 \)
\( = 9x^2 - 30x + 21 \)
Verification:
LHS = \( 9x^2 - 30x + 21 \)
RHS = \( 9x^2 - 30x + 21 \)
Since LHS = RHS, it is proved that \( (f \circ g) \circ h = f \circ (g \circ h) \).
22. This question asks for the steps of geometric construction. The solution provides a description of the steps.
- Draw the given triangle \( \triangle PQR \).
- Draw a ray QX starting from Q, making an acute angle with the side QR and on the side opposite to vertex P.
- Locate 7 points (the greater of 7 and 4) \( Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7 \) on the ray QX such that \( QQ_1 = Q_1Q_2 = \dots = Q_6Q_7 \).
- Join \( Q_4 \) (the 4th point, corresponding to the denominator) to R.
- Draw a line through \( Q_7 \) (the 7th point, corresponding to the numerator) parallel to \( Q_4R \). This line will intersect the extended line segment QR at a point R'.
- Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
- The resulting triangle, \( \triangle P'QR' \), is the required similar triangle which is an enlargement of \( \triangle PQR \).
- Draw the given triangle \( \triangle LMN \).
- Draw a ray MX starting from M, making an acute angle with the side MN and on the side opposite to vertex L.
- Locate 5 points (the greater of 4 and 5) \( M_1, M_2, M_3, M_4, M_5 \) on the ray MX such that \( MM_1 = M_1M_2 = \dots = M_4M_5 \).
- Join \( M_5 \) (the 5th point, corresponding to the denominator) to N.
- Draw a line through \( M_4 \) (the 4th point, corresponding to the numerator) parallel to \( M_5N \). This line will intersect the line segment MN at a point N'.
- Draw a line through N' parallel to LN. This line will intersect the line segment LM at a point L'.
- The resulting triangle, \( \triangle L'MN' \), is the required similar triangle which is a reduction of \( \triangle LMN \).