FIRST MID TERM TEST - 2024
MATHEMATICS
Question Paper
PART - A
1. If $A = \{x, y, z\}$ then the number of non-empty subsets of A is
2. If $B - A$ is $B$, then $A \cap B$ is
3. Which of the following is true?
4. In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?
5. Which of the following is an irrational number?
6. $0.\overline{34} + 0.3\overline{4}$ =
7. $4\sqrt{7} \times 2\sqrt{3} =$
8. The length and breadth of a rectangle plot are $5 \times 10^5$ and $4 \times 10^3$ metres respectively. Its area is
PART - B
9. Write the set of letters of the following words in Roster form:
i) ASSESSMENT ii) PRINCIPAL
10. If $S = \{\text{square, rectangle, circle, rhombus, triangle}\}$, list the elements of the following subset of S: The set of shapes which have 4 equal sides.
11. If $A = \{-3, -2, 1, 4\}$ and $B = \{0, 1, 2, 4\}$, find (i) $A - B$ (ii) $B - A$.
12. Find any three rational numbers between $-\frac{7}{11}$ and $\frac{2}{11}$.
13. Convert the decimal number $0.3\overline{5}$ in the form of $\frac{p}{q}$, where p and q are integers.
14. Write the number 625 in the form $5^n$.
15. Represent the number $0.0009000002$ in scientific notation.
16. Draw a Venn diagram and shade the region representing the set $(A - B)'$.
PART - C
17. Using the given Venn diagram, write the elements of: (i) $A \cup B$ (ii) $A \cap B$ (iii) $A - B$ (iv) $A'$ (v) $U$
18. Verify $A - (B \cup C) = (A - B) \cap (A - C)$ using Venn diagrams.
19. In a school, all students play either Hockey or Cricket or both. 300 play Hockey, 250 play Cricket and 110 play both games. Find (i) the number of students who play only Hockey (ii) the number of students who play only Cricket (iii) the total number of students in the school.
20. Represent $\sqrt{9.3}$ on a number line.
21. Represent $5.3\overline{48}$ on a number line using successive magnification.
22. Arrange in ascending order: $\sqrt[3]{2}, \sqrt{4}, \sqrt[4]{3}$.
23. Find the value of a and b if $\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b$.
24. If $A = \{b, c, e, g, h\}$, $B = \{a, c, d, g, i\}$ and $C = \{a, d, e, g, h\}$, then show that $A - (B \cap C) = (A - B) \cup (A - C)$.
Solutions
PART - A
1. The number of elements in set A is $n(A) = 3$. Total number of subsets = $2^n = 2^3 = 8$. These subsets are: $\emptyset, \{x\}, \{y\}, \{z\}, \{x,y\}, \{y,z\}, \{x,z\}, \{x,y,z\}$. The number of non-empty subsets = Total subsets - 1 (the empty set). Number of non-empty subsets = $8 - 1 = 7$. 4) 7
2. The expression $B - A$ means the set of elements that are in B but not in A. If $B - A = B$, it means that no element of A is present in B. This implies that the sets A and B are disjoint. The intersection of two disjoint sets is the empty set, $\emptyset$. 4) $\emptyset$
3. Let's evaluate the options. 1) $A - B = A \cap B'$ (by definition), not $A \cap B$. So, 1 is false. 2) $A - B$ is generally not equal to $B - A$. So, 2 is false. 3) $(A \cup B)' = A' \cap B'$ (De Morgan's Law). The option is $A' \cup B'$. So, 3 is false. 4) $(A \cap B)' = A' \cup B'$ (De Morgan's Law). This matches the option. So, 4 is true. 4) $(A \cap B)' = A' \cup B'$
4. The percentage of people who like at least one fruit is the sum of those who like only one, only two, or all three. Total percentage of people who like fruits = (Only one fruit) + (Only two fruits) + (All three fruits) $= 40\% + 35\% + 20\% = 95\%$. The percentage of people who do not like any fruit is $100\% - 95\% = 5\%$. 1) 5
5. Let's check each option. 1) $\sqrt{25} = 5$, which is a rational number. 2) $\sqrt{\frac{9}{4}} = \frac{3}{2}$, which is a rational number. 3) $\frac{7}{11}$ is in the form $\frac{p}{q}$, so it is a rational number. 4) $\pi$ is a non-terminating, non-recurring decimal, which is the definition of an irrational number. 4) $\pi$
6. The question is $0.\overline{34} + 0.\overline{34}$. Let $x = 0.\overline{34} = 0.343434...$ Then $100x = 34.343434...$ $100x - x = 34.3434... - 0.3434... \implies 99x = 34 \implies x = \frac{34}{99}$. So, the sum is $\frac{34}{99} + \frac{34}{99} = \frac{68}{99}$. Converting $\frac{68}{99}$ back to a decimal gives $0.686868... = 0.\overline{68}$. 3) $0.\overline{68}$
7. We multiply the rational parts and the irrational parts separately. $4\sqrt{7} \times 2\sqrt{3} = (4 \times 2) \times (\sqrt{7} \times \sqrt{3})$ $= 8 \times \sqrt{7 \times 3} = 8\sqrt{21}$. 2) $8\sqrt{21}$
8. Area of a rectangle = length $\times$ breadth. Length = $5 \times 10^5$ m. Breadth = $4 \times 10^3$ m. Area = $(5 \times 10^5) \times (4 \times 10^3)$ Area = $(5 \times 4) \times (10^5 \times 10^3)$ Area = $20 \times 10^{5+3} = 20 \times 10^8$ m². To write this in standard scientific notation: $20 \times 10^8 = 2 \times 10^1 \times 10^8 = 2 \times 10^9$ m².
PART - B
9. Roster form lists the unique elements of a set. i) ASSESSMENT: The unique letters are A, S, E, M, N, T. Roster form: {A, S, E, M, N, T} ii) PRINCIPAL: The unique letters are P, R, I, N, C, A, L. Roster form: {P, R, I, N, C, A, L}
10. We need to find shapes from the set $S$ that have 4 equal sides. $S = \{\text{square, rectangle, circle, rhombus, triangle}\}$ A square has 4 equal sides. A rhombus has 4 equal sides. The subset is: {square, rhombus}
11. Given $A = \{-3, -2, 1, 4\}$ and $B = \{0, 1, 2, 4\}$. (i) $A - B$ contains elements in A but not in B. $A - B = \{-3, -2, 1, 4\} - \{0, 1, 2, 4\} = \{-3, -2\}$. (i) $A - B = \{-3, -2\}$ (ii) $B - A$ contains elements in B but not in A. $B - A = \{0, 1, 2, 4\} - \{-3, -2, 1, 4\} = \{0, 2\}$. (ii) $B - A = \{0, 2\}$
12. We need three rational numbers between $-\frac{7}{11}$ and $\frac{2}{11}$. The integers between -7 and 2 are -6, -5, -4, -3, -2, -1, 0, 1. We can choose any three corresponding rational numbers with the same denominator. Three rational numbers are: $-\frac{6}{11}, 0, \frac{1}{11}$
13. Convert $0.3\overline{5}$ to the form $\frac{p}{q}$. Let $x = 0.3555...$ Multiply by 10: $10x = 3.555...$ (Equation 1) Multiply by 100: $100x = 35.555...$ (Equation 2) Subtract Equation 1 from Equation 2: $100x - 10x = 35.555... - 3.555...$ $90x = 32$ $x = \frac{32}{90}$ Simplify the fraction: $x = \frac{16}{45}$. $\frac{16}{45}$
14. We express 625 as a power of 5. $625 = 5 \times 125 = 5 \times 5 \times 25 = 5 \times 5 \times 5 \times 5 = 5^4$. $625 = 5^4$, so $n=4$.
15. To write $0.0009000002$ in scientific notation, we move the decimal point to be after the first non-zero digit. The first non-zero digit is 9. We move the decimal point 4 places to the right. $0.0009000002 = 9.000002 \times 10^{-4}$. $9.000002 \times 10^{-4}$
16. To draw the Venn diagram for $(A - B)'$:
1. First, identify the region $A - B$. This is the part of circle A that does not overlap with circle B.
2. The complement, $(A - B)'$, is everything in the universal set EXCEPT the $A - B$ region.
3. Therefore, we shade circle B and the entire area outside both circles A and B.
PART - C
17. From the Venn diagram: Universal set $U = \{1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 17\}$ Set $A = \{2, 4, 8, 10\}$ Set $B = \{4, 6, 9, 11\}$ (i) $A \cup B$ (all elements in A or B or both): $\{2, 4, 6, 8, 9, 10, 11\}$ (ii) $A \cap B$ (common elements): $\{4\}$ (iii) $A - B$ (elements in A but not B): $\{2, 8, 10\}$ (iv) $A'$ (elements in U but not A): $\{1, 3, 6, 7, 9, 11, 12, 17\}$ (v) $U$ (all elements in the diagram): $\{1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 17\}$
18. Verify $A - (B \cup C) = (A - B) \cap (A - C)$ using Venn diagrams.
LHS: $A - (B \cup C)$
1. Shade $B \cup C$ (all of B and all of C).
2. $A - (B \cup C)$ is the region of A that is NOT in the shaded $B \cup C$. This is the part of A that is completely separate from B and C.
RHS: $(A - B) \cap (A - C)$
1. Shade $A - B$ (part of A not in B).
2. Shade $A - C$ (part of A not in C).
3. The intersection is the area shaded in both cases. This is the part of A that is not in B AND not in C, which is the same region as the LHS.
Since the final shaded regions for both LHS and RHS are identical, the law is verified.
19. Let H be the set of students who play Hockey and C be the set for Cricket. Given: $n(H) = 300$, $n(C) = 250$, $n(H \cap C) = 110$. (i) Number of students who play only Hockey: $n(H \text{ only}) = n(H) - n(H \cap C) = 300 - 110 = 190$. (i) 190 students (ii) Number of students who play only Cricket: $n(C \text{ only}) = n(C) - n(H \cap C) = 250 - 110 = 140$. (ii) 140 students (iii) Total number of students in the school is $n(H \cup C)$. $n(H \cup C) = n(H) + n(C) - n(H \cap C) = 300 + 250 - 110 = 550 - 110 = 440$. (iii) 440 students
20. To represent $\sqrt{9.3}$ on a number line:
1. Draw a line and mark a point A on it.
2. Mark point B such that AB = 9.3 units.
3. From B, mark point C such that BC = 1 unit. Now AC = 10.3 units.
4. Find the midpoint of AC, let's call it O. $O = \frac{10.3}{2} = 5.15$ from A.
5. With O as the center and OA (or OC) as the radius, draw a semicircle.
6. Draw a line perpendicular to AC passing through B, intersecting the semicircle at D.
7. The length of the segment BD is equal to $\sqrt{9.3}$.
8. With B as the center and BD as the radius, draw an arc that cuts the original line at a point E. The point E represents $\sqrt{9.3}$ on the number line.
21. To represent $5.3\overline{48}$ ($=5.34848...$) on a number line using successive magnification:
Step 1: The number lies between 5 and 6. Draw a number line from 5 to 6 and divide it into 10 equal parts. The number is between 5.3 and 5.4.
Step 2: Magnify the interval [5.3, 5.4]. Divide it into 10 equal parts (5.31, 5.32, ...). The number $5.348...$ lies between 5.34 and 5.35.
Step 3: Magnify the interval [5.34, 5.35]. Divide it into 10 equal parts (5.341, 5.342, ...). The number $5.3484...$ lies between 5.348 and 5.349.
Step 4: Magnify the interval [5.348, 5.349]. Divide it into 10 equal parts. The number $5.3484...$ lies between 5.3484 and 5.3485.
This process can be continued indefinitely. The point is located by this infinite process of magnification.
22. Arrange in ascending order: $\sqrt[3]{2}, \sqrt{4}, \sqrt[4]{3}$. First, simplify $\sqrt{4} = 2$. The numbers are $\sqrt[3]{2}$, 2, and $\sqrt[4]{3}$. To compare them, we raise them to a common power. The LCM of the root indices (3, 1, 4) is 12. $(\sqrt[3]{2})^{12} = 2^{12/3} = 2^4 = 16$. $(2)^{12} = 4096$. $(\sqrt[4]{3})^{12} = 3^{12/4} = 3^3 = 27$. Comparing the results: $16 < 27 < 4096$. Therefore, the original numbers in order are $\sqrt[3]{2} < \sqrt[4]{3} < 2$. Ascending order: $\sqrt[3]{2}, \sqrt[4]{3}, \sqrt{4}$
23. Find a and b if $\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b$. We need to rationalize the denominator of the left side. Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{7}-2$. $$ \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} = \frac{(\sqrt{7}-2)^2}{(\sqrt{7})^2 - (2)^2} $$ Expand the numerator: $(\sqrt{7}-2)^2 = (\sqrt{7})^2 - 2(\sqrt{7})(2) + 2^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}$. Simplify the denominator: $7 - 4 = 3$. So, the expression becomes $\frac{11 - 4\sqrt{7}}{3} = \frac{11}{3} - \frac{4}{3}\sqrt{7}$. Now, compare this with $a\sqrt{7} + b$. We have $-\frac{4}{3}\sqrt{7} + \frac{11}{3} = a\sqrt{7} + b$. By comparing coefficients, we get $a = -\frac{4}{3}$ and $b = \frac{11}{3}$. $a = -4/3$, $b = 11/3$
24. Given $A = \{b, c, e, g, h\}$, $B = \{a, c, d, g, i\}$, $C = \{a, d, e, g, h\}$. Show that $A - (B \cap C) = (A - B) \cup (A - C)$. LHS: $A - (B \cap C)$ First, find $B \cap C$: $B \cap C = \{a, c, d, g, i\} \cap \{a, d, e, g, h\} = \{a, d, g\}$. Now, find $A - (B \cap C)$: $A - (B \cap C) = \{b, c, e, g, h\} - \{a, d, g\} = \{b, c, e, h\}$. LHS = {b, c, e, h} RHS: $(A - B) \cup (A - C)$ First, find $A - B$: $A - B = \{b, c, e, g, h\} - \{a, c, d, g, i\} = \{b, e, h\}$. Next, find $A - C$: $A - C = \{b, c, e, g, h\} - \{a, d, e, g, h\} = \{b, c\}$. Now, find the union $(A - B) \cup (A - C)$: $\{b, e, h\} \cup \{b, c\} = \{b, c, e, h\}$. RHS = {b, c, e, h} Since LHS = RHS, the identity is verified.