Algebra Class 10 - Section A Quiz

Section A

• Question 1

  • 1. Between -2 and -3
    • The number lies on the number line between -2 and -3.
  • ** 2. 10**
    • The sum of all interior angles of an n-sided polygon is (n-2) x 180 degrees.
    • Since the polygon is regular, all interior angles are equal to 140 degrees.
    • So, (n-2) x 180 = 140 x n. Solving for n, we get n = 10.
  • 3. 110 degrees
    • Let the smaller angle be x degrees. Then the larger angle is (2x - 30) degrees.
    • The sum of opposite angles of a parallelogram is 180 degrees. So, x + (2x-30) = 180.
    • Solving for x, we get x = 70 degrees. Therefore, the larger angle is 2*70 - 30 = 110 degrees.
  • 4. 2
    • A number is divisible by 3 if the sum of its digits is divisible by 3.
    • The sum of digits in 467x8 is 25 + x. So, x = 2 makes the sum divisible by 3.
  • 5. 840
    • Let the number be x. 5% of x is 42. So, 0.05x = 42
    • Solving for x, we get x = 840.
  • 6. {0, 4, 6, 8}
    • B-A means all elements in B that are not in A.
    • B is the set of even numbers. A is the set of prime numbers. So, B - A includes all even numbers except for 2, since 2 is a prime number.
  • 7. C
    • A convex polygon has all its interior angles less than 180 degrees.
  • 8. 0.3
    • the square root of 0.09 is 0.3.

• Question 2

  • A. selling price = ₹ 803
    • Discount = 6% of ₹ 850 = (6/100)*850 = ₹ 51.
    • Selling Price = Marked Price – Discount = ₹ 850 - ₹ 51 = ₹ 803.
  • B. x = 1
    • (4)^(3x+2) = (4)^(4-3x)
    • Since the bases are the same, the exponents must be equal. So, 3x+2 = 4-3x. Solving for x, we get x = 1.
  • C. A = 8, B = 3
    • From the rightmost column, we need 8 + B to give a units digit of 1. Therefore, B = 3.
    • Moving to the tens column, we have 6 + 8 + 5 = 19 + B (carry-over from previous column). Since we need the units digit to be 6, the carry-over from the hundreds column must be 7. This means A must be 8.

• Question 3

  • A. Construct a quadrilateral ABCD
    • Draw a line segment AB of length 3.5 cm.
    • At point B, construct an angle of 120 degrees.
    • Mark point C on the angle ray such that BC is 2.5 cm.
    • From point C, draw a perpendicular line segment (CD) of length 4.1 cm.
    • Join D to A. ABCD is the required quadrilateral.
  • B. X = 10
    • √900 = 30, and the equation states that √900 X = 30. Therefore, X = 1. Simplifying: 30X = 30. Therefore X = 1.
  • C. Segment AB
    • The shaded region is the line segment that connects points A and B, which is a part of the circle's circumference.
  • D. ∠x = 110 degrees
    • The angles on the straight line sum up to 180 degrees. Therefore, ∠x = 180 - 70 = 110 degrees.

Section B

• Question 4

  • A. 200 students are left out
    • The number of students in each row is equal to the number of rows, so the number of rows is √4500.
    • Since the square root of 4500 is not a whole number, we find the nearest integer that is less than √4500.
    • √4500 is approximately 67. So, 67 rows can be formed with 67*67 = 4489 students.
    • This leaves 4500 - 4489 = 200 students out.
  • B. ∠x = 90 degrees and ∠y = 55 degrees
    • Since PT is a tangent to the circle, angle OPT is 90 degrees.
    • The sum of angles in a triangle is 180 degrees. So, ∠x = 180 - 90 - 25 = 65 degrees.
    • Angle y and ∠x form a linear pair. Therefore, ∠y = 180 - ∠x = 115 degrees.

• Question 5

  • A. Five rational numbers between -3/4 and -4/5:
    • Find a common denominator for the fractions: -3/4 = -15/20 and -4/5 = -16/20.
    • Rational numbers between -15/20 and -16/20 could be: -14/20, -13/20, -12/20, -11/20, -10/20.
  • B. -1/8
    • Simplifying the expression inside the first brackets: (1/4) - (1/3) = -1/12.
    • Simplifying the expression in the second bracket: (1/2) - (1/3) = 1/6.
    • The final expression simplifies to: (-1/12) ÷ (1/6) = -1/2.

• Question 6

  • A. Find n(X∪Y), n(X∩Y), n(X’)
    • 𝜉 = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}.
    • X = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}.
    • Y = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14}.
    • X∪Y = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. So, n(X∪Y) = 16.
    • X∩Y = {10, 11, 12, 13, 14}. So, n(X∩Y) = 5.
    • X’ = {5, 6, 7, 8, 9}. So, n(X’) = 5.
  • B. 10; cube root of product is 52
    • Find the prime factorization of 10584: 2 x 2 x 2 x 1323= 2³ x 7 x 11 x 11 = 2³ x 7 x 11².
    • To make it a perfect cube, we need one more 7 and one more 11. So, multiply by 7 x 11 = 77.
    • The cube root of product is 2 x 7 x 11 = 154, which is the cube root of 10584 x 77.

• Question 7

  • A. Multiplicative inverse of 4/5 and 2^-4
    • The multiplicative inverse of a number is its reciprocal. The reciprocal of 4/5 is 5/4.
    • The multiplicative inverse of 2^-4 is 2^4 = 16. (Recall that the multiplicative inverse of x^n is x^-n).
  • B. The numbers are 43 and 34
    • Let the ten's digit be x and the unit's digit be y. The number is 10x + y.
    • The sum of digits is 7 means x+y = 7.
    • The number obtained by reversing the digits is 10y + x, and their difference is 9: 10x + y - (10y + x) = 9.
    • Simplify the equations, and solve for x and y. We get x = 4 and y = 3. So, the numbers are 43 and 34.

• Question 8

  • A. Breadth of rectangular field is 3/4 m
    • Area of rectangle = length x breadth.
    • Given that area = 10 3/12 m² = 123/12m² and length = 2m. Substitute these values and solve for breadth: 123/12 = 2 x breadth.
    • Therefore, breadth = 123/24 = 3/4 m.
  • B. Price before GST is 5000
    • Let the original price be x.
    • The price including 8% GST is 1.08x, which is equal to ₹5400: 1.08x = 5400.
    • Solving for x: x = 5400 / 1.08 = ₹ 5000.

• Question 9

  • A. Incomplete question
    • The question is incomplete and requires additional information to be solved.