Odisha B.Ed Entrance Mathematics Solutions 2023 Download PDF

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1 Odisha B.Ed Entrance Mathematics Solutions 2023

Odisha B.Ed Entrance Mathematics Solutions 2023

In Odisha B.Ed Entrance Exam there will be 20 Questions from Mathematics Section each carrying 1 mark. On correct answer you will get 1 mark and for Incorrect answer there will be negative marking of 0.25 mark. Below we have provided questions from the Mathematics section from 2023 batch, with there Answer and Solution.

For Other Batch / Year Solutions refer the table given at the last of this article.

41.

A rational number $\frac{p}{q}$ ( $q \neq 0$ ) can be expressed as a terminating decimal, if the prime factorization of $q$ is of the form:

Incorrect $2^m 3^n$ (where $m$ and $n$ are non-negative integers)
Correct $2^m 5^n$ (where $m$ and $n$ are non-negative integers)
Incorrect $3^m 5^n$ (where $m$ and $n$ are non-negative integers)
Incorrect $5^m 7^n$ (where $m$ and $n$ are non-negative integers)

Solution: The correct answer is: $2^m 5^n$ (where $m$ and $n$ are non-negative integers). A rational number can be expressed as a terminating decimal if and only if its denominator ( $q$ ) is in the form of $2^m 5^n$ , where $m$ and $n$ are non-negative integers. This ensures that all the prime factors of $q$ are either 2 or 5, leading to a finite number of digits in the decimal expansion.

42.

The difference between the two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as the remainder. Which one of the following is the number?

Incorrect360
Incorrect295
Correct270
Incorrect240

Solution: The correct answer is: (C) 270. Let’s denote the larger number as $L$ and the smaller number as $S$ . According to the given information, we have:

$L - S = 1365$

And, on dividing the larger number by the smaller number, we get:

$L = 6S + 15$

Substitute the value of $L$ from the second equation into the first equation:

$6S + 15 - S = 1365$

Solving for $S$ , we get:

$5S = 1350 \Rightarrow S = 270$

So, the correct answer is indeed 270, which is option (C).

43.

What is the number of proper subsets of the set $P \cup Q$ , where $P = \{1, 2, 5, 7\}$ and $Q = \{2, 4, 6\}$ ?

Incorrect31
Correct63
Incorrect64
Incorrect127

Solution: The correct answer is: (B) 63.

The number of proper subsets of a set with $n$ elements is $2^n - 1$ , where $n$ is the number of elements in the set.

Here, the set $P \cup Q$ is the union of sets $P$ and $Q$ , which contains all unique elements from both $P$ and $Q$ .

The number of elements in $P \cup Q$ is $|P \cup Q| = |P| + |Q| - |P \cap Q|$ .

Given that $P = \{1, 2, 5, 7\}$ and $Q = \{2, 4, 6\}$ , we have $|P| = 4$ , $|Q| = 3$ , and $|P \cap Q| = 1$ (since 2 is the only common element).

Therefore, $|P \cup Q| = 4 + 3 - 1 = 6$ .

The number of proper subsets of $P \cup Q$ is $2^6 - 1 = 64 - 1 = 63$ .

So, the correct answer is 63, which is option (B).

44.

If the set P contains 5 elements and the set Q contains 6 elements, then what is the number of one-one and onto functions from P to Q?

Incorrect 720
Incorrect 120
Incorrect 30
Correct 0

Solution : The correct answer is: (D) 0

For a function to be both one-one (injective) and onto (surjective), each element in set Q must be mapped to by exactly one element in set P.

Since set P contains 5 elements and set Q contains 6 elements, and the function must be onto, there are 6 possible elements in Q that need to be mapped to.

Now, let’s count the number of possible one-one and onto functions:

For the first element in set Q, we have 5 choices in set P to map it to.

For the second element in set Q, we have 4 choices in set P to map it to (since we can’t use the element that was already mapped to the first element in set Q).

For the third element in set Q, we have 3 choices in set P.

For the fourth element in set Q, we have 2 choices in set P.

For the fifth element in set Q, we have 1 choice in set P.

For the sixth element in set Q, we have 0 choices in set P (since all the elements in set P are already used).

Therefore, the total number of one-one and onto functions from set P to set Q is: 5 × 4 × 3 × 2 × 1 × 0 = 0.

45.

How many equivalence relations can be defined on the set S = {1, 2, 3}?

Incorrect 3
Correct 5
Incorrect 7
Incorrect 8

Solution : The correct answer is: (B) 5

An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive.

For the set S = {1, 2, 3}, let’s consider the possible equivalence relations:

1. The relation where every element is related to itself (reflexive property).

2. The relation where all elements are related to each other (symmetric property).

3. The relation where no elements are related (empty relation).

4. The relation where each element is related to itself and to one other element (reflexive and symmetric).

5. The relation where each element is related to itself and to both other elements (reflexive and symmetric).

Therefore, there are 5 possible equivalence relations that can be defined on the set S = {1, 2, 3}.

46.

Suresh, Dinesh and Ramesh became partners in a business by investing money in the ratio 3 : 6 : 8. If their investments are increased by 5%, 15% and 20% respectively, then what will be the ratio of their profits in one year?

Incorrect 7 : 46 : 64
Incorrect 19 : 46 : 64
Correct 21 : 46 : 64
Incorrect 35 : 46 : 64

Solution : The correct answer is: (C) 21 : 46 : 64

Let the original investments of Suresh, Dinesh, and Ramesh be 3x, 6x, and 8x respectively.

After increasing their investments by the given percentages, the new investments become:

Suresh: 3x + 0.05 * 3x = 3.15x

Dinesh: 6x + 0.15 * 6x = 6.9x

Ramesh: 8x + 0.20 * 8x = 9.6x

The ratio of their new investments is 3.15x : 6.9x : 9.6x, which simplifies to 21 : 46 : 64.

Since the ratio of their investments is also the ratio of their profits, the correct ratio of their profits in one year will be 21 : 46 : 64.

47.

In a bag, there are coins of 5 ps, 10 ps and 25 ps in the ratio 3 : 2 : 1. If all the coins in the bag amount to Rs. 60.00, then how many 5 ps coins are there?

Incorrect 100
Incorrect 200
Correct 300
Incorrect 400

Solution : The correct answer is: (C) 300

Let’s assume the common ratio as ‘x’.

So, the number of 5 ps coins = 3x

The number of 10 ps coins = 2x

The number of 25 ps coins = 1x

The total value of the coins is given as Rs. 60, which is 60 × 100 paisa.

So, (3x × 5) + (2x × 10) + (1x × 25) = 60 × 100

Solving the equation: 15x + 20x + 25x = 6000

60x = 6000

x = 100

Therefore, the number of 5 ps coins = 3x = 3 × 100 = 300.

Hence, there are 300 coins of 5 ps.

48.

If $\log_x y = 100$ and $\log_2 x = 10$ , then what is the value of $y$ ?

Incorrect 2
Incorrect $2^{10}$
Incorrect $2^{100}$
Correct $2^{1000}$

Solution : The correct answer is: (D) $2^{1000}$

Given $\log_x y = 100$ and $\log_2 x = 10$ .

We can rewrite $\log_x y = 100$ as $y = x^{100}$ .

Also, $\log_2 x = 10$ can be rewritten as $x = 2^{10}$ .

Substituting the value of $x$ in $y = x^{100}$ , we get $y = (2^{10})^{100} = 2^{1000}$ .

Therefore, the value of $y$ is $2^{1000}$ .

49.

What is the value of $(256)^{0.16} \times (256)^{0.09}$ ?

Incorrect 256.25
Correct 4
Incorrect 16
Incorrect 64

Solution : The correct answer is: 4 (Option B)

We can use the properties of exponents to simplify this expression. When multiplying numbers with the same base, you can add the exponents:

$(256)^{0.16} \times (256)^{0.09} = 256^{0.16 + 0.09} = 256^{0.25}$

Now, $256^{0.25}$ is the same as taking the fourth root of 256:

$256^{0.25} = \sqrt[4]{256} = 4$

So, the value of the expression is 4.

50.

If $\tan \theta = \frac{3}{4}$ for some $0 < \theta < 90^\circ$ , then what is the value of $\sin \theta$ ?

Incorrect $\frac{1}{5}$
Incorrect $\frac{2}{5}$
Correct $\frac{3}{5}$
Incorrect $\frac{4}{5}$

Solution : The correct answer is: $\frac{3}{5}$ (Option C)

We are given that $\tan \theta = \frac{3}{4}$ . In a right triangle, the tangent of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

Let’s assume the opposite side is $3x$ and the adjacent side is $4x$ . Then, we have:

$\tan \theta = \frac{3x}{4x} = \frac{3}{4}$

Using the Pythagorean theorem, we can find the length of the hypotenuse:

$(3x)^2 + (4x)^2 = h^2$

$9x^2 + 16x^2 = h^2$

$25x^2 = h^2$

$h = 5x$

The sine of the angle $\theta$ is the ratio of the length of the opposite side to the length of the hypotenuse:

$\sin \theta = \frac{3x}{5x} = \frac{3}{5}$

So, the value of $\sin \theta$ is $\frac{3}{5}$ .

51.

If $\log_3(x^4 - x^3) - \log_3(x - 1) = 3$ , then what is the value of $x$ ?

Incorrect 9
Incorrect 6
Correct 3
Incorrect 1

Solution : The correct answer is: 3 (Option C)

We are given the equation $\log_3(x^4 - x^3) - \log_3(x - 1) = 3$ .

Using the properties of logarithms, we can combine the logarithms:

$\log_3 \left( \frac{x^4 - x^3}{x - 1} \right) = 3$

Now, we can rewrite the equation as an exponential equation:

$3^3 = \frac{x^4 - x^3}{x - 1}$

$27 = \frac{x^3(x - 1)}{x - 1}$

$27 = x^3$

$x = \sqrt[3]{27} = 3$

So, the value of $x$ is 3.

52.

What is the value of the expression:

$\sin^2(1^\circ) + \sin^2(11^\circ) + \sin^2(21^\circ) + \sin^2(31^\circ) +$ $\sin^2(41^\circ) + \sin^2(45^\circ) + \sin^2(49^\circ) + \sin^2(59^\circ) +$ $\sin^2(69^\circ) + \sin^2(79^\circ) + \sin^2(89^\circ) ?$

Incorrect $4$
Incorrect $4 \frac{1}{2}$
Incorrect $5$
Correct $5 \frac{1}{2}$

Solution : The correct answer is: $5 \frac{1}{2}$ (Option D)

We have a series of sine squared terms with angles in degrees. Notice that the sum of the angles in each pair of complementary angles adds up to 90°. For example, $\sin^2(1^\circ) + \sin^2(89^\circ) = \sin^2(1^\circ) + \cos^2(1^\circ) = 1$ .

Using this property for each pair of complementary angles:

$\sin^2(1^\circ) + \sin^2(89^\circ) = 1$

$\sin^2(11^\circ) + \sin^2(79^\circ) = 1$

$\sin^2(21^\circ) + \sin^2(69^\circ) = 1$

$\sin^2(31^\circ) + \sin^2(59^\circ) = 1$

$\sin^2(41^\circ) + \sin^2(49^\circ) = 1$

$\sin^2(45^\circ) = \frac{1}{2}$

Now, adding all these equations:

$5 \times 1 + \frac{1}{2} = 5 \frac{1}{2}$

So, the value of the given expression is $5 \frac{1}{2}$ .

53.

If $S$ is the set of all distinct numbers of the form $\frac{p}{q}$ , where $p, q \in \{1, 2, 3, 4, 5, 6\}$ , then what is the total number of elements in $S$ ?

Incorrect 21
Correct 23
Incorrect 32
Incorrect 36

Solution : The correct answer is: 23 (Option B)

We need to find the number of distinct fractions of the form $\frac{p}{q}$ , where $p, q \in \{1, 2, 3, 4, 5, 6\}$ . To find the total number of distinct fractions, we need to consider all possible combinations of p and q and account for duplicates.

Here’s the approach:

For each $p$ value in the set $\{1, 2, 3, 4, 5, 6\}$ , there are 6 possible $q$ values (since $q$ can also take values from the same set).
However, there will be duplicates due to simplification. To account for this, we need to find the greatest common divisor (GCD) of $p$ and $q$ . If GCD is greater than 1, then the fraction is not in its simplest form, and we should skip it.

Let’s calculate the number of distinct fractions using this approach:

Number of distinct fractions = Total combinations of ( $p$ , $q$ ) − Duplicates.

For each $p$ value, there are 6 possible $q$ values, giving a total of $6 \times 6 = 36$ combinations.

However, there are duplicates due to simplification. To calculate duplicates, we need to find fractions that are not in their simplest form (GCD > 1).

Calculating duplicates for each $p$ value:

For $p = 1$ , the duplicates are $\frac{1}{1} \frac{2}{2} \frac{3}{3} \frac{4}{4} \frac{5}{5} \frac{6}{6}$ (total 6 duplicates).
For $p = 2$ , the duplicates are $\frac{2}{1} \frac{4}{2} \frac{6}{3}$ (total 3 duplicates).
For $p = 3$ , the duplicates are $\frac{3}{1}$ (total 1 duplicates).
For $p = 4$ , the duplicates are $\frac{4}{1}$ (total 1 duplicates).
For $p = 5$ , the duplicates are $\frac{5}{1}$ (total 1 duplicates).
For $p = 6$ , the duplicates are $\frac{6}{1}$ (total 1 duplicates).

Total number of duplicates $= 6 + 3 + 1 + 1 + 1 + 1 = 13$ .

So, the number of distinct fractions $= 36 - 13 = 23$ .

54.

The angle of elevation of a ladder leaning against a wall is 60°. If the foot of the ladder is 4.6 m away from the wall, then what is the length/height of the ladder?

Incorrect 2.3 m
Incorrect 4.7 m
Incorrect 7.8 m
Correct 9.2 m

Solution : The correct answer is (D) 9.2 m.

Let’s consider the situation: A ladder is leaning against a wall with an angle of elevation of 60°. The foot of the ladder is 4.6 m away from the wall.

We need to find the length of the ladder (height).

Using trigonometry, in a right triangle formed by the ladder, the wall, and the ground:

$\sin(60^\circ) = \frac{\text{height}}{\text{hypotenuse}}$

Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ , we can solve for the height:

$\text{height} = \frac{\sqrt{3}}{2} \times \text{hypotenuse}$

Given that the distance from the wall (base) is 4.6 m, we can use the Pythagorean theorem to find the hypotenuse (length of the ladder):

$\text{hypotenuse}^2 = \text{base}^2 + \text{height}^2$

$\text{hypotenuse}^2 = 4.6^2 + \left(\frac{\sqrt{3}}{2} \times \text{hypotenuse}\right)^2$

Solving for $\text{hypotenuse}$ , we get:

$\text{hypotenuse} = \frac{4.6}{\sqrt{1 - \frac{3}{4}}}$

$\text{hypotenuse} = \frac{4.6}{\frac{\sqrt{1}}{2}} = \frac{4.6}{\frac{1}{2}} = 9.2$

So, the length of the ladder (height) is 9.2 meters.

55.

If one of the roots of the equation $3x^2 + px + 3 = 0$ ( $p > 0$ ) is the square of the other root, then what is the value of $p$ ?

Correct 3
Incorrect 1
Incorrect 2/3
Incorrect 1/3

Solution : The correct answer is (A) 3.

Let the roots of the quadratic equation $3x^2 + px + 3 = 0$ be $\alpha$ and $\alpha^2$ , where $\alpha^2$ is the square of the other root $\alpha$ .

By Vieta’s formulas, the sum of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$ .

So, for the equation $3x^2 + px + 3 = 0$ , we have:

Sum of the roots: $\alpha + \alpha^2 = -\frac{p}{3}$ …..(1)

Product of the roots: $\alpha \cdot \alpha^2 = \frac{3}{3} = 1$

$\implies \alpha^3 = 1$ $\implies \alpha = 1$ …..(2)

Substitute $\alpha = 1$ into the equation $\alpha + \alpha^2 = -\frac{p}{3}$ :

we get, $1 + 1^2 = -\frac{p}{3}$

$\implies 2 = -\frac{p}{3}$

$\implies p = -6$ .

However, since the question specifies that $p > 0$ , this is not a valid solution.

So, Now we can go with, $\alpha^3 = 1$ ….(from $eq^n 2$ )

$\implies \alpha^3 = 1$

$\implies \alpha^3 -1 = 0$

$\implies \alpha^3 -1^3 = 0$

$\implies (\alpha-1) \times (\alpha^2 + 1 + \alpha) = 0$

$\implies \alpha = 1$ or $\alpha^2 + \alpha = -1$

We have already taken $\alpha = 1$ , but it didn’t worked. So, we can take $\alpha^2 + \alpha = -1$

$\implies \alpha^2 + \alpha = -1 = -\frac{p}{3}$ (from $eq^n 1$ ) Sum of the roots

$\implies p = 3$ (Ans)

57.

If the system of linear equations $5x + my = 10$ and $4x + ny = 8$ (where $m$ and $n$ are positive integers) have infinitely many solutions, then the minimum possible value of $m + n$ is:

Incorrect 11
Correct 9
Incorrect 7
Incorrect 5

Solution: The correct answer is (B) 9.

For a system of linear equations to have infinitely many solutions, the equations must be dependent, meaning one equation can be obtained by multiplying the other equation by a constant. In other words, the two equations must be proportional.

Comparing the given equations, we can find the condition for them to be proportional:

$\frac{5}{4} = \frac{m}{n}$ , since the coefficients of $x$ in both equations must be proportional.

Cross-multiplying, we get $5n = 4m$ , which implies $m = \frac{5}{4}n$ .

The question asks for the minimum possible value of $m + n$ , which can be achieved when $m$ and $n$ are integers with the smallest possible values. In this case, $m = 5$ and $n = 4$ , giving $m + n = 9$ .

So, the correct answer is (B) 9.

58.

A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, then its radius will be?

Incorrect 22 cm
Correct 14 cm
Incorrect 11 cm
Incorrect 7 cm

Solution: The correct answer is (B) 14 cm.

When the wire is bent to form a square with a side length of 22 cm, the total length of the wire is equal to the perimeter of the square:

Perimeter of square = 4 × side = 4 × 22 cm = 88 cm

Now, if the same wire is rebent to form a circle, the length of the wire will be the circumference of the circle:

Circumference of circle = 2πr

Given that π = 22/7 and the length of the wire is 88 cm:

2 × $\frac{22}{7}$ × r = 88

Solving for $r$ , the radius of the circle:

r = $\frac{88 \times 7}{2 \times 22}$ = 14 cm

So, the radius of the circle formed by the wire is 14 cm.

59.

Which one of the following is an obtuse angle?

Incorrect 11/21 of a right angle
Correct 8/20 of a complete rotation
Incorrect 11/21 of a complete rotation
Incorrect 8/20 of a right angle

Solution: The correct answer is (B) 8/20 of a complete rotation.

An obtuse angle is an angle greater than 90 degrees but less than 180 degrees.

A complete rotation is 360 degrees.

Let’s analyze the given options:

A. 11/21 of a right angle is less than 90 degrees, so it’s not obtuse.

B. 8/20 of a complete rotation is 8/20 × 360 degrees = 144 degrees, which is greater than 90 degrees and less than 180 degrees. This is an obtuse angle.

C. 11/21 of a complete rotation is greater than 180 degrees, so it’s not obtuse.

D. 8/20 of a right angle is 8/20 × 90 degrees = 36 degrees, which is less than 90 degrees. This is not obtuse.

Therefore, the correct answer is option (B) 8/20 of a complete rotation.

60.

PQR is a right-angled triangle right-angled at Q with PQ = 6 cm, and QR = 8 cm. What is the radius of the circle with center at O inscribed inside the triangle PQR?

Incorrect 4 cm
Incorrect 3 cm
Correct 2 cm
Incorrect 1 cm

Solution: The correct answer is (C) 2 cm.

The circle inscribed inside a right-angled triangle with the center of the circle at the point of the right angle is known as the incenter of the triangle.

The incenter is equidistant from all three sides of the triangle. Therefore, the inradius of triangle PQR will be the distance from the incenter to any of the sides.

Given that PQ = 6 cm and QR = 8 cm, the hypotenuse PR can be calculated using the Pythagorean theorem:

PR = √(PQ² + QR²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

So, the inradius (r) can be calculated using the formula:

inradius (r) = $\frac{\text{PQ} + \text{QR} - \text{PR}}{2}$ = $\frac{6 + 8 - 10}{2}$ = $\frac{4}{2}$ = 2 cm

Therefore, the radius of the circle inscribed inside triangle PQR is 2 cm.

Odisha B.Ed Entrance Mathematics Solutions 2023

A rational number () can be expressed as a terminating decimal, if the prime factorization of is of the form:

The difference between the two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as the remainder. Which one of the following is the number?

What is the number of proper subsets of the set , where and ?

If the set P contains 5 elements and the set Q contains 6 elements, then what is the number of one-one and onto functions from P to Q?

How many equivalence relations can be defined on the set S = {1, 2, 3}?

Suresh, Dinesh and Ramesh became partners in a business by investing money in the ratio 3 : 6 : 8. If their investments are increased by 5%, 15% and 20% respectively, then what will be the ratio of their profits in one year?

In a bag, there are coins of 5 ps, 10 ps and 25 ps in the ratio 3 : 2 : 1. If all the coins in the bag amount to Rs. 60.00, then how many 5 ps coins are there?

If and , then what is the value of ?

What is the value of ?

If for some , then what is the value of ?

If , then what is the value of ?