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## AP 10th Maths Answer Key 2023

AP 10th Pre final Answer Key 2023 AP SSC Maths key Sheets PDF Download 2023 . Key sheets for the 2023-2024 pre-public Examination. AP Maths 10th Class Pre-Final Answer Keys 2023: Candidates who took the pre-final examinations can review the question paper and answer key. AP SSC Pre Final Exam Maths Answer key Download 2023 Principles of Evaluation, Official key sheet download.

**1Q: LCM of 12, 15 and 21 is 420.**

12 = (2 × 2 × 3) = 2^{2} × 3^{1},

15 = (3 × 5) = 3^{1} × 5^{1}, and

21 = (3 × 7) = 3^{1} × 7^{1}

LCM (12, 15, 21) = 420

**2: Write the roster form of the set {x:x is a natural number less than 6}.**

Ans:

Natural numbers less than 6 are: 1,2,3,4,5.

This can be written as:

{1,2,3,5}

**3: The maximum number of zeroes which a quadratic polynomial can have is 2**

**The degree of a quadratic polynomial is 2.**

**So, Answer is A**

**5: **Find the roots of equation 5x^{2}–6x–2=0

5x2 – 6x – 2 = 0 ……………………………. (1)

On multiplying (1) by 5, we get 25x2 – 30x -10 = 0

(5x)2 – 2(5x)(3) +9 – 9 – 10 = 0

(5x–3)2 = 19

Taking square root on both sides, we get

5x – 3 = ±√19

5x = ±√19

5x = 3 + √19 5x = 3 – √19

x = (3 + √19)/5 x = (3 – √19)/5

**Therefore (3 + √19)/5 and (3 – √19)/5 are the roots of the given quadratic equation.**

**6: State Thale’s theorem**

Thales theorem states that:

**7: The number of tangents drawn at the end points of the diameter is 2.**

**8: The side of a cube is 4 cm long. Find its volume.**

The correct option is **D** 64cm^{3}

Given: Length of each side = 4 cm

Volume of a cube = Side^{3}

= (4cm)^{3}

= 64cm^{3}

Therefore, volume of the given cube is 64cm^{3}.

**10: “You are observing top of your school building at an angle of elevation 60° from a point which is at 20 meters distance from foot of the building”.**

**Draw a rough diagram to the above situation.**

**Solution**

Let,

BC=h be the height of the building,

The angle of elevation to the top of school is 60

and

20 mtr is the distance from the foot of the building.

11: **If P(E) = 0.05, then the probability of ‘not E’ is .**

The correct option is 0.95

P(E) = 0.05

Since P(E) + P (not E) = 1

∴ P (not E) = 1 – P(E) = 1 – 0.05 = 0.95

**12: Find the mean of the given data. 2, 3, 7, 6, 6, 3, 8**

The sum of the data is: 2 + 3 + 7 + 8 + 11 + 6 + 5 = 42

The number of data points is 7

The mean is 42 divided by 7, which is equal to 6.

So, the mean of the data 2, 3, 7, 8, 11, 6, and **5 is 6.**

**13: If A = {3, 4, 5, 6} B = {5, 6, 7, 8, 9} then illustrate A ∩ B in Venn diagram.**

**14: 6 pencils and 4 pens together cost 50 whereas 5 pencils and 6 pens together cost ₹ 46. Express the above statements in the form of Linear equations.**

**15: Check whether (x – 2) ^{2} + 1 = 2x – 3 is a quadratic equation or not.**

**16: Write the formula to find nth term of A.P. and explain the terms in it.**

Assume that a_{1}, a_{2}, a_{3},… be an arithmetic progression (AP), in which first term a_{1 }is equal to “a” and the common difference is taken as “d”, then the second term, third term, etc can be calculated as follows:

Second term, a_{2} = a+d

Third term, a_{3} = (a+d)+d = a+2d,

Fourth term, a_{4} = (a+2d)+d = a+3d, and so on.

Therefore, the nth term of an AP (a_{n}) with the first term “a” and common difference “d” is given by the formula:

**n ^{th} term of an AP, a_{n} = a+(n-1)d.**

(**Note**: The nth term of an AP (a_{n}) is sometimes called the general term of an AP, and also the last term in a sequence is sometimes denoted by “l”.)

**17: Find the distance between the two points (7, 8) and (-2, 3).**

**18: From a point Q, the length of the tangent to a circle is 24 cm, and the distance of Q from the centre is 25 cm. Find the radius of the circle.**

## Let QP be the tangent, such that, Point of contact is P.

**19 If cosA = 12 / 13 then find sinA and tanA**

**20 A die is thrown once. Find the probability of getting**

**(i) a prime number;**

**(ii) an odd number**We use the basic formula of probability to solve the problem.Number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6Number of prime numbers on dice are 1, 3 and 5 = 3(i) Probability of getting a prime number = Number of prime numbers/total number of outcomes= 3/6 = 1/2

(ii) Total number of odd numbers are 1, 3 and 5 = 3

Probability of getting a odd number = Number of odd numbers/total number of outcomes

= 3/6 = 1/2

**21: If 2 log 5 + 1/2 log 9 – log 3 = log x then find the value of x**

The given equation is

We need to find the value of x.

Using power property of logarithm, the given equation can be rewritten as

On comparing both sides we get

**22**

**If $3$ and $−3$ are two zeroes of the polynomial $(x+x−11x−9x+18),$find all the zeroes of the given polynomial.**

## Let $f(x)=x+x−11x−9x+18$

$(x+x−2)(x−9)=0$

$(x−1)(x+2)(x−3)(x+3)=0$$x=1$ or $x=−2$ or $x=3$ or $x=−3$

hence all the zeros of the given polynomial are $1,−2,3$ and $−3.$

**23 ****Solve the following system by elimination method: ****$3x+2y=1$, $2x+3y=4$**

## Given equations are

**24 Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. Write the quadratic equation to find Rohan’s present age.**

Let take Rohan’s age = x years

Hence, his mother’s age = x+26

3 years from now

Rohan’s age = x+3

Age of Rohan’s mother will =x+26+3=x+29

The product of their ages 3 years from now will be 360 so that

(x+3)(x+29)=360

⇒x2+29x+3x+87=360

⇒x2+32x+87−360=0

⇒x2+32x−273=0

⇒x2+39x−7x−273=0

⇒x(x+39)−7(x+39)=0

⇒(x+39)(x−7)=0

Either x+39=0orx−7=0

⇒x=−39 or x=7

As, age cannot be in a negative value

hence, x=7,Rohan’s age = 7 years

his mother’s age = 7+26=33 years

**25 Draw a tangent to a given circle with centre ‘O’ from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point?**

Let O be center of circle and R be a point outside it.

Draw tangent to circle at A and B through R

**26 An oil drum is in the shape of a cylinder having the following dimensions:**

Diameter is 2 m and height is 7 meters. The painter charges ₹ 3 per m² to paint the drum. Find the total charges to be paid to the painter for 10 drums.

Diameter is 2 m and height is 7 meters. The painter charges ₹ 3 per m² to paint the drum. Find the total charges to be paid to the painter for 10 drums.

Given,

Diameter of an oil drum(cylinder) = 2m

Height of the oil drum = 7m

The painter charges 5 per meter to paint the drum.

To find out,

Find the total charges to be paid to the painter for 10 drums.

Solution:

Radius of the oil drum is 1 meter.

Total surface area of a cylindrical drum = 2×π×r(r + h) = 2×22/7×1(1 + 7) = 2 ×22/7 × 8

= 352/7 = 50.28 m^2

Therefore the total surface area of cylinder is 50.28 meter square.

Painting charge per 1 m² is rupees 5.

Charge per 50.28m² is = ?

Cost of painting of 10 drums = 50.28 x 5 × 10

= 2514 rupees.

Therefore the total charges to be paid to the painter for 10 drums is 2514 rupees.

**27 Show that (1 – tan**

^{2}A)/(cot^{2}A – 1) = tan^{2}A**28. A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household.**

Family size | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |

Number of families | 7 | 8 | 2 | 2 | 1 |

**Solution**

$⇒$ Here, modal class $=3−5$