**Solutions of math(Basic) question paper of class 10 CBSE Pre-Board Exam 2021**

1. This question paper contains 36 questions divided into two parts A and B. All the questions are compulsory.

2. Part A consists of two sections-I and II. Section I has 16 questions of I mark each and section II has 4 case study-based questions. Each case-study-based questions have 5 subparts of 1 mark each.

3. Part-B consists of 16 questions- 6 questions of 2 marks, 7 questions of 3 marks, and 3 questions of 5 marks each.

4. There is no overall choice. However internal choices are provided in 5 questions of 1 mark, 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks. You have to attempt only one of the alternatives in all such questions.

5. In cash study-based questions, you have to attempt only four out of five sub- parts.

6. Use of the calculator is not permitted.

7. Please write down the serial number of the question before attempting it.

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**Q1. HCF of co-prime numbers is always**

Ans. HCF of co-prime number is 1, as an example 2 and 3 are co-prime numbers,ย factors of 2 are = 2ร1 and of 3 are = 3ร1, 1 is the highest common factor between them, therefore 1 is the HCF of co-prime numbers.

**Q2. Write a quadratic polynomial the sum of whose zeroes is -3 and product is -10**

Ans. If the zeroes of the quadratic polynomial are a and b then the quadratic polynomial is given as

xยฒ – (a+ b)x + ab

We are given a +b = -3 and ab = -10

xยฒ -(-3)x + (-10)

=xยฒ + 3x – 10

**Q3. After how many decimal places will the decimal expansion of the number**

**47 / 2 ^{3 }ร 5^{2ย }terminates.**

Ans.ย The fraction given to us is = 47/2ยณร 5ยฒ = 47/(2 ร5)ยฒ ร2 since the denominator is multiple ofย (2 ร5)ยฒ, so the decimal expansion of the given number will terminate after two places.

OR

Given that HCF (336, 54) is 6, Find LCM (336, 54).

Ans. The given numbers are 336 and 54 and their HCF is 6

The relationship between LCM ,HCF and numbers is as follows

LCM รHCF = Product of numbers

LCM ร 6 = 336 ร 54

LCM = (336 ร54)/6 =336 ร9 = 3024

**Q4. The diameter of a wheel is 1.54 m. How far will it travel in 200 revolutions?**

Ans. The diameter of a wheel is = 1.54 m

Radius of the wheel,r = 1.54/2 = 0.77 m

The distance travelled by the wheel in one revolution is = circumference of the wheel = 2ฯr = 2 (22/7) ร0.77 = 44 ร0.11 = 4.84

The distance traveled by the wheel in 200 revolutions =200ร4.84=968

**Q5. Express 234 as the product of prime numbers.**

Ans.234 ย = 2ร3ยฒร13

**Q6. Find the area of a quadrant of a circle with a radius of 14cm. (Take ฯ = 22/7)**

Ans. We are given the radius,r of the circle = 14 cm

Area of the quadrant of the circle = ฯrยฒ/4 = (22/7) (1/4)(14 ร14) =22ร7=154 cmยฒ

**Q7. A pair of linear equations of two variables has a unique solution. What kind of lines will its graph represent?**

Ans. If a pair of linear equations of two variables has a unique solution then then both lines will be intersecting lines and the co-ordinates of intersecting point is the solutions of the equations.

OR

For 3x – 7y = 10, Express y in terms of x.

Ans. The given equation is 3x – 7y = 10

– 7y = 10 -3x

y = (10 – 3x)/-7 = (3x -10)/7

โด y = (3x -10)/7

**Q8. On comparing the ratios of the coefficients, find out whether the pair of linear equations x – 2y = 0 and 3x + 4y – 20 = 0 is consistent or inconsistent.**

Ans. The condition for a pair of linear equations a_{1}xย + b_{1}y + c_{1}=0, a_{2}x +b_{2}y +c_{2}=0ย to be consistent is as follows

In the given equations x – 2y = 0 and 3x + 4y – 20 = 0

a_{1}=1ย b_{1 }=-2, c_{1}=0, a_{2}=3 ,b_{2} =4,c_{2}=-20

Therefore the pair of linear equations x – 2y = 0 and 3x + 4y – 20 = 0 is consistent

**Q9. Probability of happening of an event is 3/7. What will be the probability of not happening of that event?**

Ans. Let the event is E then the relationship between happening of probability P(E) and not happening of event isย

Since we are given P(E) = 3/7

Probability of not happening of event =4/7

**Q10. A surveyor wants to find out the height of a tower. He measures โ A as tan A = 3/4. What is the height of the tower if A is 40m from its base as shown in the figure?**

Ans. In ฮABC,we are given AB = 40 m and tan A =3/4

We have to find out height of the tower BC

Tan A =BC/AB = 3/4

BC : AB = 3 : 4

Let BC = 3x and AB = 4x

We are given AB = 40 m

4x = 40 โx = 10

โด AB = 3x = 3ร 10 = 30 m

Hence height of the tower is 30 m.

**Q11. The median of the given days with the observations in ascending order is 27.5. Find the value of x.**

**24, 25, 26, x + 2, x + 3, 30, 33, 37**

Ans.ย The median of the given data is,m = 27.5

The given data is 24, 25, 26, x + 2, x + 3, 30, 33, 37

Number of observations is = 8

Since number of observations is even number so applying the formula

2x +5 = 55

x = 25

Hence the value of x is 25

**Q12. If a line segment AB is to be divided in the ratio 5:8 internally, we draw a ray AX such that โ BAX is an acute angle. What will be the minimum number of points to be located at equal distances on ray AX?**

Ans. The minimum number of points to be located at equal distances on ray AX are = 5 + 8 = 13

**Q13. State the Pythagoras theorem.**

Ans. In a right triangle

Hยฒ = Pยฒ + Bยฒ

Where H is the hypotenuse,B is the base and P is perpendicular

OR

**If ฮABC ~ ฮDEF and โ A = 45ยฐ, โ C = 55ยฐ, then Find โ E.**

Ans. In ฮABC we are given โ A = 45ยฐ, โ C = 55ยฐ

According to angle sum property of the triangle

โ A + โ B + โ C = 180ยฐ

45ยฐ + โ B + 55ยฐ= 180ยฐ

โ B = 180ยฐ – 100ยฐ = 80ยฐ

Since ฮABC โผ ฮDEF

Therefore โ E = โ B = 80ยฐ

**Q14. A bag contains 6 red balls and 5 blue balls. Once ball is drawn at random. What is the probability of getting a blue ball.**

Ans. The total number of balls are = 6 red balls + 5 blue balls = 11 balls

Let the probabilty of drawing a blue ball is P(Blue)

Then P(Blue) = (Number of blue balls)/Total balls = 5/11

Hence probability of getting a blue ball is 5/11

OR

**A die is thrown at once. What is the probability of getting an odd number?**

Ans. If a die is thrown once then total possible outcome are =6(1,2…6)

The number of odd numbers among the total possible outcome =3(1,3 and 5)

Let the probability of getting an odd number is P(odd number)

P(odd number) = (number of odd numbers)/total outcomes= 3/6 = 1/2

**Q15. If sinฮธ = 12/13, then find cosฮธ.**

Ans.Sin ฮธ = 12/13 = Perpendicular/hypotenuse

Let the Base is Bย Perpendicular(P) is = 12x and hypotenuse(H)ย =13

Applying the Pythagoras theorem

Hยฒ = Pยฒ + Bยฒ โB = โ(Hยฒ – Pยฒ) = โ(13ยฒ -12ยฒ) =โ(169 – 144) =โ25 = 5

โดCosฮธ = B/H =5/13

**Q16. A pendulum swings through an angle of 30ยฐ and describes an arc 17.6 cm in length. Find the length of pendulum.**

Ans. The length of arc(l) subtending an angle ฮธ at the centre is given by

ฮธ= 30ยฐ, l = 17.6 cm and r is the length of pendulum

r = 4 .8 cm

The length of the pendulum is 4.8 cm

OR

**Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60ยฐ.**

Ans. We are given the angle of sector, ฮธ = 60ยฐ and radius of circle,r = 6 cm

Area of the sector ,A is given by the follwing formula

Hence area of the sector is 132/7 cmยฒ

**SECTION-II**

Question numbers 17-20 are case study-based questions. Attempt any 4 subparts from each question. Each such part carries 1 mark.

**Q17. Raman is stitching a kite-shaped patch on the cushion cover. Few questions came to his mind while stitching the patch. Give answers to his questions by looking at the figure.**

(i) Raman stitched the white thread at what angles to each other?

a) 30ยฐ

b) 60ยฐ

c) 90ยฐ

d) 60ยฐ

Ans. (i) In the kite two diagonal intersect each other at the right angle, so Raman will stitch white threads to each other with the subtention of 90ยฐ, therefore the answer is c)

(ii) Which is the correct similarity criteria applicable for smaller triangles at the upper part of this kite?

a) RHS

b) AAA

c) SSA

d) AAS

Ans.ย The upper part of the kite contains two smaller triangles formed by the intersection of two diagonal at 90ยฐย in such a way that longer diagonal bisects smaller diagonal, sides of both of the smaller triangles are equal, therefore the correct similarity criteria applicable for smaller triangles at the upper part of this kite is RHS.So, the answer is a) RHS.

(iii) Sides of two similar triangles are in the ratio 2:9. The corresponding attitudes of these triangles are in the ratio.

a) 2:3

b) 2:9

c) 81:16

d) 16:81

Ans. b) 2:9.The ratio between sides of a similar triangle is = Ratio between altitudes of the triangles = 2 : 9

(iv) Triangles switched at the tail of the kite are congruent to each other and are similar to the lower part of the kite in the ratio 2:9. If one side if the smaller triangle is of 4 cm, then the corresponding side of the kite’s lower triangle will be

a) 12 cm

b) 15 cm

c) 18 cm

d) 8 cm

(v) What is the area of the kite formed by the two perpendicular strings of length 8 and 12 cm?

a) 48 cmยฒ

b) 14 cmยฒ

c) 24 cmยฒ

d) 96 cmยฒ

**Q18. Karma went to the lab near his home for the covid 19 test along with his family members. The seats in the waiting area were as per the norms of distancing during this pandemic (as shown in the above figure). His family member took their seats surrounded by red circular areas.**

(i) Considering A as the origin, what are the coordinates of A

a) (0,1)ย ย ย b) (1,0)ย ย ย c) (0,0)ย ย ย d) (-1,-1)

Ans. c) (0,0)

(ii) What is the distance between Neetu and Karan?

a) 10 unitsย ย ย b) 2โ5 unitsย ย ย c) โ10 unitsย ย ย d) โ8 units

Ans. The coordinates ofย Neetu and Karan are (3,6) and (6,5)

The distance between Neetu and Karanย is =โ[(3-6)ยฒ +(6-5)ยฒ]= โ10

(iii) What are the coordinates ofย seat of Akash?

a) (2,3)ย ย ย ย ย b) (3,2)ย ย ย ย c) (0,3)ย ย ย d) (2,0)

Ans.a) (2,3)

(iv) What will be the coordinates of a point exactly between Akash andย Binu where a person can be seated?

(a) (3.5, 2.5)ย ย ย ย ย (b) (2.5, 5)ย ย ย ย ย (c) (10, 5)ย ย ย ย ย (d) (1.5, 0.5)

Ans.The coordinatesย of Akash andย Binu are (2,3) and (5,2)

The coordinates of the point betweenย Akash andย Binu are

= [(2+5)/2,(3+2)/2] =(3.5,2.5)

Hence answer is (a) (3.5, 2.5)

**Q19. An engineer plans to make all the pillars of the metro green with plants to contribute to a healthy environment as shown in the picture. Observe the picture and answer the questions if the dimension of one pillar is 1.5 m ร 1.5 m ร 20 m.**

(i) The shape of the pillars is

(a) Rectangleย ย ย ย ย (b) Cubeย ย ย ย ย (c) Cuboidย ย ย ย ย (d) Cylinder

Ans. ย (c) Cuboid

(ii) By using which formula he can calculate the surface area of the pillar?

(a) A = 2(lb + bh + hl)ย ย ย ย ย (b) A = 2(lb + hl)ย ย ย ย ย (c) A = 2h(l + b)ย ย ย ย ย (d) A = lb + bh + hl

Ans.A = 2h(l + b)

(iii) The lateral surface area of one pillar is

(a) 100mยฒย ย ย ย ย (b) 120mยฒย ย ย ย ย (c) 165mยฒย ย ย ย ย (d) 82.5mยฒ

Ans. The lateral surace area of theย pillar is

= 2h(l+b) = 2ร20(1.5+1.5) =40 ร3 = 120 cmยฒ

(iv) How much cement is used to fill the pillar?

(a) 44mยณย ย ย ย ย (b) 45mยณย ย ย ย ย (c) 450mยณย ย ย ย ย (d) 440mยณ

Ans. Volume of pillar = l รbรhย = 1.5 ร1.5ย ร20 =45 cmยณ

Hence answer is (c) 450mยณ

(v) Find the cost of the plantation if it costs Rs.50 per mยฒ

(a) Rs.6625ย ย ย ย ย (b) Rs.6000ย ย ย ย ย (c) Rs.5000ย ย ย ย ย (d) Rs.4100

Ans. The surface area of theย pillarย = 120 cmยฒ

Theย cost of the plantationย isย Rs.50 per mยฒ

Therefore the costย of plantation of the pillar is =50ร 120 =ย 6000

Hence the answerย is ย (b) Rs.6000

**Q20. **

**A park has swings made of rubber and iron chain. Sachin who is studying in class X has noticed that this is a mathematical shape, he has learned in maths class. Following questions raised in his mind. Answer the questions by observing both pictures:**

(i) Name the shape in which the wire is bent.

(a) Spiralย ย ย ย ย (b) elipseย ย ย ย ย (c) linearย ย ย ย ย (d) Parabola

Ans. ย (d) Parabola

(ii) How many zeroes are there for the polynomial (shape of wire)?

(a) 2ย ย ย ย (b) 3ย ย ย ย (c) 1ย ย ย ย (d) 0

Ans.(a) 2

(iii) The zeroes of the polynomial are

(a) -1, 4ย ย ย ย (b) -1, 3ย ย ย ย (c) 3, 5ย ย ย ย ย (d) -4, 2

Ans.(a) -1, 4

(iv) What will be the expression of the polynomial?

(a) xยฒ-3x -4 ย ย ย (b) xยฒ – 2x + 3ย ย ย ย ย (c) xยฒ – 2x – 3ย ย ย ย ย (d) xยฒ + 2x + 3

Ans. The zeroesย ofย the polynomial are =-1 and 4

Therefore the polynomial is

(x+1)(x -4) =xยฒ-3x -4

Hence the answer is (a) xยฒ-3x -4

(v) What is the value of the polynomial if x = 1?

(a) -4ย ย ย ย ย (b) 5ย ย ย ย ย (c) -5ย ย ย ย ย (d) -6

Ans. The polynomial is ย xยฒ-3x -4

Value of polynomial at x =1 is 1ยฒ-3ร1 -4ย ย =1-3-4=1-7=-6

Hence answer is ย ย (d) -6

**Part -B**

Question No.21 to 26 are very short answers type questions of 2 marks each.

**Q21.Find the value of a if the distance between the points (-3,-14) and B(a,-5) is 9 units.**

Ans. The distance between two pointsย (x_{1},y_{1}) and (x_{2,}y_{2}) is given by

We are given the distance between the points (-3,-14) and B(a,-5) is 9 units.

aยฒ +6a +9 +81 = 81

aยฒ +6a +9 =ย 0

aยฒ +3a+3a + 9 =0

a(a +3) +3(aย +3) = 0

(a + 3)(a +3) = 0

a = ยฑ3

Hence the value of a is ยฑ3

OR

**Find a relation between x and y such that the point (x,y) is equidistant from the point (3,6) and (-3,4).**

Ans. According to question

The distance between (x,y) and (3,6) = The distance between (x,y) and (-3,4)

Applying the distance formula

xยฒ + 9 -6x +ย yยฒ +36 -12y = xยฒ +9 +6xย + yยฒย +16 -8y -6x

-6x +36 -12yย =16 -8y

-6x +36-16= -8y +12y

4y = 20 -6x

**Q22. In the given figure DE||BC, find the value of x.ย **

In ฮABC,ย DE||BC

Applying BPT theorem

x(3x -19) = (xย +3)(3xย -4)

3xยฒ -19x = 3xยฒ -4x +9x -12

-19x =ย -4x +9x -12

24 x = 12 โ x = 1/2

**Q23.The sides a,b,c of a right triangle, where c is the hypotenuse, are circumscribing a circle. Prove that the radius rย of the circle is given by r = (a+b-c)/2.**

Ans.

In right triangle ฮABC, AC=c, BC=a andย ABย =b

Quadrilatral DEBD will be a square ,soย DB =r

AD = b – r

AD = AF= b – r…(i)[Tangents drawn from anย external pointย to the circle are equal]

BEย =ย r

CE =BC-BE =a -r

CEย =CF = a -r …(ii)[Tangents drawn from anย external pointย to the circle are equa]

AF = ACย – CF =c -(a -r) =c -a+ r from (ii)

b-r = c -a+ r fromย (i)

2r = a + b – c

r = (a + b – c)/2, Hence proved

**Q24. Draw a line segment of length 7.6 cm and divide it internally in the ratio 3: 2. Measure the two parts.ย **

Ans.

**Steps of construction**

(i) Draw a line segment AB of the length 7.6 cm.

(ii) Draw a ray AX forming an acute angle with AB.

(iii) Draw arcs on AX such thatย AA_{1ย }=A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4} =A_{4}A_{5}

(iv) Join A_{5}B and draw a line A_{3}C||A_{5}B

We get AC : CBย = 3 : 2

**Q25. If cos A =7/25, find the value of tan A + cot A.ย **

Ans. cos A =7/25 = base/hypotenuse

Perpendicur = โ(25ยฒ -7ยฒ) = โ(625 – 49) = โ576 = 24

tanย Aย = perp./base =24/7 andย cot A =base/perp. = 7/24

tan A + cot A = 24/7 + 7/24 = (576+49)/168 = 625/168

OR

**If 5x = sec ฮธ and 5/x = tan ฮธ,thenย findย the value of 5(xยฒ – 3/xยฒ).ย **

Ans. We are given 5x = sec ฮธ โ x = sec ฮธ/5….(i)

5/x = tan ฮธย โ 1/x =ย tan ฮธ/5……(ii)

5(xยฒ – 3/xยฒ)โ 5[(sec ฮธ/5)ยฒย -3ร(tan ฮธ/5)ยฒ]

= 1/5(secยฒ ฮธ -3tanยฒ ฮธ) = 1/5(1 + tanยฒ ฮธ -3tanยฒ ฮธ)ย =1/5(1 -2tanยฒ ฮธ)

**Q26.How many terms of the AP 24,21,18……must be taken so that their sum is 78?**

Ans. The given AP is 24,21,18……

Let there are n terms to make the sum 78.

Now, we have S_{n}= 78, a = 24, d = 21 – 24 = -3

n(48 -3n+ 3) = 156

48n -3nยฒ +3n = 156

-3nยฒ + 51n – 156 = 0

3nยฒ – 51 n + 156 = 0

nยฒ – 17 n + 52 = 0

nยฒ – 13 n -4n + 52 = 0

n(n -13) – 4(n -13) =0

(n -13)(n -4) = 0

n = 13, 4

Therefore, either the sum of the first 13 terms or the sum of the first 4 terms of the series is 78

**Q27.In figure two tangentsย TP and TQ are drawn to a circle with centre O from an external point P. Prove that โ PTQ = 2โ QPQ.**

Ans.

Ans.

**GIVEN:** TP and TQ are the tangents drawn from T to the circle.

**TO PROVE:** โ PTQ = 2โ QPQ

**PROOF:** PO is the radius of the circle

TP = TQ (tangents drawn from an external point to the circle are equal)

โ TPQ = โ TQP (angles opposite to equal sides in a ฮ)

โ OPT = 90ยฐ (angle between radius and tangent)

โ TPQ =ย 90ยฐ- โ QPQ

According to the angle sum property of the ฮ

โ PTQ + โ TPQ + โ TQP = 180ยฐ

โ PTQ + โ TPQ + โ TPQ = 180ยฐ

โ PTQ + 2โ TPQ = 180ยฐ

โ PTQ +2(90ยฐ- โ QPQ) = 180ยฐ

โ PTQ + 180ยฐ – 2โ QPQ = 180ยฐ

โ PTQ = 2โ QPQ,Hence proved

**Q28. Sunita has some notes of Rs.50 and Rs.100 amounting to a total of Rs.15,500. If the total number of notes is 200, then Find how many notes of Rs.50 and Rs.100 each she has?**

Ans. Let the number of Rs.50 notes are x and the number of Rs.100 note are y

According to first condition of the question

50x + 100y = 15,500

x + 2y = 310…..(i)

According to second condition of the question

x + y = 200……(ii)

Subtracting equation (ii) from equation (i)

y = 110

Putting the value of y = 110 in equation (ii)

x +110 = 200

x = 200 – 110 = 90

Hence she has 90 notes of Rs 50 and 11o notes of Rs 100

**Q29. Prove that โ2 is an irrational number.**

Ans. Let โ2 is a rational number

โ2 = a/b (where a and b are co-prime numbers)

bโ2 = a

2bยฒ = aยฒ…….(i)

We get from equation (i) that aยฒ is divisible by 2

โด a will also be divisible by 2

Now, we can write a = 2c (where c is another positive integer)

Putting a = 2c in equation number (i)

2bยฒ = (2c)ยฒ= 4cยฒ

bยฒ = 2cยฒ……(ii)

From (ii) it is clear that bยฒ is divisible by 2

Then b will also be divisible by 2

It is evident from equations (i) and (ii) that 2 is a common factor between a and b, so a and b can’t be co-prime. Hence our assumption for โ2 as a rational number is wrong, obviously, โ2 is an irrational number.

**Q30. 90 cards numbered from 1 to 90 are placed in a box. If one card is drawn at random from the box find the probability that it is:**

(i) a two-digit number

(ii) a perfect square

(iii) a number divisible by 5

Ans. (i) Total numbers from 1 to 90 are =90โTotal possible O/C =90

The two digit number from 1 to 90 are =90-9 = 81, Favourable O/C =81

P(two-digit number) = Favourable O/C/Total possible O/C=81/90 =9/10

(ii) The outcomes of card drawn is of complete square number are=4,9,16,25,36,49,64,81 โFavourable outcomes =8

P(a perfect square) = 8/90 = 4/45

(iii) A number divisible by 5 are =5,10,15,20,25,30,35…..90(total numbers 18)โ Favourable outcomes =18

P(a number divisible by 5) = 18/90 = 1/5

OR

**Red Queen and a black jack are removed from a pack of 52 playing cards. Find the probability that the card drawn from the remaining cards is:**

(i) a red card

(ii) neither a jack nor a king

(iii) either a king or a queen

Ans. (ii)ย A red queen and a black jack is removed from 52 cards then the total remaining cards are =52 -2 = 50 cardsโTotal possible outcomes =50

Total number of remaining red cards =2-1= 1โFavourable outcomes =1

P(a red card) =1/50

(ii) Total number of possible outcomes =50

The number of jacks and kings =3 +4 = 7

The number of cards other than jacks and king = 50 -7 = 43โFavourable outcomes =43

P(neither a jack nor a king) = 43/90

(iii) Total number of possible outcomes =50

The number of queens and kings =3 +4 = 7โFavourable outcomes =7

P(either a king or a queen) = 7/50

**Q31. A well of diameter 3m is dug 14m deep. The earth taken out of it has been spread evenly all-around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.**

Ans. Let the height of the embankment is = h, the radius of the well,r =3/2=1.5 m, the width of the embankment is =4m, inner radius of the embanment,r = 1.5 m and outer radius of embankment,R = 4+1.5 = 5.5 m,depth of the well,H = 14 m

Since the embankment is created by the earth taken out from the well

Therefore, The volume of well = The volume of the embankment

ฯrยฒH = ฯ(Rยฒ-rยฒ)h

1.5ยฒร14 = (5.5ยฒ -1.5ยฒ)h

(30.25-2.25)h = 31.5

28 h = 31.5

h = 31.5/28 = 1.125

Therefore,height of embankment is =1.125 m

Ans.ย We have to prove that

Dividing numerator and denominator of LHSย by sin ฮธ

Rearranging it

Multiplying numerator and denominator by the conjugate of (cot ฮธ – cosec ฮธ)i.e =(cot ฮธ + cosec ฮธ)

**Q33. A train traveling a distance of 360 km at a uniform speed would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.**

Ans. Letย the original speed of the train is = xย km/h

The distance traveled by train = 360 km

Time taken by the train with original speed = distance/time=360/x

If the train’s speed changed by the increase of 5 km/h, then it’s speed becomes =(x + 5) km/h.

Time taken by train with modified speed = 360/(x+5) h

48 minutes = 48/60 = 4/5 =0.8 h

According to question

360/x – 360/(x+5) = 0.8

360[1/x – 1/(x +5)] = 0.8

360[(x+5-x)/(x+5)x] =0.8

1800 = 0.8x(x+5)

0.8xยฒ + 4x – 1800 = 0

0.2xยฒ +x – 450 = 0

2xยฒ +10x -4500 = 0

Applying the quadratic equation

x = (-10 +190)/4 = 180/4 = 45 or x = (-10 -190)/4 = -200/4 = -50

Since speed of the train can’t be negative,so the original speed of the train is 45 km/h.

Q34. A statue which is x m tall stands on the top of a 100 m pedestal on the ground. From a point on the ground, the angle of elevation of the top of the statue is 60ยฐ and from the same point, the angle of elevation of the top of the pedestal is 45ยฐ. Find the height of the statue.

Ans.

In the figure, BD is the pedestal of height 100 m and DA is the statute of the height x

We are given the angle of elevation of the top of the statue from a point C on the ground =60ยฐ

The angle of elevation of the top of the pedestal from the same point C =45ยฐ

tan 45ยฐ = BD/BC = 100/BC

1 = 100/BCโ BC = 100 m

tan 60ยฐ=AB/BC = (100 +x)/100

(100 +x)/100 = โ3

100 +x = 100โ3

x = 100โ3 – 100 = 100(โ3 – 1)

Hence, height of the statue is 100(โ3 – 1) m

**Q35.Find the number of terms in an AP 18,15,12……. -48 and also find the sum of all of its terms.**

Ans.ย Let there are n terms in the AP, the given AP is 18,15,12……. -48

The n^{th} term of AP is given by

a_{n}= a + (n-1)d, where a = 18, d = 15 -18 = -3 and a_{n }ย = -48

18+(n-1)(-3) = -48

-3n +3 = -48 -18 = -66

-3n = -69

n = 23

Applying the following formula for getting the sum of n terms of an AP

Putting the value of n =23,a = 18, d = 15 -18 = -3

= 23ร18 -23ร66

= 414 – 1518 = -1104

Hence,sum of all terms of the given AP is -1104

**Q36. Find the missing frequenciesย f _{1 }ย and f_{2} if the mean of all observations given below is 38.2.**

Ans. We are given the mean of all observations is 38.2

Arranging the table for evaluating mean of the given data

Class Interval | Frequency(f) | Class mark(x) | fx |

0 – 10 | 4 | 5 | 20 |

10 – 20 | 4 | 15 | 60 |

20 -30 | f_{1} | 25 | 25 f_{1} |

30 -40 | 10 | 35 | 350 |

40 -50 | f_{2} | 45 | 45 f_{2} |

50 -60 | 8 | 55 | 440 |

60 -70 | 5 | 65 | 325 |

N=50 | โfx | =1195+25 f_{1}+45 |

We have

f_{1}+ f_{2}+31 = 50โf_{1}+ f_{2}= 19……(i)

Applying the direct method for getting the mean

25** f _{1}**+45

**f**=ย 1910- 1195

_{2ย }25** f _{1}**+45

**f**= 715……(ii)

_{2}Substituting **ย f _{1ย }**=( 19-

**f**) equation (i) in equation (ii)

_{2ย }25** ( 19- f _{2ย } )**+45

**f**= 715

_{2}475 –** 25f _{2ย }** + 45

**f**= 715

_{2}20** f _{2}**= 715 – 475 = 340

**ย f _{2}**= 17

Putting the value of **f _{2}**= 17 in equation ( i)

**f _{1}**= 19-17 = 2

Hence the value of **f _{1}** is 2 and of

**f**ย is 17

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