Assignment SA-2 Class X Topic : Height And Distance

1.The angle of elevation of a ladder leaning against a wall is 60^o and the foot of the ladder is 9.5 meter away from  the wall. Find the length of the ladder. [ 19m ]

2. If the length of the shadow cast by a pole be  times the length of the pole, find the angle of elevation of the sun. [ 30^o ]

3.   A tree is broken by the wind. The top stuck the ground at an angle of 30^o and at a distance of 30 m from the root.   Find the total height of the tree.

4.     A circus artist is climbing from the ground along a rope stretched from the top of vertical pole and tied at the ground level 30^o. Calculate the distance covered by the artist in climbing to the top of the pole. [ 24 m ]

5. A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60^o. When he was 40 m away from the bank he finds that the angle of elevation to be 30^o.  Find: - 

(i) The height of the tee, 

(ii) The width of the river, correct up to two decimal places.[(i)34.64m (ii) 20m]

6.  An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the  angles of elevations of two planes to a same point on the ground are 60^o and 45^o respectively.  Find the vertical distance between the aeroplanes at that at that instant. [ 1693.34 m ]

7.  The angle of elevation of the top of the hill at the foot of a tower is 60^o and the angle of elevation of the top of tower from the foot of hill is 30^o. If the tower is 50 m high, what is the height of the hill. [ 150 m ]

8.    There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. Let P and Q be       points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top     the tree from P and Q respectively are 30^o and 45^o, find the height of the tree. 

9.        Two pillars of equal heights are on either sides of a roadway, which is 150 m wide. The angles of elevation of       the top of pillars are 60^o  and 30^o at a point on the roadway between the pillars. Find the position of the point    between the pillars and the height of each pillar.       (64.95m)

10.        At the foot of mountain, the elevation of its peak is 45^o. After ascending 1 km towards the mountain up an inclination of 30^o, the elevation changes to 60^o. Find the height of mountain. (1.366 km)

11.        From the top of the building 15m high, the angle of elevation of the top of a tower is found to be 30^o. From the       bottom of the same building, the angle of elevation of the top of tower is found to be 30^o. Find the height of the     tower and the distance between the tower and the building. (22.5m, 12.975m)

12.        A fire in a building B is reported on the telephone to two fire stations P and Q, 10 km apart from each other on a    straight road. P observes that the fire is at angle of 60^o to the road and Q observe that it is an angle of 45^o to the       road. Which station should send its team and how much this team has to travel? (P, 7.32km)

13.        The shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sunrays meet the ground  at an angle of 60^o. Find the angle between the sunrays and the ground at the time of long shadow. ( 30^o)

14.        From a point in the cricket ground, the angle of elevation of a vertical tower is found to be θ at a distance of  200m from the tower. On walking 125 m towards the tower the angle of elevation becomes 2θ. Find the height of   tower. (100m)

15.        A boy standing on the ground and flying a kite with 75 m of string at an elevation of 45^o. Another boy is standing on the roof of 25 m high building and is flying his kite at an elevation of 30^o. Both the boys are on the opposite side of the two kites. Find the length of the string that the second boy must have, so that the kites meet.(56.05 m)

16.        As observed from the top of light house, 100m high above the sea level, the angle of depression of a ship, sailing directly towards it, changes from 30^o to 45^o. Determine the distance traveled by the ship during the period of observation.  ( 73.2m)

17.        An aeroplane at an altitude of 200 m observes the angle of depression of opposite points on two banks of a river  to be 45^o and 60^o. Find the width of the river.  ( 315.4m)

18.        From the top of a cliff 150m high, the angles of depression of two boats are 60^o and 30^o. Find the distance  between the boats, if the  boats are (i) on the side of cliff. (ii) on the opposite sides of the cliff.     [ (i) 173.2m (ii) 346.4m ]

19.        A man standing on the deck of a ship, which is 10m above the water level, observe the angle of elevation of the  top of a hill as 60^o and the angle of depression of the base of the hill as 30^o.Calculate the distance of the hill from the ship and the height of the hill.   [17.3m, 40m].

20.        The angle of elevation and depression of the top and the bottom of a light house from the top of the building, 60m high, are 30^o and 60^o respectively. Find (i) The difference between the heights of the light house and the  building (ii) Distance between the light house and the building. [ (i) 20m, (ii) 34.64m]

21.A pole 5m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from Point ‘A’ on the ground is 60^o and the angle of depression of the point ‘A’ from the top of tower is 45^o. Find the height of tower.  [ 6.83m]

22.        Man on a cliff observes a boat at an angle of depression of 30^o which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is  found to be 60^o. Find the time taken by the boat to reach the shore. [ 9 minutes]

23.  A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly       towards it. If it takes 12 minutes for the angle of depression to change from 30^o to 45^o, how soon after this will       the car reach the observation tower. Give your answer correct to nearest seconds.  [16 min. 24 sec.]

                                                     JSUNIL TUTORIAL CBSE MATHS & SCIENCE

CBSE10th Class Solved Paper - Arithmetic Progression

1. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution: 12 = a + 2d

106 = a + 49d

So, 106-12 = 47d

Or, 94 = 47d

Or, d = 2

Hence, a = 8

And, n₂₉ = 8 + 28x2 = 64

2. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution: -8 = a + 8d

4 = a + 2d

Or, -8 – 4 = 6d

Or, -12 = 6d

Or, d = -2

Hence, a = -8 + 16 = 8

0 = 8 + -2(n-1)

Or, 8 = 2(n-1)

Or, n-1 = 4

Or, n = 5

3. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution: n₇ = a + 6d

And, n₁₀ = a + 9d

Or, a + 9d – a – 6d = 7

Or, 3d = 7

Or, d = 7/3

4. Which term of the AP: 3. 15, 27, 39, … will be 132 more than its 54th term?

Solution: d = 12,

132/12 = 11

So, 54 + 11 = 65th term will be 132 more than the 54th term.

5.  How many three digit numbers are divisible by 7?

Solution: Smallest three digit number divisible by 7 is 105

Greatest three digit number divisible by 7 is 994

Number of terms

= {(last term – first term )/common difference }+1

= {(994-105)/7}+1

= (889/7)+1=127+1=128

6. How many multiples of 4 lie between 10 and 250?

Solution: Smallest number divisible by 4 after 10 is 12,

The greatest number below 250 which is divisible by 4 is 248

Number of terms: {(248-12)/4}+1

{236/4}+1 = 59+1 = 60

7. For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?

Solution: In the first AP       a = 63 and d = 2

In the second AP                 a = 3 and d = 7

As per question,

63+2(n-1) = 2+ 7(n-1)

Or, 61 = 5 (n-1)

Or, n-1 = 61/5

As the result is not an integer so there wont be a term with equal values for both APs.

8. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution: As the 7th term exceeds the 5th term by 12, so the 5th term will exceed the 3rd term by 12 as well

So, n₃ = 16

n₅ = 28

n₇ = 40

n₄ or n₆ can be calculated by taking average of the preceding and next term

So, n₄ = (28+16)/2 = 22

This gives the d = 6

AP: 4, 10, 16, 22, 28, 34, 40, 46, ……..

9. Find the 20th term from the last term of the AP: 3, 8, 13, ……, 253.

Solution: a = 3, d = 5

253 = 3 + 5(n-1)

Or, 5(n-1) = 250

Or, n-1 = 50

Or, n = 51

So, the 20th term from the last term = 51 – 19 = 32nd term

Now, n₃₂ = 3 + 5x31 = 158

10. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th terms is 44. Find the first three terms of the AP.

Solution: a + 3d + a + 7d = 24

Or, 2a + 10d = 24

Similarly, 2a + 14d = 44

So, 44 – 24 = 4d

Or, d = 5

2a + 10x5 = 24

Or, a + 25 = 12

Or, a = -13

So, first three terms of AP: -13, -8, -3,

11. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reached Rs. 7000.?

Solution: 7000 = 5000 + 200(n-1)

Or, 200(n-1) = 2000

Or, n-1 = 10

Or, n = 11

12. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her savings become Rs. 20.75, find n.

Solution: 20.75 = 5 + 1.75(n-1)

Or, 1.75(n-1) = 15.75

Or, n-1 = 9

Or, n = 10

For more papers visit  CBSE: 10th Class Sample Test Paper- Arithmetic Progression

10th probability illustrative solved examples

ILLUSTRATIVE EXAMPLES

Example 1. An unbiased die is thrown. What is the probability of getting :

(i) an odd number (ii) a multiple of 3 (iii) a perfect square number (iv) a number less than 4.

Solution.

An unbiased die is thrown we may get 1, 2, 3,4,5,6

So,  total number of all possible outcomes= 6

(i) Favourable outcomes for an odd number are 1, 3, 5.

So, no. of favourable outcomes = 3

P (an odd number) = No. of favourable outcomes/ Total no. of possible outcome= 3/6=1/2

(ii) favourable outcomes for a multiple of 3  are 3 and 6.

So, no. of favourable outcomes = 2

P (a multiple of 3) = 2/6= 1/3

(iii) favourable outcomes for ) a perfect square number  are 1 and 4.

So, no. of favourable outcomes = 2

 P (a perfect square number)  =2/6= 1/3

(iv) favourable outcomes for a number less than 4. are 1, 2 and 3.

So, no. of favourable outcomes = 3

P (a number less than 4) = 3/6 =1/2

2. Three unbiased coins are tossed together. Find the probability of getting :

(i) all heads (ii) two heads (iii) one head (iv) at least two heads

Solution. When three unbiased coins are tossed together, possible outcomes are

HHH, HHT, HTH, HTT, THH, THT, TTH and  TTT.       
So, total no. of possible outcomes = 8

(i) favourable outcome = HHH

So, No. of favourable outcome = 1

P (all heads)  =no. of favourable outcomes/Total no. possible outcomes= 1/8

(ii) favourable outcomes are HHT, THH and HTH.

So, no. of favourable outcomes = 3

P (two heads) = 3/8

(iii) favourable outcomes are HTT, THT and TTH.

 So, no. of favourable outcomes = 3       P (one head) =  3/8

(iv) favourable outcomes are HHH, HHT, HTH and THH.

So, no. of favourable outcomes = 4

P (at least two heads) 4/8 =1/2

Example 3. Find the probability that a leap year selected at random will contain 53 Sundays.

Solution. In a leap year, there are 366 days. But  366 days = 52 weeks + 2 days.

Thus, a leap year has always 52 sundays.

The remaining 2 days can be :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Clearly, there are seven elementary events associated with this random experiment.

Let E be the event that a leap year has 53 sundays. Clearly, the event E will happer if the last two

days of the leap year are either Sunday and Monday or Saturday and Sunday.

Favourable no. of elementary events = 2

Hence, required probability = 2/7 

Example 4. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn.

Find the probability that the card drawn is :

(i) an ace (ii) either red or king  (iii) a face card (iv) a red face card

Solution. here, total no. of possible outcomes = 52.

(i) There are 4 ace cards in a pack of 52 cards. One ace can be chosen in 4 ways.

So, favourable no. of outcomes = 4

P (an ace) no. = of favourable outcomes/Total no. of possible outcomes= 4/52=1/13

(ii) There are 26 red cards, including 2 red kings. Also, there are 4 kings, two red and two black.

card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings)

So, favourable no. of outcomes = 28

        P(either red or king) =28/52= 7/13

(iii) Kings, queens and jacks are the face cards.

So, favourable no. of outcomes = 3 × 4 = 12

      P(a face card) = 12/52 =3/13

(iv)  There are 6 red face cards, 3 each from diamonds and hearts.

So, favourable no. of outcomes = 6

                      P(a red face card) = 6/52= 3/26

Example 5. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is :

(i) an even number                                     (ii) a number less than 14 

(iii) a number which is a perfect square (iv) a prime number less than 20.

Solution.

From 2 to 101, these are (101–2) + 1 = 100 numbers.

So, total no. of possible outcomes = 100.

(i) From 2 to 101, the even numbers are 2, 4, 6, ...., 100 which are 50 in number.

So, number of favourable outcomes = 50

     P(an even number) = no. of favourable outcomes/Total no. of possible outcomes

                                       = 50/100 = 1/2

(ii) From 2 to 101, the numbers less than 14 are 2, 3, ...., 13 which are 12 in number.

So, no. of favourable outcomes    = 12

              P(a number less than 14) = 12/100 = 3/25

(iii) From 2 to 101, the perfect squares are 4, 9, 16, ..... 100, which are 9 in number.

              So, no. of favourable outcomes = 9

 P (a number which is a perfect square) = 9/100

(iv) From 2 to 101, the prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19 which are 8 in

number.

So, no. of favourable  outcomes  = 8

        P (a prime no. less than 20) = 8/100=2/25

Example 6. A bag contains 3 red balls and 5 black balls.A ball is drawn at random from a bag. What is the

probability that the ball drawn is :  (i) red (ii) not red

Solution. Total number of balls = 3 + 5 = 8

(i) P (red ball) = no. of red balls /Total no.of balls=3/8

(ii) P (not red ball) = 1 – P(red ball) =1 – 3/8 =5/8

Example  7. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 . Find the number of blue marbles in the jar.

Solution. Total number of elementary events = 24.

Let there be x green marbles.

P (green marbles is drawn) = x/24

but, P(green marbles is drawn)= 2/3 (given)

So, x / 24  = 2/3         x=24x2/3                       x   =   16

 Number of green marbles = 16

Number of blue marbles = 24 – 16 = 8 Ans.

Example 8. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?

 Solution. Total area of rectangular region = 3 m × 2 m = 6 m2

Area of the circle = p r²  = p (1/2)² m²        =  p/ 4 m²

P (die to land inside the circle) = p/ 4 m² ÷  6 = p/24 

Example 9. A bag contains 12 balls out of which x are white.

(i) If one ball is drawn at random, what is the probability that it will be a white ball?

(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be

double than that in (i). Find x.

Solution. (i) Total number elementary events = 12.

There are x white balls out of which one can be chosen in x ways.

So, favourable number of elementary events = x

 p1 = P (white ball) = no. of favourable outcomes / Total no. of possible outcomes = x /12

 (ii) If 6 more white balls are put in the bag, then total number of balls in the bag =12 + 6 = 18

and, no. of white balls in the bag = (x + 6)

P₂ = P (getting a white ball) = ( x + 6 )/ 18

It is given that, p₂  =  2 p₁

( x + 6 )/ 18 = 2 ( x / 2)

 x = 3

Example 10. Two dice are thrown simultaneously. Find the probability of getting :

(i) a doublet i.e. same number on both dice. (ii) the sum as a prime number.

Solution. Possible outcomes associated to the random experiment of throwing two dice are :

(1, 1), (1, 2), ......., (1, 6)

(2, 1), (2, 2), ......., (2, 6)

.......................................

.......................................

(6,1) , (6, 2), ......., (6, 6)

Total number of possible outcomes = 6 × 6 = 36

(i) The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Total no. of favourable outcomes = 6

So, P(a doublet) = no. of favourable outcomes/Total no. of possible outcomes = 6/36= 1/6

(ii) Here, favourable sum (as a prime number) are 2, 3, 5, 7 and 11.

So, favourable outcomes are (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2),

(3, 4), (4,3), (6, 5) and (5, 6).

no. of favourable outcomes = 15

P (the sum as a prime number) = 15/36 = 5/12