Vedanta Excel in Mathematics Book 10 TG Final

Note:
(i) Two events are said to be independent if the occurrence of an event affects the

occurrence of another event.
(ii) If A and B are be independent events then P (A and B) = P (AÇB) = P (A) ´ P( B)
(iii) If A, B and C are three independent events then P (AÇBÇC) = P (A)´ P (B)´ P (C)
14. Demonstrate the probability of depend and independent events in tree diagram and
encourage the students to solve the related problems.

Solution of selected questions from Vedanta Excel in Mathematics

1. From a well-suffled pack of 52 cards, a card is drawn at random. What is the prob-

ability of getting neither a black jack or a red queen.

Solution:

Let, J and Q denote the events of getting a black jack and queen respectively.

Then, n(S) = 52, n(J) = 2, n(Q) = 4

Now,

Probability of getting either black jack or queen = P(J or Q) = P(J∪Q)

= P(J) + P(Q)

= n(J) + n(Q) = 2 + 2 = 1
n(S) n(S) 52 52 13
1 12
Probability of getting neither black jack not red queen = P( J ∪ Q) = 1 – 13 = 13

2. Two unbiased dice are rolled. Find the probability that the sum of the number on
the two facer is either divisible by 5 or divisible by 6.

Solution:

LtievtelDy5. and D6 denote the event of getting the sum of number be divisible by 5 and 6 respec-

Then, (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

S = (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

∴ n(S) = 36

D5 = {(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)} ∴n(D5) = 7
D6 = {(1,5), (2,4), (3,3), (4,2), (5,1), (6,6)} ∴n(D6) = 6
7 6 13
Now p(D5 or D6) = p(D5) + p(D6) = 36 + 36 = 36

Hence, the probability that the sum of number on the two faces is either divisible by 5 or
13
divisible by 6 is 36

3. A number card is drawn randomly from the set of numbered cards, numbered from
6 to 39. Find the probability that the card may be a prime number or cubed number.

Solution:

Let P and C denote the event of getting a prime and cubed numbered cards respectively.

Then, S = {6, 7, 8, .........., 39}

P = {7, 11, 13, 17, 19, 23, 29, 31,37} ∴ n(P) = 9

C ={8, 27} ∴ n(C) = 2

Now, probability of getting prime or cubed numbered card

P(P or C) = P(P) + P(C) = 9 + 2 = 11 11
So, 34 a 34 34 34
the probability that the card is prime number or cubed number is

151 Vedanta Excel in Mathematics Teachers' Manual - 10

4. A natural number is chosen at random from the first 30 natural numbers. What is
the probability that the number chosen is either multiple of 3 or a multiple of 4 ?

Solution:

Let M3 and M4 denote the events of getting a multiple of 3 and multiple of 4
respectively.

Then, S = {1, 2 , 3, ........, 30} ∴ n(S) = (30 − 1) + 1 = 30

M3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} ∴ n(M3) = 10
M4 = {4, 8, 12, 16, 20, 24, 28} ∴ n(M4) = 7
n(M3∩M4) = {12, 24} ∴ n(M3∩M4) = 2

Now, P(M3 or M4) = P(M3) + P(M4) − P(M3∩M4)

= n(M3) + n(M4) − n(M3∩M4) = 10 + 7 − 2 = 15 = 1
n(S) n(S) n(S) 30 30 30 30 2

So, the probability that the number chosen is either multiple of 3 or a multiple of 4 is 12.
5. If a card is drawn at random from a pack of 52 cards and at the same time a marble

is drawn at random from a bag containing 2 red marbles and 3 blue marbles. Find

the probability of getting a blue marble and a king.

Solution:

Let K, B and R denote the events of getting king card, blue and red marbles respectively.
n(K) 4 1
Then, For drawing a card: n(S1) = 52, n(K) = 4 ∴P(K) = n(S) = 52 = 13

FSoinr cderathweinegveanmtsaarrbelei:nnd(eSp2e)n=de2n+t 3 = 5, n(B) = 3 ∴ P(B) = n(B) = 3
n(S) 5

∴ P(K and B) = P(K) × P(B) = 1 × 3 = 3
13 5 65
Hence the required probability is 635.
6. A bag contains 5 black, 7 blue and 4 yellow balls. A ball is drawn at random and it

is replaced, then another ball is drawn. Find the probability that:

i) the first is blue and the second is black

ii) both of them are yellow

iii) both of them are of the same colour

iv) the first is black and the second is yellow

v) both of them are not black

Solution:

∴LTehTteoBntak,,lnBn(eBuakm)n=bdeYr5,odnfe(bnBaoel)tles=tnh7(eSa)env=den5nt(+Yo7f) getting a black, blue and yellow balls respectively.
=4 16
+4 =

Since the ball which is draw at first is replaced to draw another ball, it is an

independent event. n(Be) n(BK)
n(S) n(S)
i) P(Be and Bk) = P(Be ∩ Bk) = P1(Be) × P2(Bk) = × = 7 nn×((YS15))6 ==146235×56146
ii) P(Both are yellow) = 16
n(Y) 1
P(YY) = P(Y∩Y) = P1(Y) × P2(Y) = n(S) × = 16

iii) P(Both are of same colour) = P(Be Be or Bk Bk or YY) = P(Be Be) + P( Bk Bk) + P(YY)

= 7 × 7 + 5 × 5 + 4 × 4 = 49 + 25 + 16 = 90 45 = 45
16 16 16 16 16 16 256 256 128 128

Vedanta Excel in Mathematics Teachers' Manual - 10 152

iv) P(Bk and Y) = P(Bk) × P(Y) = 5 × 4 = 5
16 16 64

v) P(Bk ∪ Bk) = P1(Bk) × P2(Bk) = [1 − P(Bk)] [1 − P(Bk)] = 1 − 5 1 − 5 = 12
16 16 256

7. A can solve 90% of the problems given in the exercise of probability and B can solve

70%. What is the probability that at least one of them will solve a problem selected

at random from the exercise ?

Solution: 90 9 70 7
100 10 100 10
Here, p(A) = = and p(B) = =

Since, the event of solving problem is independent.
9 7
So, p(A and B) = p(A ∩ B) = p(A) × p(B) = 10 × 10 = 63
100
Also, the problem may be solved by both the students A and B.

So, the events are non - mutually exclusive.

Now, p(solving problem by at least one of them) = p(A ∪ B) = p(A) + p(B) − p(A ∩ B)

= 7 + 7 − 63 = 63 = 0.97.
10 10 100 100

8. Three children were born in a family. By drawing a tree diagram, find the following

probabilities.

i) at least two daughters ii) all of them are boys iii) at least a boy

Solution:

Let S and D denote the events of having son and daughter respectively.

Drawing a probability tree diagram: 1
8
P(S1S2S3) =

) = 12

S P(S 3

1 2 D P(D3) = 1 P(S1S2D3) = 1
= 2 8
)
P(S
2

S P(S3) = 1 1
2 8
D S P(S1D2S3) =
P(D ) = 1 D P(S1D2D3) = 1
=12 2 2 8

P(S ) P(D ) = 1
3 2
1

S1 1
D ) = 2 8
P(S P(D1S2S3) =
P(D S 3
1 D
) =21 P(S2) = 1 P(D ) 1 1
2 3 2 8
S = P(D1S2D3) = 1
P(D1D2S3) = 8
D P(D 1
2 2
) =21 P(S3) =

S P(D
D 3

) = 21 P(D1D2D3) = 1
8

Now, 1 1 1 1 1
8 8 8 8 2
i) P(at least two daughter) = P(SDD) + P(DSD) + P(DDS) + P(DDD) = + + + =
ii)
P(all are boys) = P(SSS) = 1
8

153 Vedanta Excel in Mathematics Teachers' Manual - 10

iii) P(at least a boys) = P(SSS) + P(SSD) + P(SDS) + P(SDD) + P(DSS) + P(DSD) + P(DDS)

are 3 red=, 518b+lu81e +a18nd+ 1 y+e18llo+w18 i+de81nt=ica87l pencils in a box. Two pencils are
9. There 28

drawn at random and without replacing the first one.

(i) Find the probability that both of them are of blue colour.

(ii) Find the probability that both of them are not red colour.

(iii) Find the probability that at least one is yellow.

(iv) Find the probability that none of them are of blue and yellow.

(v) Find the probability that all of them are of same colour.

(vi) Show all these probabilities in a tree diagram.

Solution

Let R1, B1 and Y1 denote the events of getting red, blue and yellow pencils in the
first trial respectively and R2, B2 and Y2 denote the events of getting red, blue and
yellow pencils in the second trial respectively.

Then, no. of possible outcomes = 3 + 5 + 2 = 10

Since, the pencil which is drawn in the first trial is replaced to draw another

pencil. So, it is a dependent event.

(i) The probability that both of them are of blue colour,

P (B1B2) = P (B1) × P (B2) = 5 × 4 = 2
10 9 9
(ii) The probability that both of them are not red colour,

P (B1B2) + P (B1Y2) + P (Y1B2) + P (Y1Y2)

= 5 × 4 + 5 × 2 + 2 × 5 + 2 × 1 = 2 + 1 + 1 + 1 = 21 = 7
10 9 10 9 10 9 10 9 9 9 9 45 45 15
Alternatively,

The probability that both of them are not red colour,

P (R1 ∩ Y2) = P (R1 ∪ Y2)

= [1 - P (R1)] × [1 - P (R2/ (B1Y1)]

= (1 – 130) × (1 – 39) = 7 × 2 = 7
10 3 15
(iii) The probability that at least one is yellow,

P (R1Y2) + P (B1Y2) + P (Y1R2) + P (Y1B2) + P (Y1Y2)

= 3 × 2 + 5 × 2 + 2 × 3 + 2 × 5 + 2 × 1
10 9 10 9 10 9 10 9 10 9

= 1 + 1 + 1 × 1 + 1 = 17
15 9 15 9 45 45
(iv) The probability that none of them are blue and yellow

= Probability of getting both red balls = P (R1R2) = 3 × 2 = 1
10 9 15

(v) The probability that all of them are same colour, = P (R1R2) + P (B1B2) + P (Y1Y2)

= 3 × 2 + 5 × 4 + 2 × 1 = 1 + 2 + 1 = 14
10 9 10 9 10 9 15 9 45 45

Vedanta Excel in Mathematics Teachers' Manual - 10 154

(vi) 2 P(R1 R2) = 1)
=
3R 9 15
5B
2Y P(R ) P(R2) = 5 P(R1 B2) = 1
9 6
2

103 2R5B2Y P(Y ) = 2 P(R1 Y2) = 1
2 9 15

=

) 39 1
6
P(R ) = P(B1 R2) =
1
2
P(B1) = 5 3R P(R
10
4B 4 P(B1 B2) = 2
2Y P(B2) = 9 9
P(Y = 92
P(Y 2 ) P(B1 Y2) = 1
9
)

1

=210 5B3R P(R2) = 3
1Y 9
P(Y1 R2) = 1
15
P(B
2 ) = 5
9
P(Y
P(Y1 B2) = 1
) 9

2

=19

P(Y1 Y2) = 1
45

10. A card is drawn from a pack of numbered cards numbered from 1 to 20. If the

second card is drawn without replacing the first one, find the probability of the

following events,

(i) Both of them are of the multiple of 5.

(ii) The first is of even number and second prime number.

(iii) None of them are multiples of 3.

(iv) Show all these probabilities in a tree diagram.

Solution

(i) LreestpMec1taivnedlyM. 2 denote the events of getting multiple of 5 in the first and second drawing
∴No. tohfecparrdobsaobfimlituylttihpaltesbootfh5o, fnth(Mem1) =are4oafnmd unlt(iMpl2e) = 3
Now, of 5,
ngP(Meu1tmdt1i)enb×ngeorePteev(xetMchnee2np)eut=vm2e2n4bi0tnesr×otehfx1e3gc9eeft=iptritsn9t23g5t,reeivavelennrensupprmeimcbteievrneeluxycmeabpnetdr2E2, e2a,vnTedn2

2LPe(atMnEd11M, pT2r)1i=amnedP prime number
(ii) the events of panridmPe 2nduemnboeter

except 2 in the second trial respectively. Also, R denotes the remaining numbered
cards. Then,

No. of cards of even numbers except 2, n (E) = 9

No. of cards of even prime number 2, n (T) = 1

No. of cards of prime numbers except 2, n (P) = 7

No. of cards of remaining numbers, n (R) = 3

Now,

The probability that the first is of even and second is prime numbers,

P (EP) + P (TP) = P (E1) × P (P2) + P (T1) × P (P2)
9 7 1 7 70 7
= 20 × 19 + 20 × 19 = 380 = 38

155 Vedanta Excel in Mathematics Teachers' Manual - 10

(iii) Let M1 and M2 denote the events of getting multiple of 3 in the first and second
drawing respectively.

∴No. of cards of multiples of 3, n (M1) = 6 and n (M2) = 5
Now, the probability that none of them are of multiple of 3,

P (M1M2)= 1 – P (M1) × P (M2) = 1 – 6 × 5 = 1 – 3 = 35
20 19 38 38
(iv) Required tree-diagrams
3
) = 319 P(M1M2) = 95

P(M 2

3M 16
95
(i) = 420 16M P(M ) = 1196 P(M1M2) =
2
4M P(M )

16M 1

P(M ) =2106 4M P(M2) = 149 P(M1M2) = 16
1 15M 95

P(M ) = 1915 12
19
2

P(M1M2) =

(ii) 8E P(E 2)PPP(=((RP2T8)212)=)9=1=93P111(79PE9(1ERPP12E(()EE2)=11PT=223))287==01985336898300
1T
7P
3R

209 ) = 919 P(T1E2) = 9
7
= 19 P(T1P2) 380
139
) 9E
7P P(T1R2)
P(E P(E 1 1) = = 7
1 3R 380
P(P
120
9E P(T ) = P(R ) = 3
1T 1 380
1
=

7P P(P ) = 207 9E P(E 2)PPP(=((RP2T9)212)=)9=1=9P311(169P9P1(RPPP21()(EPP21=1)TP22=3))89==306833012879100
3R 1 1T
6P
P(R 3R

)

1

=

320

) = 919 P(R1E2) = 27
119 P(R1T2)
9E 380
1T
P(E 2 2) = = 3
380
P(T
P(P
7P P(R 2 ) = 179
2R 2 =192 P(R1P2) = 21
) 380

P(R1R2) = 3
190

Vedanta Excel in Mathematics Teachers' Manual - 10 156

) = 519 P(M1M2) = 3
38
P(M 2

5M 21
95
= 620 14M P(M ) = 1194 P(M1M2) =
2
P(M )
(iii)
1

6M P(M2) = 169

14M P(M P(M1M2) = 42
1 190
) =2104 6M
13M
P(M
) = 1913 91
2 190

P(M1M2) =

Extra questions

1. A box contains the lottery tickets numbered from 3 to 32. If a ticket is drawn at random,
[Ans: 51]
what is the probability of the ticket bearing square or cube number?

2. An ace of diamond is lost from a deck of 52 playing cards and a card is drawn at random.
[Ans: 137]
What is the probability of getting black faced card or ace?

3. A dice is rolled and a coin is tossed at the same time, find the probability of occurring
[Ans: 14]
even number on the dice and head on the coin.

4. A card is draw from a well-shuffled pack of 52 cards and at the same time a marble is

drawn at random from a bag containing 2 green and 3 red marbles of same shape and
[Ans: 110]
size. Find the probability of getting a spade and green marble.

5. A bag contains 1 yellow, 1 black and 1 green balls of same shape and size. Two balls are
drawn at randomly one after another without replacement; show the probabilities of all
possible outcomes in a tree-diagram.

6. A coin is tossed thrice successively. Show the probabilities of all the possible outcomes
1
in a tree diagram and find the probability of getting all heads. [Ans: 8 ]

157 Vedanta Excel in Mathematics Teachers' Manual - 10

SEE Model Question Set-2077 Maximum Marks: 100
Compulsory Mathematics

Time: 3 hrs.

Candidates are required to answer in their own words as far as practicable. Credit shall be

given to originality in expression, creativity and neatness in hand, not to rote learning.

Attempt all the questions

Group- A 6×1 = 6

1. (a) If the selling price of a calculator after discount is Rs x and the selling price with

VAT is Rs y, what is the rate of VAT?

(b) Write down the formula to calculate the surface area of a sphere whose radius is

p cm. 3N − c.f .
4
2. (a) What is the order of the surd 4 3 ? f

(b) What does ‘L’ represent in the formula Q3 = L + ×h? E

A B

3. (a) In the given figure, which triangle is equal in area to ∆ABC?

(b) Define cyclic quadrilateral. DC

Group- B 17 × 2= 34

4. (a) If US $ 1 = NRs 114.50 and GBP £ 1 = NRs 148.85, convert US $ 1001 into

pound.

(b) The population of a town in beginning of B.S. 2077 was 110000. If the population

of the town increased 1% by birth and 2% by immigration every year, what will

be the population of the town at the end of B.S. 2078?

5. (a) How many cubic meters of earth must be dug out to construct a cylindrical well

which is 20 m deep and the radius of the base is 50 cm? (Use π = 3.14)

(b) The diameter of volleyball is 20 cm. Find its surface area. (Use π = 3.14)

A'

(c) Find the area of rectangular surfaces of the triangular prism A 4.5cm

given alongside. 3cm
6. (a) Find the HCF of a2 – 4 and a3 – 8 B' C'
B 3.5cm C 5cm
(b) Find the LCM of z4 + z2 + 1 and z3 – z2 + z

7. (a) Simplify: xa(b−c) ÷  xb c
x b(a −c)  xa 
 

(b) Solve: 2 x – 3 ×3 x – 2 = 3

(c) Simplify: 3 128 − 3 16
32
8. (a) In the given figure, WXYZ is a parallelogram, WU and XU UZ Y

are joined to meet YZ produced at U and V is any point on

the side WX. If the area of ∆UWX is 35 sq. cm, what is the W VX

area of ∆VYZ? Find it.

Vedanta Excel in Mathematics Teachers' Manual - 10 158

(b) In the given figure, O is the centre of circle. If ∠OQ = QR, find

the value of ∠PTS. P OS

QR

(c) In the figure, O is centre of circle, TAN the tangent and T
B

M

BM // ON. If∠ONA = 300. Find the value of ∠NOA and BAM. O

TA 30° N

9. (a) In the adjoining figure, area of ∆ABC is 10 sq. cm, AB = 5cm, A

∠A = 800 and ∠C = 700, find the measurement of BC. 5cm 70° C
In a continuous series, if Σfm = 850 + 35p, N = 30 + p and 80°
(b) arithmetic mean (X) = 31, find the value of p and N.
B

10. (a) From a pack of 52 playing cards, a card is drawn at random. What is the probability

of getting a black king or a red queen?

(b) A bag contains a dozen of pencils of same shape and size. Among them 7 are red

and the rest are blue. Two pencils are drawn randomly in succession without

replacement. Show the probability of all possible outcomes in a tree diagram.

Group- C 10×4= 40

11. In a group of 100 non-veg people, 40 like chicken but not fish and 30 like fish but

not chicken. If the number of people who do not like any of the two non-veg items is

double the number of people who like both items, find the number of people who like

at most one non-veg item by using a Venn-diagram.

12. After allowing 20% discount on the marked price and then levying 13% value added

tax, a cycle was sold. If the discount amount was Rs 2,200, how much value added tax

was levied on the price of the cycle? Find it.

13. An umbrella is made up by stitching 6 isosceles triangular pieces of cloth. The

measurement of the base of each triangular piece is 28 cm and the lengths of two

equal sides are 50 cm each. How much cloth is required for the umbrella? If the rate

of cost of the cloth is 25 paisa per sq. cm, find the total cost of the cloth to make the

umbrella. x −1 3 1
+ 3x +x − 3x
14. Simplify: 2x 2 − 2 + x 2 − 2 − 2x2 + 1

15. Solve: 25x – 6 × 5x + 1 + 125 = 0
16. Prove that the area of triangle BCD is equal to half of the area of a parallelogram ABCE

on the same base BC and between the same parallels AD and BC.

17. Construct a triangle ABC in which AB = 4.7 cm, ∠ABC = 450 and ∠ACB = 600. Also,
construct a parallelogram having one side 6.5 cm and equal in area to the triangle

ABC.

18. Explore experimentally the relationship between the inscribed angles of a circle
standing on the same arc. (Two circles with radii at least 3 cm are necessary)

159 Vedanta Excel in Mathematics Teachers' Manual - 10

19. A man 1.5 m tall standing at a distance of 54 m from a tree observes the angle of

elevation of the top of the tree to be 450. Find the height of the tree.

20. The median of the following distribution is 32, find the value of p.

Marks obtained 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65

No. of students 5 8 p 9 7 1

Group- D 4×5 = 20
21. Nepal Bank Limited slightly changes its policy and decides to pay half-yearly interest

at the following rates for the deposits in the Fixed Account.

Deposit Time Rate of interest p.a. Minimum balance required

Up to 6 Months 6.75% Rs 25,000

Above 6 Months up to 1 Year 7.25% Rs 25,000

Above 1 Year up to 5 Years 7.5% Rs 25,000

Dhurmus gets the above information. He borrows a loan of Rs 2,00,000 from Manoj at
the rate of 5% simple interest for 2 years and immediately deposites the same sum for
the same duration of time in his Fixed Account of Nepal Bank Limited.

(i) Calculate the amount that he accumulated in 2 years.
(ii) Calculate the amount that he had to pay to Manoj?
(iii) Find his gain at the end of 2 years.
22. A cylindrical tin of 40cm high and base radius 14 cm is

completely filled with the cement. For the cement work of wall,
a mason pours the cement on the ground from the tin and finds
a conical heap of height 30 cm.
(i) Calculate the base radius of the heap of the cement.
(ii) Find the surface area of heap.
(iii) If 1 cm3 equals to 2.5 g, find the mass of the cement.
23. From the figure given below, find the heights of the table and the cat.


24. Five students Ambika, Binaya, Chhiring, Dawa and Imran are plying in the park.
(i) Ambika, Binaya, Chhiring and Dawa are standing on the circumference of a

circular track such Dawa is equidistance from Ambika and Chhiring.
(ii) Imran is standing outside the circular track but Binaya, Chhiring and Imran are

on the same straight line.
(iii) The distance between Ambika and Binaya and the distance between Chhiring

and Imran are same.
Use the above conditions to prove Binaya, Imran and Dawa act as the vertices of an

isosceles triangle.
The End

Vedanta Excel in Mathematics Teachers' Manual - 10 160