chapter : 1 - arithmetic progression and geometric...

MT EDUCARE LTD. ALGEBRA

SCHOOL SECTION 339

HOTS PROBLEMS

CHAPTER : 1 - ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION

1. If 3rd term of an A.P. is ‘p’ and the 4th term is ‘q’. Find its nth term andhence find its 10th term. (5 marks)

Given : t3 = pt4 = q

To find : tn and t10

Sol. tn = a + (n – 1) dt3 = a + (3 – 1) dp = a + 2d

a + 2d = p ........(i)t4 = a + (4 – 1) d

q = a + 3d a + 3d = q ........(ii)

Subtracting (ii) from (i),a + 2d = pa + 3d = q

(– ) (– ) (–)– d = p – q

Multiplying both sides by – 1d = q – p

Substituting d = q – p in (i),a + 2 (q – p) = p

a + 2q – 2p = p a = p + 2p – 2q a = 3p – 2q

tn = a + (n – 1)d

tn = 3p – 2q + (n – 1) (q – p) t10 = 3p – 2q + (10 – 1) (q – p) = 3p – 2q + 9 (q – p)

= 3p – 2q + 9q – 9p

t10 = 7q – 6p

2. Find three consecutive terms in a G.P. such that their sum is 38 andproduct is 1728. (5 marks)

Sol. Let the three consecutive terms of G.P. be a

r, a and ar.

From first given condition,a

r + a + ar = 38 .......(i)

From second given condition,a

r × a × ar = 1728

a3 = 1728 a = 12 [Taking cube roots on both sides]

Substituting a = 12 in (i),12

r + 12 + 12r = 38

Multiplying throughout by r,

EXTRA HOTS SUMS

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SCHOOL SECTION340

12 + 12r + 12r2 = 38r 12r2 + 12r – 38r + 12 = 0 12r2 – 26r + 12 = 0

Dividing throughout by 2,6r2 – 13r + 6 = 0

6r2 – 9r – 4r + 6 = 0 3r (2r – 3) – 2 (2r – 3) = 0 (2r – 3) (3r – 2) = 0 2r – 3 = 0 or 3r – 2 = 0 2r = 3 or 3r = 2

r =3

2or r =

2

3

If r =3

2 then If r =

2

3 then

a

r=

1232

a

r=

1223

=12 2

3

=

12 3

2

a

r= 8

a

r= 18

ar =3

122

ar =2

123

ar = 18 ar = 8

The three consecutive terms of a G.P. are 8, 12, 18 or 18, 12, 8.

3. The 5th, 8th and 11th term of a G.P. are p, q and s. Show that q2 = ps. (4 marks)

Given : For a G.P.t5 = pt8 = qt11 = s

To find : q2 = psProof : For a G.P.

tn = arn–1

t5 = ar5–1

p = ar4 .......(i)t8 = ar8–1

q = ar7 .......(ii)t11 = ar11–1

s = ar10 ......(iii)q = ar7 [From (ii)L.H.S. = q2

= (ar7)2

= a2r14 ......(iv)R.H.S. = ps

= ar4 . ar10

= a2r4+10

= a2r14 .......(v)From (iv) and (v),L.H.S. = R.H.S.

q2 = ps


SCHOOL SECTION 341

4. If each term of A.P. is doubled, is the resulting sequence also an A.P. ? Ifit is write its first term, common difference and nth term. (4 marks)

Sol. Let the A.P. be a, a + d, a + 2d, a + 3d .......... .If each term of the A.P. is doubled the new sequence will be2a, 2a + 2d, 2a + 4d, 2a + 6d [Multiplying each term by 20]t1 = 2at2 = 2a + 2dt3 = 2a + 4dt4 = 2a + 6dt2 – t1 = 2a + 2d – 2a

= 2d ........(i)t3 – t2 = 2a + 4d – (2a + 2d)

= 2a + 4d – 2a – 2d= 2d .......(ii)

t4 – t3 = 2a + 6d – (2a + 4d)= 2a + 6d – 2a – 4d= 2d ......(iii)

The difference between two consecutive term is 2d remaining constantthe new sequence is an A.P. withFirst term (A) = 2aCommon difference (D) = 2dnth term (Tn) = A + (n – 1)D

= 2a + (n – 1) 2d= 2 [a + (n – 1) d

Tn = 2tn

5. Three consecutive terms in G.P. are such that their sum is 26 and sumof their squares is 364. Find the three consecutive terms in G.P.(5 marks)

Sol. Let the three consecutive terms in G.P. be a

r, a, ar

According to the first condition,a

r + a + ar = 26 .......(i)

According to the second condition,

a

r

2

2 + a2 + a2r2 = 364 ......(ii)

From (i) a

r + ar = 26 – a

Squaring both sides,

aar

r

2

= (26 – a)2

a a

2 ar (ar)rr

2

22 = 262 – 2 × 26 × a + a2

a

2a a rr

2

2 2 22 = 676 – 52a + a2

a

2a – a a rr

2

2 2 2 22 = 676 – 52a

a

a a rr

2

2 2 22 = 676 – 52a ......(iii)

Substituting (ii) in (iii),364 = 676 – 52a

52a = 676 – 364


SCHOOL SECTION342

52a = 312

a =312

52 a = 6

Substituting a = 6 in (i) we get,

6

r + 6 + 6r = 26

Multiplying throughout by r6 + 6r + 6r2 = 26r

6r2 + 6r – 26r + 6 = 0 6r2 – 20r + 6 = 0

Dividing through by 2,3r2 – 10r + 3 = 0

3r2 – 9r – r + 3 = 0 3r (r – 3) – 1 (r – 3) = 0 (r – 3) (3r – 1) = 0 r – 3 = 0 or 3r – 1 = 0

r = 3 or r =1

3

If r = 3 If r = 1

3

a

r =

6

3= 2

a

r=

613

= 6 × 3 = 18

ar = 6 × 3 = 18 ar = 1

63

= 2

The three consecutive terms in a G.P. are 2, 6, 18 or 18, 6, 2.

6. If sum of m terms is ‘n’ and sum of n terms is ‘m’ then show that sum of(m + n) terms is – (m + n). (5 marks)

Given : Sm = nSn = m

To find : Sm+n = – (m + n)

Proof : Sm =m

2 [2a + (m – 1) d]

n =m

2 [2a + (m – 1) d] .......(i)

Sn =n

2 [2a + (n – 1) d]

m =n

2[2a + (n – 1) d] .......(ii)

Subtracting (ii) from (i),

n – m =m

2 [2a + (m – 1) d] –

n

2 [2a + (n – 1) d]

Multiplying throughout by 2,

2 (n – m) = m [2a + (m – 1) d] – n [2a + (n – 1) d]

2 (n – m) = m [2a + md – d] – n [2a + nd – d]


SCHOOL SECTION 343

2 (n – m) = 2am – 2an + m2d – n2d – md + nd 2 (n – m) = 2a (m – n) + d (m2 – n2) – d (m – n) – 2 (m – n) = 2a (m – n) + d (m – n) (m + n) – d (m – n)

Dividing throughout by m – n– 2 = 2a + d (m + n) – d

– 2 = 2a + (m + n – 1) d ......(iii)

Now, Sm+n =m n

2

[2a + (m + n – 1) d]

Sm+n =m n

2

[– 2]

Sm+n = m + n (– 1)

Sm+n = – (m + n)

7. 200 logs of wood are stacked in the following manner 20 logs in thebottom row, 19 in the next row, 18 in the row next to it and so on. Inhow many rows 200 logs are placed and how many logs are there in thetop row. (5 marks)

Sol. There are 20 logs in the first row, 19 in the second row and 18 in therow next to itThis arrangement of logs20, 19, 18, ............. forms an A.P. witha = 20, d = – 1Let 200 logs be arranged in n rows

Sn = 200We know,

Sn =n

2[2a + (n – 1) d]

200 =n

2 [2 (20) + (n – 1) (– 1)]

400 = n [40 – n + 1] 400 = n (41 – n] 400 = 41n – n2

n2 – 41n + 400 = 0 n2 – 25n – 16n + 400 = 0 n (n – 25) – 16 (n – 25) = 0 n – 25 = 0 or n – 16 = 0 n = 25 or n = 16

If n = 25 tn = a + (n – 1) d

t25 = 20 + (25 – 1) (– 1) t25 = 20 + 24 (– 1) t25 = 20 – 24 t25 = – 4 No. of logs in the 25th row cannot be negative n = 25 is not acceptable n = 16

tn = a + (n – 1) d t16 = 20 + (16 – 1) (– 1) t16 = 20 + 15 (– 1) t16 = 20 – 15 t16 = 5

200 logs are placed in 16 rows and there are 5 logs in the top row.


SCHOOL SECTION344

CHAPTER : 2 - QUADRATIC EQUATIONS

1. Two pipes running together can fill a cistern in 1

313

minutes. If one

pipe takes 3 minutes more than the other to fill it, find the time inwhich each pipe would fill the cistern. (5 marks)

Sol. Let the time taken by other pipe to fill the cistern be ‘x’ minutes Time taken by first pipe to fill the cistern = (x + 3) minutes

Portion of cistern filled by other pipe in one minute = 1

x

Portion of cistern filled by first pipe in one minute = 1

x 3Portion of cistern filled by both the pipes in one minute

=

11

313

=

14013

=13

40From the given condition,

1 1

x x 3

= 13

40

x 3 x

x (x 3)

=

13

40

2x 3

x 3x

2 =

13

40 40 (2x + 3) = 13 (x2 + 3x) 80x + 120 = 13x2 + 39x 13x2 – 41x – 120 = 0 13x2 – 65x + 24x – 120 = 0 13x (x – 5) + 24 (x – 5) = 0 (x – 5) (13x + 24) = 0 x – 5 = 0 or 13x + 24 = 0

x = 5 or x = –24

13

x = –24

13 is not acceptable because time cannot be negative.

x = 5 x + 3 = 5 + 3 = 8

Time taken by two pipes to fill the cistern separately are5 minutes and 8 minutes.

2. If the equation (1 + m2) x2 + 2m cx + (c2 – a2) = 0 has equal roots, provethat c2 = a2 (1 + m2). (4 marks)

Sol. (1 + m2) x2 + 2mcx + (c2 – a2) = 0Comparing with Ax2 + Bx + C = 0 we get, A = 1 + m2, B = 2mc, C = c2 – a2

The equation has equal roots, B2 – 4AC = 0 (2mc)2 – 4(1 + m2) (c2 – a2) = 0 4m2c2 – 4(1 (c2 – a2) + m2 (c2 – a2)] = 0


SCHOOL SECTION 345

Dividing throughout by 4,m2c2 – [c2 – a2 + m2c2 – m2a2] = 0

m2c2 – c2 + a2 – m2c2 + m2a2 = 0 – c2 + a2 + m2a2 = 0 a2 + m2 a2 = c2

c2 = a2 (1 + m2)Hence proved.

3. Out of a group of swans, 72

times the square root of the number are

playing on the shore of a tank. The two remaining one are playing withamorous fight in the water. What is the total no. of swans ? (5 marks)

Sol. Let total no. of swans be x then,

No. of swans playing on the shore of the tank = 7

x2

No. of swans with amorous fight = 2From the given condition,

x =7

x 22

Multiplying throughout by 2 we get,

2x =7

x 22

2x – 7 x – 4 = 0

2 x2 – 7 x – 4 = 0

Substituting x = m

2m2 – 7m – 4 = 0 2m2 – 8m + 1m – 4 = 0 2m (m – 4) + 1 (m – 4) = 0 (m – 4) (2m + 1) = 0 (m – 4) (2m + 1) = 0 m – 4 = 0 or 2m + 1 = 0

m = 4 or m = –1

2Resubstituting m = x we get,

x = 4 or x = –1

2Squaring throughout,

x = 42 or x = –1

2

2

x = 16 or x = 1

4

x = 1

4 is not acceptable because no. of swans cannot be a fraction.

x = 16

Total no. of swans is 16.

4. Two trains leave a railway station at the same time. The first traintravels due west and the second train due north. The first train travels5km/hr faster than second train. If after two hours, they are 50 kmapart. Find the speed of each train. (5 marks)

Sol. Let speed of second train be x km/hr Speed of first train = (x + 5) km/hr


SCHOOL SECTION346

Distance = Speed × timeDistance covered by second train = x × 2 = (2x) kmDistance covered by first train = (x + 5)2 = (2x + 10) kmIn ABC,m ABC = 90ºBy Pythagoras theorem,

AB2 + BC2 = AC2

(2x)2 + (2x + 10)2 = (50)2

4x2 + (2x)2 + 2 + 2x × 10 + 102 = 2500 4x2 + 4x2 + 40x + 100 – 2500 = 0 8x2 + 40x – 2400 = 0

Dividing throughout by 8,x2 + 5x – 300 = 0

x2 + 20x – 15x – 300 = 0 x (x + 20) – 15 (x + 20) = 0 (x + 20) (x – 15) = 0 x + 20 = 0 or x – 15 = 0 x = – 20 or x = 15

x = – 20 is not acceptable because speed cannot be negative x = 15 x + 5 = 15 + 5 = 20

The speed of the trains are 15 km/hr and 20 km/hr.

5. There is a square field whose side is 44 m. A square flower bed is preparedin its center leaving a gravel path all round the flower bed. The totalcost of laying flower bed and gravelling the path at Rs. 2.75 and Rs.1.50 per square meter respectively is Rs. 4904. Find the with of thegravel path. (5 marks)

Sol. Length of the square field = 44 m Area of the square field = 44 × 44 = 1936 sq.m.

Let width of the gravel path be ‘x’ m Length of the square flower bed = (44 – 2x)m

Area of the square flower bed = (44 – 2x)2 sq.m.Area of the gravel path

= Area of the field – Area of the square flower bed= 1936 – (44 – 2x)2

= 1936 – [442 – 2 × 44 × 2x + (2x)2]= 1936 – (1936 – 176x + 4x2)= 1936 – 1936 + 176x – 4x2

= (176x – 4x2) sq.m.Cost of laying the flower bed

= (Area of flower bed) × (Rate per sq. m.)

= [442 – 2 × 44 × 2x + (2x)2] × 275

100

= [1936 – 176x + 4x2] × 275

100

= [484 – 44x + x2] × 4 × 275

100= [484 – 44x + x2] 11= Rs. (5324 – 484x + 11x2)

Cost of laying the gravel path= (Area of gravel path) × (Rate per sq. m.)= (176x – 4x2) (1.50)

A

C B2x + 10

2x50 km

44 – 2x

44 –

2x

FlowerBed

x

x

x

x


SCHOOL SECTION 347

= (176x – 4x2) × 150

100

= 2 (88x – 2x2) × 3

2= Rs. (264x – 6x2)

As per the given condition,5324 – 484x + 11x2 + 264x – 6x2 = 4904

5x2 – 220x + 420 = 0Dividing throughout by 5,

x2 – 44x + 84 = 0 x2 – 42x – 2x + 84 = 0 x (x – 42) – 2 (x – 42) = 0 (x – 42) (x – 2) = 0 x = 42 or x – 2 = 0 x = 42 or x = 2

x = 42 is not acceptable because side of field is 44 m. x = 2

The width of the gravel path is 2 m.

6. Solve : 2x4 – 9x3 + 14x2 – 9x + 2 = 0 (5 marks)Sol. 2x4 – 9x3 + 14x2 – 9x + 2 = 0

Dividing throughout by x2 we get,2x 9x 14x 9x 2

– –x x x x x

4 3 2

2 2 2 2 2 = 0

2x2 – 9x + 14 – 9

x +

2

x2 = 0

2x2 + 2

x2 – 9x – 9

x + 14 = 0

1 1

2 x – 9 x 14xx

22 = 0

Substituting x + 1

x= m

Squaring both sides 1

xx

2

= m2

x2 + 2 × x × 1

x +

1

x

2

= m2

x2 + 2 + 1

x2 = m2

x2 + 1

x2 = m2 – 2

2 (m2 – 2) – 9m + 14 = 0 2m2 – 4 – 9m + 14 = 0 2m2 – 9m + 10 = 0 2m2 – 4m – 5m + 10 = 0 2m (m – 2) – 5(m – 2) = 0 (m – 2) (2m – 5) = 0 m – 2 = 0 or 2m – 5 = 0 m = 2 or 2m = 5

m = 2 or m = 5

2

Resubstituting m = 1

xx


SCHOOL SECTION348

1x

x = 2 ......(i)

1x

x =

5

2. ......(ii)

1x 2

x [From (i)]

Multiplying throughout by ‘x’,x2 + 1 = 2x

x2 – 2x + 1 = 0 (x – 1)2 = 0 x – 1 = 0 x = 1

1x

x =

5

2[From (ii)]

Multiplying throughout by 2x we get,2x2 + 2 = 5x

2x2 – 5x + 2 = 0 2x2 – 4x – 1x + 2 = 0 2x (x – 2) – 1 (x – 2) = 0 (x – 2) (2x – 1) = 0 x – 2 = 0 or 2x – 1 = 0

x = 2 or x = 1

2

Solution set = 1

1, 2 ,2

CHAPTER : 3 - LINEAR EQUATIONS IN TWO VARIABLES

1. Solve : 2x + 3y = 17, 2x+2 – 3y+1 = 5. (4 marks)Sol. 2x + 3y = 17 ......(i)

2x+2 – 3y+1 = 5 2x . 22 – 3y . 31 = 5 4.2x – 3.3y = 5 .....(ii)

Substituting 2x = a and 3y = b in (i) and (ii),a + b = 17 ....(iii)

4a – 3b = 5 .....(iv)Multiplying (iii) by 3

3a + 3b = 51 ......(v)Adding (iv) and (v),

4a – 3b = 53a + 3b = 51

7a = 56 a = 8Substituting a = 8 in (iii),

8 + b = 17 b = 17 – 8 b = 9

Resubstituting the values of a and ba = 2x

8 = 2x

23 = 2x

x = 3 [ Bases are equal, powers are also equal]


SCHOOL SECTION 349

b = 3y

9 = 3y

32 = 3y [ Bases are equal, powers are also equal] y = 2

x = 3 and y = 2 is the solution.

2. Solve : (4 marks)Sol. (a – b)x + (a + b)y = a2 – 2ab – b2 .......(i)

(a + b) (x + y) = a2 + b2

(a + b)x + (a + b)y = a2 + b2 ......(ii)Subtracting (ii) from (i),(a – b)x + (a + b)y = a2 – 2ab – b2

(a + b)x + (a + b)y = a2 + b2

(–) (–) (–) (–)[a – b – (a + b)]x = – 2ab – 2b2

(a – b – a – b) x = – 2b (a + b) – 2bx = – 2b (a + b)

x =–2b (a b)

– 2b

x = (a + b)Substituting x = a + b in (i),(a – b) (a + b) + (a + b)y = a2 – 2ab – b2

a2 – b2 + (a + b)y = a2 – 2ab – b2

(a + b) y = – 2ab

y =–2ab

a b

x = a + b and y = –2ab

a b is the solution of given simultaneous equations.

3. Find the area of the region bounded by the following lines and X-axis.4x – 3y + 4 = 0, 4x + 3y – 20 = 0. (5 marks)

Sol. 4x – 3y + 4 = 0 4x + 4 = 3y

4x 4

3

= y

y = 4x 4

3

x 2 5 – 4

y 4 8 – 4

(x, y) (2, 4) (5, 8) (– 4, – 4)

4x + 3y – 20 = 0 3y = 20 – 4x

y = 20 – 4x

3x 2 – 1 5

y 4 8 0

(x, y) (2, 4) (–1, 8) (5, 0)


SCHOOL SECTION350

A (ABC) =1

b h2

= 1

6 42

A (ABC) = 12 sq. units

4. Places A and B 100 km apart on the highway. One car starts from A andanother from B at the same time. If the cars travel in the same direction atdifferent speed, they meet in 5 hours. If they travel towards each other,they meet in 1 hour. What are the speeds of the two cars ? (5 marks)

Sol. Let the speed of first car, starting from A be x km/hrLet the speed of second car, starting from B be y km/hrDistance = Speed × Time

Y

Scale : 1 cm = 1 uniton both the axes

Y

41 2 3 5 X-5 -4 -3 -2X

-1

-2

-3

4

5

3

2

-4

4x –

3y

+ 4

= 0

(5, 8)

0

6

8

7

1

-1

(– 1, 8)

(2, 4)A

(– 4, – 4)

4x + 3y – 20 = 0


SCHOOL SECTION 351

Distance travelled by first car in 5 hours = 5x kmDistance travelled by second car in 5 hours = 5y km

AC – BC = ABFrom first given condition,5x – 5y = 100Dividing throughout by 5,x – y = 20 ......(i)Distance travelled by first car in 1 hour = 1x = x kmDistance travelled by second car in 1 hour = 1y = y km

AC + BC = ABFrom second given condition,x + y = 100 ......(ii)Adding (i) and (ii),x – y = 20x + y = 1002x = 120

x = 60Substituting x = 60 in (ii),60 + y = 100

y = 100 – 60 y = 40

The speed of cars are 60 km/hr and 40 km/hr.

5. Find the values of p and q for which the following system of equationshas infinite solutions. (4 marks)

Sol. 2x + 3y = 7(p + q) + (2p – q) = 21Comparing with a1x + b1y = ca1 = 2, b1 = 3, c1 = 7Comparing with a2x + b2y = c2

a2 = p + q, b2 = 2p – q, c2 = 21 The equation have infinite solutions

a

a1

2 =

b

b1

2 =

c

c1

2

2

p q = 3

2p q = 7

21

2

p q = 7

21 42 = 7 (p + q) p + q = 6 ........(i)

3

2p – q = 3

21 3 × 21 = 3 (2p – q)

3 21

3

= 2p – q

2p – q = 21 ......(ii)

A B C100 km5y km

5x km

A B C100 km

x km y km


SCHOOL SECTION352

Adding (i) and (ii), p + q = 62p – q = 213p = 27

p = 9Substituting p = 9 in (i),9 + q = 6

q = 6 – 9 q = – 3

p = 9 and q = – 3.

6. The ratio of incomes of two persons is 9 : 7 and the ratio of theirexpenditures is 4 : 3. If each of them saves Rs. 200 per month. Findtheir monthly incomes. (4 marks)

Sol. Let monthly incomes of two persons be Rs. x and Rs. yFrom first given condition,x

y = 9

7 7x = 9y 7x – 9y = 0 ......(i) Each one of them saves Rs. 200

Expenditure of first person = Rs. (x – 200)Expenditure of second person = Rs. (y – 200)From second given condition,x – 200

y – 200 = 4

3 3 (x – 200) = 4 (y – 200) 3x – 600 = 4y – 800 3x – 4y = – 200 ......(ii)

D = 7 – 9

3 – 4 = (7 × – 4) – (– 9 × 3) = – 28 + 27 = – 1

Dx = 0 – 9

– 200 – 4 = (0 × – 4) – (– 9 × – 200) = 0 – 1800 = – 1800.

Dy = 7 0

3 – 200 = (7 × – 200) – (0 × 3) = – 1400 – 0 = – 1400

By Cramer’s rule,

x = D

Dx =

–1800

–1 = 1800

y = D

Dy

= –1400

–1 = 1400

The income of two person are Rs. 1800 and Rs. 1400.

CHAPTER : 4 - PROBABILITY

1. A coin is tossed twice. If the second toss results in a tail a die is thrown.Write the sample space.

Sol. S = { HH, TH, HT1, HT2, HT3, HT4, HT5, HT6,TT1, TT2, TT3, TT4, TT5, TT6 }

n (S) = 14


SCHOOL SECTION 353

2. Two dice are thrown together. What is the probability that sum of thenumbers on two dice is 5 or number on the second die is greater than orequal or equal to the number on the first die. (4 marks)

Sol. When two dice are thrownS = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)(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) = 36Let A be the event that the sum of the numbers on the two dice is 5A = { (1, 4), (2, 3), (3, 2), (4, 1) }n (A) = 4

P (A) =n (A)

n (S)

P (A) =4

36Let B be the event that number on the second die is greater than orequal to the number on the first dieB = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 2), (2, 3), (2, 4), (2, 5), (2, 6)(3, 3), (3, 4), (3, 5), (3, 6),(4, 4), (4, 5), (4, 6)(5, 5), (5, 6),(6, 6) }

n (B) = 21

P (B) =n (B)

n (S)

P (B) =21

36A B is the event that sum of the numbers on the two dice is 5 andthe number on the second die is greater than or equal to the numberon the first die.A B = { (1, 4), (2, 3) }n (A B) = 2

P (A B) =(A B)

n (S)

P (A B) =2

36

A B is the event that sum of the number on two dice is 5 or number onthe second die is greater than or equal to the number on the first die.P (A B) = P (A) + P (B) – P (A B)

=4 21 2

–36 36 36

=4 21 – 2

36

P (A B) =23

36


SCHOOL SECTION354

3. On the disc show below, a player spins the arrow twice. The fraction ab

is formed where a is the number of the sector where the arrow stopsafter the first spin and b is the number of the sector where the arrowstops after the second spin. On every spin each of the numbered sectorhas are equal probability of being the sector on which the arrow stops. What

is the probability that the fraction ab

is greater than 1 ? (4 marks)

Sol. Since the arrow can stop in any one of thesix sectors. So a and b both car assumevalues from 1 to 6.Thus, the ordered pain (a, b) can be as follows :S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

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

For the fraction a

b to be greater than 1, a should be greater than B.

Let A be the event that the fraction a

b A = { (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3),

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

n (A) = 15

P (A) =n (A)

n (S)

P (A) =15

36

P (A) =5

12

4. In the adjoining figure,a dart thrown lands in the interior ofthe circle. What is the probability thatthe dart will land in the shaded region ? (4 marks)

Sol. We have l (AB) = l (CD) = 8l (BC) = l (AD) = 6ABCD is a rectangleIn ABC,m ABC = 90ºBy Pythagoras theorem,AC2 = AB2 + BC2

AC2 = 82 + 62 AC2 = 64 + 36 AC2 = 100 AC = 10

Area of circle = (OA)2

= (5)2

= 25= 25 × 3.14= 78.5 sq. units

2

1

6

5

4

3

O

8

6

8

6

D

A

C

B


SCHOOL SECTION 355

Area of ABCD = AB × BC= 8 × 6= 48 sq. units

Area of shaded reagin = Area of circle – Area of rectangle= 78.5 – 48= 30.5 sq. units

Let A be the event that the dart lands in the shaded region

P (A) =Area of shaded region

Area of circle

=30.5

78.5

=305

785

P (A) =61

157

5. A letter is chosen at random, from the letter in the word ‘ASSASSINATION’.Find the probability that the letter chosen is a (i) vovel (ii) consonant.

(3 marks)Sol. There are 13 letters in the word ‘ASSASSINATION’ out of which one

letter can be chosen in n (S) = 13(i) Let A be the event that the letter chosen is a vowel

There are 6 vowels n (A) = 6

P (A) =n (A)

n (S)

P (A) =6

13(ii) A is the event that the letter chosen is a consonant.

P (A) + P (A) = 1 P (A) = 1 – P (A)

P (A) = 6

1 –13

P (A) = 13 – 6

13

P (A) = 7

13

6. A number ‘x’ is selected from the numbers 1, 2, 3 and then a secondnumber ‘y’ is selected from the numbers 1, 4, 9. What is the probabilitythat the product xy of the two numbers will be less than 9. (3 marks)

Sol. Number ‘x’ can be selected in three ways and corresponding to eachsuch way there are three ways of selecting number ‘y’.

Two numbers can be selected in 9 ways as listed below :S = { (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (2, 9),

(3, 1), (3, 4), (3, 9) }n (A) = 5

P (A) =n (A)

n (S)

P (A) =5

9


SCHOOL SECTION356

CHAPTER : 5 - Statistics - I1. The mean of the following frequency distribution is 50. Find the value

of f : (5 marks)

Class interval 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100Frequency 17 f 32 24 19

Sol. Class Class marks Frequency fixiinterval (xi) (f i)

0 - 20 10 17 17020 - 40 30 f 30f40 - 60 50 32 160060 - 80 70 24 168080 - 100 90 19 1710

Total 92 + f 5160 + 30f

Mean x =

i i

i

f x

f

50 =5160 30

92

f

f 50 (92 + f) = 5160 + 30f 4600 + 50f = 5160 + 30f 50f – 30f = 5160 – 4600 20f = 560

f =560

20

f = 28

2. An incomplete frequency distribution is given as follows :Classes interval Frequency

10 - 20 1220 - 30 3030 - 40 ?40 - 50 6550 - 60 ?60 - 70 2570 - 80 18

Total 229Given that median value is 46, determine the missing frequencies usingthe medians formula. (5 marks)

Sol. Median is 46 it lies in the class 40 - 50 and the corresponding frequencyis 65.

Classes Frequency Cumulative frequencyc.f.

10 - 20 12 1220 - 30 30 4230 - 40 f1 42 + f1

40 - 50 65 f 42 + f1 + 65 = 107 + f1

50 - 60 f2 107 + f1 + f2

60 - 70 25 107 + f1 + f2 + 25 = 132 + f1 + f2

70 - 80 18 131 + f1 + f2 + 18 = 150 + f1 + f2 = 229

Total 229


SCHOOL SECTION 357

From the last c.f.150 + f1 + f2 = 229

f1 + f2 = 229 – 150 f1 + f2 = 79 f2 = 79 – f1 ......(i)

Median =N h

L – c. .2

ff

Where L = 40, N = 229, c.f. = 42 + f1, h =10, f = 65

Median =229 10

40 – (42 )2 65

1f

46 =229 2

40 – 42 –2 13

1f

46 – 40 =229 2

– 42 –2 13

f1

13

62

=229

– 42 –2 1f

39 =229

– 42 –2 1f

78 = 229 – 84 – 2f1

2f1 = 229 – 84 – 78 2f1 = 67

f1 =67

2 f1 = 33.5 34

f2 = 79 – f1 [From (i)] f2 = 79 – 34 f2 = 45

Hence f1 = 34 and f2 = 45.

3. The following data gives the information on the observed life time(in hour) of 225 electrical components :

Lifetime (in hour) Frequency0 - 20 10

20 - 40 3540 - 60 5260 - 80 61

80 - 100 38100 - 120 29

Determine the modal life time of the components. (5 marks)Sol. The class having maximum frequency is 60 - 80 is the modal class.

Lifetime (in hour) Frequency

0 - 20 1020 - 40 3540 - 60 52 f1

60 - 80 61 fm

80 - 100 38 f2

100 - 120 29

The modal class is 60 - 80 and L = 60, f1 = 52, fm = 61, f2 = 38, h = 20


SCHOOL SECTION358

Mode =–

L h2 – –

m 1

m 1 2

f f

f f f

=61 – 52

60 202 (61) – 52 – 38

=9

60 20122 – 90

=9

60 2032

Mode = 60 + 5.625

Mode = 65.625 hours.

CHAPTER : 6 - Statistics - II

1. The following pie-diagram shows the percentage distribution of theexpenditure incurred in publishing a book study the diagram and answerthe questions :(a) If for certain quantity of books, the

publisher has to pay Rs. 30,600as printing, then what will beamount of royalty to be paid forthese books.

(b) What is the central angle of thesector corresponding to the expenditureincurred on Royalty. (3 marks)

Sol.

(a) Let the total cost for certain quantity of books be Rs. ‘x’Printing cost = 20% of x

30600 =20

x100

30600 100

20

= x

x = 153000 Royalty paid for these book = 15% of total cost

=15

153000100

= 15 × 1530

= 22950

(b) Measure of central angle for Royalty =15

360100

= 54º

2. Draw histogram and frequency polygon both in one figure sharing thefollowing information in a city. (3 marks)

Age (years) 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50

No. of doctors 26 58 52 36 20

Printingcost20%

Trasporatio

n

20%

Promotion cost

10%Royalty15%

Binding20% Paper cost

25%


SCHOOL SECTION 359

4. The annual profits earned by 30 shops if a shopping complex in a localitygive rise to the following distribution.

Profit (in lakhs of Rs.) No. of shops (frequency)

More than or equal to 5 30More than or equal to 10 28More than or equal to 15 16More than or equal to 20 14More than or equal to 25 10More than or equal to 30 7More than or equal to 35 3 (5 marks)

25

30

35

40

45

Scale : On X = axis : 1 cm = 2.5 yearsOn Y = axis : 1 cm = 5 doctors

50

10

15

20

5

Age in years

X

Y

0 3025 35 X40 45 50

YN

o.

of

do

cto

rs

55

60

•

•

•

•

•

•

•

65

70

75


SCHOOL SECTION360

Draw both ogive curves for the above data and hence obtain the median.Sol. Profit in lakh Classes No. of shops frequency c.f. less

of Rs. typeMore than orequal to 5 5 - 10 30 2 2

More than orequal to 10 10 - 15 28 12 14






Classes (Profit in lakhs of Rs.)Y

10

12

14

16

18

Cu

mu

lati

ve

fre

qu

en

cy

(N

o.

of

sh

op

s)

20

22

Y

4

6

8

2

X X0

(5, 0)

5 10 15 2 0 25 30 35 40

24

26

28

30

•

•

•

•

•

•

•

•

•

•

•

••

•

•

•

•

(10, 2)

(15, 14)

(15, 16) (20, 16)

(20, 14)

(25, 10)

(30, 7)

(35, 3)

(40, 0)

(25, 20)

(30, 23)

(35, 27)

(40, 30)(20, 28)

(5, 30)

Median = Rs. 17.5 lakhs

Scale : On X = axis : 1 cm = Rs. 5 lakhsOn Y = axis : 1 cm = 2 shops


SCHOOL SECTION 361

3. Prove that the total area of the rectangles in a histogram is equal to thetotal area bounded by the corresponding frequency polygon and theX-axis. (5 marks)

Proof : Consider the following histogram and the corresponding frequency polygon.Let the total area of the rectangles of the histogram be A1 sq. units andthe area enclosed by the frequency polygon and the X-axis be A2 sq. units.Then, we find the area of triangles numbered 1, 2, ....., 10Let us denote these triangles as 1, 2, ........., 10

Then, 2, 4, 6, 8 and 10 are included in are A1 but not in area A2

.......(i)1, 3, 5, 7 and 9 are included in area A2 but not in area A1 ......(ii)The remaining area is common to both the figures, ......(iii)Now, consider 1, i.e., ABC and 2, i.e., EDCIn these triangles,B D [Each is a right angle]seg AB = seg DE [Each of length 5 units]ACB ECD [Vertically opposite angles]

ABC EDC [A-S-A test for congruency] A (ABC) = A (EDC) [Congruent triangles are of equal area]

i.e., A (1) = A (2) ......(iv)

Similarly, we can prove thatA (3) = A (4), A (5) = A (6), A (7) = A (8) and A (9) = A (10)

........(v) From (iv) and (v),

A (1) + A (3) + A (5) + A (7) + A (9) = A (2) + A (4) + A (6) +A (8) + A (10) .......(vi)

From (i), (ii), (iii) and (vi)Area of the histogram = area of the frequency polygon with the X-axis.

Fre

qu

en

cy

25

30

35

40

45

Scale : 1 cm = 5 units on Y axis

10

15

20

5

ClassX

Y0 2 0 30 X40 50

Y

10

D E 3

2

C

BA 1

4

5

6 10

8

9

7•

•

•

•

•

•

chapter : 1 - arithmetic progression and geometric...

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