(a) CO3-2 – 1+3=4

(b) PO4-3 – 1+4=5

(c) P2O5 –  2+5=7

(d) CO –  1+1=2

It is known that the sum of the opposite angles of a cyclic quadrilateral is 180o

Thus, we have

C +A = 180

4y + 20- 4x = 180

- 4x + 4y = 160

x - y = - 40 ……………(1)

And, B + D = 180

3y - 5 - 7x + 5 = 180

- 7x + 3y = 180 ………..(2)

Multiplying 3 to equation (1), we get

3x - 3y = - 120 ………(3)

Adding equation (2) to equation (3), we get

- 7x + 3x = 180 – 120

- 4x = 60

x = -15

Substituting this value in equation (i), we get

x - y = - 40

-y-15 = - 40

y = 40-15

= 25

A = 4y + 20 = 20+4(25) = 120°

B = 3y - 5 = - 5+3(25) = 70°

C = - 4x = - 4(- 15) = 60°

D = 5-7x

D= 5- 7(-15) = 110°

Hence, all the angles are measured

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying p to equation (1)  and q to equation (2), we get

p2x + pqy = p2 - pq ………… (iii)

q2x - pqy = pq + q2 ………… (iv)

Adding equation (iii) and equation (iv),we get

p2x + q2 x = p2  + q2

(p2 + q2 ) x = p2 + q2

x = (p2 + q2)/ p2 + q2 = 1

From equation (i), we get

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying a to equation (i) and  b to equation (ii), we obtain

a2x + aby = ac ………………… (iii)

b2x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a2 – b2) x = ac - bc– b

x = (ac - bc– b)/ (a2 – b2)

x = c(a-b) –b / (a2+b2)

From equation (i), we obtain

ax +by = c

a{c(a-b)-b)/ (a2 – b2)} +by=c

ac(a-b)-ab/ (a2 – b2)+by=c

by=c–ac(a-b)-ab/(a2 – b2)

by=abc – b2 c+ab/a2-b2

y = c(a-b)+a/a2-b2

(iii) x/a – y/b = 0

ax + by = a2 + b2

x/a – y/b = 0

=> bx - ay = 0 ……. (i)

ax + by = a2 + b2 …….. (ii)

Multiplying a and b to equation (i) and (ii) respectively, we get

b2x - aby = 0 …………… (iii)

a2x + aby = a 3 + ab3 …… (iv)

Adding equations (iii) and (iv), we get

b2x + a2x = a 3 + ab2

x (b2 + a2) = a (a2 + b2) x = a

Using equation (i), we get

b(a) - ay = 0

ab - ay = 0

ay = ab,

y = b

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(a + b) y + (a – b) x = a2- 2ab - b2 …………… (i)

(x + y)(a + b)  = a 2 + b2

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

Subtracting equation (ii) from equation (i), we get

(a - b) x - (a + b) x = (a 2 - 2ab - b 2) - (a2 + b2)

x(a - b - a - b) = - 2ab - 2b2

- 2bx = - 2b (b + a)

x = b + a

Substituting this value in equation (i), we get

(a + b)(a - b)  +y (a + b)  = a2- 2ab – b2

a2 - b2 + y(a + b)  = a2- 2ab – b2

(a + b) y = - 2ab

y = -2ab/(a+b)

(v) 152x - 378y = - 74

76x - 189y = - 37

x =(189y-137)/76……………..…(i)

- 378x + 152y = - 604

- 189x + 76y = - 302 ………….. (ii)

Using the value of x in equation (ii), we get

-189(189y-37/76)+76y=-302

- (189)2y + 189 × 37 + (76)2 y = - 302 × 76

189 × 37 + 302 × 76 = (189)2 y - (76)2y

6993 + 22952 = (189 - 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x = (189-37)/76

x = 152/76 = 2

Given,

5x – y = 5

=> y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

From the above graph we can see that the triangle formed is ?ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) ,  C(0,-5) and B(0,-3).

Let the number of rows be A and the number of students in a row be B.

Total number of students = Number of rows x Number of students in a row

=AB

Using the information, that is given,

First Condition:

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3 – – – – – – – – – – – – – (1)

Second condition:

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6 – – – – – – – – – (2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9+3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class = AB = 4 x 9 = 36

Let the speed of the train be A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.

Speed of the train = Distance travelled by train / Time taken to travel that distance

A = N (distance)/ X (time)

Or, N = AX – – – – – – – – – – – (1)

Using the information that is given, we get:

(A+10) = X/(N-2)

(A + 10) (N – 2) = X

AN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 – – – – – – – – – – (2)

(A-10) = X/(N+3)

(A – 10) (N + 3) = X

AN – 10N + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30 – – – – – – – – – (3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

(-2) x (50) + 10N = 20

-100 +10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get,

Distance travelled by the train, X = AN

= 50 x 12

= 600 km

Hence, the distance covered by the train is 600km.

Let Sangam have Rs A with him and Reuben have Rs B with him.

Using the information that is given we get,

A + 100 = 2(B – 100) ? A + 100 = 2B – 200

Or A – 2B = -300 – – – – – – – (1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70 – – – – – – (2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140 – – – – – – – (3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440

 A = 440/11 = 40

Using A =40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340

B = 170

Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju sister Ann B/2 yrs

By using the information that is given,

Case (i)

When Ani is older than that of Biju by 3 yrs then A – B = 3 – – – – – – – – (1)

2A-B/2 = 30

4A – B = 60 – – – – – – – – – – – (2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

Case (ii)

When Biju is older than Ani,

B – A = 3 – – – – – – – – – (1)

2A - B/2 = 30

4A – B = 60 – – – – – – – – – (2)

Adding the equation (1) and (2) we get,

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given,

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore,

Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60

And y = 1/n

y = 80

Therefore,

Speed of the train = 60 km/h

Speed of the bus = 80 km/h