1. The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Answer:

The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

Let the age of Ani and Biju be x and y years respectively.

Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

Biju’s sister Cathy And age of = years

By using the information given in the question,

Case (I) When Ani is older than Biju by 3 years, x – y = 3 (i)

2x –= 30

4x – y = 60 (ii)

Subtracting (i) from (ii), we obtain 3x = 60 – 3 = 57

x –= 19

Therefore, age of Ani = 19 years

And age of Biju = 19 – 3 = 16 years

Case (II) When Biju is older than Ani, y – x = 3 (i)

2x –= 30

4x – y = 60 (ii)

Adding (i) and (ii), we obtain 3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

Answer:

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = – 300        … (1)

6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70            … (2)

Multiplying equation (2) by 2, we obtain:
12x – 2y = 140        … (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

We know that, Speed =

x =

Or, d = xt (i)

Using the information given in the question, we obtain

(x + 10) =

(x+10) (t-2) = d

Xt+10t-2x-20 = d

By using equation (i),

we obtain – 2x + 10t = 20 (ii)

(x-10) =

(x-10) (t+3) = d

Xt-10t+3x-30 = d

By using equation (i), we

obtain 3x – 10t = 30 (iii)

Adding equations (ii) and (iii),

we obtain x = 50

Using equation (ii), we

obtain ( – 2) × (50) + 10t = 20

– 100 + 10t =

10t = 120

t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km

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

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:

Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row
= xy


According to the question,
Total number of students = (x – 1) (y + 3)
xy = (x – 1) (y + 3)

= xy – y + 3x – 3
3x – y – 3 = 0
3x – y = 3 … (1)

Total number of students = (x + 2) (y – 3)
xy = xy + 2y – 3x – 6
3x – 2y = -6 … (2)

Subtracting equation (2) from (1), we obtain:
y = 9

Substituting the value of y in equation (1), we obtain:
3x – 9 = 3
3x = 9 + 3 = 12
x = 4

Number of rows = x = 4
Number of students in a row = y = 9

Total number of students in a class = xy = 4 x 9 = 36

5. In a ∆ ABC, ∠C = 3∠B = 2(∠A + ∠B). Find three angles.

Answer:

C = 3B = 2(A + B)

Taking 3B = 2(A + B)

B = 2A

2 –B = 0 ….(i)

We know that the sum of the measures of all angles of a triangle is 180°.

A + B + C = 

A + B + 3B = 

A + 4B = …….(2)

Multiplying equation (1) by 4, we obtain:

8A – 4B = 0 …….(3)

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

9A = 

A = 

From eq. (2), we get,

+ 4B = 

4B = 

B = 

C = 3B

3* = 

Hence the measures of A, B and C are respectively.

6. Draw the graphs of the equations 5x – y =5 and 3x – y = 3.Determine the co-ordinate of the vertices of the triangle formed by these lines and the y- axis.

Answer:

5x – y =5

= y = 5x – 5

Three solutions of this equation can be written in a table as follows:

3x – y = 3 or,
y = 3x – 3
The solution table will be as follows.

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is ΔABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, – 3), C (0, – 5).

7. Solve the following pair of linear equations:

(i) px + py = p – q

x – p = py + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x/a – y/b = 0

ax + b = a² + b²

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

( a + b )( x + y ) = a² + b²

(v) 152x – 378y = -74

-378x + 152y = -604

Answer:

(i) px + qy = p – q … (1)
qx – py = p + q … (2)

Multiplying equation (1) by p and equation (2) by q,

we obtain p2x + pqy = p2 – pq … (3)
q2x – pqy = pq + q2 … (4)
Adding equations (3) and (4),
we obtain p2x + q2 x = p2 + q2
(p2 + q2) x = p2 + q2

x == 1

From equation (1),

we obtain p (1) + qy = p – q
qy = – q,so, y = – 1

(ii) ax + by = c … (1) bx + ay = 1 + c … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
… (3)
… (4)
Subtracting equation (4) from equation (3),

From equation (1), we obtain ax + by = c

(iii) 

……..(1)

……..(2)

Multiplying equation (1) and (2) by b and a respectively, we obtain:

……..(3)

……..(4)

Adding equations (3) and (4), we obtain:

By using (1), we obtain b (a) – ay = 0
ab – ay = 0 ay = ab
y = b

(iv)… (1)

……..(2)
Subtracting equation (2) from (1), we obtain:

Using equation (1) ,we obtain,

(v)152x – 378y = –74… (1)
–378x + 152y = –604 … (2)
Adding the equations (1) and (2), we obtain:
–226x – 226y = –678

Subtracting the equation (2) from equation (1), we obtain:
530x – 530y = 530

Adding equations (3) and (4), we obtain:
2x = 4

x = 2
Substituting the value of x in equation (3), we obtain:
y = 1

8. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Answer:

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is.

Therefore,

4y + 20 – 4x = 180 – 4x + 4y = 160

x – y = – 40 (i)

Also

3y – 5 – 7x + 5 = 180 – 7x + 3y = 180 (ii)

Multiplying equation (1) by 3, we obtain:
3x – 3y = – 120 (iii)

Adding equations (2) and (3), we obtain:
– 7x + 3x = 180 – 120
– 4x = 60
x = -15

By using equation (i), we obtain x – y = – 40
-15 – y = – 40
y = -15 + 40 = 25

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2