Enhanced Curriculum Support

SecA

1.If the quadratic equation x2+4x + k = 0 has real and equal roots then,

2. Graphically the pair of equations given by 6x-3y+10 = 0 , 2x-y+9=0 represents two lines which are?

3. Write the nature of roots of the quadratic equation 9x2-6x-2=0?

4. If a pair of linear equations is consistent, then the lines will be?

5. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is?

6. The pair of equations x = a and y = b graphically represents lines which are?

7. For what value of k, do the equations 3x – y +8 =0 and 6x – ky = - 16 represent coincident lines ?

8. One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be?

9. If x = a, y = b is the solution of the equations x – y= 2 and x + y = 4, then the values of a and b are,respectively.

10. Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹2 coins are, respectively.

Sol

1. k=4

2. Parallel

3. 2 distinct real roots

4. intersecting or coincident

5. 154

6. intersecting at (a, b)

7. 2

8. 10x – 14y = − 4

9. 3 and 1

10. y = 25 and x = 25

SecB

1. Solve the following system of equations by substitution method: 2x + 3y=8, 3x-2y=7

2. Determine whether the following system of equations is consistent or inconsistent: 5x-y=1 ,10x-2y=2

3. The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not.

Sol

1. x=2 y=1

2. Consistent (they represent the same line)

3. The given statement is not true as the line represented by x = 7 is of the from x = a. The graph of the equation is a line parallel to the y-axis.

SecC

1.Find the point of intersection of the lines represented by the equations x+y=5 and 2x-3y=4

2. Solve the following system of equations using the elimination method: 2x+3y=9, 4x+6y=18

3. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

4.The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

5. Given that 3 is irrational, prove that 5 + 23 is irrational.

Sol

1. Point of intersection:(3,2)

2. Infinitely many solutions (they represent the same line)

3. Let the two-digit number = 10x + y

Case I : Multiplying the sum of the digits by 8 and then subtracting 5 = two-digit number

⇒ 2x - 7y = -5 ...(i)

Case II : Multiplying the difference of the digits by 16 and then adding 3 = two-digit number

⇒ 6x - 17y = -3 ...(ii)

Now, multiplying in Eq. (i) by 3 and then subtracting from Eq. (ii), we get y = 3

Now, put the value of y in eq. (i), we get x = 8

Hence, the required two-digit number = 10x + y = 10 × 8 + 3 = 80 + 3 = 83

4. Let the present age (in years) of father and his two children be x, y and z years respectively.

Now by given condition,

x = 2(y+z) ...(i)

And after 20 years, (x+20) = (y+20) + (z+20)

⇒ y + z +40 = x +20

⇒ y + z = x -20

On putting the value of (y + z) in Eq. (i) and get the present age of father,

x = 2(x-20)

x = 2x -40

2x - x = 40

∴ x = 40

Hence, the father’s present age is 40 years.

SecD

1. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 15 days, pays Rs 850 as hostel charges. Find the fixed charges and the cost of food per day.

2.Two trains are moving in the same direction at speeds of 56 km/hr and 29 km/hr. The faster train crosses a man sitting in the slower train in 10 seconds. Find the length of the faster train.

3.The sum of a two-digit number and the number obtained by reversing the digits is 99. If the digits of the number differ by 3, find the number.

4. Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of ₹400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of ₹ 4 for 5 bananas, his total collection would have been ₹ 460. Find the total number of bananas he had.

Sol

1. Fixed charges = Rs 400, Cost of food per day = Rs 30

2. Length of the faster train = 75 meters

3. The number is 63 or 36

4. Case I : Cost of first lot at the rate of ₹2 for 3 bananas + Cost of second lot at the rate of ₹ 1 per banana = Amount received

⇒ 23x + y = 400

⇒ 2x + 3y = 1200

Case II : Cost of first lot at the rate of ₹1 per banana + Cost of second lot at the rate of ₹4 for 5 bananas = Amount received

⇒ x + 45 y = 460

⇒ 5x + 4y = 2300

On multiplying in Eq. (i) by 4 and Eq. (ii) by 3 and then subtracting them, we get

∴ x = 300

Now, put the value of x in Eq. (i), we get

∴ y = 200

∴ Total number of bananas = Number of bananas in lot A + Number of bananas in lot B

= x + y = 300 + 200 =

Hence, he had 500 bananas.

Problem1

Situation:

A theatre sells tickets for a show in two categories: adult and child. The total number of tickets sold is 120. The revenue from the tickets is $700, with adult tickets costing $10 each and child tickets costing $5 each. Find the number of adult and child tickets sold.

1.From the problem, we have: x + y = 120 10x + 5y = 700

Sol

Solution

Equations:

Let the number of adult tickets be x and the number of child tickets be y.

Simplify the second equation by dividing by 5:

2x + y = 140

Now we have the system of equations:

1. x + y=120

2. 2x + y = 140

Subtract the first equation from the second equation:

(2x+y)-(x+y) = 140 -120

x = 20

Substitute x = 20 back into the first equation:

20 + y = 120

y= 100

So, 20 adult tickets and 100 child tickets were sold.

Problem 2

Situation:

The sum of the ages of a father and his son is 50 years. Five years ago, the father's age was four times the son's age. Find their current ages.

1.From the problem, we have: 1. x+y=50, 2.x-5=4(y-5)

Sol

Solution

Equations:

Let the father's age be x years, and the son's age be y years.

Substitute 𝑥=50−𝑦 into the second equation:

(50-y) - 5 = 4(y-5)

45 - y = 4y-20

45 + 20 = 5y

65 = 5y

y=13

Substitute y =13 back into the first equation:

x + 13 = 50

x = 37

So, the father is 37 years old, and the son is 13 years old.

Problem 3

Situation:

John went to a stationery shop to buy some notebooks and pens. He bought 5 notebooks and 3 pens for $23. Another customer bought 3 notebooks and 2 pens for $14. Find the cost of one notebook and one pen.

1. 5x+3y=23 2.3x+2y=14

Sol

Solution

To solve this system of equations, we can use the elimination method.

First, multiply the second equation by 3 to make the coefficients of x the same in both equations:

3(3x+2y) = 3(14)

9x+6y = 42

Next, multiply the first equation by 2:

2(5x+3y)=2(23)

10x+6y = 46

Now, subtract the second modified equation from the first modified equation:

10x+6y -9x -6y = 46-42

x = 4

Substitute x = 4 back into one of the original equations to find y:

5(4) + 3y = 23

20 + 3y =23

3y = 3

y = 1

So, the cost of one notebook is $4, and the cost of one pen is $1.

Q1

1. Two friends, Raj and Simran, start running from the same point on a circular track at the same time. Raj runs at a speed of 10 km/h and Simran at 12 km/h. If they run in the same direction, they meet after 3 hours. If they run in opposite directions, they meet after 1 hour. Determine the length of the circular track.

Sol

Solution:

When they run in the same direction:

Relative speed = 12 - 10 = 2 kmh

Distance covered = 2 x 3 = 6km

When they run in the opposite direction:

Relative speed = 12 + 10 = 22 kmh

Distance covered = 22 x 1 = 22km

Since both represent the same track length, we equate the distances:

6km = 22km

The length of the circular track is 22 km.

Q2

2. The sum of the digits of a two-digit number is 9. If the digits are reversed, the number obtained is 27 less than the original number. Find the original number.

Sol

Solution:

Let the tens digit be x and the units digit be y.

x + y = 9 ............(1)

The original number is = 10x + y

The reversed number is = 10y + x

10x + y - 27 = 10y + x

9x - 9y =27

x - y = 3.......(2)

Add Equations 1 and 2:

(x + y) (x - y) = 9 + 3

2x = 12

x = 6

Substitute x=6 into Equation 1:

6 + y = 9

y = 3

The original number is 10x + y = 10 x 6 + 3 = 63.

Q3

3. A farmer has a field in the shape of a rectangle. The length of the field is 5 meters more than twice its width. If the perimeter of the field is 230 meters, find the dimensions of the field.

Sol

Solution:

Let the width be w meters.

Perimeter P = 2(l + w);

2(2w + 5 + w) = 230

2(3w + 5) = 230

6w + 10 = 230

6w = 230

w = 36.67

Length l = 2(36.67) + 5

l = 78.34

The dimensions of the field are approximately 36.67 meters by 78.34 meters.

Q1

1. Aruna has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Re 1 and Rs 2 coins are, respectively.

Sol

Given, Aruna has Re. 1 and Rs. 2 coins with her.

We have to find the number of Re.1 and Rs.2 coins.

Let the number of Re.1 coins be x.

Let the number of Rs.2 coins be y.

Given, x + y = 50 ------------- (1)

Given, the amount of money is Rs. 75

So, x + 2y = 75 --------------- (2)

On solving the linear equations (1) and (2),

First subtracting (2) and (1),

(x + 2y) - (x + y) = 75 - 50

x + 2y - x - y = 25

By grouping,

(x - x) + (2y - y) = 25

y = 25

Put y = 25 in (1),

x + 25 = 50

x = 50 - 25

x = 25.

Therefore, the number of Re.1 and Rs.2 coins are 25 and 25.

Q2

2.The cost of 4 pens and 4 pencil boxes is ₹ 100. Three times the cost of a pen is ₹ 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Sol

Consider the cost of a pen as ₹x.

Consider the cost of a pencil box as ₹y.

According to the given condition,

4x + 4y = 100

x + y = 25-------(1)

3x = y + 15

3x - y = 15-------(2)

(1) and (2) are the pair of linear equations in two variables.

Add (1) and (2),we get,

4x = 40

x = 10.

Substituting x = 10 in (1),we get.

y = 25 - 10

y = 15.

Therefore, the cost of a pen and a pencil box are ₹10 and ₹15, respectively.

Q3

3. A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Sol

Consider v kmh as the speed of the stream.

Given,

Speed of a person rowing in still water = 5 kmh.

Speed of a person rowing in downstream = (5 + ν ) kmh.

Speed of a person has rowing upstream = (5 - ν ) kmh.

Time taken by the person to cover 40 km downstream,

t₁ = 405+v hours.

Time taken by the person to cover 40km upstream,

t₂ = 405-v hours.

According to the given condition,

t₂ = t₁ x 3.

405-v = 405+v x 3.

15-v = 35+v

Let us solve this linear equation.

5+ v = 15 - 3v = 4v = 10

v = 104

v = 2.5kmh.

Therefore, the speed of the stream is 2.5 km/h.

Q4

4. A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Sol

Let us consider the speed of the motorboat in still water to be u kmhr.

Let us consider the speed of the stream to be ν kmh.

downstream speed of motorboat = (u + ν) kmh.

upstream speed of motorboat = (u - ν) kmh.

Time taken to travel 30 km upstream,

t₁ = 30u-v hours

Time taken to travel 28km downstream,

t₂ = 28u+v hours.

As per the first condition,

t₁ + t₂ = 7 hours

30u-v + 28u+v = 7-----(1)

Time taken to travel 21 km upstream,

t₃ = 21u-v hours.

Time taken to travel 21 km downstream,

t₄ = 21u+v hours.

As per the second condition,

t₄ + t₃ = 5 hours

21u+v + 21u+v = 5----(2)

Let us consider,

x = 1u+v

y = 1u-v

Rearranging (1) and (2), we get,

30x + 28y = 7------(3)

21x + 21y = 5-----(4)

x + y = 5/21.

Solve the linear equations (3) and (4)

Multiplying equation (4) by 28 and then subtracting from (3), we get,

(30x - 28y) - (28x + 28y) = 7 - 14021

2x = 7 - 203

2x = 13

x = 16.

Substituting the value of x in(4), we get,

16 + y = 521

y = 521 = 16 = 10 - 742 = 342

y = 114.

x = 1u+v = 16

u + v = 6-----(5)

y = 1u-v = 114

u-v = 14----(6)

Adding (5) and (6), we get,

2u = 20

u = 10.

Substituting the value of u in (5),we get,

10 + ν = 6

ν = -4.

u = 10.

ν = -4.

Therefore, the speed of the motorboat in still water is 10 kmh and the speed of the stream 4 kmh.

Q5

5. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Sol

Explanation:

Let the numbers be 5x and 6x.

Given that, if 8 is subtracted from each of the numbers, the ratio becomes 4 : 5.

(5x - 8) : (6x - 8) = 4 : 5

(5x - 8)(6x - 8) = 45

5(5x - 8) = 4(6x - 8)

25x - 40 = 24x - 32

25x - 24x = 40 - 32

x = 8

the first number = 5x = 5 × (8) = 40

the second number = 6x = 6 × (8) = 48

Thus, the numbers are 40 and 48 which are in the ratio 5 : 6.

Question

Mathematics teacher of a school took the standard 10 students to see the painting exhibition which was held at ART COLLEGE OF EDUCATION, Bangalore. It is the part of art integration of Mathematics. The teacher and students had interest in painting as well. Students were eager to see the above paintings. The teacher explained that the above paintings are based on concept of a pair of linear equations of two variables.

Based on your understanding of the above case study, answer all the five questions below:

1. If the speed of boat is 5 km/hr and speed of stream is 2 km/hr. What is the speed of the boat in Downstream?

2.If the speed of boat is 5 km/hr and speed of stream is 2 km/hr. What is the speed of the boat in Upstream?

3. A boat goes 21 km downstream. What is the time required to cover it?

4. A boat goes 12 km Up stream. What is the time required to cover it?

5. If speed of boat and stream be x km/hr and y km/hr respectively. What is the distance covered by down stream boat in ‘t’hours?

Solution

Solution

1. The speed of the boat in downstream is the sum of the speed of the boat and the speed of the stream.

Speed downstream =Speed of boat + Speed of stream

= 5 kmhr + 2 kmhr = kmhr

2. The speed of the boat in upstream is the difference between the speed of the boat and the speed of the stream.

Speed upstream =Speed of boat + Speed of stream

= 5 kmhr - 2 kmhr = kmhr

3. The time required to cover a certain distance downstream can be calculated using the formula:

Time = distanceSpeeddownstream

Given the distance is 21 km and the speed downstream is 7 kmhr:

Time = 21km7kmhr = hours.

4. The time required to cover a certain distance downstream can be calculated using the formula:

Time = distanceSpeed upstream

Given the distance is 12 km and the speed upstream is 3 kmhr:

Time = 12km3km = hours.

5. Distance = (x+y)t km