Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 6 Pairs Of Equations are provided here with simple step-by-step explanations. These solutions for Pairs Of Equations are extremely popular among Class 9 students for Maths Pairs Of Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 84:

#### Question 1:

When five pens were bought together, the total price was reduced by three rupees and only seventeen rupees had to be paid. Had they been bought one by one, how much would have been the price of each pen?

#### Answer:

Let the price of one pen be Rs.*x*.

Number of pens bought = 5

Cost of five pens = 5 × Rs.*x* = Rs.5*x*

Amount paid for the five pens = Rs.17

Actual amount of the five pens = Rs (17 + 3) = Rs.20

∴ 5*x* = 20

⇒ *x* =

⇒ *x* = 4

Thus, the price of one pen is Rs.4.

#### Page No 84:

#### Question 2:

Four added to half of a number gives hundred. What is the number?

#### Answer:

Let the required number be *x*.

According to the question:

Thus, the required number is 192.

#### Page No 84:

#### Question 3:

Two persons divided hundred rupees between them and one got ten rupees more than the other. How much did each get?

#### Answer:

Total money with the two people = Rs.100

Let the amount received by the first person be Rs.*x*.

The amount received by the other person = Rs (*x *+ 10)

According to the question,

⇒ *x* + (*x* + 10) = 100

⇒ 2*x* = 100 − 10

⇒ 2*x* = 90

⇒ *x* =

⇒ *x* = 45

*x* + 10 = 45 + 10 = 55

Therefore, the amounts received by the first and the second person are Rs.45 and Rs.55 respectively.

#### Page No 84:

#### Question 4:

A hundred rupee note was changed to ten rupee notes and twenty rupee notes, eight notes in all. How many tens and how many twenties?

#### Answer:

Total amount of money = Rs.100

Total number of ten and twenty rupee notes = 8

Let the total number of ten rupee notes be *x*.

The total number of twenty rupee notes = 8 − *x*

Total amount made by ten rupee notes = *x* × Rs.10 = Rs.10*x*

Total amount made by twenty rupee notes = (8 − *x*) × Rs.20

= Rs.20(8 − *x*)

= Rs (160 − 20*x*)

∴ 10*x* + 160 − 20*x* = 100

⇒ −10*x* = 100 − 160

⇒ −10*x* = −60

⇒ *x* =

⇒ *x* = 6

8 − *x *= 8 − 6 = 2

Thus, the total number of ten and twenty rupee notes is 6 and 2 respectively.

#### Page No 84:

#### Question 5:

Raji’s age is three times that of her brother Rajan. After two years, Raji’s age would be two times that of Rajan. How old are they now?

#### Answer:

Let Rajan’s age be *x *years.

Raji’s age = 3 × *x *years = 3*x* years

After two years:

Rajan’s age = (*x *+ 2) years

Raji’s age = (3*x* + 2) years

According to the question,

3*x* + 2 = 2(*x* + 2)

⇒ 3*x* + 2 = 2*x *+ 4

⇒ 3*x* − 2*x** *= 4 − 2

⇒ *x* = 2

3*x* = 6

Thus, Rajan’s age is 2 years and Raji’s age is 6 years.

#### Page No 89:

#### Question 1:

The number of girls in a class is 4 more than the number of boys. On a day when only 8 boys were absent, the numbers of girls were double the number of boys. How many girls and boys are there in that class?

#### Answer:

Let the number of boys in the class be *x* and the number of girls in the class be *y*.

The number of girls in the class is four more than the number of boys.

∴ *y* = *x* + 4 … (1)

Number of boys absent on a particular day = 8

∴ Number of boys present on that day = *x* − 8

According to the question:

⇒ 2(*x *− 8) = *y*

⇒ 2*x* − 16 = *y*

⇒ 2*x* − *y* = 16

Putting the value of *y *from equation (1):

2*x* − (*x* + 4) = 16

⇒ 2*x* − *x* − 4 = 16

⇒ *x* = 16 + 4

⇒ *x* = 20

From equation (1), we have:

*y* = 20 + 4 = 24

Thus, there are 20 boys and 24 girls in the class.

#### Page No 89:

#### Question 2:

The price of a table and a chair together is 2500 rupees. The price of 3 tables and 4 chairs of this kind is 8000 rupees. What is the price of each?

#### Answer:

Let the price of table be Rs.*x* and the price of chair be Rs.*y*.

Total price of the chair and the table = Rs.2500

⇒ *x *+ *y* = 2500

⇒ *y* = 2500 − *x * … (1)

Total price of three tables and four chairs = Rs.8000

⇒ 3*x** *+ 4*y* = 8000

⇒ 3*x* + 4(2500 − *x*) = 8000 (From equation (1))

⇒ 3*x* + 10000 − 4*x* = 8000

⇒ −*x* = 8000 − 10000

⇒ −*x* = −2000

⇒ *x* = 2000

From equation (1), we have:

*y* = 2500 − 2000 = 500

Thus, the price of the table and the chair is Rs.2000 and Rs.500 respectively.

#### Page No 89:

#### Question 3:

Manu is 4 years younger than Mridula. 4 years back, Mridula’s age was one and a half times Manu’s age. What are their ages now?

#### Answer:

Let Manu’s age be *x* years and Mridula’s age be *y *years.

Manu is four years younger than Mridula.

⇒ *x* = *y* − 4 … (1)

Four years back:

Manu’s age = (*x* − 4) years

Mridula’s age = (*y *− 4) years

According to the question:

× (*x* − 4) = (*y* − 4)

⇒(*x* − 4) = (*y* − 4)

⇒ 3*x* − 12 = 2*y* − 8

⇒ 3*x* − 2*y* = −8 + 12

⇒ 3*x* − 2*y* = 4

Putting the value of *y *from equation (1):

⇒ 3(*y *− 4)* *− 2*y* = 4

⇒ 3*y* − 12 − 2*y* = 4

⇒ *y* = 4 + 12

⇒ *y* = 16

From equation (1), we have:

*x* = 16 − 4 = 12

Thus, Manu’s age is 12 years and Mridula’s age is 16 years.

#### Page No 89:

#### Question 4:

Thomas deposited 7000 rupees in all in two banks. The rate of interest is 8% in one bank and 11% in the other. After a year, he got 680 rupees as interest from both banks together. How much did he deposit in each bank?

#### Answer:

Let money deposited by Thomas in the first and the second bank be Rs.*x* and Rs.*y *respectively.

Total money deposited = Rs.7000

⇒ *x* + *y* = 7000

⇒ *x* = 7000 − *y* … (1)

Rate of interest in the first bank = 8%

Rate of interest in the second bank = 11%

Time period = 1 year

Simple interest received from both the banks together = Rs.680

Simple interest =

Simple interest from the first bank

Simple interest from the second bank

Adding equation (2) and equation (3):

From equation (1), we have:

*x* = 7000 − 4000* =* 3000

Thus, the money deposited by Thomas in the first and the second bank is Rs.3000 and Rs.4000 respectively.

#### Page No 94:

#### Question 1:

Some autorikshaws and bikes are parked in a place. There are 19 in all. The total number of wheels is 45. How many of each vehicle are there?

#### Answer:

Let the number of auto rickshaws and bikes parked at the place be *x* and *y* respectively.

Total number of vehicles = 19

⇒ *x* + *y* = 19 … (1)

Number of wheels in a bike = 2

Number of wheels in an auto rickshaw = 3

Total number of wheels = 45

⇒ 3*x* + 2*y* = 45 ... (2)

Multiplying equation (1) by 3:

3*x* + 3*y* = 57 ... (3)

Subtracting equation (2) from equation (3):

*y *= 12

Putting the value of *y* in equation (1):

*x* + 12 = 19

⇒ *x* = 19 − 12 = 7

Thus, there were 7 auto rickshaws and 12 bikes parked at the place.

#### Page No 94:

#### Question 2:

Four times a number added to thrice another number gives 43. Twice the second number, subtracted from thrice the first gives 11. What are the numbers?

#### Answer:

Let the first number be *x* and the second number be *y*.

According to the question:

4*x* + 3*y* = 43 … (1)

Also,* *3*x* − 2*y* = 11 ... (2)

Multiplying equation (1) by 3 and equation (2) by 4:

12*x* + 9*y* = 129 ... (3)

12*x* − 8*y* = 44 ... (4)

Subtracting equation (4) from equation (3):

⇒ 17*y** *= 85

⇒ *y* =

⇒ *y* = 5

Putting the value of *y* in equation (1):

4*x* + 15 = 43

⇒ 4*x* = 28

⇒ *x* = 7

Thus, the first number is 7 and the second number is 5.

#### Page No 94:

#### Question 3:

The sum of the digits of a two-digit number is 11. If the digits of the number are interchanged, we get a number which 27 more than the original number. What is the number?

#### Answer:

Let the digits at tens and units place of a two-digit number be *x* and *y* respectively.

Sum of the digits of the two-digit number = 11

⇒ *x* + *y *= 11 … (1)

We know that any two digit number is represented by 10*x* + *y*.

If the digits are interchanged then the two-digit number is represented by 10*y* + *x*.

According to the question:

10*y* + *x* = 10*x* + *y* + 27

⇒ 10*y* + *x* − 10*x* − *y* = 27

⇒ 9*y* − 9*x* = 27

⇒ *y* − *x* = 3 ... (2)

Adding equation (1) and (2):

2*y** *= 14

⇒ *y* =

⇒ *y* = 7

From equation (1), we get:

*x* = 11 − 7 = 4

Number = 10*x* + *y* = 40 + 7 = 47

Thus, the two digit number is 47.

#### Page No 94:

#### Question 4:

From the middle of a square, a small square is cut off. The area of the resulting figure is 65 square centimetres and the sum of the lengths of all its edges is 52 centimetres. What is the length of a side of the original square and the square cut off?

#### Answer:

Let the side of the original square be *x* cm and of the small square which is cut off from the original square be *y *cm.

Sum of the length of all the edges of the original square = 52 cm

⇒ 4*x* = 52

⇒ *x* = 13

∴ Length of each side of the original square = 13 cm

According to the question:

Area of resulting figure = 65 cm^{2}

⇒ *x*^{2} − *y*^{2} = 65

⇒ (13)^{2} − *y*^{2} = 65

⇒ 169 − *y*^{2} = 65

⇒ *y*^{2} = 104

⇒ *y *=

∴ Length of each side of the small square = cm

#### Page No 94:

#### Question 5:

A particle travelling along a straight line starts with a velocity of metres per second and its speed increases per second by a metres. The particle travels 10 metres is 2 second and 28 metres is 4 seconds. Find *u* and *a*.

(If the distance travelled in *t* seconds is *s *metres, then)

#### Answer:

Initial velocity of the particle = *u* m/sec

Rate of increase in speed = *a* m/sec

We know that distance travelled, *s* in time, *t* is given by:

Distance travelled in 2 seconds = 10 m

Distance travelled in 4 seconds = 28 m

Subtracting equation (1) from equation (2):

*a* = 2

From equation (1), we get:

*u* = 3

∴ *u* = 3 and *a* = 2

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