Exercise 3.5 Pair Of Linear Equations In Two Variables Class 10th Mathematics KC Sinha Solution

Exercise 3.5

  1. The sum of the two numbers is 18. The sum of their reciprocals is 1/4. Find the…
  2. The sum of two numbers is 15 and sum of their reciprocals is 3/10 . Find the…
  3. Two numbers are in the ratio of 5 : 6. If 8 is subtracted from each of the numb,…
  4. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the…
  5. Two positive numbers differ by 3 and their product is 54. Find the numbers.…
  6. Two numbers are in the ratio of 3 : 5. If 5 is subtracted from each of the…
  7. Two numbers are in the ratio of 3 : 4. If 8 is added to each number, they become…
  8. Two numbers differ by 2 and their product is 360. Find the numbers.…
  9. Two numbers differ by 4 and their product is 192. Find the numbers.…
  10. Two numbers differ by 4 and their product is 96. Find the numbers.…
  11. The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly…
  12. Scooter charges consist of fixed charges and the remaining depending upon the…
  13. A part of monthly hostel charges in a college is fixed and the remaining depend…
  14. Taxi charges in a city consist of fixed charges per day and the remaining…
  15. A part of monthly hostel charges in a college are fixed and the remaining…
  16. The total expenditure per month of a household consists of a fixed rent of the…
  17. The car rental charges in a city comprise a fixed charge together with the…
  18. The sum of a two - digit number and the number formed by interchanging the…
  19. A two - digit number is 4 times the sum of its digits. If 18 is added to the…
  20. A number consists of two digits. When it is divided by the sum of its digits,…
  21. The sum of the digits of a two - digit number is 12. The number obtained by…
  22. A two - digit number is 3 more than 4 times the sum of its digits. If 18 is…
  23. A number consisting of two digits is seven times the sum of its digits. When 27…
  24. The sum of the digits of a two - digit number is 15. The number obtained by…
  25. The sum of the numerator and denominator of a fraction is 3 less than twice the…
  26. The sum of the numerator and denominator of a fraction is 4 more than twice the…
  27. The sum of the numerator and denominator of a fraction is 8. If 3 is added to…
  28. The numerator of a fraction is one less than its denominator. If 3 is added to…
  29. The age of the father is 3 years more than 3 times the son's age. 3 years here,…
  30. Two years ago, a man was five times as old as his son. Two years later his age…
  31. Father's age is three times the sum of ages of his two children. After 5 years,…
  32. Five years ago, A was thrice as old as B and ten years later, A shall be twice…
  33. Ten years hence, a man's age will be twice the age of his son. Ten years ago,…
  34. Find a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)…
  35. Find the four angles of a cyclic quadrilateral ABCD in which ∠A = (2x — 3)°, ∠B…
  36. In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles. ∠A = 20°, ∠B = 40°,…
  37. In a ΔABC, ∠A = x°, ∠B = (3x)° and ∠C = y°. If 3y — 5x = 30, show that the…
  38. The area of a rectangle gets reduced by 8 m^2 , when its length is reduced by…
  39. The length of a room exceeds its breadth by 3 metres. If the length is…
  40. Two places A and B are 120 km apart from each other on a highway. A car starts…
  41. A train travels a distance of 300 km at a constant speed. If the speed of SE…
  42. A plane left 30 minutes later than the scheduled time and in order to reach the…
  43. A man travels 600 km partly by train and partly by car. If he covers 400 km by…
  44. Places A and B are 80 km apart from each other on a highway. One car starts…
  45. A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12…
  46. A man travels 370 km, partly by train and partly by car. If he covers 250 km by…

Exercise 3.5
Question 1.

The sum of the two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x + y = 18 …(i)


 …(ii)


Eq. (ii) can be re - written as


 …(iii)


On putting the value of x + y = 18 in Eq. (iii), we get



⇒ xy = 72



On putting the value of  in Eq. (i), we get



⇒ 72 + y2 = 18y


⇒ y2 – 18y + 72 = 0


⇒ y2 – 12y – 6y + 72 = 0


⇒ y(y – 12) – 6(y – 12) = 0


⇒ (y – 6)(y – 12) = 0


⇒ y = 6 and 12




Hence, the two numbers are 6 and 12.



Question 2.

The sum of two numbers is 15 and sum of their reciprocals is . Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x + y = 15 …(i)


 …(ii)


Eq. (ii) can be re - written as


 …(iii)


On putting the value of  in Eq. (iii), we get



⇒ xy = 50



On putting the value of  in Eq. (i), we get



⇒ 50 + y2 = 15y


⇒ y2 – 15y + 50 = 0


⇒ y2 – 10y – 5y + 50 = 0


⇒ y(y – 10) – 5(y – 10) = 0


⇒ (y – 5)(y – 10) = 0





Hence, the two numbers are 5 and 10.



Question 3.

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


Answer:

Let the two numbers be .

According to the question,



 …(i)


Also, 


⇒ 5(x – 8) = 4(y – 8)


⇒ 5x – 40 = 4y – 32


⇒ 5x – 4y = 8 …(ii)


On putting the value of  in Eq. (ii), we get




⇒ x = 40


On putting the value of x = 40 in Eq. (i), we get



Hence, the two numbers are 40 and 48.



Question 4.

The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x + y = 16 …(i)


 …(ii)


Eq. (ii) can be re - written as


 …(iii)


On putting the value of x + y = 16 in Eq. (iii), we get



xy = 48



On putting the value of  in Eq. (i), we get



⇒ 48 + y2 = 16y


⇒ y2 – 16y + 48 = 0


⇒ y2 – 12y – 4y + 48 = 0


⇒ y(y – 12) – 4(y – 12) = 0


⇒ (y – 4)(y – 12) = 0


⇒ y = 4 and 12




Hence, the two numbers are 4 and 12.



Question 5.

Two positive numbers differ by 3 and their product is 54. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x – y = 3 …(i)


Also, x×y = 54


 …(ii)


On putting the value of  in Eq. (i), we get



⇒ 54 – y2 = 3y


⇒ y2 + 3y – 54 = 0


⇒ y2 + 9y – 6y – 54 = 0


⇒ y(y + 9) – 6(y + 9) = 0


⇒ (y – 6)(y + 9) = 0


⇒ y = – 9 and 6


But y = – 9 can’t be the one number as it is given that the numbers are positive.



Hence, the two numbers are 9 and 6.



Question 6.

Two numbers are in the ratio of 3 : 5. If 5 is subtracted from each of the number they become in the ratio of 1 : 2. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,



 …(i)


Also, 


⇒ 2(x – 5) = (y – 5)


⇒ 2x – 10 = y – 5


⇒ 2x – y = 5 …(ii)


On putting the value of  in Eq. (ii), we get




⇒ x = 15


On putting the value of x = 15 in Eq. (i), we get



Hence, the two numbers are 15 and 25.



Question 7.

Two numbers are in the ratio of 3 : 4. If 8 is added to each number, they become in the ratio of 4 : 5. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,



 …(i)


Also, 


⇒ 5(x + 8) = 4(y + 8)


⇒ 5x + 40 = 4y + 32


⇒ 5x – 4y = – 8 …(ii)


On putting the value of  in Eq. (ii), we get




⇒ x = 24


On putting the value of x = 24 in Eq. (i), we get



Hence, the two numbers are 24 and 32.



Question 8.

Two numbers differ by 2 and their product is 360. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x – y = 2 …(i)


Also, x×y = 360


 …(ii)


On putting the value of  in Eq. (i), we get



⇒ 360 – y2 = 2y


⇒ y2 + 2y – 360 = 0


⇒ y2 + 20y – 18y – 360 = 0


⇒ y(y + 20) – 18(y + 20) = 0


⇒ (y – 18)(y + 20) = 0


⇒ y = – 20 and 18


But y = – 20 can’t be the one number as it is given that the numbers are positive.



Hence, the two numbers are 20 and 18.



Question 9.

Two numbers differ by 4 and their product is 192. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x – y = 4 …(i)


Also, x×y = 192


 …(ii)


On putting the value of  in Eq. (i), we get



⇒ 192 – y2 = 4y


⇒ y2 + 4y – 192 = 0


⇒ y2 + 16y – 12y – 192 = 0


⇒ y(y + 16) – 12(y + 16) = 0


⇒ (y – 12)(y + 16) = 0


⇒ y = – 16 and 12


But y = – 16 can’t be the one number as it is given that the numbers are positive.



Hence, the two numbers are 16 and 12.



Question 10.

Two numbers differ by 4 and their product is 96. Find the numbers.


Answer:

Let the two numbers be x and y.

According to the question,


x – y = 4 …(i)


Also, x×y = 96


 …(ii)


On putting the value of  in Eq. (i), we get



⇒ 96 – y2 = 4y


⇒ y2 + 4y – 96 = 0


⇒ y2 + 12y – 8y – 96 = 0


⇒ y(y + 12) – 8(y + 12) = 0


⇒ (y – 8)(y + 12) = 0


⇒ y = – 8 and 12


But y = – 8 can’t be the one number as it is given that the numbers are positive.



Hence, the two numbers are 8 and 12.



Question 11.

The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly expenditures are in the ratio of 7 : 5. If each saves Rs. 3000 per month, find the monthly income of each.


Answer:

Given ratio of incomes = 5:4

And the ratio of their expenditures = 7:5


Saving of each person = Rs. 3000


Let incomes of two persons = 5x and 4x


And their expenditures = 7y and 5y


According to the question,


5x – 7y = 3000 …(i)


4x – 5y = 3000 …(ii)


On multiplying Eq. (i) by 4 and Eq. (ii) by 5 to make the coefficients of x equal, we get


20x – 28y = 12000 …(iii)


20x – 25y = 15000 …(iv)


On subtracting Eq. (iii) from (iv), we get


20x – 25y – 20x + 28y = 15000 – 12000


⇒ 3y = 3000


⇒ y = 1000


On putting the y = 1000 in Eq. (i), we get


5x – 7y = 3000


⇒ 5x – 7(1000) = 3000


⇒ 5x = 10000


⇒ x = 2000


Thus, monthly income of both the persons are 5(2000) and 4(2000), i.e. Rs. 10000 and Rs. 8000



Question 12.

Scooter charges consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a person travels 12 km, he pays Rs. 45 and for travelling 20 km, he pays Rs. 73. Express the above statements in the form of simultaneous equations and hence, find the fixed charges and the rate per km.


Answer:

Let fixed charge = Rs x


and charge per kilometer = Rs y


According to the question,


x + 12y = 45 …(i)


and x + 20y = 73 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 20y – x – 12y = 73 – 45


⇒ 8y = 28



On putting the value of y = 3.5 in Eq. (i), we get


x + 12(3.5) = 45


⇒ x + 42 = 45


⇒ x = 45 – 42 = 3


Hence, monthly fixed charges is Rs. 3 and charge per kilometer is Rs. 3.5



Question 13.

A part of monthly hostel charges in a college is fixed and the remaining depend on the number of days one has taken food in the mess. When a student A, takes food for 22 days, he has to pay Rs. 1380 as hostel charges, whereas a student B, who takes food for 28 days, pays Rs. 1680 as hostel charges. Find the fixed charge and the cost of food per day.


Answer:

Let fixed hostel charge (monthly) = Rs x

and cost of food for one day = Rs y


In case of student A,


x + 22y = 1380 …(i)


In case of student B,


x + 28y = 1680 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 28y – x – 22y = 1680 – 1380


⇒ 6y = 300


⇒ y = 50


On putting the value of y = 50 in Eq. (i), we get


x + 22(50) = 1380


⇒ x + 1100 = 1380


⇒ x = 1380 – 1100 = 280


Hence, monthly fixed charges is Rs. 280 and cost of food per day is Rs. 50



Question 14.

Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels 110 km, he pays Rs. 690, and for travelling 200 km, he pays Rs. 1050. Find the fixed charges per day and the rate per km.


Answer:

Let fixed charge = Rs. x

and charge per kilometer = Rs. y


According to the question,


x + 110y = 690 …(i)


and x + 200y = 1050 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 200y – x – 110y = 1050 – 690


⇒ 90y = 360


⇒ y = 40


On putting the value of y = 40 in Eq. (i), we get


x + 110(40) = 690


⇒ x + 440 = 690


⇒ x = 690 – 440 = 250


Hence, monthly fixed charges is Rs. 250 and charge per kilometer is Rs. 40



Question 15.

A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs. 1750 as hostel charges whereas a student a d who takes food for 28 days, pays Rs. 1900 as hostel charges. Find the fixed charges and the cost of the food per day.


Answer:

Let fixed hostel charge (monthly) = Rs x

and cost of food for one day = Rs y


In case of student A,


x + 25y = 1750 …(i)


In case of student B,


x + 28y = 1900 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 28y – x – 25y = 1900 – 1750


⇒ 3y = 150


⇒ y = 50


On putting the value of y = 50 in Eq. (i), we get


x + 25(50) = 1750


⇒ x + 1250 = 1750


⇒ x = 1750 – 1250 = 500


Hence, monthly fixed charges is Rs. 500 and cost of food per day is Rs. 50



Question 16.

The total expenditure per month of a household consists of a fixed rent of the house and the mess charges, depending upon the number of people sharing the house. The total monthly expenditure is Rs. 3,900 for 2 people and Rs. 7,500 for 5 people. Find the rent of the house and the mess charges per head per month.


Answer:

Let fixed rent of the house = Rs. x

And the mess charges per hesd per month = Rs. y


According to the question,


x + 2y = 3900 …(i)


and x + 5y = 7500 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 5y – x – 2y = 7500 – 3900


⇒ 3y = 3600


⇒ y = 1200


On putting the value of y = 1200 in Eq. (i), we get


x + 2 (1200) = 3900


⇒ x + 2400 = 3900


⇒ x = 1500


Hence, fixed rent of the house is Rs. 1500 and the mess charges per head per month is Rs. 1200.



Question 17.

The car rental charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 13 km, the charge paid is Rs. 96 and for a journey of 18 km, the charge paid is Rs. 131. What will a person have to pay for travelling a distance of 25 km?


Answer:

Let fixed charge = Rs. x

and charge per kilometer = Rs. y


According to the question,


x + 13y = 96 …(i)


and x + 18y = 131 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


x + 18y – x – 13y = 131 – 96


⇒ 5y = 35


⇒ y = 7


On putting the value of y = 7 in Eq. (i), we get


x + 13 (7) = 96


⇒ x + 91 = 96


⇒ x = 5


Hence, monthly fixed charges is Rs. 5 and charge per kilometer is Rs. 7


Now, amount to be paid for travelling 25 km


= Fixed charge + Rs 7 ×25


= 5 + 175


= Rs. 180


Hence, the amount paid by a person for travelling 25km is Rs. 180



Question 18.

The sum of a two - digit number and the number formed by interchanging the digits is 132. If 12 is added to the number, the new number becomes 5 times the sum of the digits. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


After interchanging the digits, New number = x + 10y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,


(10x + y) + (x + 10y) = 132


⇒ 11x + 11y = 132


⇒ 11(x + y) = 132


⇒ x + y = 12 …(i)


and 10x + y + 12 = 5(x + y)


⇒ 10x + y + 12 = 5x + 5y


⇒ 10x – 5x + y – 5y = – 12


⇒ 5x – 4y = – 12 …(ii)


From Eq. (i), we get


x = 12 – y …(iii)


On substituting the value of x = 12 – y in Eq. (ii), we get


5(12 – y) – 4y = – 12


⇒ 60 – 5y – 4y = – 12


⇒ – 9y = – 12 – 60


⇒ – 9y = – 72


⇒ y = 8


On putting the value of y = 8 in Eq. (iii), we get


x = 12 – 8 = 4


So, the Original number = 10x + y


= 10×4 + 8


= 48


Hence, the two digit number is 48.



Question 19.

A two - digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the two digit number = 10x + y


The sum of the digit = x + y


According to the question,


(10x + y) = 4(x + y)


⇒ 10x + y = 4x + 4y


⇒ 10x – 4x + y – 4y = 0


⇒ 6x – 3y = 0


⇒ 2x – y = 0


⇒ y = 2x …(i)


After interchanging the digits, New number = x + 10y


and 10x + y + 18 = x + 10y


⇒ 10x + y + 18 = x + 10y


⇒ 10x – x + y – 10y = – 18


⇒ 9x – 9y = – 18


⇒ x – y = – 2 …(ii)


On substituting the value of y = 2x in Eq. (ii), we get


x – y = – 18


⇒ x – 2x = – 2


⇒ – x = – 2


⇒ x = 2


On putting the value of x = 2 in Eq. (i), we get


y = 2×2 = 4


So, the Original number = 10x + y


= 10×2 + 4


= 20 + 4


= 24


Hence, the two digit number is 24.



Question 20.

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,



⇒ 10x + y = 6(x + y)


⇒ 10x + y = 6x + 6y


⇒ 10x + y – 6x – 6y


⇒ 4x – 5y = 0 …(i)


The reverse of the number = x + 10y


and 10x + y – 9 = x + 10y


⇒ 10x + y – 9 = x + 10y


⇒ 10x – x + y – 10y = 9


⇒ 9x – 9y = 9


⇒ x – y = 1


⇒ x = y + 1 …(ii)


On substituting the value of x = y + 1 in Eq. (i), we get


4x – 5y = 0


⇒ 4(y + 1) – 5y = 0


⇒ 4y + 4 – 5y = 0


⇒ 4 – y = 0


⇒ y = 4


On substituting the value of y = 4 in Eq. (ii), we get


x = y + 1


⇒ x = 4 + 1


⇒ x = 5


So, the Original number = 10x + y


= 10×5 + 4


= 50 + 4


= 54


Hence, the two digit number is 54.



Question 21.

The sum of the digits of a two - digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,


x + y = 12 …(i)


After interchanging the digits, the number = x + 10y


and 10x + y + 18 = x + 10y


⇒ 10x + y + 18 = x + 10y


⇒ 10x – x + y – 10y = – 18


⇒ 9x – 9y = – 18


⇒ x – y = – 2 …(ii)


On adding Eq. (i) and (ii) , we get


x + y + x – y = 12 – 2


⇒ 2x = 10


⇒ x = 5


On substituting the value of x = 5 in Eq. (i), we get


x + y = 12


⇒ 5 + y = 12


⇒ y = 7


So, the Original number = 10x + y


= 10×5 + 7


= 50 + 7


= 57


Hence, the two digit number is 57.



Question 22.

A two - digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,


10x + y = 3 + 4(x + y)


⇒ 10x + y = 3 + 4x + 4y


⇒ 10x + y – 4x – 4y = 3


⇒ 6x – 3y = 3


⇒ 2x – y = 1 …(i)


The reverse number = x + 10y


and 10x + y + 18 = x + 10y


⇒ 10x + y + 18 = x + 10y


⇒ 10x – x + y – 10y = – 18


⇒ 9x – 9y = – 18


⇒ x – y = – 2 …(ii)


On subtracting Eq. (i) from Eq. (ii) , we get


x – y – 2x + y = – 2 – 1


⇒ – x = – 3


⇒ x = 3


On substituting the value of x = 3 in Eq. (i), we get


2(3) – y = 1


⇒ 6 – y = 1


⇒ – y = 1 – 6


⇒ – y = – 5


⇒ y = 5


So, the Original number = 10x + y


= 10×3 + 5


= 30 + 5


= 35


Hence, the two digit number is 35.



Question 23.

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,


10x + y = 7(x + y)


⇒ 10x + y = 7x + 7y


⇒ 10x + y – 7x – 7y = 0


⇒ 3x – 6y = 0


⇒ x – 2y = 0


⇒ x = 2y …(i)


The reverse number = x + 10y


and 10x + y – 27 = x + 10y


⇒ 10x + y – 27 = x + 10y


⇒ 10x – x + y – 10y = 27


⇒ 9x – 9y = 27


⇒ x – y = 3 …(ii)


On substituting the value of x = 2y in Eq. (ii), we get


x – y = 3


⇒ 2y – y = 3


⇒ y = 3


On putting the value of y = 3 in Eq. (i), we get


x = 2(3) = 6


So, the Original number = 10x + y


= 10×6 + 3


= 60 + 3


= 63


Hence, the two digit number is 63.



Question 24.

The sum of the digits of a two - digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.


Answer:

Let unit’s digit = y

and the ten’s digit = x


So, the original number = 10x + y


The sum of the number = 10x + y


The sum of the digit = x + y


According to the question,


x + y = 15 …(i)


After interchanging the digits, the number = x + 10y


and 10x + y + 9 = x + 10y


⇒ 10x + y + 9 = x + 10y


⇒ 10x – x + y – 10y = – 9


⇒ 9x – 9y = – 9


⇒ x – y = – 1 …(ii)


On adding Eq. (i) and (ii) , we get


x + y + x – y = 15 – 1


⇒ 2x = 14


⇒ x = 7


On substituting the value of x = 5 in Eq. (i), we get


x + y = 15


⇒ 7 + y = 15


⇒ y = 8


So, the Original number = 10x + y


= 10×7 + 8


= 70 + 8


= 78


Hence, the two digit number is 78.



Question 25.

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.


Answer:

Let the numerator = x

and the denominator = y


So, the fraction 


According to the question,


Condition I:


x + y = 2y – 3


⇒ x + y – 2y = – 3


⇒ x – y = – 3 …(i)


Condition II:



⇒ 2(x – 1) = y – 1


⇒ 2x – 2 = y – 1


⇒ 2x – y = 1 …(ii)


On subtracting Eq. (i) from Eq. (ii), we get


2x – y – x + y = 1 + 3


⇒ x = 4


On putting the value of x in Eq. (i), we get


4 – y = – 3


⇒ y = 7


So, the numerator is 4 and the denominator is 7


Hence, the fraction is 



Question 26.

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, then are in the ratio 2 : 3. Determine the fraction.


Answer:

Let the numerator = x

and the denominator = y


So, the fraction 


According to the question,


Condition I:


x + y = 2x + 4


⇒ x + y – 2x = 4


⇒ – x + y = 4


⇒ y = 4 + x …(i)


Condition II:



⇒ 3(x + 3) = 2( y + 3)


⇒ 3x + 9 = 2y + 6


⇒ 3x – 2y = – 3 …(ii)


On putting the value of y in Eq.(ii) , we get


3x – 2(4 + x) = – 3


⇒ 3x – 8 – 2x = – 3


⇒ x = 5


On putting the value of x in Eq. (i), we get


y = 4 + 5


⇒ y = 9


So, the numerator is 5 and the denominator is 9


Hence, the fraction is 



Question 27.

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both 3 the numerator and the denominator, the fraction becomes 3/4. Find the fraction.


Answer:

Let the numerator = x

and the denominator = y


So, the fraction 


According to the question,


Condition I:


x + y = 8


⇒ y = 8 – x …(i)


Condition II:



⇒ 4(x + 3) = 3( y + 3)


⇒ 4x + 12 = 3y + 9


⇒ 4x – 3y = – 3 …(ii)


On putting the value of y in Eq.(ii) , we get


4x – 3(8 – x) = – 3


⇒ 4x – 24 + 3x = – 3


⇒ 7x = 21


⇒ x = 3


On putting the value of x in Eq. (i), we get


y = 8 – 3


⇒ y = 5


So, the numerator is 3 and the denominator is 5


Hence, the fraction is 



Question 28.

The numerator of a fraction is one less than its denominator. If 3 is added to each of the numerator and denominator, the fraction is increased by 3/28, find the fraction. 28


Answer:

Let the denominator = x

Given that numerator is one less than the denominator


⇒ numerator = x – 1


So, the fraction 


According to the question,






⇒ 28{(x2 + 2x) – (x2 –x + 3x – 3)} = 3 (x2 + 3x)


⇒ 28x2 + 56x – 28x2 – 56x + 84 = 3x2 + 9x


⇒ 3x2 + 9x – 84 = 0


⇒ x2 + 3x – 28 = 0


⇒ x2 + 7x – 4x – 28 = 0


⇒ x(x + 7) – 4(x + 7) = 0


⇒ (x – 4) (x + 7) = 0


⇒ x = 4 and – 7


But x is a natural number


Hence, x = 4


So, the fraction is 



Question 29.

The age of the father is 3 years more than 3 times the son's age. 3 years here, the age of the father will be 10 years more than twice the age of the son. Find their present ages.


Answer:

Let the age of father = x years

And the age of his son = y years


According to the question,


x = 3 + 3y ...(i)


Three year here,


Father’s age = (x + 3) years


Son’s age = (y + 3) years


According to the question,


(x + 3) = 10 + 2(y + 3)


⇒ x + 3 = 10 + 2y + 6


⇒ x = 2y + 13 …(ii)


From Eq. (i) and (ii), we get


3 + 3y = 13 + 2y


⇒ 3y – 2y = 13 – 3


⇒ y = 10


On putting the value of y = 7 in Eq. (i), we get


x = 3 + 3(10)


⇒ x = 3 + 30


⇒ x = 33


Hence, the age of father is 33 years and the age of his son is 10 years.



Question 30.

Two years ago, a man was five times as old as his son. Two years later his age will be 8 more than three times the age of the son. Find the present ages of man and his son.


Answer:

Let the age of a man = x years

And the age of his son = y years


Two years ago,


Man’s age = (x – 2) years


Son’s age = (y – 2) years


According to the question,


(x – 2) = 5(y – 2)


⇒ x – 2 = 5y – 10


⇒ x = 5y – 10 + 2


⇒ x = 5y – 8 ...(i)


Two years later,


Father’s age = (x + 2) years


Son’s age = (y + 2) years


According to the question,


(x + 2) = 8 + 3(y + 2)


⇒ x + 2 = 8 + 3y + 6


⇒ x = 3y + 12 …(ii)


From Eq. (i) and (ii), we get


5y – 8 = 3y + 12


⇒ 5y – 3y = 12 + 8


⇒ 2y = 20


⇒ y = 10


On putting the value of y = 11 in Eq. (i), we get


x = 5(10) – 8


⇒ x = 50 – 8


⇒ x = 42


Hence, the age of man is 42 years and the age of his son is 10 years.



Question 31.

Father's age is three times the sum of ages of his two children. After 5 years, his age will be twice the sum of ages of two children. Find the age of father.


Answer:

Let the age of two children be x and y

So, the father’s present age = 3(x + y)


After five years,


Age of two children = (x + 5) + (y + 5) years


= ( x + y + 10) years


So, the age of father after five years = 3(x + y) + 5


= 3x + 3y + 5


According to the question,


3x + 3y + 5 = 2(x + y + 10)


⇒ 3x + 3y + 5 = 2x + 2y + 20


⇒ 3x – 2x + 3y – 2y = 20 – 5


⇒ x + y = 15


So, the age of two children = 15 years


And the age of father = 3(15) = 45years


Hence, the age of father is 45 years and the age of his two children is 15 years.



Question 32.

Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?


Answer:

Let the age of A = x years

And the age of B = y years


Five years ago,


A’s age = (x – 5) years


B’s age = (y – 5) years


According to the question,


(x – 5) = 3(y – 5)


⇒ x – 5 = 3y – 15


⇒ x = 3y – 10 …(i)


Ten years later,


A’s age = (x + 10)


B’s age = (y + 10)


According to the question,


(x + 10) = 2(y + 10)


⇒ x + 10 = 2y + 20


⇒ x = 2y + 10 …(ii)


From Eq. (i) and (ii), we get


3y – 10 = 2y + 10


⇒ 3y – 2y = 10 + 10


⇒ y = 20


On putting the value of y = 20 in Eq. (i), we get


x = 3(20) – 10


⇒ x = 50


Hence, the age of person A is 50years and Age of B is 20years.



Question 33.

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages.


Answer:

Let the age of a man = x years

And the age of his son = y years


Ten years hence,


Man’s age = (x + 10) years


Son’s age = (y + 10) years


According to the question,


(x + 10) = 2(y + 10)


⇒ x + 10 = 2y + 20


⇒ x = 2y + 20 – 10


⇒ x = 2y + 10 ...(i)


Ten years ago,


Father’s age = (x – 10) years


Son’s age = (y – 10) years


According to the question,


(x – 10) = 4(y – 10)


⇒ x – 10 = 4y – 40


⇒ x = 4y – 30 …(ii)


From Eq. (i) and (ii), we get


2y + 10 = 4y – 30


⇒ 2y – 4y = – 30 – 10


⇒ – 2y = – 40


⇒ y = 20


On putting the value of y = 20 in Eq. (i), we get


x = 2y + 10


⇒ x = 2(20) + 10


⇒ x = 50


Hence, the age of man is 50 years and the age of his son is 20 years.



Question 34.

Find a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10) and ∠D = (4x — 5)°. Find the four angles.


Answer:


We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°


A + C = 180° and B + D = 180°


⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180


⇒ 2x + 2y = 180 – 14 and 4x + y – 2 = 180


⇒ x + y = 83 and 4x + y = 182


So, we get pair of linear equation i.e.


x + y = 83 …(i)


4x + y = 182 …(ii)


On subtracting Eq.(i) from (ii), we get


4x + y – x – y = 182 – 83


⇒ 3x = 99


⇒ x = 33


On putting the value of x = 33 in Eq. (i) we get,


33 + y = 83


⇒ y = 83 – 33 = 50


On putting the values of x and y, we calculate the angles as


A = (2x + 43)° = 2(33) + 4 = 66 + 4 = 70°


B = (y + 3)° = 50 + 3 = 53°


C = (2y + 10)° = 2(50) + 10 = 100 + 10 = 110°


and D = (4x — 5)° = 4(33) – 5 = 132 – 5 = 127°


Hence, the angles are A = 63°, B = 57°, C = 117°, D = 123°



Question 35.

Find the four angles of a cyclic quadrilateral ABCD in which ∠A = (2x — 3)°,

∠B = (y + 7)°, ∠C = (2y + 17)° and ∠D = (4x — 9)°.

∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°.


Answer:

We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°

A + C = 180° and B + D = 180°


⇒ 2x – 3 + 2y + 17 = 180 and y + 7 + 4x – 9 = 180


⇒ 2x + 2y + 14 = 180 and 4x + y – 2 = 180


⇒ 2x + 2y = 180 – 14 and 4x + y = 182


⇒ x + y = 83 and 4x + y = 182


So, we get pair of linear equation i.e.


x + y = 83 …(i)


4x + y = 182 …(ii)


On subtracting Eq.(i) from (ii), we get


4x + y – x – y = 182 – 83


⇒ 3x = 99


⇒ x = 33


On putting the value of x = 33 in Eq. (i) we get,


33 + y = 83


⇒ y = 83 – 33 = 50


On putting the values of x and y, we calculate the angles as


A = (2x — 3)° = 2(33) – 3 = 66 – 3 = 63°


B = (y + 7)° = 50 + 7 = 57°


C = (2y + 17)° = 2(50) + 17 = 100 + 17 = 117°


and D = (4x — 9)° = 4(33) – 9 = 132 – 9 = 123°


Hence, the angles are A = 63°, B = 57°, C = 117°, D = 123°



Question 36.

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

∠A = 20°, ∠B = 40°, ∠C = 120°.


Answer:

We know that, in a triangle , the sum of three angles is 180°

∴ A + B + C = 180° …(a)


According to the question,



On taking II and III, we get


⇒ 3B = 2 (A + B)


⇒ 3B = 2 A + 2 B


⇒ B = 2 A …(i)


Now, on taking I and II, we get


C = 3 B


⇒ C = 3(2 A) (from eq. (i))


⇒ C = 6 A …(ii)


On substituting the value of B and C in Eq. (a), we get


A + 2A + 6A = 180°


⇒ 9A = 180°


⇒ A = 20°


On puuting the value of A = 20° in Eq. (i) and (ii), we get


B = 2 A = 2(20) = 40°


C = 6 A = 6(20) = 120°


Hence, the angles areA = 20°, B = 40°, C = 120°



Question 37.

In a ΔABC, ∠A = x°, ∠B = (3x)° and ∠C = y°.

If 3y — 5x = 30, show that the triangle is right – angled.


Answer:

We know that, in a triangle , the sum of three angles is 180°

∴ A + B + C = 180°


According to the question,


x + 3x + y = 180


⇒ 4x + y = 180


⇒ y = 180 – 4x …(i)


Given that 3y — 5x = 30 …(ii)


On substituting the value of y in Eq. (ii), we get


3(180 – 4x) – 5x = 30


⇒ 540 – 12x – 5x = 30


⇒ 540 – 17x = 30


⇒ – 17x = 30 – 540


⇒ – 17x = – 510


⇒ x = 30


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


⇒ y = 180 – 4(30)


⇒ y = 60


On putting the value of x and y, we calculate the angles


A = x° = 30°


B = (3x)° = 3(30) = 90°


and C = y° = 60°


Here, we can see that B = 90° , so triangle is a right angled.



Question 38.

The area of a rectangle gets reduced by 8 m2, when its length is reduced by 5m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m2. Find the length and the breadth of the rectangle.


Answer:

Let the length of a rectangle = x m

and the breadth of a rectangle = y m


Then, Area of rectangle = xy m2


Condition I :


Area is reduced by 8m2, when length = (x – 5) m and breadth = (y + 3) m


Then, area of rectangle = (x – 5)×(y + 3) m2


According to the question,


xy – (x – 5)×(y + 3) = 8


⇒ xy – (xy + 3x – 5y – 15) = 8


⇒ xy – xy – 3x + 5y + 15 = 8


⇒ – 3x + 5y = 8 – 15


⇒ 3x – 5y = 7 …(i)


Condition II:


Area is increased by 74m2, when length = (x + 3) m and breadth = (y + 2) m


Then, area of rectangle = (x + 3)×(y + 2) m2


According to the question,


(x + 3)×(y + 2) – xy = 74


⇒ (xy + 3y + 2x + 6) – xy = 74


⇒ xy + 2x + 3y + 6 – xy = 74


⇒ 2x + 3y = 74 – 6


⇒ 2x + 3y = 68 …(ii)


On multiplying Eq. (i) by 2 and Eq. (ii) by 3, we get


6x – 10y = 14 …(iii)


6x + 9y = 204 …(iv)


On subtracting Eq. (i) from Eq. (ii), we get


6x + 9y – 6x + 10y = 204 – 14


⇒ 19y = 190


⇒ y = 10


On putting the value of y = 10 in Eq. (i), we get


3x – 5 (10) = 7


⇒ 3x – 50 = 7


⇒ 3x = 57


⇒ x = 19


Hence, the length of the rectangle is 19m and the breadth of a rectangle is 10m



Question 39.

The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.


Answer:

Let the breadth of a room = x m

According to the question,


Length of the room = x + 3


Then, Area of room = (x + 3)× (x) m2


= x2 + 3x


Condition II:


Area remains same,


when length = (x + 3 + 3) m = (x + 6) m


and breadth = (x – 2) m


According to the question,


x2 + 3x = (x + 6)(x – 2)


⇒ x2 + 3x = x2 – 2x + 6x – 12


⇒ 3x = 4x – 12


⇒ 3x – 4x = – 12


⇒ x = 12


So, length of the room = (x + 3) = 12 + 3 = 15m


Hence, the length of the room is 15m and the breadth of a room is 12m



Question 40.

Two places A and B are 120 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 6 hours, and if they move in opposite directions, they meet in 1 hour 12 minutes. Find the speed of each car.


Answer:

Let the speed of car I = x km/hr

And the speed of car II = y km/hr


Car I starts from point A and Car II starts from point B.


Let two cars meet at C after 6h.


Distance travelled by car I in 6h = 6x km


Distance travelled by car II in 6h = 6y km


Since, they are travelling in same direction, sign should be negative


6x – 6y = 120


⇒ x –y = 20 …(i)


Now, Let two cars meet after 1hr 12min


1hr 12min 


Since they are travelling in opposite directions, sign should be positive.



⇒ 6x + 6y = 120 × 5


⇒ x + y = 100 …(ii)


On adding (i) and (ii) , we get


x – y + x + y = 20 + 100


⇒ 2x = 120


⇒ x = 60


Putting the value of x = 60 in Eq. (i), we get


60 – y = 20


⇒ y = 40


So, the speed of the two cars are 60km/h and 40 km/hr respectively.



Question 41.

A train travels a distance of 300 km at a constant speed. If the speed of SE the 2O train is increased by 5 km an hour, the journey would have taken 2 hours less. Find the original speed of the train.


Answer:

Total distance travelled = 300km

Let the speed of train = x km/hr


We know that,



Hence, time taken by train 


According to the question,


Speed of the train is increased by 5km an hour


∴ the new speed of the train = (x + 5)km/hr


Time taken to cover 300km 


Given that time taken is 2hrs less from the previous time




⇒ 300x + 1500 – 300x = 2x (x + 5)


⇒ 1500 = 2x2 + 10x


⇒ 750 = x2 + 5x


⇒ x2 + 5x – 750 = 0


⇒ x2 + 30x – 25x – 750 = 0


⇒ x (x + 30) – 25 (x + 30) = 0


⇒ (x – 25) (x + 30) = 0


⇒ (x – 25) = 0 or (x + 30) = 0


∴ x = 25 or x = – 30


Since, speed can’t be negative.


Hence, the speed of the train is 25km/hr



Question 42.

A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.


Answer:

Let the usual time taken by the aeroplane = x km/hr

Distance to the destination = 1500km


We know that,



Hence, speed 


According to the question,


Plane left 30min later than the scheduled time


30min 


Time taken by the aeroplane 


∴ the speed of the plane 


Given that speed has to increase by 250 km/hr






⇒ 6(2x – 2x + 1) = 2x2 – x


⇒ 6 = 2x2 – x


⇒ 2x2 –x – 6 = 0


⇒ 2x2 – 4x + 3x – 6 = 0


⇒ 2x (x – 2) + 3 (x – 2) = 0


⇒ (2x + 3) (x – 2) = 0


⇒ (2x + 3) = 0 or (x – 2) = 0


∴  or x = 2


Since, time can’t be negative.


Hence, the time taken by the aeroplane is 2hrs and the speed is 750km/hr



Question 43.

A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.


Answer:

Let the speed of a train = x km/hr

And the speed of a car = y km/hr


Total distance travelled = 600km


According to the question,


If he covers 400km by train and rest by car i.e. (600 – 400) = 200km


Time take = 6hrs 30min 


If he travels 200km by train and rest by car i.e. (600 – 200) = 400km


He takes half hour longer i.e. 7 hours


So, total time = train time + car time


We know that,



 …(i)


 …(ii)



400u + 200v = 6.5 …(iii)


and 200u + 400v = 7 …(iv)


On multiplying Eq. (iii) by 2 and Eq. (iv) by 4, we get


800u + 400v = 13 …(a)


800u + 1600v = 28 …(b)


On subtracting Eq. (a) from Eq. (b), we get


800u + 1600v – 800u – 400v = 28 – 13


⇒ 1200v = 15




On putting the value of v in Eq. (iv), we get



⇒ 200u + 5 = 7


⇒ 200u = 2



So, we get  and 


⇒ x = 100 and y = 80


Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.



Question 44.

Places A and B are 80 km apart from each other on a highway. One car starts from A and another from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find speed of the cars.


Answer:

Let the speed of car I = x km/hr

And the speed of car II = y km/hr


Car I starts from point A and Car II starts from point B.


Let two cars meet at C after 8h.


Distance travelled by car I in 8h = 8x km


Distance travelled by car II in 8h = 8y km


Since, they are travelling in same direction, sign should be negative


8x – 8y = 80


⇒ x –y = 10 …(i)


Now, Let two cars meet after 1hr 20 min


1hr 20min 


Since they are travelling in opposite directions, sign should be positive.



⇒ 4x + 4y = 240


⇒ x + y = 60 …(ii)


On adding (i) and (ii) , we get


x – y + x + y = 10 + 60


⇒ 2x = 70


⇒ x = 35


Putting the value of x = 25 in Eq. (i), we get


35 – y = 10


⇒ y = 25


So, the speed of the two cars are 35km/h and 25 km/hr respectively.


Question 45.

A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream in the same time. Find the speed of the boat in still water and that of the stream.


Answer:

Let speed of the boat in still water = x km/hr

and speed of the stream = y km/hr


Then, the speed of the boat downstream = (x + y)km/hr


And speed of the boat upstream = (x – y)km/hr


According to the question


ConditionI: When boat goes 16 km upstream, let the time taken be t1.


Then,



When boat goes 24 km downstream, let the time taken be t2.


Then,



But total time taken (t1 + t2) = 6 hours


 …(a)


Condition II: When boat goes 12 km upstream, let the time taken be T1.


Then,



When boat goes 36 km downstream, let the time taken be T2.


Then,



But total time taken (T1 + T2) = 6 hours


 …(b)


Now, we solve tis pair of linear equations by elimination method


 …(i)


And  …(ii)


On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients equal of first term, we get the equation as


 …(iii)


 …(iv)


On substracting Eq. (iii) from Eq. (iv), we get





⇒ x – y = 12 …(a)


On putting the value of x – y = 12 in Eq. (i), we get





⇒ x + y = 4 …(b)


Adding Eq. (a) and (b), we get


⇒ 2x = 16


⇒ x = 8


On putting value of x = 8 in eq. (a), we get


8 – y = 12


⇒ y = – 4 but speed can’t be negative


⇒ y = 4


Hence, x = 8 and y = 4 , which is the required solution.


Hence, the speed of the boat in still water is 8km/hr and speed of the stream is 4km/hr



Question 46.

A man travels 370 km, partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.


Answer:

Let the speed of a train = x km/hr

And the speed of a car = y km/hr


Total distance travelled = 370km


According to the question,


If he covers 250km by train and rest by car i.e. (370 – 250) = 120km


Time take = 4hrs


If he travels 130km by train and rest by car i.e. (370 – 130) = 240km


He takes 18min longer i.e. 


So, total time = train time + car time


We know that,



 …(i)


 …(ii)



250u + 120v = 4 …(iii)


and 130u + 240v = 4.3 …(iv)


On multiplying Eq. (iii) by 2


500u + 240v = 8 …(v)


On subtracting Eq. (iv) from Eq. (v), we get


500u + 240v – 130u – 240v = 8 – 4.3


⇒ 370u = 3.7




On putting the value of v in Eq. (iv), we get



⇒ 1.3 + 240v = 4.3


⇒ 240v = 3



So, we get  and 


⇒ x = 100 and y = 80


Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.


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