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# RS Aggarwal Solutions: Exercise 3E - Linear Equations in two variables Notes | EduRev

## Class 10 : RS Aggarwal Solutions: Exercise 3E - Linear Equations in two variables Notes | EduRev

``` Page 1

Linear equations in two variables 3E
32. 5600, while 4 chairs and 3 tables together cost
4340. Find the cost of each chair and that of each table.
Sol:
5x + 4y = 5600
Multiplying (i) by 3 and (ii) by 4, we get
15x 16x = 16800 17360
-x = -560
Page 2

Linear equations in two variables 3E
32. 5600, while 4 chairs and 3 tables together cost
4340. Find the cost of each chair and that of each table.
Sol:
5x + 4y = 5600
Multiplying (i) by 3 and (ii) by 4, we get
15x 16x = 16800 17360
-x = -560

x = 560
Substituting x = 560 in (i), we have
5 × 560 + 4y = 5600
4y = 5600 2800
y = = 700
Hence, the cost of a chair and that a table are respectively
33. 23 spoons and 17 forks cost Rs.1770, while 17 spoons and 23 forks cost Rs.1830. Find the
cost of each spoon and that of a fork.
Sol:
Let the cost of a spoon be Rs.x and that of a fork be Rs.y. Then
Adding (i) and (ii), we get
40x + 40y = 3600
Now, subtracting (ii) from (i), we get
6x 6y = -60
x y = -
Adding (iii) and (iv), we get
2x = 80 x = 40
Substituting x = 40 in (iii), we get
40 + y = 90 y = 50
Hence, the cost of a spoon that of a fork is Rs.40 and Rs.50 respectively.
34. A lady has only 50-paisa coins and 25-paisa coins in her purse. If she has 50 coins in all
totaling Rs.19.50, how many coins of each kind does she have?
Sol:
Let x and y be the number of 50-paisa and 25-paisa coins respectively. Then
i)
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y = 50 - 39
y = = 22
Subtracting y = 22 in (i), we get
x + 22 = 50
x = 50 22 = 28
Hence, the number of 25-paisa and 50-paisa coins is 22 and 28 respectively.
Page 3

Linear equations in two variables 3E
32. 5600, while 4 chairs and 3 tables together cost
4340. Find the cost of each chair and that of each table.
Sol:
5x + 4y = 5600
Multiplying (i) by 3 and (ii) by 4, we get
15x 16x = 16800 17360
-x = -560

x = 560
Substituting x = 560 in (i), we have
5 × 560 + 4y = 5600
4y = 5600 2800
y = = 700
Hence, the cost of a chair and that a table are respectively
33. 23 spoons and 17 forks cost Rs.1770, while 17 spoons and 23 forks cost Rs.1830. Find the
cost of each spoon and that of a fork.
Sol:
Let the cost of a spoon be Rs.x and that of a fork be Rs.y. Then
Adding (i) and (ii), we get
40x + 40y = 3600
Now, subtracting (ii) from (i), we get
6x 6y = -60
x y = -
Adding (iii) and (iv), we get
2x = 80 x = 40
Substituting x = 40 in (iii), we get
40 + y = 90 y = 50
Hence, the cost of a spoon that of a fork is Rs.40 and Rs.50 respectively.
34. A lady has only 50-paisa coins and 25-paisa coins in her purse. If she has 50 coins in all
totaling Rs.19.50, how many coins of each kind does she have?
Sol:
Let x and y be the number of 50-paisa and 25-paisa coins respectively. Then
i)
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y = 50 - 39
y = = 22
Subtracting y = 22 in (i), we get
x + 22 = 50
x = 50 22 = 28
Hence, the number of 25-paisa and 50-paisa coins is 22 and 28 respectively.

35. The sum of two numbers is 137 and their differences are 43. Find the numbers.
Sol:
Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137
x -
On adding (i) and (ii), we get
2x = 180 x = 90
On substituting x = 90 in (i), we get
90 + y = 137
y = (137 90) = 47
Hence, the required numbers are 90 and 47.
36. Find two numbers such that the sum of twice the first and thrice the second is 92, and four
times the first exceeds seven times the second by 2.
Sol:
Let the first number be x and the second number be y.
Then, we have:
4x -
On multiplying (i) by 7 and (ii) by 3, we get
12x -
On adding (iii) and (iv), we get
26x = 650
x = 25
On substituting x = 25 in (i), we get
2 × 25 + 3y = 92
50 + 3y = 92
3y = (92 50) = 42
y = 14
Hence, the first number is 25 and the second number is 14.
37. Find the numbers such that the sum of thrice the first and the second is 142, and four times
the first exceeds the second by 138.
Sol:
Let the first number be x and the second number be y.
Then, we have:
Page 4

Linear equations in two variables 3E
32. 5600, while 4 chairs and 3 tables together cost
4340. Find the cost of each chair and that of each table.
Sol:
5x + 4y = 5600
Multiplying (i) by 3 and (ii) by 4, we get
15x 16x = 16800 17360
-x = -560

x = 560
Substituting x = 560 in (i), we have
5 × 560 + 4y = 5600
4y = 5600 2800
y = = 700
Hence, the cost of a chair and that a table are respectively
33. 23 spoons and 17 forks cost Rs.1770, while 17 spoons and 23 forks cost Rs.1830. Find the
cost of each spoon and that of a fork.
Sol:
Let the cost of a spoon be Rs.x and that of a fork be Rs.y. Then
Adding (i) and (ii), we get
40x + 40y = 3600
Now, subtracting (ii) from (i), we get
6x 6y = -60
x y = -
Adding (iii) and (iv), we get
2x = 80 x = 40
Substituting x = 40 in (iii), we get
40 + y = 90 y = 50
Hence, the cost of a spoon that of a fork is Rs.40 and Rs.50 respectively.
34. A lady has only 50-paisa coins and 25-paisa coins in her purse. If she has 50 coins in all
totaling Rs.19.50, how many coins of each kind does she have?
Sol:
Let x and y be the number of 50-paisa and 25-paisa coins respectively. Then
i)
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y = 50 - 39
y = = 22
Subtracting y = 22 in (i), we get
x + 22 = 50
x = 50 22 = 28
Hence, the number of 25-paisa and 50-paisa coins is 22 and 28 respectively.

35. The sum of two numbers is 137 and their differences are 43. Find the numbers.
Sol:
Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137
x -
On adding (i) and (ii), we get
2x = 180 x = 90
On substituting x = 90 in (i), we get
90 + y = 137
y = (137 90) = 47
Hence, the required numbers are 90 and 47.
36. Find two numbers such that the sum of twice the first and thrice the second is 92, and four
times the first exceeds seven times the second by 2.
Sol:
Let the first number be x and the second number be y.
Then, we have:
4x -
On multiplying (i) by 7 and (ii) by 3, we get
12x -
On adding (iii) and (iv), we get
26x = 650
x = 25
On substituting x = 25 in (i), we get
2 × 25 + 3y = 92
50 + 3y = 92
3y = (92 50) = 42
y = 14
Hence, the first number is 25 and the second number is 14.
37. Find the numbers such that the sum of thrice the first and the second is 142, and four times
the first exceeds the second by 138.
Sol:
Let the first number be x and the second number be y.
Then, we have:

4x -
On adding (i) and (ii), we get
7x = 280
x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
y = (142 120) = 22
y = 22
Hence, the first number is 40 and the second number is 22.
38. If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21
is subtracted from twice the smaller number, it results in the greater number. Find the
numbers.
Sol:
Let the greater  number be x and the smaller number be y.
Then, we have:
25x 45 = y or 2x
2y - 21 = x or
On multiplying (i) by 2, we get:
4x -
On adding (ii) and (iii), we get
3x = (90 + 21) = 111
x = 37
On substituting x = 37 in (i), we get
2 × 37 - y = 45
74 - y = 45
y = (74 - 45) = 29
Hence, the greater number is 37 and the smaller number is 29.
39. If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient
and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the
quotient and 5 as the remainder. Find the numbers.
Sol:
We know:
Dividend = Divisor × Quotient + Remainder
Let the larger number be x and the smaller be y.
Page 5

Linear equations in two variables 3E
32. 5600, while 4 chairs and 3 tables together cost
4340. Find the cost of each chair and that of each table.
Sol:
5x + 4y = 5600
Multiplying (i) by 3 and (ii) by 4, we get
15x 16x = 16800 17360
-x = -560

x = 560
Substituting x = 560 in (i), we have
5 × 560 + 4y = 5600
4y = 5600 2800
y = = 700
Hence, the cost of a chair and that a table are respectively
33. 23 spoons and 17 forks cost Rs.1770, while 17 spoons and 23 forks cost Rs.1830. Find the
cost of each spoon and that of a fork.
Sol:
Let the cost of a spoon be Rs.x and that of a fork be Rs.y. Then
Adding (i) and (ii), we get
40x + 40y = 3600
Now, subtracting (ii) from (i), we get
6x 6y = -60
x y = -
Adding (iii) and (iv), we get
2x = 80 x = 40
Substituting x = 40 in (iii), we get
40 + y = 90 y = 50
Hence, the cost of a spoon that of a fork is Rs.40 and Rs.50 respectively.
34. A lady has only 50-paisa coins and 25-paisa coins in her purse. If she has 50 coins in all
totaling Rs.19.50, how many coins of each kind does she have?
Sol:
Let x and y be the number of 50-paisa and 25-paisa coins respectively. Then
i)
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y = 50 - 39
y = = 22
Subtracting y = 22 in (i), we get
x + 22 = 50
x = 50 22 = 28
Hence, the number of 25-paisa and 50-paisa coins is 22 and 28 respectively.

35. The sum of two numbers is 137 and their differences are 43. Find the numbers.
Sol:
Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137
x -
On adding (i) and (ii), we get
2x = 180 x = 90
On substituting x = 90 in (i), we get
90 + y = 137
y = (137 90) = 47
Hence, the required numbers are 90 and 47.
36. Find two numbers such that the sum of twice the first and thrice the second is 92, and four
times the first exceeds seven times the second by 2.
Sol:
Let the first number be x and the second number be y.
Then, we have:
4x -
On multiplying (i) by 7 and (ii) by 3, we get
12x -
On adding (iii) and (iv), we get
26x = 650
x = 25
On substituting x = 25 in (i), we get
2 × 25 + 3y = 92
50 + 3y = 92
3y = (92 50) = 42
y = 14
Hence, the first number is 25 and the second number is 14.
37. Find the numbers such that the sum of thrice the first and the second is 142, and four times
the first exceeds the second by 138.
Sol:
Let the first number be x and the second number be y.
Then, we have:

4x -
On adding (i) and (ii), we get
7x = 280
x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
y = (142 120) = 22
y = 22
Hence, the first number is 40 and the second number is 22.
38. If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21
is subtracted from twice the smaller number, it results in the greater number. Find the
numbers.
Sol:
Let the greater  number be x and the smaller number be y.
Then, we have:
25x 45 = y or 2x
2y - 21 = x or
On multiplying (i) by 2, we get:
4x -
On adding (ii) and (iii), we get
3x = (90 + 21) = 111
x = 37
On substituting x = 37 in (i), we get
2 × 37 - y = 45
74 - y = 45
y = (74 - 45) = 29
Hence, the greater number is 37 and the smaller number is 29.
39. If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient
and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the
quotient and 5 as the remainder. Find the numbers.
Sol:
We know:
Dividend = Divisor × Quotient + Remainder
Let the larger number be x and the smaller be y.

Then, we have:
3x = y × 4 + 8 or 3x
5y = x × 3 + 5 or
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get
3x - 4 × 13 = 8
3x = (8 + 52) = 60
x = 20
Hence, the larger number is 20 and the smaller number is 13.
40. If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is
subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.
Sol:
Let the required numbers be x and y.
Now, we have:
=
By cross multiplication, we get:
2x + 4 = y + 2
2x y = -
Again, we have:
=
By cross multiplication, we get:
11x 44 = 5y 20
11x
On multiplying (i) by 5, we get:
10x 5y = -10
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 y = -2
68 y = -2
y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.
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