Interactive OP Malhotra Class 9 Solutions Chapter 5 Simultaneous Linear Equations in Two Variables Ex 5(B) engage students in active learning and exploration.

## S Chand Class 9 ICSE Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variables Ex 5(B)

Question 1.
If one number is thrice the other and their sum is 16, find the numbers.
Solution:
Sum of two numbers = 16
Let first number = A
and second number = y
∴ x + y = 16 … (i)
and A = 3y
⇒ x – 3y = 0 … (ii)
Subtracting (ii) from (i)
4y = 16 ⇒ y = $$\frac { 16 }{ 4 }$$ = 4
∴ x = 3y = 3 x 4 = 12
Hence numbers are 4, 12 or 12, 4

Question 2.
The sum of two numbers is 6 and their difference is 4. Find the numbers.
Solution:
Sum of two numbers = 6
and difference = 4
Let first number = x
and second number = y
∴ x + y = 6 … (i)
and x – y = 4 … (ii)
2x = 10 ⇒ x = $$\frac { 10 }{ 2 }$$ = 5
Subtracting, we get
2y = 2 ⇒ y = $$\frac { 2 }{ 2 }$$ = 1
∴ Numbers are 5, 1 Question 3.
In a two digit number, the sum of digits is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number ?
Solution:
Sum of two digit of a two digit number = 13
Let units digit = x
and tens digit = y
∴ Number = x + 10y
and number by reversing the digit = y + 10x
According to the conditions,
x + y = 13 … (i)
and (y + 10x) – (x + 10y) = 45
⇒ y + 10x – x – 10y = 45
⇒ 9x – 9y = 45
⇒ x – y = 5 … (ii)
(Dividing by 9)
2x = 18 ⇒ x = $$\frac { 18 }{ 2 }$$ = 9
and subtracting (ii) from (i).
2y = 8 ⇒ y = $$\frac { 8 }{ 2 }$$ = 4
Number = x + 10y = 9 + 10 x 4
= 9 + 40 = 49

Question 4.
A number consists of two digits whose sum is 5. When the digits are reversed, the number becomes greater by 9. Find the number.
Solution:
Sum of two digits of a two digit number = 5
Let units digit = x
and tens digit = y
then number = x + 10y
By reversing the digits,
the New number = y + 10x
According to the conditions, x + y = 5 … (i)
and y + 10x – (x + 10y) = 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 … (ii)
(Dividing by 9)
2x = 6 ⇒ x = $$\frac { 6 }{ 2 }$$ = 3
and subtracting (ii) from (i)
2y = 4 ⇒ y = $$\frac { 4 }{ 2 }$$ = 2
∴ Number = x + 10y
= 3 + 10 x 2
= 3 + 20
= 32

Question 5.
The sum of a two digit number and the number obtained by reversing the order of its digits is 121 and the two digits differ by 3. Find the numbers.
Solution:
Difference of two digits of a two digit number = 3
Let units digit = x
and tens digit = y
∴ Number = x + 10y
By reversing the digits, the new number
= y – 10x
According to the conditions.
x – y = 3 … (i)
x + 10y + y + 10x = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 … (ii)
(Dividing by 11)
2x = 14 ⇒ x = $$\frac { 14 }{ 2 }$$ = 7
Subtracting (i) from (ii)
2y = 8 ⇒ y = $$\frac { 8 }{ 2 }$$ = 4
∴ Number = x + 10y
= 7 + 10 x 4
= 7 + 40 = 47
OR y + 70x = 4 + 10 x 7 = 4 + 70 = 74 Question 6.
Seven times a given two digit number is equal to four times the number obtained by reversing the order of digits. The sum of the digits of the number is 3. Find the number.
Solution:
In two digits number,
Sum of its digits = 3
Let ones digit = x
and tens digit = y
∴ Number = x + 10y
and new number by reversing the digit,
= y + 10x
According to the conditions,
x + y = 3 … (i)
and 7 (x + 10y) = 4 (y + 10x)
7x + 70y = 4y + 40x
⇒ 40x – 7x + 4y – 70y = 0
⇒ 33x – 66y = 0
⇒ x – 2y = 0 … (ii)
(Dividing by 33)
⇒ x = 2y
Substituting the value of x in (i)
2y + y = 3 ⇒ 3y = 3 = y = $$\frac { 3 }{ 3 }$$ = 1
and x = 2y = 2 x 1 = 2
Number = x + 10y = 2 + 10 x 1
= 2 + 10 = 12

Question 7.
If three times, the larger of the two numbers is divided by the smaller one. we get 4 as quotient and 3 as remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as quotient and 1 as remainder. Find the numbers.
Solution:
Let larger number = x
and smaller number = y
According to the conditions given
$$\frac { 3x }{ y }$$ = 4 x y + 3 ⇒ 3x = 4y + 3
⇒ 3x – 4y = 3 … (i)
7y = x × 5 + 1
⇒ 7y = 5x + 1
⇒ 5x – 7y = – 1 … (ii)
Multiplying (i) by 7 and (ii) by 4, From (i)
3 x 25 – 4y = 3 ⇒ 75 – 4y = 3
⇒ 4y = 75 – 3 = 72 ⇒ y = $$\frac { 72 }{ 2 }$$ = 18
∴ Larger number = 25
and smaller number = 18

Question 8.
Divide 36 into four parts so that if 2 is added to the first part, 2 is subtracted from the second part, the third part is multiplied by 2, and the fourth part is divided by 2, then the resulting number is the same in each case.
Solution:
Total = 36
Let the result in each case = x
∴ First number = x – 2
Second number = x + 2
Third number = $$\frac { x }{ 2 }$$
and fourth number = 2x
According to the conditions.
x + 2 + x + 2 + $$\frac { x }{ 2 }$$ + 2x = 36
⇒ 4$$\frac { 1 }{ 2 }$$ = 36 ⇒ $$\frac { 9 }{ 2 }$$x = 36
⇒ x = $$\frac{36 \times 2}{9}$$ = 8
∴ First number = 8 – 2 = 6
Second number = 8 + 2 = 10
Third number = $$\frac { 8 }{ 2 }$$ = 4
and fourth number = 8 x 2 = 16

Question 9.
If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes $$\frac { 5 }{ 8 }$$ and if the numerator and the denominator of the same fraction are each increased by 1, the fraction becomes equal to $$\frac { 1 }{ 2 }$$. Find the fraction.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = $$\frac { x }{ y }$$
According to the conditions given,
$$\frac{x+2}{y+1}=\frac{5}{8}$$ ⇒ 8x + 16 = 5y + 5
⇒ 8x – 5y = 5 – 16
⇒ 8x – Sy = -11 … (i)
and $$\frac{x+1}{y+1}=\frac{1}{2}$$⇒ 2x + 2 = y + 1
⇒ 2x – y = 1 – 2 = – 1 … (ii)
From (ii) y = 2 x + 1
Substituting the value on y in (i)
8x – 5y = – 11
⇒ 8x – 5 (2x + 1) = – 11
⇒ 8x – 10x – 5 = – 11
⇒ – 2x = – 11 -5 = – 6
⇒ x = $$\frac { -6 }{ -2 }$$ = 3
∴ y = 2x + 1 = 2 x 3 + 1 = 6 + 1 = 7
and fraction = $$\frac{x}{y}=\frac{3}{7}$$ Question 10.
If 1 is added to the denominator of a fraction, the fraction becomes $$\frac { 1 }{ 2 }$$. If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.
Solution:
Let numerator of a fraction = x
and denominator = y
∴ Fraction = $$\frac { x }{ y }$$
According to the condition,
$$\frac{x}{y+1}=\frac{1}{2}$$ ⇒ 2x = y + 1
⇒ 2x – y = 1 … (i)
and $$\frac { x+1 }{ y }$$ = 1 ⇒ x + 1 = y
⇒ x – y = – 1 … (ii)
Subtracting (ii) from (i)
x = 2
From (ii) 2 – y = – 1 ⇒ – y = – 1 – 2 = – 3
∴ y = 3
∴ Fraction = $$\frac { 2 }{ 3 }$$

Question 11.
When die numerator of a fraction is increased by 4, the fraction increases by $$\frac { 2 }{ 3 }$$. What is the denominator of the fraction?
Solution:
Let the numerator of a fraction = x
and its denominator = y
∴ Fraction = $$\frac { x }{ y }$$
According to the condition,
$$\frac{x+4}{y}=\frac{x}{y}+\frac{2}{3} \Rightarrow \frac{x+4}{y}-\frac{x}{y}=\frac{2}{3}$$
$$\frac{x+4-x}{y}=\frac{2}{3} \Rightarrow \frac{4}{y}=\frac{2}{3}$$
⇒ y = $$\frac{4 \times 3}{2}$$ = 6
∴ Denominator = 6

Question 12.
Father is six times as old as his son. Four years hence he will be four times as old as his son at that time. Find their present ages.
Solution:
Let the age of father = x years
and age of son = y years
∴ x = 6y … (i)
4 years hence,
Age of father will be = x + 4
and age of son = y + 4
∴ (x + 4) = 4 (y + 4)
⇒ x + 4 = 4y + 16
⇒ x = 4y + 16 – 4
⇒ x = 4y + 12 … (ii)
From (i) and (ii)
6y = 4y + 12
⇒ 6y – 4y = 12 ⇒ 2y = 12
⇒ y = 6
∴ x = 6y = 6 x 6 = 36
Present age of father = 36 years
and age of son = 6 years

Question 13.
The age of the father is 3 years more than three times the age of the son. Three years hence father’s age will be 10 years more than twice the age of the son. Determine their present ages.
Solution:
Let the age of son = x years
and age of father = y years
∴ y = 3x + 3 … (i)
3 years hence,
Age of son will be = (x + 3) years
and age of father = (y + 3) years
∴ y + 3 = 2(x + 3) + 10
⇒ y + 3 = 2x + 6 + 10
⇒ y = 2x + 6 + 10 – 3
⇒ y = 2x + 13 … (ii)
From (i) and (ii),
3x + 3 = 2x + 13 ⇒ 3x – 2x = 13 – 3
⇒ x = 10
and y = 3x + 3 = 3 x 10 + 3
= 30 + 3 = 33
∴ Present age of son = 10 years
and age of father = 33 years

Question 14.
The age of a man is three times the sum of the ages of his two children and five years hence his age will be double the sum of their ages. Find his present age.
Solution:
Let the age of the man = x years .
and sum of ages of his two children = y years
∴ x = 3y … (i)
5 years hence,
Age of the man will be = (x + 5) years
and sum of ages of two children = (y + 10) years
∴ x + 5 = 2 (y + 10)
⇒ x + 5 = 2y + 20
⇒ x = 2y + 20 – 5
⇒ x = 2y + 15 … (ii)
From (i) and (ii)
3y = 2y+ 15
⇒ 3y – 2y = 15
⇒ y = 15
and x = 3y = 3 x 15 = 45
∴ The age of the man = 45 years

Question 15.
Ram’s father is 4 times as old as Ram. Five years ago, his father was 9 times as old as he was then. What are their present ages ?
Solution:
Let age of Ram, = x years
and age of Ram’s father = y years
∴ y = 4x … (i)
5 years ago,
Age of Ram was = (x – 5) years
and age of his father = (y – 5) years
∴ y – 5 = 9 (x – 5)
⇒ y – 5 = 9x – 45 ⇒ y = 9x – 45 + 5
⇒ y = 9x – 40 … (ii)
From (i) and (ii),
9x – 40 = 4x ⇒ 9x – 4x = 40
⇒ 5x = 40 ⇒ x = $$\frac { 40 }{ 2 }$$ = 8
∴ y = 4x = 4 x 8 = 32
Hence present age of father = 32 years
and age of Ram = 8 years

Question 16.
At the time of marriage a man was 6 years older than his wife, but 12 years after his marriage, his age was $$\frac { 6 }{ 5 }$$ th of the age of his wife. What were their ages at the time of marriage ?
Solution:
At the time of marriage
Let age of a man = x years
and age of his wife = y years
⇒ x = y + 6 … (i)
12 years after,
Age of a man = (x + 12) years
and age of his wife = (y + 12) years
∴ x + 12 = $$\frac { 6 }{ 5 }$$ (y + 12)
⇒ 5x + 60 = 6y + 72 (By cross multiplication)
⇒ 5x – 6y = 72 – 60 = 12 … (ii)
From (i) x = y + 6
∴ 5 (y + 6) – 6y = 12
5y + 30 – 6y = 12
⇒ – y = 12 – 30 = – 18
∴ y = 18
and x = y + 6 = 18 + 6 = 24
∴ At the time of their marriage
Age of man = 24 years
and age of his wife = 18 years Question 17.
The present ages of Ram and Shyam are in the ratio 5 : 6. Five years ago, the ratio was 4:5. Find their present ages.
Solution:
Ratio in the present age of Ram and Shyam = 5 : 6
Let age of Ram = x years
and age of Shyam = y years
∴ $$\frac{x}{y}=\frac{5}{6}$$ ⇒ 6x = 5y
⇒ x = $$\frac { 5 }{ 6 }$$y … (i)
5 yeas ago,
Age of Ram was = (x – 5) years
and age of Shyam = (y – 5) years
∴ $$\frac{x-5}{y-5}=\frac{4}{5}$$
⇒ 5x – 25 = 4y – 20
⇒ 5x – 4y = – 20 + 25 = 5
⇒ 5x – 4y = 5 … (ii)
From (ii)
5 x $$\frac { 5 }{ 6 }$$y – 4y = 5 ⇒ $$\frac { 25 }{ 6 }$$y – 4y = 5
⇒ 25y – 24y = 30 ⇒ y = 30
∴ x = $$\frac { 5 }{ 6 }$$y = $$\frac { 5 }{ 6 }$$ x 30 = 25
∴ Present age of Ram = 25 years
and age of Shyam = 30 years

Question 18.
The cost of 5 pencils and 6 erasers is Rs. 1.80 whereas that of 3 pencils and 2 erasers is 92 paise. Find the cost of each of a pencil and an eraser.
Solution:
Let cost of one pencil = x paise
and cost of one eraser = y paise
According to the conditions,
5x + 6y = 1.80 = 180 … (i)
3x + 2y = 92 … (ii)
Multiplying (i) by 1 and (ii) by 3, we get x = $$\frac { -96 }{ -4 }$$ = 24
From (i)
5 x 24 + 6y = 180 ⇒ 120 + 6y = 180
⇒ 6y= 180- 120 = 60 ⇒ y = $$\frac { 60 }{ 6 }$$ = 10
∴ Cost of one pencil = 24 paise
and cost of one eraser = 10 paise

Question 19.
9 chairs and 5 tables cost Rs. 90 while 5 chairs and 4 tables cost Rs. 61. Find the price of 6 chairs and 3 tables.
Solution:
Let cost of one chair = Rs. x
and cost of one table = Rs. y
According to the conditions,
9x + 5y = 90 … (i)
5x + 4y = 61 … (ii)
Multiplying (i) by 4 and (ii) by 5, ⇒ x = $$\frac { 55 }{ 11 }$$ = 5
From (i)
9 x 5 + 5y = 90
⇒ 45 + 5y = 90
⇒ 5y = 90 – 45 = 45
= y = $$\frac { 45 }{ 5 }$$ = 9
∴ Cost of 1 chair = Rs. 5 and cost of 1 table = Rs. 9 Now cost of 6 chairs and 3 tables = 6 x 5 + 9 x 3 = 30 + 27 = Rs. 57

Question 20.
A horse and two cows together cost Rs. 680. If a horse cost Rs. 80 more than a cow, find the cost of each.
Solution:
Let cost of one horse = Rs. x
and cost of one cow = Rs. y
According to the conditions,
x – 2y = 680 … (i)
x = y + 80
∴ From (i)
y + 80 + 2y = 680
⇒ 3y = 680 – 80 = 600
⇒ y = $$\frac { 600 }{ 3 }$$ = 200
and x = y + 80 = 200 + 80 = 280
∴ Cost of one horse = Rs. 280
and cost of one cow = Rs. 200

Question 21.
If one kg of sweets and 3 kg of apples cost Rs. 33 and 2 kg of sweets and one kg of apples cost Rs. 31, what is the cost per kg of each ?
Solution:
Let cost of 1 kg sweets = Rs. x
and cost of 1 kg apples = Rs. y
According to the conditions,
x + 3y = 33 … (i)
2x + y = 31 … (ii)
Multiplying (i) by 1 and (ii) by 3, we get ⇒ x = $$\frac { -60 }{ -5 }$$ = 12
From (i)
12 + 3y = 33 ⇒ 3y = 33 – 12 = 21
⇒ y = $$\frac { 21 }{ 3 }$$ = 7
∴ Price of 1 kg sweets = Rs. 12
and price of 1 kg apples = Rs. 7

Question 22.
“A pen costs Rs. 3.50 more than a pencil and the cost of 3 pens and 2 pencils is Rs. 13”. Taking x and y as the cost (in Rs) of a pen and a pencil respectively, write two simultaneous equations in x and y which satisfy the statement given above. (Do not solve the equations). Solution:
Let cost of one pen = Rs. x
and cost of one pencil = Rs. y
∴ x = y + 3.5
x – y = 3.5 … (i)
and 3x + 2y = 13 … (ii)
⇒ 3 (y + 3.5) + 2y = 13
⇒ 3y + 10.5 + 27 = 13
⇒ 5y = 13- 10.5 = 2.5
y = $$\frac { 2.5 }{ 5 }$$ = 0.5 = 50 paise
and x = y + 3.5 = 0.50 + 3.50 = 4.0
Cost of one pen = Rs. 4
and cost of one pencil = 50 paise Question 23.
A man buys postage stamps of denominations 3 paise and 5 paise, for Re 1.00. He buys 22 stamps in all. Find the number of 3 paise stamps bought by him.
Solution:
Total stamps = 22
Let number of stamps of 3 paise = x
Number of stamps of 5 paise = y
According to the conditions given,
x + y = 22 … (i)
and $$\frac{x \times 3}{100}+\frac{y \times 5}{100}$$ = 1
⇒ 3x + 57 = 100 … (ii)
From (i) y = 22 – x
Substituting the value of y in (ii)
3x + 5 (22 – x) = 100
3x + 110 – 5x = 100 – 2x
= 100 – 110 = – 10
x = $$\frac { -10 }{ – 2 }$$ = 5
∴ No. of stamps of 3 paise = 5

Question 24.
There were 2500 persons who bought tickets to see a village fair. The adults paid 75 paise each for their admission tickets but the children paid 25 paise each. If the total receipts amounted to Rs. 1503, using an equation method, find how many children saw the fair?
Solution:
Total persons = 2500
Cost of adult ticket = 75 paise
and cost of child ticket = 25 paise
Total amount of sold tickets = Rs. 1503
= 150300 paise
Let number of children = x and number of adults = y
According to the condition,
25x + 75y = 150300 … (i)
x + y = 2500 … (ii)
From (ii) y = 2500 – x,
Substituting the value of y in (i)
25x + 75 (2500 – x) = 150300
⇒ 25x + 187500 – 75x = 150300
⇒ – 50x = 150300 – 187500
⇒ – 50x = – 37200
x = $$\frac{-37200}{-50}$$ = 744
∴ No. of children = 744
and no. of adults = 2500 – 744 = 1756

Question 25.
In a triangle, the sum of the two angles is equal to the third. If the difference between them is 50°, determine the angles.
Solution:
Sum of three angles of a triangle = 180°
Sum of two angles = third angle
and difference between them = 50
Let first angle = x
Second angle = 7
∴ Third angle = x + 7
Now x + 7 + x + 7 = 180
⇒ 2x + 2y = 180
⇒ x + y = 90 … (i)
and x – y = 50 … (ii)
2x – 140 ⇒ x = $$\frac { 140 }{ 2 }$$ = 70
and subtracting,
2y = 40 ⇒ y = $$\frac { 40 }{ 2 }$$ = 20
∴ First angle = 70°
Second angle = 20°
and third angle = 70° + 20° = 90°

Question 26.
In a parallelogram, one angle is $$\frac { 2 }{ 5 }$$ th of the adjacent angles. Determine the angles of this parallelogram.
Solution:
In a parallelogram, one angle = $$\frac { 2 }{ 5 }$$th of the adjacent angle
Let first angle = x
Second angle = y
According to the condition,
x + y = 180° … (i)
(co-interior angle)
x = $$\frac { 2 }{ 5 }$$y … (ii)
∴ $$\frac { 2 }{ 5 }$$y + y = 180° ⇒ 2y + 5y = 900
⇒ 7y = 900 ⇒ y = $$\frac { 900 }{ 7 }$$
∴ x = $$\frac { 1 }{ 2 }$$
∵ Opposite angles of the parallelogram are equal
∴ Angles of the parallelogram are
$$\frac{900^{\circ}}{7}, \frac{360^{\circ}}{7}, \frac{900^{\circ}}{7}, \frac{360^{\circ}}{7}$$

Question 27.
If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.
Solution:
Let length of rectangle (l) = x
∴ Area = l x b = xy m²
In first case,
Length = x + 1
and breadth = y + 1
∴ (x + 1) (y + 1) = xy + 21
⇒ xy + x + y + 1 = xy + 21
⇒ x + y = 21 – 1 = 20 … (i)
and in second case,
Length = x + 1
Then area,
(x + 1) (y – 1) = xy – 5
⇒ xy – x + y – 1 = xy – 5
⇒ – x + y = – 5 + 1 = – 4
⇒ x – y = 4 … (ii)
2x + 24 ⇒ x = $$\frac { 24 }{ 2 }$$ = 12
and subtracting (iii) from (i)
2y = 16 ⇒ y = $$\frac { 16 }{ 2 }$$ = 8
∴ Length = 12 m
Now perimeter = 2 (l + b)
= 2 (12 + 8) = 2 x 20 = 40 m Question 28.
A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.
Solution:
Distance = 8 km
Time taken downstream = 40 minutes = $$\frac { 40 }{ 60 }$$hr.
and time taken upstream = 60 min. = 1 hr.
Let speed of sailor in still water = x km/hr
and speed of the current = y km/hr.
According to the conditions,
$$\frac{8}{x+y}=\frac{40}{60}=\frac{2}{3} \Rightarrow x+y=\frac{8 \times 3}{2}$$ = 12 … (i)
and $$\frac{8}{x-y}$$ = 1 ⇒ x – y = 8 … (ii)
2x = 20 ⇒ x = 10
and subtracting 2y = 4 ⇒ y = 2
∴ Speed of the sailor = 10 kin/hr.
and speed of current = 2 km/hr.

Question 29.
The boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream, in 6$$\frac { 1 }{ 2 }$$hours. Find the speed of the boat in still water and also the speed of the stream.
Solution:
In 6 hours,
Upstream distance = 24 km
and downstream distance = 28 km
and in 6$$\frac { 1 }{ 2 }$$ hours
Upstream distance = 30 km
Downstream distance = 31 km
Let the speed of boat = x km/hr
and speed of stream = y km/hr
Now according to the conditon, and subtracting, we get
2y = 8 ⇒ y = $$\frac { 8 }{ 2 }$$ = 4
∴ Speed of boat = 10 km/hr.
and speed of stream = 4 km/hr.

Question 30.
Satish and Ashok start at the same time from two places 21 km apart. If they walk in the same direction Satish overtakes Ashok in the 21 hours but if they walk in opposite directions, they meet in 3 hours. Find the rate at which each of them walks.
Solution:
Distance between two places = 21 km Let speed of A = x km/hr.
and speed of B = y km/ hr.
and x > y
In first case, they meet in 21 hours
and in second case, they meet in 3 hours
In first case,
Distance travelled by A = x × 21 km
and by B = y x 21 km
∴ 2x – 21y = 21 ⇒ x – y = 1 … (i)
In second case,
Distance travelled by A = 3x km
and by B = 3y km
∴ 3x + 3y = 21 ⇒ x + y = 7 … (ii)
Adding (i) and (ii), we get
and subtracting (i) from (ii),
2y = 6 ⇒ y = $$\frac { 6 }{ 2 }$$ = 3
∴ Speed of A = 4 km/hr. and speed of B = 3 km/hr.

Question 31.
A boy walks to a picnic spot from his house in 6 hours, but he can travel the same distance on his cycle in 2 hours. If his average cycling speed is 7 kilometers per hour faster than his average walking speed, find his average walking speed and his average cycling speed.
Solution:
Distance covered by walking in = 6 hours
and by cycling in = 2 hours
Average speed of cycling is greater than by walking by = 7 km/hr.
Let the speed of walking = x km/hr
and speed of cycling = y km/hr.
∴ y – x = 7
⇒ y = x + 7 … (i)
and distance covered by walking = x × 6 km
and by cycling = y x 2
∴ 6x = 2y ⇒ y = $$\frac { 6 }{ 2 }$$ = 3x … (ii)
From (i) and (ii),
x + 7 = 3x ⇒ 3x – x = 7
⇒ 2x = 7 ⇒ x = $$\frac { 7 }{ 2 }$$ = 3.5
and y = 3x = 3 x 3.5 = 10.5
∴ Speed of a man by walking = 3.5 km/hr.
and speed by cycling =10.5 km/hr.

Question 32.
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.
Solution:
Total distance covered = 600 km
Let the speed of train = x km/hr.
and speed of car = y km/hr.
In first case distance travelled by train = 400 km
and by car = 600 – 400 = 200 km
In second case, distance travelled by train = 200 km
and by car = 600 – 200 = 400 km
According to the conditions, ∴ Speed of train = 100 km/hr.
and speed of car = 80 km/hr.