Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 1 Pairs of Equations Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 1 Solutions Pairs of Equations
Pairs of Equations Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 1 Pairs of Equations Solutions Questions and Answers
Class 9 Maths Chapter 1 Kerala Syllabus – Mental Math And Algebra
Intext Questions and Answers
Question 1.
There are 100 beads, black and white, in a box; 10 more black than white. How many black and white beads?
Answer:
If we remove the extra 10 black beads in the box then the number of white and black beads in the box will be the same.
That is,
Black bead + white bead = 90
Since they are equal in number.
So, the number of white beads or black beads = \(\frac{90}{2}\) = 45
But the number of black beads is 10 more than the white beads.
Thus,
Number of White beads = 45
Number of black beads = 45 + 10 = 55
Textual Questions and Answers
Now try these problems in any way you like: as mental math, as an equation with a single letter or as a pair of equations with two letters:
Question 1.
Priya bought a bag and a pair of slippers for 1100 rupees. The bag costs 300 rupees more than the slippers. What is the price of the slippers? And the price of the bag?
Answer:
The total price for a bag and a pair of slippers is 1100 rupees.
The bag costs 300 rupees more than the slippers.
Then if this extra 300 rupees is removed, the price of the bag and shoes will be equal.
Thus, each cost 400 rupees.
N ow, if we add the extra rupees 300 to the bag,
Price of a bag = 700 rupees
Price of a slipper = 400 rupees Algebraically,
Let x be the price of a slipper, then
Price of a bag = x + 300 rupees
That is,
x + x + 300 = 1100
2x = 1100 – 300 = 800
x = 400
Thus,
Price of a slipper = 400 rupees
Price of a bag = 400 + 300 = 700 rupees
Question 2.
The sum of two numbers is 26 and their difference is 4. What are the numbers?
Answer:
Sum = 26, Difference = 4
Larger number = \(\frac{\text { Sum }+ \text { Difference }}{2}=\frac{30}{2}\) = 15
Smaller number = \(\frac{\text { Sum }- \text { Difference }}{2}=\frac{22}{2}\) = 11
Question 3.
The perimeter of a rectangle is 40 centimetres, and one side is 8 centimetres longer than the other. Calculate the lengths of the sides.
Answer:
Let x be the length of one side and x + 8 be the length of another side.
Perimeter = 40 cm
2 (x + x + 8) = 40
x + x + 8 = 20
2x= 12
x = 6
Thus, the length of one side = 6 cm
Length of another side = 6 + 8 = 14 cm
Question 4.
A wire three and a half metres long is to be cut into two pieces, with one piece bent into a square and the other into an equilateral triangle. The lengths of the sides of both must be the same. How should the wire be cut?
Answer:
Let x be the length of the side of both square and équilateral triangle then the sum of their perimeter is equal to the length of the wire.
That is,
4x + 3x = 3.5 m
7x = 3.5
⇒ x = \(\frac{3.5}{7}\) = 0.5
0.5 = 2 m
4x = 4
3x = 3 × 0.5 = 1.5 m
Thus, the wire will be cut into 2 metres and 1.5 metres.
Question 5.
In a class, there are 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice the number of boys. How many boys and girls are there in the class?
Answer:
Let the number of boys = x, then
Number of girls = x + 4
The day when 8 boys were absent then,
Number of boys = x – 8
That is,
2 (x – 8) = x + 4
2x – 16 = x + 4
2x – x = 4 + 16
x = 20
Thus,
Number of boys = 20
Number of girls = 24
Question 6.
A fraction simplified after adding 1 to its numerator becomes \(\frac{1}{3}\). If instead, 1 is added to the denominator and then simplified, it becomes \(\frac{1}{4}\). What is the fraction?
Answer:
Let the decimal be \(\frac{x}{y}\) then,
\(\frac{x+1}{y}=\frac{1}{3}\) ⇒ y = 3(x + 1) ….(1)
\(\frac{x}{y+1}=\frac{1}{4}\) ⇒ 4x = y + 1 …. (2)
Substitute the value of y from equation (1) into (2) we get,
3(x+ 1) + 1 = 4x
3x + 3 + 1 = 4x
3x + 4 = 4x
x = 4
Substitute the value of x equation in (1) we get,
y = 3(x + 1) = 3(4 + 1)
= 3 × 5 = 15
Thus, the fraction = \(\frac{4}{15}\)
Question 7.
A person invested 100000 rupees in two schemes, with interest rates 7% and 6%. After one year, they got 6750 rupees as interest from both these together. How much did they invest in each scheme?
Answer:
Total amount invested = 100000 rupees
Let x be the amount invested in the first scheme and 100000 – x be the amount invested in the second scheme then,
\(\frac{x \times 7}{100}+\frac{(100000-x) 6}{100}\) = 6750
\(\frac{7 x+600000-6 x}{100}\) = 6750
7x – 6x = 675000 – 600000
x = 75000
Thus, the amount invested in the first scheme = 75000 rupees
Amount invested in the second scheme 25000 rupees
Question 8.
An object starts with a speed of u m/s and travels along a straight line. If the speed increases at the rate of a m/s every second, the speed at time t seconds is u + at. The speed at one second is 5 m/s and at five seconds, 13 m/s. What is the rate at which speed is increasing? What was the starting speed?
Answer:
Speed in t seconds = u + at
Speed at one second = 5 m/s
u + a × 1 = 5
u + a = 5 …(1)
Speed at five seconds = 13 m/s
u + a × 5 = 13
u + 5a = 13 …(2)
(2) – (1) ⇒ 4a = 8
a = 2
From (2) u = 5 – 2 = 3
Thus, the starting speed, u = 3 m/s
The rate at which speed is increased, a = 2 m/s
Class 9 Maths Kerala Syllabus Chapter 1 Solutions – Two Equations
Textual Questions And Answers
Question 1.
The price of 4 pens and 3 pencils is 66 rupees. The price of 7 pens and 3 pencils is 111 rupees. What is the price of a pen? The price of a pencil?
Answer:
Let x be the price of the pen and y be the price of the pencil then,
4x + 3y = 66
7x + 3y = 111
Here, the number of pencils is the same in both cases.
Therefore, the price of the excess three pens = 111 – 66 = 45
Thus, the price of one pen = \(\frac{45}{3}\) = 15
Price of four pens = 4 × 15 = 60 rupees
Price of remaining 3 pencil = 66 – 60 = 6 rupees
Therefore, price of one pencil = \(\frac{6}{3}\) = 2 rupees
Question 2.
The perimeter of a rectangle is 26 centimetres. Another rectangle with twice the length and thrice the breadth has perimeter 62 centimetres. What are the length and breadth of the first rectangle?
Answer:
Let x and y be the length and breadth of the rectangle respectively.
Perimeter of the first rectangle = 26 centimetres
Therefore, 2( x + y ) = 26
x + y = 13 …(1)
Perimeter of the second rectangle = 62 centimetres
Therefore, 2 (2x + 3y) = 62
2x + 3y = 31 …(2)
(1) × 2 → 2x + 2y = 26 …(3)
(1) — (3) → y = 5
Substituting the value of y in equation (1) we get,
x = 13 – y = 13 – 5 = 8
Thus, the length and breadth of the first rectangle is 8 cm and 5 cm respectively.
Question 3.
The price of two kilograms of orange and three kilograms of apple is 520 rupees. The price of three kilograms of orange and two kilograms of apple is 480 rupees. What is the price of each?
Answer:
Let x andy be the price of one kilogram orange and one kilogram apple respectively then,
2x + 3y = 520 …….(1)
3x + 2y = 480 …….(2)
Here, the weight of the orange in the first equation is the same as that of the weight of the apple in the second equation. Also, the weight of the apple in the first equation is the same as that of the weight of the orange in the second equation.
Thus, in this type of problem, we can form equations by adding and subtracting the above equations to make it easier.
Therefore,
(1) + (2) → 5x + 5y= 1000
x + y = 200 ……(3)
(1) – (2) → -x + y = 40 …(4)
(3) + (4) → 2y = 240
y = 120
Thus, x = 200 – y = 200 – 120 = 80
Therefore, Price of one kilogram of orange = 80 rupees
Price of one kilogram apple = 120 rupees
Question 4.
A wire one metre long is cut into two pieces, one of which is bent into a square and the other into an equilateral triangle. Three times the side of the square and two times the side of the equilateral triangle makes 71 centimetres. What are the lengths of the pieces?
Answer:
Length of the wire = 1 metre = 100 centimetres
Let x and y be the length of the side of the square and equilateral triangle respectively, and the sum of the perimeter of the square and equilateral triangle is 1 metre (100 centimetres) then,
4x + 3y = 100 …(1)
And also, 3x + 2y = 71 ……(2)
(1) × 3 → 12x + 9y = 300 …(3)
(2) × 4 → 12x + 8y = 284 …(4)
(3) – (4) → y= 16
Therefore, 4x = 100 -3y= 100 – 3 × 16 = 52
x = \(\frac{52}{4}\) = 13
Thus, the length of the wire pieces is 4x and 3x.
Therefore, the lengths of the pieces is 52 cm and 48 cm.
Question 5.
Four years ago, Rahim’s age was three times the age of Ramu. After two years, this would become two times. What are their ages now?
Answer:
Let Rahim’s present age = x and
Ramu’s present age = y
Four years ago:
3(y – 4 ) = x – 4
3y-x = -4 + 12= 8 …(1)
After two years:
2(y + 2) = x + 2
2y + 4 = x + 2
2y – x = 2 – 4 = -2 ……(2)
(1) – (2) → y = 8 – (-2) = 10
Thus, from equation (1) we get,
3 × 10 – x = 8
x = 30 – 8 = 22
Therefore, Present age of Rahim = 22
Present age of Ramu =10
Question 6.
The difference of the two smaller angles of a right triangle is 20°. Calculate all three angles.
Answer:
x + y = 90°
x – y = 20°
(1) + (2) → 2x =110
x = \(\frac{110}{2}\) = 55°
(1) – (2) → 2y = 90 – 20
y = \(\frac{90-20}{2}\) = 35°
Therefore, the three angles are 90°, 55°, 35°
Question 7.
When a larger number is divided by a smaller number, the quotient and remainder are both 2. When 5 times the smaller is divided by the larger, the quotient and remainder are still both 2. What are the numbers?
Answer:
Let x be the larger number and y be the smaller number then,
That is, 5y = 2(2y + 2) + 2
5y = 4y + 4 + 2
y = 6
x = 2(6) + 2 = 14
Thus, the numbers are 14 and 6.
Question 8.
The sum of the digits of a two-digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
Let 10x + y be the two-digit number then
x + y = 11 …(1)
If the number is interchanged, then the digit will be 10y + x
Thus,
10y + x = (10x + y) + 27
9y – 9x = 27
y – x = 3 ……..(2)
(1) + (2) → 2y = 14
⇒ y = 7
∴ x = 4
Thus, the two-digit number is 47.
Question 9.
The price of 17 trophies and 16 medals is 2180 rupees. The price of 16 trophies and 17 medals is 2110 rupees. What is the price of each?
Answer:
Let x be the price of the trophies and y be the price of the medals then,
17 x + 16y = 2180 …(1)
16x + 17y = 2110 …(2)
(1) + (2) → 33x + 33y = 4290
x + y = 130 …(3)
(1) – (2) → x – y = 70 ………..(4)
x = \(\frac{130+70}{2}\) = 100
y = \(\frac{130-70}{2}\) = 30
Thus, the price of a trophy =100 rupees
and Price of a medal = 30 rupees
Question 10.
An object starts with a speed of u m/s and travels along a straight line. If the speed increases at the rate of a m/s every second, the distance travelled in time t seconds is ut + at2. The distance travelled in 2 seconds is 10 metres and the distance travelled in 4 seconds is 28 metres. What was the starting speed? What is the rate at which speed is increasing?
Answer:
Distance in t sec = ut + at²
When t = 2 then,
u × 2 + \(\frac{1}{2}\) a × 2² = 10
2 u + 2a = 10
u + a = 5 ………..(1)
When t = 4 then,
u × 4 + – a × 42 = 28
4u + 8a = 28
u + 2a = 7 ……..(2)
(2) – (1) → a = 2
u = 5 – 2 = 3
Thus, the starting speed = 3 m/s
The rate at which speed is increased = 2 m/s
Question 11.
A two-digit number is equal to 6 times the sum of its digits. The number got by interchanging the digits is 9 more than 4 times the sum of the digits. What is the number?
Answer:
Let 10x + y be the two-digits number then,
10x + y = 6(x + y)
4x – 5y = 0 …(1)
10y + x = 4(x + y) + 9
10y + x – 4x – 4y = 9
6y – 3x = 9 …………(2)
(1) × 3 → 12x – 15y = 0 …(3)
(2) × 4 → -12x + 24y = 36 …(4)
(3) + (4) → 0 + 9 y = 36
y = 4
Substituting the value in equation (1) we get,
4x – 5(4) = 0
4x = 20
x = 5
Thus, the number is 54.
Question 12.
11 added to a number gives twice another number, 20 added to the second number gives twice the first number. What are the numbers?
Answer:
Let the number are x, y.
When 11 added to a number gives twice another number then,
2y = x + 11
x = 2y – 11
When 20 added to the second number gives twice the first number then,
2x = y + 20
That is, 2 (2y— 11) = y + 20
4y – 22 = y + 20
3y = 42
y = 14
Therefore, x = 2 × 14 – 11 = 17
Thus, the numbers are 17 and 14.
Intext Questions And Answers
Question 1.
The price of 2 pens and 3 notebooks is 110 rupees. The price of 2 pens and 5 notebooks is 170 rupees. What is the price of a pen? And a notebook?
Answer:
Let x be the price of a pen, and y be the price of a notebook
When the price of 2 pens and 3 notebooks is Rs. 110, then
2x + 3y = 110 ……….(1)
The price for 2 pens and 5 notebooks is 170, then
2x + 5y = 170 ……….(2)
(2) – (1) 2y = 60 ⇒ y = 30
(1) → 2x + 3(30) = 110
⇒ 2x = 110 – 90 = 20
⇒ x = 10
Thus,
Price of a pen =10 rupees
Price of a notebook = 30 rupees
Question 2.
The price of 3 pencils and 4 pens is 66 rupees and the price of 6 pencils and 3 pens is 72 rupees. What is the price of each?
Answer:
Let x be the price of the pencil and y be the price of the pen
When the price of 3 pencils and 4 pens is 66 rupees, then
3x + 4y = 66 (1)
When the price of 6 pencils and 3 pens is 72 rupees, then
6x + 3y = 72 ….(2)
(1) × 2 ⇒ 6x + 8y = 132 ….(3)
(3) – (2) ⇒ 5y = 60
⇒ y = 12
(1) ⇒ 3x + 4(12) = 66
⇒ x = 6
Thus,
Price of a pencil = 6 rupees
Price of a pen = 12 rupees
Question 3.
When a small vessel was filled and e|p|tied five times and a big one two times into a bucket, it contains 20 litres. Instead when this was done twice with the small vessel and thrice with the big, it contained only 19 litres. How much can each vessel hold?
Answer:
Let x litres be filled in a small vessel and y litres be filled in a large vessel.
When a small vessel was filled and emptied five times and a big one two times into a bucket, it contained 20 litres then,
5x + 2y = 20 ….(1)
When this was done twice with the small vessel and thrice with the big, it contained only 19 litres then,
2x + 3y = 19 ….(2)
(1) × 2 → 10x + 4y = 40 ….(3)
(2) × 5 → 10x + 15y = 95 ….(4)
(4) – (3) → 11 y = 55
⇒ y = 5
Substituting this in (1) we get,
5x + 2(5) = 20
5x – 20 = 10
= 10
⇒ x – 2
Thus, the small vessel can hold 2 litres and large vessel can hold 5 litres.
Question 4.
Four times a number added to three times another number gives 43. Two times the second number subtracted from three times the first gives 11. What are the numbers?
Answer:
Let x be the first number, and y be the second number.
When four times a number added to three times another number gives 43, then
4x + 3y = 43 …..(1)
When two times the second number subtracted from three times the first gives 11, then
3x – 2y = 11….(2)
(1) × 3 → 12 + 9y = 129 ….(3)
(2) × 4 → 12x – 8y = 44 ….(4)
(3) – (4) → 17y = 85
⇒ y = 5
(1) → 4x + 3(5) = 43
⇒ 4x = 43 – 15 = 28
⇒ x = 7
Thus, the first number = 7
Second number = 5
Question 5.
The sum of two numbers is 28 and their difference is 12. What are the numbers?
Answer:
Let x be the first number and y be the second number.
When sum of two numbers is 28 then,
x + y = 28 ……(1)
When difference is 12 then,
x – y = 12 ….(2)
(1) + (2) → 2x = 40
⇒ x = 20 (1) → 20 + y = 28
⇒ y = 28 — 20 = 8
Thus, the numbers are 20 and 8.
Pairs of Equations Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
There are some sheep and shepherds in a field. There are a total of 20 heads and 56 legs. Find the number of sheep and the number of shepherds?
Answer:
Let number of sheep = x
Number of shepherds = y
x + y = 20 …(1)
4x + 2y = 56 ……(2)
(1) × 4 → 4x + 4y = 80 …(3)
(3) – (2) → 2y = 24
y = 12
x = 20 – 12 = 8
Thus, Number of sheep = 8
Number of shepherds = 12
Question 2.
A total of 21 vehicles including bikes and autorickshaw were parked in a cinema theatre. If the total number of wheels is 49, what is the number of autorickshaws and bike each?
Answer:
Let the number of bikes = x and
Number of autorickshaws = y ‘
x + y = 21 …(1)
Since bikes have 2 wheel and autorickshaw have 3 wheels then,
2x + 3y = 49 …(2)
(1) × 3 → 3x + 3y = 63 …(3)
(3) – (2) → x 4 14
Therefore, y = 21 – 14 = 7
Thus, number of bikes =14
Number of autorickshaws = 7
Question 3.
If the sum of two numbers is 30 and the difference is 4, then what are the numbers?
Answer:
Let x and y be the two numbers then x + y = 30 …(1)
x – y = 4 …(2)
Adding (1) and (2) we get,
x = \(\frac{30+4}{2}\) = 17
Subtracting (1) from (2) we get,
y = \(\frac{30-4}{2}\) = 13
Thus, the numbers are 17 and 13.
Question 4.
The perimeter of a rectangle is 100 cm and the larger side is 8 cm longer than the smaller side. Find the length of the larger side and the smaller side?
Answer:
et x and y be the length of larger side and smaller side respectively then,
2x + 2y = 100
x + y = 50 …(1)
x – y = 8 …(2)
(1) + (2) → 2 x = 58
x = 29
Substituting the value in equation (1) we get,
y = 21
Thus, the length of the larger side = 29
Length of the smaller side = 21
Question 5.
The difference between the acute angles of an isosceles triangle is 10 degrees. Find the degree of each angles?
Answer:
Here, x > y
x – y = 10° …(1)
x + y = 90° …(2)
Adding (1) and (2) we get,
x = \(\frac{10+90}{2}\) = 50°
Subtracting (1) and (2) we get,
y = \(\frac{90-10}{2}\) = 40°
Thus, the angles are 50° and 40°.
Question 6.
The difference of two integers is 11 and -th of their sum is 5. What are the numbers?
Answer:
Here x > y then
x – y = 11 …(1)
\(\frac{x+y}{5}\) = 5
⇒ x + y = 25 …(2)
Adding equation (1) and (2) we get,
x = \(\frac{25+11}{2}\) = 18
Subtracting equation (2) from (1) we get,
y = \(\frac{25-11}{2}\) = 7
Thus, the number are 18 and 7
Question 7.
In a box there are 17 coins including ten rupee coins and five rupee coins. If the total value of these is 105 rupees, how many coins each?
Answer:
Let the number of ten rupee coins = x
Number of five rupee coins = y
x + y = 17 …(1)
10x + 5y = 105 …(2)
(1) × 10 → 10x + 10y = 170 …(3)
(3) – (2) → 5 y = 65
y = 13
Therefore, x = 17 – 13 = 4
Thus, the number of ten rupee coins = 4
Number of five rupee coins =13
Question 8.
The sum of the digits of a two-digit number is 13 and the number obtained by reversing the digits is 45 more than the first number. Find the numbers?
Answer:
Let x be in the ones place and y be in the tens place then,
x + y = 13 …(1)
(10y + x) – (10x + y) = 45
9y – 9x = 45
y – x = 5 …(2)
(1) + (2) – 2y = 18
y = 9
Therefore, x = 13 – 9 = 4
Thus, number = 49
Number obtain by reversing the digits = 94
Question 9.
The difference of the areas of two squares is 56 and the sum of their perimeters is 56. Find the side length of each square?
Answer:
x² – y² = 56 …….(1)
4x + 4y = 56
x + y = \(\frac{56}{4}\) = 14 ……..(2)
(1) → (x + y)(x – y) = 56
14(x – y) = 56
x – y = 4 ……….(3)
(2) + (3) →
x = \(\frac{14+4}{2}\) = 9
Therefore, y = 5
Question 10.
The larger side of a rectangle is 7 cm longer than the smaller side. The diagonal is 13 cm. Find the length of the sides?
Answer:
Here, x > y
x – y = 7 …(1)
x² + y² = 13²
(x -² y)² = x² + y² — 2xy
7² = 13² – 2xy
xy = \(\frac{13^2-7^2}{2}=\frac{120}{2}\) = 60
(x + y)² = x² + y² + 2xy
= 169 + 120 = 289
x + y = \(\sqrt{289}\) = 17
(1) + (2) → 2x = 24
x = 12cm
(2) – (1) → 2y = 10
y = 5 cm