EXERCISE E Chapter 2 solutions CA foundation maths
Chapter 2 Equations exercise E CA foundation maths solutions are given.
You should study the textbook lesson Equations very well.
You should observe the given below solutions and try them in your own method.
Solutions for exercise E chapter 2 Equations CA foundation maths.
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CA foundation maths solutions
Chapter 2
Exercise E
1. Monthly incomes of two persons are in the ratio 4:5 and their monthly expenses are in the ratio 7:9. If each save Rs. 50 per month find their monthly incomes.
Solution:
Let monthly incomes of two persons are in the ratio 4x:5x
Let monthly expenses are in the ratio 7y:9y
Each save per month = Rs. 50
Then, 4x – 7y = 50 ………… (1)
5x – 9y = 50 ………… (2)
Multiply equation (1) by 5, we get 20x – 35y = 250 …… (3)
Multiply equation (2) by 4, we get 20x – 36y = 200 …… (4)
Equation (3) – (4), we get
y = 50
Substitute y vale in equation (1), we get
4x – 7(50) = 50
4x – 350 = 50
4x = 350 + 50
4x = 400
x = 100
4x = 400
5x = 500
Therefore, their monthly incomes are Rs. 400 and Rs. 500
2. Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are increased by 12.
Solution:
let x/y be the fraction and it is equal to 1/2
Both numerator and denominator are increased by 2
(x + 2)/ (y + 2) = 1/2 ……. (1)
let x/y be the fraction and it is equal to 3/4
Both numerator and denominator are increased by 12
(x + 12)/ (y + 12) = 3/4 ……. (2)
From (1), we get 2x + 4 = y + 2
2x – y = – 2 …….. (3)
From (2), we get 4x + 48 = 3y + 36
4x – 3y = – 12 ……. (4)
Multiply (3) by 2, we get 4x – 2y = – 4 ……. (5)
Equation (5) – (4), we get y = 8
Substitute y value in (3), we get 2x – 8 = – 2
2x = 6
x = 3
Therefore fraction = x/y = 3/8
3. The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age.
Solution.
Let present age of a person be y and sum of ages of his two sons be x
The age of the person is twice the sum of the ages of his two sons
y = 2x
Five years ago, age of the person was y – 5
The sum of ages of two sons will be x – 10
From the problem
y – 5 = 3(x – 10)
y – 3x = – 25
x = 25
Substitute x value in (1)
y = 2 (25) = 50
Therefore, the present age of the person is 50 years.
4. A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number.
Solution.
From option
A number between 10 and 100 is five times the sum of its digits
4 + 5 = 9
45 is five times of 9
If 9 be added to it the digits are reversed
45 + 9 = 54
Therefore, the number is 45
5. The wages of 8 men and 6 boys amount to Rs.33. If 4 men earn Rs. 4.50 more than 5 boys determine the wages of each man and boy.
Solution.
Let wage of man be Rs. x and wage of boy be y
The wages of 8 men and 6 boys amount to Rs.33
8x + 6y = 33 ……. (1)
4 men earn Rs. 4.50 more than 5 boys
4x = 5y + 4.5
4x – 5y = 4.5 …… (2)
Multiply (2) by 2, we get 8x – 10y = 9 …… (3)
Equation (1) – (3), we get 16y = 24
y = 1.5
Substitute y value in (1), we get
8x + 6(1.5) = 33
8x = 33 – 9
8x = 24
x = 3
Therefore, The wage of man x = Rs. 3 and the wage of boy y = Rs. 1.5
6. A number consisting of two digits is 4 times the sum of its digits and if 27 be added to it the digits are reversed. The number is
Solution.
Let x be the unit place and y be at 10th place
The number is 10y + x
The number consisting of two digits is 4 times the sum of its digits
10y + x = 4(x + y)
10y + x = 4x + 4y
3x – 6y = 0
x – 2y = 0 …… (1)
If 27 be added to it the digits are reversed
10y + x + 27 = 10x + y
9y – 9x = 27
y – x = 3 …… (2)
From (1) and (2)
y = 3 and x = 6
Therefore, the number is 10(3) + 6 = 36
7. Of two number 1/5th of the greater is equal to 1/3rd of the smaller and their sum is 16. The numbers are
Solution.
Let smaller number be x and greater number be y
1/5th of the greater is equal to 1/3rd of the smaller
1/5 y = 1/3 x
5x + 3y = 0 …… (1)
Their sum is 16
x + y = 16 ……. (2)
Multiply (2) by 3, we get 3x + 3y = 48 …… (3)
Add (1) and (3), we get 8x = 48 or x = 6
Substitute x value in (1)
5(6) – 3y = 0 or 10 – y = 0 or y = 10
Therefore, the numbers are 6, 10.
8. Y is older than X by 7 years, 15 years back age of X was 3/4 of age of Y. Their present ages are
Solution.
Let ages of two persons be X and Y
Y -is older than X by 7 years
Y = X + 7 …… (1)
15 years back age of X was 3/4 of age of Y.
Then age of Y is Y – 15 and age of X is X – 15
X – 15 = 3/4 (Y – 15)
4X – 60 = 3Y – 45
4x – 60 = 3(X + 7) – 45 [ from (1)]
4X – 60 = 3X + 21 – 45
X = 36
Substitute X value in (1), we get
Y = 36 + 7
Y = 43
Therefore, their present ages are 36 years and 43 years.
9. The sum of the digits in a three-digit number is 12. If the digits are reversed the number is increased by 495 but reversing only the tens and unit digits increases the number by 36. The number is
Solution.
Let the digit at unit, tens and hundredth place be x. y and z
The sum of the digits in a three-digit number is 12
x + y + z = 12 ……. (1)
If the digits are reversed the number is increased by 495
Original number 100x + 10y + z
On reversing digits new number is 100z + 10y + x
100x + 10y + z = 100z + 10y + x + 495
99z – 99x = 495
x – z = 5
z = x – 5 …. (2)
Reversing only of tens and unit digits increases the number by 36
10x + y = 10y + x + 36
9x – y = 36
x – y = 4
y = x – 4 …… (3)
Substitute y and z values in (1), we get
x + x – 4 + x – 5 = 12
3x = 21
x = 7
From (2)
z = 7 – 5 = 2
From (3)
y = 7 – 4 = 3
x = 7, y = 3 and z = 2
Therefore, the original number is 237
10. Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3rd of the smaller and 1/5th of the greater number is together 21. The numbers are
Solution.
Let greater number and smaller number be x and y
Two numbers are such that twice the greater number exceeds twice the smaller one by 18
2x = 2y + 18
x – y = 9 ……(1)
1/3rd of the smaller and 1/5th of the greater number is together 21
x/5 + y/3 = 21
3x + 5y = 315 …….. (2)
Multiply (1) by 5, we get
5x – 5y = 45 ……. (3)
Add (2) and (3), we get
8x = 360
x = 45
Substitute x value in (1), we get
45 – y = 9
y = 36
Therefore, the numbers are 45 and 36
11. The demand and supply equations for a certain commodity are 4q + 7p = 17 and p = q/3 + 7/4 respectively where p is the market price and q is the quantity then equilibrium price and quantity are
Solution.
4q + 7p = 17 …… (1)
p = q/3 + 7/4
4q – 12p = – 21 …….. (2)
Subtract (2) from (1), we get
19p = 38 or p = 2
Substitute p value in (1), we get
4q + 7(2) = 17
4q = 17 – 14 = 3
4q = 3
q = 3/4
Therefore, p = 2 and q = 3/4
Note: Observe the given solutions and try them in your own way.
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CA foundation maths solutions