Selina Concise Class 9 Maths Chapter 6 Simultaneous Equations Exercise 6F Solutions
EXERCISE – 6F
(Q1) Five years ago, A’s age was four times the age of B. Five years hence, A’s age will be twice the age of B. Find their present ages.
Solution:
Let present age of A be x years and let present age B be y years.
From first condition,
Five years ago, A’s age was four times the age of B.
x – 5 = 4 (y – 5)
x – 5 = 4y – 20
x – 4y = -20+5
x – 4y = -15 —— (i)
From second condition,
Five years hence, A’s age will be twice the age of B,
x + 5 = 2 (y + 5)
x + 5 = 2y + 10
x + 5 = 2y + 10
x – 2y = 10 – 5
x – 2y = 5 —- (ii)
Subtracting equation (i) — (ii), we get,
x – 4y = -15
x – 2y = 5
(-) (+) (-)
___________
-2y = -20
y = -20/-2
y = 10
Put y = 10 in equation (i), we get,
x – 4 (10) = -15
x – 40 = -15
x = -15 + 40
x = 25
∴Present age of A = 25 years and present age of B = 10 years.
(Q2) A is 20 years older than B, 5 years ago, A was 3 times as old as B. Find their present ages.
Solution:
Let the present age of A is x years and let the present age of B is y years.
From first condition,
A is 20 years older than B.
x = y + 20
x – y = 20 —– (i)
From second condition,
5 years ago, A was 3 times as old as B.
x – 5 = 3 (y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
x – 3y = -10 —- (ii)
Subtracting equation (i) —- (ii), we get,
x – y = 20
x – 3y = -10
(-) (+) (+)
______________
2y = 30
y = 30/2
y = 15
Put y = 5 in equation (i), we get,
x – 15 = 20
x = 20 + 15
x = 35
∴Present age of A = 35 and present age of B = 15 years.
(Q3) Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of mother and her daughter.
Solution:
Let present age of mother be daughter and let present age of daughter be y years.
From first condition,
Four years ago, a mother was four times as old as her daughter.
x – 4 = 4 (y – 4)
x – 4 = 4y – 16
x – 4y = -16 + 4
x – 4y = -12 —— (i)
From second condition,
Six years later, the mother will be two and a half times as old as her daughter at that time.
x + 6 = 2 1/2 (y + 6)
x + 6 = 4+1/2 (y + 6)
x + 6 = 5/2 (y + 6)
2 (x + 6) = 5 (y + 6)
2x + 12 = 5y + 30
2x – 5y = 30 – 12
2x – 5y = 18 —– (ii)
Multiply equation (i) by 2,
2x – 2 × 4y = -12×2
2x – 8y = -24 —– (iii)
On solving equation (ii) and (iii),
Subtracting equation (ii) and (iii), we get,
2x – 5y = 18
2x – 8y = -24
(-) (+) (-)
______________
3y = 42
y = 42/3
y = 14
Put y = 14 in equation (i), we get,
x – 4y = – 12
x – 4 (14) = -12
x – 56 = -12
x = -12 + 56
x = 44
∴The present age of mother is 44 years and the present age of daughter is 14 years.
(Q4) The age of a man is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.
Solution:
Let present age of man be x years and the present age of two children be y years.
From first condition,
The age of a man is twice the sum of the ages of his two children
x = 2y —- (i)
From second condition,
After 20 years, his age will be equal to the sum of the ages of his children at that time.
x + 20 = y + 20 + 20 (There are two children so, we have to add 20 + 20)
x + 20 = y + 40
x – y = 40 – 20
x – y = 20 —- (ii)
From equation (i) and (ii),
x – y = 20
2y – y = 20
y = 20
Put y = 20 in equation (i), we get,
x = 2×20
x = 40
∴The present age of man is 40 years and the present age of two children is 20 years.
(Q5) The annual incomes of A and B are in the ratio 3:4 and their annual expenditure are in the ratio 5:7. If each RS 5000; find their annual incomes.
Solution:
Let annual income of A = 3x and annual income of B = 4x
Let annual expenditure = 5y and annual expenditure = 7y
Case (I) 3x – 5y = 5000 —— (i)
Case (II) 4x – 7y = 5000 —– (ii)
Multiply equation (i) by 4, we get,
4×3x – 4×5y = 4×5000
12x – 20y = 20000 ——- (iii)
Multiply equation (ii) by 3, we get,
3×4x – 3×7y = 3×5000
12x – 21y = 15000 —— (iv)
Subtracting equation (iii) and (iv), we get,
12x – 20y = 20000
12x – 21y = 15,000
(-) (+) (-)
____________________
y = 5000
Put y = 5000 in equation (i), we get,
3x – 5y = 5000
3x – 5 × 5000 = 5000
3x – 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 30000/3
x = 10000
∴Annual income of A = 3x
= 3 × 10000
= 30000
and annual income of B = 4x
= 4 × 10000
= 40000
(Q6) In an, examination, the ratio of passes to failures was 4:1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5:1. Find the number of students who appeared for the examination.
Solution:
Let number of passed students x and number of fail students = y
Case (I) = the ratio of passes to failures was 4:1
x/y = 4/1
x = 4y —– (i)
Case (II): Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5:1.
Total student = x + y – 30
Pass students = x – 20
Fail students = (x + y – 30) – (x – 20)
= x + y – 30 – x + 20
= y – 10
∴ x-20/y-10 = 5/1
1 × (x – 20) = 5 × (y – 10)
x – 20 = 5y – 50
x – 5y = -50 + 20
x – 5y = -30 — (ii)
From equation (i) and (ii)
x – 5y = -30
4y – 5y = -30 (∵from (i))
-y = -30
y = 30
Put y = 30 in equation (i), we get
x = 4 × 30
x = 120
Pass students = 120 and fail students = 30
∴Total students = 120 + 30
= 150
(Q7) A and B both the have some pencils. If A gives 10 pencils to B, then B will have twice as many as A. And if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have?
Solution:
Let A has x pencils and let B has y pencils.
Case I:
If A gives 10 pencils to B, then B will have twice as many as A
2 (x – 10) = y + 10
2x – 20 = y + 10
2x – y = 10 + 20
2x – y = 30 —– (i)
Case II:
If B gives 10 pencils to A, then they will have the same number of pencils.
x + 10 = y – 10
x – y = -10 – 10
x – y = -20 —– (ii)
Subtracting equations (i) and (ii),
2x – y = 30
x – y = -20
(-) (+) (+)
_____________
x = 50
Put x = 50 in equation (i), we get
2x – y = 30
2×50 – y = 30
100 – y = 30
-y = 30 – 100
-y = -70
y = 70
∴ A has 50 pencils and B has 70 pencils.
(Q8) 1250 persons went to see a circus-show each adult paid RS. 75 and each child paid RS. 25 for the admission ticket, find the number of adults and number of children, if the total collection from them amounts to Rs. 61,250.
Solution:
Let number of adults be x and
Let number of children be y
Case I: x + y = 1250 —– (ii)
Case II: 75x + 25y = 61250
25 (3x + y) = 61250
3x + y = 61250/25
3x + y = 2450 —– (ii)
Subtracting equations first from second,
3x + y = 2450
x + y = 1250
(-) (-) (-)
_________________
2x = 1200
x = 1200/2
x = 600
Put x = 600 in equation (i),
we get,
600 + y = 1250
y = 1250 – 600
y = 650
∴The number of adults is 60 and the number of children is 650
(Q9) Two article A and B are sold for RS 1167 making 5% profit on A and 7%. Profit on B. If the two articles are sold for RS 1165, a profit of 7% is made on A and a profit of 5% is made on B. Find the cost prices of each articles.
Solution:
Let cost price of A = x and cost price of B = y
Case I:
Selling price of A = x + x × 5%
= x + x × 5/100
= x + 5x/100
= 100x + 5x/100
= 105x/100
Selling price of B = y + y × 7%
= y + y × 7/100
= y + 7y/100
= 100y + 7y/100
= 107y/100
∴105x/100 + 107y/100 = 1162
105x + 107y/100 = 1162
105x + 107y = 1162×100
105x + 107y = 116200 —— (i)
Case II:
107x/100 + 105/100 y = 1165
107x + 105y/100 = 1165
107x + 105y = 1165 × 100
107x + 105y = 116500 —— (ii)
Now, adding equations (i) and (ii),
105x + 107x = 116700
107x + 105x = 116500
_____________________
212x + 212y = 233200
212 (x + y) = 233200
x + y = 233200/212
x + y = 1100 —– (iii)
Now, Subtracting equation (i) and (ii),
105x + 107y = 116700
107 + 105y = 116500
(-) (+) (-)
______________________
-2x + 2y = 200
2 (-x + y) = 200
– x + y = 200/2
-x +y = 100 —— (iv)
Now, Adding equation (iii) and (iv), we get,
x + y = 1100
-x + y = 100
____________
2y = 1200
y = 1200/2
y = 600
Put y = 600 in equation (iv), we get,
-x + 600 = 100
-x = 100 – 600
– x = -500
x = 500
∴The cost price of A is 500 and the cost price of B is 600.
(Q10) Pooja and Ritu and do a piece of work in 17 1/7 days. If one day work of Pooja be three fourth of one day work of Ritu find in how many days each will do the work alone.
Solution:
Let Pooja can do work alone in x days. 1 day work of pooja 1/x and Let Ritu can do work alone in y days. 1 day work of Ritu = 1/y
Pooja and Ritu can do work in = 17.1/7
= 17×7+1/7
= 119+1/7
= 120/7
1 day work of Pooja and Ritu = 7/120 = y/x
Case I:
1/x + 1/y = 7/120 —– (i)
Case II:
1/x = 3/4 × 1/y
1/x = 3/4y
x = 4y/3
Put x = 4y/3 in equation (i), we get,
1/(4y/3) + 1/y = 7/120
3/4y + 1/y = 7/120
3 + 4/4y = 7/120
7/4y = 7/120
4y = 120
y = 120/4
y = 30
Put y = 30 in equation (ii),
We get,
x = 4/3 × (30)
x = 120/3
x = 40
∴ Pooja can do work alone in 40 days and Ritu can do work alone in 30 days.
Here is your solution of Selina Concise Class 9 Maths Chapter 6 Simultaneous Equations Exercise 6F
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