Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5C Goyal Brothers ICSE Maths Solutions Ch-5. In this article you will learn how to Solve word problems easily. Visit official Website CISCE for detail information about ICSE Board Class-9.
Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5C Goyal Brothers ICSE Maths Solutions Ch-5
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-5 | Simultaneous Linear Equations |
| Writer | RS Aggrawal |
| Topics | how to Solve word problems |
| Academic Session | 2024-2025 |
How to Solve Word Problems on Simultaneous Linear Equations
- Select variables to represent the unknown quantities.
- Using the given information, write a system of two linear equations relating the two variables.
- Solve the system of linear equations using either substitution or elimination
Page- 86,87
Exercise- 5C
(Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5C Goyal Brothers ICSE Maths Solutions Ch-5)
Que-1: The sum of two numbers is 53 and their difference is 25. Find the numbers.
Sol: We have, Let x+y = 53 ………..(1)
x−y = 25 ………..(2)
Additional number,
2x = 78
2x = 78
x = 78/2
x = 39
x value in equation (1) to get,
∴ x+y = 53
39+y = 53
y = 53−39
y = 14
∴ 39 and 14 are the number.
Que-2: The sum of two numbers exceeds thrice the smaller by 2. If the difference between them is 19, find the numbers.
Sol: We have, Let x, y
taking x>y
∴ (1) x+y = 3y+2
(2) x−y = 19
From the (2) equation we can say that,
x = 19+y ………..(3)
x value in equation to get,
19+2y = 3y+2
y = 17
Y value in equation (3) to get,
x = 19+y
x = 19+17
x = 36
Que-3: The sum of two numbers is 51. If the larger is doubled and the smaller is tripled, the difference is 12. Find the numbers.
Sol: According to the question,
2x – 3(51 – x) = 12
2x – 153 + 3x = 12
5x = 12 + 153
5x = 165
x = 165/5
x = 33
Numbers are
x = 33
(51 – x) = 51 – 33 = 18
Que-4: Find two numbers such that the sum of twice the first and thrice the second is 103 and four time the first exceeds seven times the second by 11.
Sol: Let the numbers be x, y.
According to the question,
2x + 3y = 103 …eqn(1)
4x = 7y+11
=> 4x – 7y = 11 …. Eqn(2)
Eqn(1) * 2 – eqn(2)
4x + 6y = 206
4x -7y = 11
13y = 195
y = 15 .
Substitute in eqn(1)
2x+3(15) = 103
2x + 45 = 103
2x = 103 – 45
2x = 58
x = 29 .
The numbers are 29 , 15
Que-5: Find two 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 us consider,
First number =𝑥 and
Second number =𝑦
As per the statement,
3𝑥+𝑦 = 142 …(1)
4𝑥−𝑦 = 138 …(2)
Using elimination method
Add (1) and (2), we get
7𝑥 = 280
𝑥 = 40
From (1):
3×40+𝑦 = 142
120+𝑦 = 142
𝑦 = 142−120 = 22.
Que-6: Of the two numbers, 4 times the smaller one is less than 3 times the larger one by 6. Also, the sum of the numbers is larger than 6 times their difference by 5. Find the numbers.
Sol: Let the Smaller number = x
Let the larger Number = y
According to the question,
3y – 4x = 6 …..(i)
Also, It is given ,
(x+y) – 6(y-x) = 5
= x + y -6y + 6x = 5
= -5y + 7x = 5 ……(ii)
Multiply by 5 in eqn (i) and by 3 in eqn (ii)
15y – 20x = 30
-15y + 21x = 15
On solving (i) and (ii),
x = 45
Substituting x = 45 in (1)
⇒ 3y – (4 ×45) = 6
⇒ 3y -180 = 6
⇒ 3y = 180+ 6
⇒ 3y = 186
⇒ y = 186/3
⇒ y = 62
Que-7: If from twice the greater of the two numbers, 45 is subtracted, the result is the other number. If from twice the smaller number, 21 is subtracted, the result is 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 – y = 45 ……….(i)
2y – 21 = x or –x +2y = 21 ………(ii)
On multiplying (i) by 2, we get:
4x – 2y = 90 ………..(iii)
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
Que-8: If three times the larger of the two number is divided by the smaller, then the quotient is 4 and remainder is 5. If 6 times the smaller is divided by the larger, the quotient is 4 and the remainder is 2. 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 + 5 or 3x – 4y = 5 ……….(i)
6y = x × 4 + 2 or –4x +6y = 2 ………(ii)
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get
3x – 4 × 13 = 5
⇒ 3x = (5 + 52) = 57
⇒ x = 19
Hence, the larger number is 19 and the smaller number is 13.
Que-9: If two is added to each of the given numbers, then 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:
(𝑥+2)/(𝑦+2) = 1/2
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2x – y = -2 ……(i)
Again, we have:
(𝑥−4)/(𝑦−4) = 5/11
By cross multiplication, we get:
11x – 44 = 5y – 20
⇒11x – 5y = 24 ……(ii)
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.
Que-10: The difference between two numbers is 12 and the difference between their squares is 456. Find the numbers.
Sol: Let the required number x and y
Now, the number we have :
x-y = 12 …….(i)
x²-y² = 456 …….(ii)
Taking eqn (ii) we have,
(x+y)(x-y) = 456
(x+y) 12 = 456
(x+y) = 456/12
(x+y) = 38 …….(iii)
Now adding eqn (i) and eqn (iii) we get,
2x = 50
x = 25
Putting the value of x in eqn (i),
25-y = 12
y = 25-12 = 13.
Que-11: Find the fractions which becomes 1/2 when its numerator is increased by 6 and is equal to 1/3 when its denominator is increased by 7.
Sol: Let numerator be x
And denominator be y
According to question.
x+(6/y) = 1/2
Cross multi the eq
It become ……. 2x+12 = y …..eq 1
x/(y+7) =1/3
It become ……..y+7 = 3x…….eq 2
y = 2x+12 we will substitute the value of y in eq 2
2x+12+7 = 3x
19+2x = 3x
19 = 3x-2x
19 = x
Now we put x= 19 in
y = 2x + 12
y = 2× 19 +12
y = 38 +12
y = 50
So the fraction is 19/ 50.
Que-12: A fraction becomes 1/2 when 1 is subtracted from its numerator and 1 is added to its denominator. Also, it becomes 1/3 when 6 is subtracted from its numerator and 1 from the denominator. Find the original fraction.
Sol: When 1 is subtracted from the numerator and 1 is added to its denominator, we get:
(x-1)/(y+1) = 1/2
2(x-1) = y+1
2x-2 = y+1
2x-y-3 = 0 …… (i)
(2) When 6 is subtracted from the numerator and 1 is subtracted from its denominator, we get:
(x-6)/(y-1) = 1/3
3(x-6) = y-1
3x-18 = y-1
3x-y-17 = 0 …… (ii)
Now, Subtracting eq. (i) from (ii), we get
3x – y – 17 = 0
2x – y – 3 = 0
we get,
x -14 = 0
∴ x = 14
Substituting x = 14 in eq. (i), we get
2(14) – y – 3 = 0
⇒ 28 – y – 3 = 0
⇒ 25 – y = 0
⇒ y = 25
Now, the original fraction is 14/25.
Que-13: The denominator of a fraction is greater tham its numerator by 9. If 7 is subtracted from both, its numerator and denominator, the fraction becomes 2/3. Find the original fraction.
Sol: y = x + 9 (The denominator is greater than the numerator by 9)
(x-7)/(y-7) = 2/3 (The fraction becomes 2/3)
[(x-7)/(x+9-7)] = 2/3
[(x-7)/(x+2)] = 2/3
Next, we can cross-multiply:
3(x-7) = 2(x+2)
Expanding both sides:
3x – 21 = 2x + 4
Now, we can solve for x:
3x – 2x = 4 + 21
x = 25
Substituting the value of x back into the first equation to find y:
y = x + 9
y = 25 + 9
y = 34
Therefore, the original fraction is 25/34.
Que-14: A number consists of 2 digits, the difference of whose digits is 3. If 4 times the number is equal to 7 times the number obtained by reversing the digits, find the number.
Sol: Let numbers be x at ones place and y at tens place so
10y+x is that digit
now reversed digit is 10x+y
according to question
7 (10y+x) = 4(10x+y)
x = 2y …… (i)
now,
given x-y = 3
from eq. (i) 2y = x
2y-y = 3
y = 3
so,
x = 63
Que-15: A number consists of 2 digits, the difference of whose digits if 5. If 8 times the number is equal to 3 times the number obtained by reversing the digits, find the number.
Sol: let number is xy, where x is ten’s place and y is unit place
so it can be written as (10x+y)
given that |x−y| = 5
that is x−y = +5 …..eq(1)
or x−y = −5 …..eq(2)
Now,
8(10x+y) = 3(10y+x)
80x+8y = 30y+3x
77x−22y = 0
7x−2y = 0 ……eq(3)
solving eq(1) and eq(3)
7x−2y = 0
−(2x−2y = 10)
it gives 5x = −10
hence x = −2
put in eq(1)
−2−y = 5
y = −7
so number,10x+y = 10(−2)+(−7)
N = −27
solving eq(2) and eq(3)
7x−2y = 0
−(2x−2y = −10)
it gives, 5x = 10
x = 2
put in eq(1)
2−y = −5
y = 7
so,
N = 10x+y = 10∗2+7
N = 27
Que-16: The result of dividing a number of two digits by a number with digits reversed is 1*(3/4). If the sum of the digits is 12, find the number.
Sol: Let the two digits of no. be x & y
Hence; number in → 10x+y
reverse of number is → 10y+x
ATQ; (10x+y)/(10y+x) = 7/4
⇒ 40x+4y = 70y+7x
⇒ 33x = 66y
hence → x = 2y −−−−(i)
Also; x+y = 12
⇒ 2y+y = 12 from (i)
⇒ y = 4 & x = 8
Hence number in 84
Que-17: When a two-digit number is divided by the sum of its digits, the quotient is 8. On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1. Find the number.
Sol: Let’s assume the two-digit number is represented as 10x + y, where x is the tens digit and y is the units digit.
According to the given condition, when the number is divided by the sum of its digits (x + y), the quotient is 8. This can be expressed as:
(10x + y) / (x + y) = 8
x – 3y = 1
We now have a system of two equations:
(10x + y) / (x + y) = 8
x – 3y = 1
To solve this system, we can substitute x from the second equation into the first equation:
(10(3y + 1) + y) / (3y + 1 + y) = 8
Simplifying the equation further:
(31y + 10) / (4y + 1) = 8
Cross-multiplying gives:
31y + 10 = 8(4y + 1)
Expanding and simplifying:
31y + 10 = 32y + 8
Subtracting 31y from both sides:
10 = y + 8
y = 2
Now substituting the value of y into the second equation:
x – 3(2) = 1
x – 6 = 1
x = 7
Therefore, the number is 10x + y = 10(7) + 2 = 70 + 2 = 72.
So, the number is 72.
Que-18: A number of two digits exceeds four times the sum of its digit by 6 and the number increased by 9 on reversing the digits. Find the number.
Sol: Let’s consider the digits at ten’s place as x and let the digit at unit’s place be y.
Number = 10 × x + y = 10x + y,
On reversing digits the number is = 10 × y + x = 10y + x.
According to first condition, we have
⇒ 10x + y – [4(x + y)] = 6
⇒ 10x + y – 4x – 4y = 6
⇒ 10x – 4x + y – 4y = 6
⇒ 6x – 3y = 6
⇒ 2x – y = 2 ……(i)
According to second condition, we have
⇒ 10x + y + 9 = 10y + x
⇒ 10x – x + y – 10y = -9
⇒ 9x – 9y = -9
⇒ x – y = -1
⇒ y – x = 1 ……(ii)
Adding eq. (i) and (ii) we get,
⇒ 2x – y + (y – x) = 2 + 1
⇒ x = 3.
Substituting value of x in (ii) we get,
⇒ y – 3 = 1
⇒ y = 4.
Number = 10 × x + y = 10 × 3 + 4 = 30 + 4 = 34.
Que-19: The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 12 less than twice the original number. Find the original number.
Sol: Let the two digit number be 10x+y
Sum of digit : x+y = 12 __(i)
A.t.Q : 2(10x+y)−(10y+x) = 12
⇒ 20x+2y−10y−x = 12
⇒ 19x−8y = 12
⇒ 19x−8y−12 = 0 __(ii)
Put eq. (i) in eq. (ii)
⇒19(12−y)−8y−12 = 0
⇒ 228−19y−8y−12 = 0
⇒ 27y = 216
⇒ y = 8
x+y = 12
x = 12−8
x = 4
Number : 48
Que-20: If 11 pens and 19 pencils together cost Rs502; while 19 pens and 11 pencils together cost Rs758, how much 3 pens and 6 pencils cost together?
Sol: let the cost of 1 pen be x
let the cost of i pencil be y
= 11x + 19y = rs 502 -equation 1
=19x + 11y = rs 758 -equation 2
adding and subtracting the above equations
after adding- 30x + 30y = 1260
= x + y = 42 = equation 3
after subtracting = 8x – 8y = 256
= x-y = 32 = equation 4
now subtract equation 4 from equation 1
= x+y = 42
= x-y = 32
we get,
= 2y = 10
= y = 5
now substitute y=5 in equation 3
= x+5 = 42
= x = 42-5
= x = 37
= 3(x)+6(y)
= 3(37)+6(5)
= 111+30
= rs 141 is the answer.
Que-21: 5kg sugar and 7kg rice together cost Rs258, while 7kg sugar and 5kg rice together cost Rs246. Find the total cost of 8kg sugar and 10kg rice.
Sol: Make sugar = x
rice = y
5x + 7y = 258……Eq(1)
7x + 5y = 246……Eq(2)
Multiply Eq(1) by 5
Multiply Eq(2) by 7
25x + 35y = 1290….Eq(1)
49x + 35y = 1722…Eq(2)
Subtract Eq(1) from Eq(2)
24x = 432
x = 18
Substitute x = 18 into Eq(1)
5x + 7y = 258……Eq(1)
5(18) + 7y = 258
90 + 7y = 258
7y = 258 – 90
7y = 168
y = 24
x = 18; y = 24
6kg of sugar
8(18) = 144
10kg of rice
10(24) = 240
Total price = 384
Que-22: One year ago a man was four times as old as his son. After 6 years, his age exceeds twice his son’s age by 9 years. Find their present ages.
Sol: Let the present age of father be ‘f’ and that of son be ‘s’.
According to question:
One year ago, a father was four times as old as his son.
(f – 1) = 4 (s – 1)
=> f – 1 = 4s – 4
=> f = 4s – 3 ——- (i)
After six years his age exceeds twice his son’s age by 9 years.
(f + 6) = 2(s + 6) + 9
If we put the value of equation (i), in the above equation, we get:
4s – 3 + 6 = 2s + 12 + 9
=> 4s + 3 = 2s + 21
=> 4s – 2s = 21 – 3
=> 2s = 18
=> s = 9
If we put the value of ‘s’ in equation (i), we get:
f = 4 × 9 – 3
=> f = 33
Que-23: 5 year ago, A has thrice as old as B and 10 years later, A shall be twice as old as B. What are the present ages of A and B.
Sol: Let the age of A be a, and age of B be b
Now, a−5 = 3(b−5)
a = 3b−10 (1)
After 10 years,
a+10 = 2(b+10)
a = 2b+10 (2)
From 1 and 2,
3b−10 = 2b+10
b = 20
And, a = 2b+10 = 2×20+10 = 50 years
Thus, age of A = 50 years and age of B = 20 years.
Que-24: The monthly incomes of A and B are in the ratio 7:5 and their expenditures are in the ratio 3:2. If each saves Rs1500 per month, find their monthly incomes.
Sol: Let 7x is the monthly income of A and 5x is the monthly income of B. Also, let their expenditures be 3y and 2y.
Write two equations for the saving amount of A and B.
7x-3y = 1500 ……(1)
5x-2y = 1500 ………..(2)
To solve equations (1) and (2) find the value of from the first equation.
7x = 1500 +3y
x = (1500+3y)/7
Substitute (1500+3y)/7 for x into the equation (2) and simplify to find y.
5{(1500+3y)/7} – 2y = 1500
{(7500+15y)/7} – 2y = 1500
7500 + 15y – 14y = 10500
y = 3000
Substitute 3000 for y into the equation and find x.
x = {(1500+3.3000)/7}
x = {(1500+9000)/7}
x = 10500/7
x = 1500
Substitute 1500 for x into 7x and find the monthly income of A.
7 (1500) = 10500
Substitute 1500 for y into 5x and find the monthly income of B.
5(1500) = 7500
Que-25: A 90% acid solution is mixed with a 97% acid solution to obtain 21 litres of a 95% solution. Find the quantity of each of the solutions to get the resultant mixtures.
Sol: Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
0.90x + 0.97y = 21 × 0.95
⇒ 0.90x + 0.97y = 21 × 0.95 ……….(i)
And
x + y = 21
From (ii), substitute y = 21 – x in (i) to get
0.90x + 0.97(21 – x) = 21 × 0.95
⇒ 0.90x + 0.97× 21 – 0.97x = 21 × 0.95
⇒ 0.07x = 0.97× 21 – 21 × 0.95
⇒𝑥 = (21×0.02)/0.07 = 6
Now, putting x = 6 in (ii), we have
6 + y = 21 ⇒ y = 15
Hence, the request quantities are 6 litres and 15 litres.
Que-26: There are two examinations hall A and B. If 12 pupils are sent from A to B, the number of pupil in each room becomes the same. If 11 pupil are sent from room B to room A, then the number of pupils in A is double their number in B. Find the number of pupils in each room.
Sol: 12 pupils sent from Hall A to Hall B and the number of pupils becomes the same:
⇒ x – 12 = y + 12
⇒ x = y + 24 ——- [ 1 ]
11 pupils sent from Hall B to Hall A, the number of pupils in Hall A doubled
⇒ x + 11 = 2(y – 11) ——- [ 2 ]
Solve x and y:
Substitute { 1 ] into [ 2 ]:
y + 24 + 11 = 2(y – 11)
y + 35 =2y – 22
y = 57 —— Substitute into [ 1 ]
x = y + 24
x = 57 + 24
x = 81
Hall A = x =81
Hall B = y = 57
Que-27: A and B each have a certain number of marbles. A say to B, “If you give 30 to me, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many marbles does each have ?
Sol: A+30 = 2(B-30)
A+30 = 2B-60
A-2B = -90 …….(i)
3(A-10) = B+10
3A-30 = B+10
3A-B = 40 ……. (ii)
Multiply by 2 in eqn (ii) we get,
6A-2b = 80 …….(iii)
On adding eqn (i) and (iii) we get,
-5A = -170
A = 34
Putting the A = 34 in eqn (i) we get,
34-2B = -90
2B = 90+34
B = 124/2
B = 62.
Que-28: The present age of a man is 3 years more than three times the age of his son. Three years hence, the man’s age will be 10 years more than twice the age of his son. Determine their present age.
Sol: Let the son’s age be x.
Then, according to the condition given, father’s age = 3x + 3
3 years later
Son’s age= x + 3
Father’s age = 3x + 3 + 3 = 3x + 6
According to the condition given,
3x + 6 = 10 + 2 ( x + 3 )
3x + 6 = 10 + 2x + 6
x = 10
Thus, the son’s present age is 10 years and the father’s age is 3(10) + 3 = 33 years.
Que-29: The length of a room exceeds its breadth by 3 metres. If the length is increased by 3m and breadth is decreased by 2 metres, the area remains the same. Find the length and breadth of the room.
Sol: Let the length of the room be x meters and he breadth of the room be y meters.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
⇒ x – y = 3 …….(i)
And, (x + 3) (y – 2) = xy
⇒ xy – 2x + 3y – 6 = xy
⇒ 3y – 2x = 6 ……..(ii)
On multiplying (i) by 2, we get:
2x – 2y = 6 ……….(iii)
On adding (ii) and (iii), we get:
y = (6 + 6) = 12
On substituting y = 12 in (i), we get:
x – 12 = 3
⇒ x = (3 + 12) = 15
Hence, the length of the room is 15 meters and its breadth is 12 meters.
Que-30: The area of a rectangle gets reduced by 8m², if its length is reduced by 5m and breadth increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74m². Find the length and breadth of a rectangle.
Sol: When the length is reduced by 5m and the breadth is increased by 3 m:
New length = (x – 5) m
New breadth = (y + 3) m
∴ New area = (x – 5) (y + 3) sq.m
∴ xy – (x – 5) (y + 3) = 8
⇒ xy – [xy – 5y + 3x – 15] = 8
⇒ xy – xy + 5y – 3x + 15 = 8
⇒ 3x – 5y = 7 ………(i)
When the length is increased by 3 m and the breadth is increased by 2 m:
New length = (x + 3) m
New breadth = (y + 2) m
∴ New area = (x + 3) (y + 2) sq.m
⇒ (x + 3) (y + 2) – xy = 74
⇒ [xy + 3y + 2x + 6] – xy = 74
⇒ 2x + 3y = 68 ………(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x – 15y = 21 ……….(iii)
10x + 15y = 340 ………(iv)
On adding (iii) and (iv), we get:
19x = 361
⇒ x = 19
On substituting x = 19 in (iii), we get:
9 × 19 – 15y = 21
⇒171 – 15y = 21
⇒15y = (171 – 21) = 150
⇒y = 10
Hence, the length is 19m and the breadth is 10m.
Que-31: A motorboat takes 6hrs to cover 100km downstream and 30km upstream. If the motorboat goes 75km downstream and return back to its starting point in 8hrs, find the speed of the motorboat in still water and the rate of the stream.
Sol: Let the rate of flow of the stream be y km/h
Speed of boat upstream = (x – y) km/h.
Speed of boat downstream = (x + y)km/h.
we know time = distance/speed.
now , ATQ,
Time for 100 km downstream and 30 km upstream
100/(x + y) + 30/(x – y)
And it takes 6 hrs to cover downstream and upstream. Then
100/(x + y) + 30/(x – y) = 6
Time for 75 km downstream and returning (means 75 km upstream)
= 75/(x + y) + 75/(x – y)
Given that the time taken is 8 hours
75/(x + y) + 75/(x – y) = 8
Put 1/(x+y) = p and 1/(x-y) = q
now the equation should be .
100p + 30q = 6
50p + 15q = 3————( 1 )
75p + 75q = 8———-( 2 )
multiply by ( 5 ) in —–( 1 )
250p + 75q = 15 —— ( 3 )
On adding eqn (2) and (3) we get,
175p = 7
p = 1/25 [ put in ——-( 1 ) ]
50(1/25) + 15q = 3
2 + 15q = 3
q = 1/15 = 1/(x – y)
x – y = 15————( 3 )
p = 1/25 = 1/(x + y)
x + y = 25———( 4 )
On adding eqn (3) and (4) we get,
2x = 40
x = 20 [ put in ——( 3 ) ]
x – y = 15
20 – y = 15
y = 20 – 15
y = 5 , x = 20
Hence, the speed of the motor boat in still water is 20 km/h and rate of flow of the stream is 5 km/h.
Que-32: A man sold a chair and a table for Rs2178, thereby making a profit of 12% on the chair and 16% on the table. By selling them for Rs2154, he gains 16% on the chair and 12% on the table. Find the cost price of each.
Sol: Let the cost price of a chair and table be x and y respectively.
According to the question,
x + (12x/100) + y + 16y/100 = 2178
112x + 116y = 217800 ….(i)
Also
x + (16x/100) + y + (12y/100) = 2154
116x + 112y = 215400 …(ii)
Solving (i) and (ii) we get x = Rs. 650 and y = Rs. 1250
Que-33: A man travel 600km partly by train and partly by car. If he covers 120km by train and the rest by car, it takes him 8hrs. But, if he travels 200km by the train and the rest by the car, he takes 20min longer. Find the speed of the car and that of the train.
Sol: Let the speed of the train be x km/h
And the speed of the car be y km/h
⇒ 120/x + 480/y = 8
⇒ 120(1/x + 4/y) = 8
⇒ 1/x + 4/y = 1/15 …i)
In the second condition
⇒ Total time = 8 + 20/60 = 25/3 hr
∴ 200/x + 400/y = 25/3
⇒ 200(1/x + 2/y) = 25/3
⇒ 1/x + 2/y = 1/24 …ii)
After solving equation (i) and (ii)
(By substracting equation 2 from equation 1)
⇒ x = 60 km/h
⇒ y = 80 km/h.
Que-34: 6 mens and 8 boys can finish a piece of work in 14 days while 8 mens and 12 boys can do it in 10 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Sol: Let the time taken by man be x days
time taken by boys be y days
work done by 1 man in one day = 1/x
work done by 1 boy in one day = 1/y
(8/x)+(12/y) = 1/10…………eq 1
(6/x)+ (8/y) = 1/14……….eq2
Let u = 1/x & v = 1/y
8u+ 12y = 1/10………eq3
6u+8v = 1/14………..eq4
Multiplying eq3 by 2 & eq4 by 3
16u+ 24y = 2/10………eq5
18u+24v = 3/14………eq6
subtract eq 5 & 6
-2u = -3/14+2/10
-2u = (-3×10 +2×14)/140
-2u = ( -30+28)/140
-2u = -2/140
u = 1/140
put this value in eq 5
16u+ 24y = 2/10
16×(1/140) +24y = 1/5
(4/35)+24y = 1/5
24y = (1/5)-(4/35)
24y = (1×7 -4)/35
24y = (7-4)/35
24y = 3/35
y = 3/35×24
y = 1/280
u = 1/x
1/140 = 1/x
x = 140
v = 1/y
1/280 = 1/y
y = 280
A man complete the work in 140 days
A boy can complete the work in 280 days
Que-35: A lady has 25-P and 50-P coins in her purse. If in all she has 80 coins totalling Rs25, how many coins of each kind does she have?
Sol: Let number of 25p coins be x .
No. of 50p coins = 80-x.
Now,
25*x + 50(80-x) = 25*100
25x + 4000 – 50x = 2500
25x = 1500
x = 1500/25 = 60.
Thus, No. of 25 paise coins is 60 and
No. of 50 paise coins will be 80-60 = 20 .
Que-36: A and B together can do a piece of work in 6 days. If A’s one day’s work be 1*(1/2)times the one day’s work of B, find in how many days, each alone can finish the work.
Sol: One day work of A is 1/x
One day work of B is 1/y
Total one day work of A and B is (1/x)+(1/y) which is equal to 1/6 .
Then,
1/x+1/y = 1/6 −(i)
It is also mentioned in the question that, A’s one day work is one and a half times the one day’s work of B.
Hence, 1/x = 3/2×1/y −(ii)
Now comparing (i) and (ii) equation we get,
(1/x)+(1/y) = 1/6
(3/2y)+(1/y) = 1/6
[1/y(3/2+1)] = 1/6
1/y(5/2) = 1/6
y = 6×5/2
y = 15
Now,
1/x = 3/2×1y
1/x = 3/2×1/15
x = (15×2)/3
x = 10
Hence , A requires 10 days to finish the work alone and B requires 15 days to finish the work.
– : End of Simultaneous Linear Equations Class 9 RS Aggarwal Exe-5C Goyal Brothers ICSE Maths Solutions :–
Return to : – RS Aggarwal Solutions for ICSE Class-9 Mathematics
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