ML Aggarwal Problems on Simultaneous Linear Equations Exe-6 Class 9 ICSE Maths APC Understanding Solutions. Solutions of Exercise-6. This post is the Solutions of ML Aggarwal Chapter 6- Problem on Simultaneous Linear Equations for ICSE Maths Class-9. APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-6. Problem on Simultaneous Linear Equations for ICSE Board Class-9. Visit official website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Problems on Simultaneous Linear Equations Exe-6 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 9th |
| Chapter-6 | Problems on Simultaneous Linear Equations |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-6 Questions |
| Edition | 2021-2022 |
Exe-6 Solutions of ML Aggarwal for ICSE Class-9 Ch-6, Problems on Simultaneous Linear Equations
Note:- Before viewing Solutions of Chapter -6 Problem on Simultaneous Linear Equations Class-9 of ML Aggarwal Solutions . Read the Chapter Carefully. Then solve all example given in Exercise-6, MCQS and Chapter Test.
Problems on Simultaneous Linear Equations Exe-6
ML Aggarwal Class 9 ICSE Maths Solutions
Page 138
Question 1. The sum of two numbers is 50 and their difference is 16. Find the numbers.
Answer :
Let’s assume the two numbers to be x and y
Then according to the given conditions, we have
x + y = 50 … (i) and
x – y = 16 … (ii)
Now, adding (i) and (ii) we get
2x = 66
x = 33
On substituting the value of x in (i), we get
33 + y = 50
y = 50 – 33
y = 17
Therefore, the two numbers are 33 and 17.
Question 2. The sum of two numbers is 2. If their difference is 20, find the numbers.
Answer :
Let’s assume the two numbers to be x and y
Then according to the given conditions, we have
x + y = 2 … (i) and
x – y = 20 … (ii)
Now, adding (i) and (ii) we get
2x = 22
x = 11
On substituting the value of x in (i), we get
11 + y = 2
y = 2 – 11
y = -9
Therefore, the two numbers are 11 and -9.
Question 3. The sum of two numbers is 43. If the larger is doubled and the smaller is tripled, the difference is 36. Find the two numbers.
Answer :
Let’s assume the two numbers to be x and y such that x > y
Then according to the given conditions, we have
x + y = 43 … (i) and
2x – 3y = 36 … (ii)
Now, multiplying (i) by 3 and adding with (ii) we get
3x + 3y = 129
2x – 3y = 36
—————–
5x = 165
x = 165/5 = 33
On substituting the value of x in (i), we get
33 + y = 43
y = 43 – 33
y = 10
Therefore, the two numbers are 33 and 10.
Question 4. The cost of 5 kg of sugar and 7 kg of rice is Rs. 153, and the cost of 7 kg of sugar and 5 kg of rice is Rs. 147. Find the cost of 6 kg of sugar and 10 kg of rice.
Answer :
Let’s assume the cost of 1 kg of sugar = Rs x
And, let the cost of 1 kg of rice = Rs y
Then according to the given conditions, we have
5x + 7y = 153 … (i) and
7x + 5y = 147 … (ii)
Multiplying (i) by 7 and (ii) by 5, we have
35x + 49y = 1071 … (iii)
35x + 25y = 735 … (iv)
(-)—(-)——(-)————- Subtracting (iv) from (iii), we get
24y = 336
y = 336/24
y = 14
On substituting the value of y in (i), we get
5x + 7(14) = 153
5x + 98 = 153
5x = 153 – 98
5x = 55
x = 55/5 = 11
So, the cost of 1 kg of sugar is Rs 11 and the cost of 1 kg of rice is Rs 14.
Now,
Cost of 6 kg of sugar = Rs 11 x 6 = Rs 66
Cost of 10 kg of rice = Rs 14 x 10 = Rs 140
Thus, the cost of 6 kg of sugar and 10 kg of rice = Rs 66 + Rs 140 = Rs 206.
Question 5. The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 84 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweets was Rs. 2800, find how much sweets of each kind they purchased.
Answer :
Let the quantity of sweet costing Rs 70 be x
And, the quantity of sweet costing Rs 84 be y
Given, total quantity of sweets purchased is 34 kg
x + y = 34 … (i)
Also given, the total money spent is Rs 2800
70x + 84y = 2800 … (ii)
Multiplying (i) by 70 and subtracting with (ii), we get
70x + 70y = 2520
70x + 84y = 2800
(-)—(-)—–(-)——
-14y = -280
y = -280/-14
y = 20
On substituting the value of y in equation (i), we get
x + 20 = 36
x = 36 – 20
x = 16
Therefore, the quantities of sweets purchased are 16 kg which costs Rs 70 per kg and 20 kg which costs Rs 84 per kg.
Question 6. If from twice the greater of two numbers 16 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is still half the other number. What are the numbers.
Answer :
Let’s consider the greater number to be x and the smaller number to be y.
Then, according to the given conditions we have
2x – 16 = y/2 ⇒ 4x – 32 = y ⇒ 4x – y = 32 … (i)
And,
x/2 – 1 = y/2 ⇒ x – 2 = y ⇒ x – y = 2 … (ii)
Now, subtracting (ii) from (i) we get
4x – y = 32
x – y = 2
(-)–(+)—(-)—
3x = 30
x = 10
On substituting the value of x in (i), we get
2(10) – 16 = y/2
20 – 16 = y/2
4 = y/2
y = 8
Therefore, the two numbers are 10 and 8.
Question 7. There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is Rs. 8.50, how many of each are there?
Answer :
Let the number of 20 paise coins be x
And let the number of 25 paise coins be y.
Then, according to the given conditions we have
x + y = 38 … (i) and
20x + 25y = 850 … (ii) [Since, 8.50 Rs = 850 paise]
Performing (ii) – 20 x (i), we get
20x + 25y = 850
20x + 20y = 760
(-)–(-)—–(-)—–
5y = 90
y = 90/5
y = 18
Substituting the value of y in (i), we have
x + 18 = 38
x = 38 – 18
x = 20
Therefore, the number of 20 paise coins are 20 and the number of 25 paise coins are 18.
Question 8. A man has certain notes of denominations Rs. 20 and Rs. 5 which amount to Rs. 380. If the number of notes of each kind is interchanged, they amount to Rs. 60 less as before. Find the number of notes of each denomination.
Answer :
Let’s consider the number of 20 rupee notes to be x
And the number of 5 rupee notes be y
Then, according to the given conditions we have
20x + 5y = 380 … (i) and
5x + 20y = 380 – 60
⇒ 5x + 20 y = 320 … (ii)
Now, multiplying (i) by 4 and subtracting with (ii) we get
80x + 20y = 1520
5x + 20 y = 320
(-)–(-)—-(-)——-
75x = 1200
x = 1200/75
x = 16
On substituting the value of x in (i), we get
20(16) + 5y = 380
320 + 5y = 380
5y = 380 – 320 = 60
y = 60/5
y = 12
Therefore, number of 20 rupee notes = 16 and number of 5 rupee notes = 12
Problem on Simultaneous Linear Equations Exe-6
ML Aggarwal Class 9 ICSE Maths Solutions
Page 139
Question 9. The ratio of two numbers is 2/3. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Answer :
Let’s assume the two numbers to be x and y.
Given that the ratio of the numbers = 2/3
Then,
x/y = 2/3
3x = 2y
⇒ 3x – 2y = 0 … (i)
Also given, if 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio
(x – 2)/(y – 8) = 3/2
2 (x – 2) = 3 (y – 8)
2x – 4 = 3y – 24
⇒ 2x – 3y = -20 … (ii)
Now, performing 3 x (i) – 2 x (ii) we get
9x – 6y = 0
4x – 6y = -40
(-)–(+)—-(+)—
5x = 40
x = 40/5
x = 8
On substituting the value of x in (i), we get
3(8) – 2y = 0
24 = 2y
y = 24/2 = 12
Therefore, the numbers are 8 and 12.
Question 10. If 1 is added to the numerator of a fraction, it becomes 1/5; if 1 is taken from the denominator, it becomes 1/7, find the fraction.
Answer :
Let the fraction be x/y
Then according to the given conditions, we have
(x + 1)/y = 1/5
5 (x + 1) = y
5x + 5 = y
⇒ 5x – y = -5 … (i)
And,
x/(y – 1) = 1/7
7x = y – 1
7x – y = -1 … (ii)
Now, subtracting (i) from (ii) we get
7x – y = -1
5x – y = -5
(-)–(+)-(+)—
2x = 4
x = 2
On substituting the value of x in (i), we get
5(2) – y = -5
10 – y = -5
y = 10 + 5
y = 15
Therefore, the fraction is 2/15.
Question 11. If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to 5/8 and if the numerator and denominator are each diminished by 1, the fraction becomes equal to ½, find the fraction.
Answer :
Let the fraction be x/y
Then according to the given conditions, we have
(x + 2)/ (y + 1) = 5/8
8 (x + 2) = 5 (y + 1)
8x + 16 = 5y + 5
8x – 5y = 5 – 16
8x – 5y = -11 … (i)
And,
(x – 1)/ (y – 1) = ½
2 (x – 1) = (y – 1)
2x – 2 = y – 1
2x – y = 1 … (ii)
Now, performing (i) – 4 x (ii) we get
8x – 5y = -11
8x – 4y = 4
(-)—(+)—(-)
-y = -15
y = 15
On substituting the value of y in (ii), we get
2x – 15 = 1
2x = 1 + 15
x = 16/2 = 8
Therefore, the fraction is 8/15.
Question 12. Find the fraction which becomes ½ when the denominator is increased by 4 and is equal to 1/8, when the numerator is diminished by 5.
Answer :
Let the fraction be x/y.
Then according to the given conditions, we have
x/ (y + 4) = ½
2x = y + 4
2x – y = 4 … (i)
And,
(x – 5)/ y = 1/8
8 (x – 5) = y
8x – 40 = y
8x – y = 40 … (ii)
Now, subtracting (i) from (ii) we get
8x – y = 40
2x – y = 4
(-)–(+)—(-)—
6x = 36
x = 36/6
x = 6
On substituting the value of x in (i), we get
2(6) – y = 4
12 – y = 4
y = 12 – 4 = 8
Therefore, the fraction is 6/8.
Question 13. In a two-digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit’s digit of the original number, find the number.
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
Then according to the first condition, we have
x + y = 7 … (i)
Also, the number = 10 × x + y × 1 = 10y + x
Now, according to the second condition we have
10y + x = 2y + 28
x + 8y = 28 … (ii)
Subtracting (i) from (ii), we get
x + 8y = 28
x + y = 7
(-)—(-)—(-)
7y = 21
y = 21/7
y = 3
Substituting the value of y in (i), we get
x + 3 = 7
x = 7 – 3
x = 4
Therefore, the number is 10 x 4 + 3 = 43.
Question 14. A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
The number is 10 × x + y × 1 = 10x + y
So, reversing the number = 10 × y + x × 1 = 10y + y
Then according to the first condition, we have
10x + y = 4 (x + y) + 6
10x – 4x + y – 4y = 6
6x – 3y = 6
2x – y = 2 … (i)
And according to the second condition, we have
10x + y + 9 = 10y + x
10x – x + y – 10y = -9
9x – 9y = -9
x – y = -1 … (ii)
Now, subtracting (ii) from (i) we have
2x – y = 2
x – y = -1
(-)–(+)–(+)—
x = 3
Substituting the value of x in (i), we get
2(3) – y = 2
6 – y = 2
y = 6 – 2
y = 4
Therefore, the number is 10 x 3 + 4 = 30 + 4 = 34.
Question 15. When a two-digit number is divided by the sum of its digits the quotient is 8. If the ten’s digit is diminished by three times the unit’s digit the remainder is 1. What is the number?
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
The number is 10 × x + y × 1 = 10x + y
Then according to the first condition, we have
(10x + y)/ (x + y) = 8
10x + y = 8 (x + y)
10x + y = 8x + 8y
10x – 8x + y – 8y = 0
2x – 7y = 0 … (i)
And according to the second condition, we have
x – 3y = 1 … (ii)
Performing (i) – 2 x (ii), we get
2x – 7y = 0
2x – 6y = 2
(-)—(+)—(-)
-y = -2
y = 2
Substituting the value of y in (i), we get
2x – 7(2) = 0
2x = 14
x = 14/2
x = 7
Therefore, the number is = 10 x 7 + 2 = 70 + 2 = 72.
Question 16. The result of dividing a number of two digits by the number with digits reversed is 1¾. If the sum of digits is 12, find the number.
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
The number is 10 × x + y × 1 = 10x + y
So, reversing the number = 10 × y + x × 1 = 10y + y
Then according to the first condition given in the problem, we have
(10x + y)/ (10y + x) = 1¾
(10x + y)/ (10y + x) = 7/4
4 (10x + y) = 7 (10y + x)
40x + 4y = 70y + 7x
40x – 7x – 70y + 4y = 0
33x – 66y = 0
x – 2y = 0 … (i)
And according to the second condition of the given problem, we have
x + y = 12 … (ii)
From equation (i) and (ii), we have
x + y = 12
x – 2y = 0
(-)—(+)—(-)
3y = 12
y = 12/3
y = 4
On substituting the value of y in (i), we have
x – 2(4) = 0
x = 8
Therefore, the number is 10 x 8 + 4 = 84.
Question 17. The result of dividing a number of two digits by the number with the digits reversed is 5/6. If the difference of digits is 1, find the number.
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
The number is 10 × x + y × 1 = 10x + y
So, reversing the number = 10 × y + x × 1 = 10y + y
Then according to the first condition given in the problem, we have
(10x + y)/ (10y + x) = 5/6
6 (10x + y) = 5 (10y + x)
60x + 6y = 50y + 5x
60x – 5x – 50y + 6y = 0
55x – 44y = 0
5x – 4y = 0 … (i) [Dividing by 11 on both sides]
And, according to the second condition in the given problem, we have
y – x = 1
-x + y = 1 … (ii)
Performing (i) + 5 x (ii), we get
5x – 4y = 0
-5x + 5y = 5
——————
y = 5
On substituting the value of y in equation (i), we have
5x – 4(5) = 0
5x – 20 = 0
5x = 20
x = 20/5
x = 4
Therefore, the number is 10 x 4 + 5 = 40 + 5 = 45.
Question 18. A number of three digits has the hundred digit 4 times the unit digit and the sum of three digits is 14. If the three digits are written in the reverse order, the value of the number is decreased by 594. Find the number.
Answer :
Let’s consider the digit at tens place as x
And let the digit at unit place be y
Then, the digit at hundred place = 4y
Hence, the number is 100 x 4y + 10 × x + y × 1 = 400y + 10x + y = 401y + 10x
So, reversing the number = 100 × y + 10 × x + 4y × 1 = 100y + 10x + 4y = 104y + 10x
Then according to the first condition given in the problem, we have
x + y + 4y = 14
x + 5y = 14 … (i)
And according to the second condition given in the problem, we have
401y + 10x = 104y + 10x + 594
401y – 104y = 594
297y = 594
y = 594/297
y = 2
On substituting the value of y in equation (i), we have
x + 5(2) = 14
x + 10 = 14
x = 14 – 10
x = 4
Therefore, the number is = 10x + 401y = 10(4) + 401(2) = 40 + 802 = 842.
Question 19. Four years ago Marina was three times old as her daughter. Six years from now the mother will be twice as old as her daughter. Find their present ages.
Answer :
Let take the present age of Marina as x years and
The present age of Marina’s daughter as y years.
Now, four years ago
Age of Marina will be (x – 4) years
Age of Marina’s daughter will be (y – 4) years
According to the first condition given in the problem, we have
(x – 4) = 3 (y – 4)
x – 4 = 3y – 12
x – 3y = -8 … (i)
Now, the age of Maria after 6 years will be (x + 6) years
And the age of Maria’s daughter will be (y + 6) years
Then, according to the second condition given in the problem, we have
x + 6 = 2 (y + 6)
x + 6 = 2y + 12
x – 2y = 6 … (ii)
From equations (i) and (ii), we get
x – 2y = 6
x – 3y = -8
(-)—(+)—(+)
y = 14
On substituting the value of y in (i), we get
x – 3 x 14 = -8
x – 42 = -8
x = -8 + 42
x = 34
Therefore, the age of Marina is 34 years and the age of her daughter is 14 years.
Question 20. On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains Rs. 70. If he sells the tea set at 5% gain and lemon set at 10% gain, he gains Rs. 130. Find the cost price of the lemon set.
Answer :
Given,
Loss on tea set = 5% and
Gain on lemon set = 15%
Let’s assume the C.P of tea set = Rs x and
The C.P of lemon set = Rs y
Then according to the given conditions, we have
(15y/100)-(5x/100)=70
15y – 5x = 7000
-x + 3y = 1400 … (i)
And,
(5x/100)+(10y/100)=130
5x + 10y = 13000
x + 2y = 2600 … (ii)
Now, adding (i) and (ii) we get
-x + 3y = 1400
x + 2y = 2600
——————
5y = 4000
y = 4000/5
y = 800
Therefore, the Cost price of lemon set = Rs 800.
Question 21. A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of Rs, 1300. If he had interchanged the amounts, he would have received Rs. 40 more as yearly interest. How much did he invest at different rates?
Answer :
Let the amount invested at S.I be Rs x at rate = 12% p.a.
And another investment at S.I = Rs y at rate = 10% p.a.
Then according to the given conditions, we have
(12x/100)+(10y/100)=1300
12x + 10y = 130000
6x + 5y = 65000 … (i)
And,
(10x/100)+(12y/100)=1340
10x + 12y = 134000
5x + 6y = 67000 … (ii)
Multiplying (i) by 6 and (ii) by 5, we have
36x + 30y = 390000
25x + 30y = 335000
(-)—(-)—–(-)——–
11x = 55000
x = 55000/11
x = 5000
On substituting the value of x in (i), we get
6(5000) + 5y = 6500
30000 + 5y = 6500
5y = 65000 – 3000
y = 35000/5 = 7000
Therefore, the investment at 12% is Rs 5000 and the investment at 10% is Rs 7000.
Question 22. A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting Rs. 1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got Rs. 20 more. Find the cost price of the table and the list price of the chair.
Answer :
Given,
Profit on table = 8%
Discount on chair = 10%
Let C.P. of table = Rs x and C.P. of chair = Rs y
Then according to the given condition, we have
110x + 92y = 102800
55x + 46y = 51400 … (ii)
Now, multiplying (i) by 55 and (ii) by 6, we have
330x + 275y = 308000
330x + 276y = 308400
(-)—-(-)——-(-)——–
-y = -400
y = 400
On substituting the value of y in equation (i), we get
6x + 5(400) = 5600
6x = 5600 – 2000
x = 3600/6
x = 600
Therefore, the C.P. of table = Rs 600 and the C.P. of chair = Rs 400.
Question 23. A and B have some money with them. A said to B, “if you give me Rs. 100, my money will become 75% of the money left with you.” B said to A” instead if you give me Rs. 100, your money will become 40% of my money. How much money did A and B have originally?
Answer :
Let’s assume A has money = x
And B has money = y
Then according to the given conditions, we have
x – 100 = (y – 100) x (75/100)
x – 100 = (y – 100) x (3/4)
4x – 400 = 3y – 300
4x – 3y = 400 – 300
4x – 3y = 100 … (i)
Also,
x – 100 = (y + 100) (40/100)
x – 100 = (y + 100) (2/5)
5x – 500 = 2y + 200
5x – 2y = 200 + 500
5x – 2y = 700 … (ii)
Now, multiplying (i) by 2 and (ii) by 3, we have
8x – 6y = -1400
15x – 6y = 2100
(-)—(+)—(-)—-
-7x = -3500
x = -3500/ -7
x = 500
On substituting the value of x in (i), we get
4(500) – 3y = -700
2000 – 3y = -700
3y = 2000 + 700
y = 2700/3
y = 900
Therefore, A has money Rs 500 and B has money Rs 900.
Problem on Simultaneous Linear Equations Exe-6
ML Aggarwal Class 9 ICSE Maths Solutions
Page 140
Question 24. The students of a class are made to stand in (complete) rows. If one student is extra in a row, there would be 2 rows less, and if one student is less in a row, there would be 3 rows more. Find the number of students in the class.
Answer :
Let the number of students in one row be taken as x
And let the number of rows be taken as y
Then the total number of students = xy
Then according to the first condition given in the problem, we have
(x + 1) (y – 2) = xy
xy – 2x + y – 2 = xy
-2x + y = 2 … (i)
And, according to the second condition given in the problem, we have
(x – 1) (y + 3) = xy
xy + 3x – y – 3 = xy
3x – y = 3 … (ii)
Adding equations (i) and (ii), we get
-2x + y = 2
3x – y = 3
————–
x = 5
On substituting the value of x in equation (i), we have
-2(5) + y = 2
-10 + y = 2
y = 2 + 10
y = 12
Therefore, the number of students = xy = 5 x 12 = 60.
Question 25. A jeweler has bars of 18-carat gold and 12- carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams? (Pure gold is 24 carat)
Answer :
Let’s assume the quantity of 18 carat gold as x gm and 12 carat gold as y gm.
Then according to the first condition given in the problem, we have
x + y = 120 … (i)
Pure gold is 24 carat
So, the purity of 18 carat gold = (18/24) x 100%
= (¾) x 100
= 75%
Purity of 12 carat gold = (12/24) x 100%
= ½ x 100%
= 50%
And, the purity of 16 carat gold = (16/24) x 100%
= (2/3) x 100%
= 200/3%
Now, according to the second condition given in the problem, we have
75x + 50y = 200/3 x 120
75x + 50y = 200 x 40
75x + 50y = 8000
3x + 2y = 320 … (ii)
Performing (ii) – 2 x (i), we get
3x + 2y = 320
2x + 2y = 240
(-)—(-)—(-)—
x = 80
On substituting the value of x in equation (i), we get
80 + y = 120
y = 120 – 80
y = 40
Therefore, the jeweler requires 80 gm of 18 carat gold and 40 gm of 12 carat gold to obtain a bar of 16 carat gold weighing 120 gm.
Question 26. A and B together can do a piece of work in 15 days. If A’s one day work is 1½ times the one day’s work of B, find in how many days can each do the work.
Answer :
Let A’s one day work be x and B’s one day work be y.
Then according to the first condition given in the problem, we have
x = (3/2)y
2x = 3y
2x – 3y = 0 … (i)
Also given, in 15 days: A and B together can do a piece of work
So, according to this condition we have
x + y = 1/15
15 (x + y) = 1
15x + 15y = 1 … (ii)
Multiplying equation (i) by 5, we get
10x – 15y = 0
15x + 15y = 1
——————
25x = 1
x = 1/25
On substituting the value of x in equation (i), we get
2(1/25) – 3y = 0
2/25 = 3y
y = 2/75
Therefore, Man A will do the work in 1/x days = 1/(1/25) = 25 day and Man B will do the work in (1/y) days = 1/(2/75) = 75/2 = 37½ days.
Question 27. 2 men and 5 women can do a piece of work in 4 days, while one man and one woman can finish it in 12 days. How long would it take for 1 man to do the work?
Answer :
Let’s assume that 1 man takes x days to do work and y days for a women.
So, the amount of work done by 1 man in 1 day = 1/x
And the amount of work done by 1 woman in 1 day = 1/y
Now,
The amount of work done by 2 men in 1 day will be = 2/x
And the amount of work done by 5 women in 1 day = 5/y
Then according to the given conditions in the problem, we have
2/x + 5/y = ¼ … (i)
1/x + 1/y = 1/12 … (ii)
Multiplying equation (ii) by 5, we get
5/x + 5/y = 5/12
2/x + 5/y = ¼
(-)—(-)—(-)
3/x = 5/12 – ¼
3/x = (5 – 3)/12
3/x = 2/12 = 1/6
x = 18
Therefore, 1 man can do the work in 18 days.
Question 28. A train covered a certain distance at a uniform speed. If the train had been 30 km/hr faster, it would have taken 2 hours less than the scheduled time. If the train were slower by 15 km/hr, it would have taken 2 hours more than the scheduled time. Find the length of the journey.
Answer :
Let the actual speed of the train be x km/hr and the scheduled time be y hours.
Then, the distance of the journey = speed x time = xy
According to the given conditions in the problem, we have
(x + 30) (y – 2) = xy
xy – 2x + 30y – 60 = xy
-2x + 30y = 60 … (i)
And,
(x – 15) (y + 2) = xy
xy – 15y + 2x – 30 = xy
2x – 15y = 30 … (ii)
From equations (i) and (ii), we have
-2x + 30y = 60
2x – 15y = 30
——————
15y = 90
y = 90/15
y = 6
On substituting the value of y in (i), we have
-2x + 30(6) = 60
-2x + 180 = 60
-2x = 60 – 180 = -120
x = -120/-2
x = 60
Therefore, the distance of the journey = 60 x 6 = 360 km.
Question 29. A boat takes 2 hours to go 40 km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.
Answer :
Let’s assume the speed of the boat in still water be x km/hr
And the speed of the stream = y km/hr
So, the speed of the boat in downstream = (x + y) km/hr
The speed of the boat in upstream = (x – y) km/hr
We know,
Distance = Speed x time
Now, according to the given conditions in the problem, we have
40 = (x + y) × 2
x + y = 20 … (i)
And,
40 = (x – y) × 4
x – y = 10 … (ii)
Adding (i) and (ii), we have
x + y = 20
x – y = 10
————-
2x = 30
x = 15
On substituting the value of x in equation (i), we have
15 + y = 20
y = 20 – 15
y = 5
Therefore, speed of the boat in still water = 15 km/hr and speed of the stream = 5 km/hr.
Question 30. A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48 minutes longer to cover the same distance against the current. Find the speed of the boat in still water and the speed of the current.
Answer :
Let’s assume the speed of the boat in still to be x km/hr
And the speed of the current = y km/hr
Speed of the boat in the direction of current = (x + y) km/hr
Speed of the boat against the current = (x – y) km/hr
We know, distance = speed x time
Then according to the given conditions in the problem, we have
44 = (x + y) x 4
x + y = 44/4
x + y = 11 … (i)
And,
x – y = 5 … (ii)
Now, adding equations (i) and (ii) we have
x + y = 11
x – y = 5
————-
2x = 16
x = 16/2
x = 8
On substituting the value of x in (i), we get
8 + y = 11
y = 11 – 8
y = 3
Therefore, the speed of the boat in still water = 8 km/hr and speed of the current = 3 km/hr.
Question 31. An aeroplane flies 1680 km with a head wind in 3.5 hours. On the return trip with same wind blowing, the plane takes 3 hours. Find the plane’s air speed and the wind speed.
Answer :
Let’s assume the speed of the plane = x km/hr
And let the speed of wind = y km/hr
So, the speed of the aeroplane in the direction of wind = (x + y) km/hr
Speed of the aeroplane in the opposite direction of wind = (x – y) km/hr
We know, distance = speed x time
Then according to the given conditions, we have
1680 = (x – y) x 3.5
x – y = 1680/3.5
x – y = 480 … (i)
And,
1680 = (x + y) x 3
x + y = 1680/3
x + y = 560 … (ii)
From equation (i) and (ii), we get
x – y = 480
x + y = 560
—————
2x = 1040
x = 1040/2 = 520
Substituting the value of x in equation (i), we get
520 – y = 480
y = 520 – 480
y = 40
Therefore, the speed of aeroplane = 520 km/hr and the speed of wind = 40 km/hr.
Question 32. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When Bhawana takes food for 20 days, she has to pay Rs. 2600 as hostel charges; whereas when Divya takes food for 26 days, she pays Rs. 3020 as hostel charges. Find the fixed charges and the cost of food per day.
Answer :
Let the fixed charges = Rs x and
The charges per day = Rs y
Then according to the given conditions, we have
x + 20y = 2600 … (i)
x + 26y = 3020 … (ii)
(-)—(-)—(-)——–
On subtracting, we get
-6y = -420
y = 420/6
y = 70
Substituting the value of y in (i), we get
x + 20 x 70 = 2600
x = 2600 – 1400
x = 1200
Therefore, the fixed charges = Rs 1200 and daily charges = Rs 70.
— : End of ML Aggarwal Problems on Simultaneous Linear Equations Exe – 6 Class 9 ICSE Maths Solutions :–
Return to :- ML Aggarawal Maths Solutions for ICSE Class-9
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