RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables Exercise  3E Class 10 Maths
Chapter Name  RS Aggarwal Chapter 3 Linear Equations in Two Variables 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 3E Solutions
1. 5 chairs and 4 tables together cost ₹5600, while 4 chairs and 3 tables together cost ₹ 4340. Find the cost of each chair and that of each table.
Solution
Let the cost of a chair be ₹ x that of a table be ₹ y, then
5x + 4y = 5600 ...(i)
4x + 3y = 4340 ...(ii)
Multiplying (i) 3 and (ii) by 4, we get
15x – 16x = 16800 – 17360
⇒  x =  560
⇒ x = 560
Substituting x = 560 in (i), we have
5×560 + 4y = 5600
⇒ 4y = 5600  2800
⇒ y = 2800/4 = 700
Hence, the cost of a chair and that a table are respectively ₹560 and ₹ 700.
2. 23 spoons and 17 forks cost Rs 1770, while 17 spoons and 23 forks cost Rs 1830. Find the cost of each spoon and that of a fork.
Solution
Let the cost of a spoon be Rs x and that of a fork be Rs y. Then,
23x + 17y = 1770 ...(i)
17x + 23y = 1830 ...(ii)
Adding (i) and (ii), we get
40x + 40y = 3600
⇒ x + y = 90 ...(iii)
Now, subtracting (ii) from (i), we get
6x – 6y =  60
⇒ x – y =  10 ...(iv)
Adding (iii) and (iv), we get
2x = 80
⇒ x = 40
Substituting x = 40 in (iii), we get
40 + y = 90
⇒ y = 50
Hence, the cost of a spoon that of a fork is Rs 40 and Rs 50 respectively.
3. A lady has only 50paisa coins in her purse. If she has 50 coins in all totaling Rs 19.50, how many coins of each kind does she have?
Solution
Let x and y be the number of 50paisa and 25paisa coins respectively. Then
x + y = 50 ...(i)
0.5x + 0.25y = 19.50 ...(ii)
Multiplying (ii) by 2 and subtracting it from (i), we get
0.5y = 50 – 39
⇒ y = 11/0.5 = 22
Subtracting y = 22 in (i), we get
x + 22 = 50
⇒ x = 50 – 22 = 28
Hence, the number of 25paisa and 50paisa coins is 22 and 28 respectively.
4. The sum of two numbers is 137 and their differences are 43. Find the numbers
Solution
Let the larger number be x and the smaller number be y.
Then we have:
x + y = 137 ...(i)
x – y = 43 ...(ii)
On adding (i) and (ii), we get
2x = 180
⇒ x = 90
On substituting x = 90 in (i), we get
90 + y = 137
⇒ y = (137 – 90) = 47
Hence, the required numbers are 90 and 47.
5. Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.
Solution
Let the first number be x and the second number be y.
Then, we have:
2x + 3y = 92 ...(i)
4x – 7y = 2 ...(ii)
On multiplying (i) by 7 and (ii) by 3, we get
14x + 21y = 644 ...(iii)
12x – 21y = 6 ...(iv)
On adding (iii) and (iv), we get
26x = 650
⇒ x = 25
On substituting x = 25 in (i), we get
2×25 + 3y = 92
⇒ 50 + 3y = 92
⇒ 3y = (92 – 50) = 42
⇒ y = 14
Hence, the first number is 25 and the second number is 14.
6. Find the numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.
Solution
Let the first number be x and the second number be y.
Then, we have:
3x + y = 142 ...(i)
4x – y = 138 ...(ii)
On adding (i) and (ii), we get
7x = 280
⇒ x = 40
On substituting x = 40 in (i), we get:
3×40 + y = 142
⇒ y = (142 – 120) = 22
⇒ y = 22
Hence, the first number is 40 and the second number is 22.
7. If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.
Solution
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
Hence, the greater number is 37 and the smaller number is 29.
8. If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.
Solution
We know:
Dividend = Divisor × Quotient + Remainder
Let the larger number be x and the smaller be y.
Then, we have:
3x = y×4 + 8 or 3x – 4y = 8 ...(i)
5y = x×3 + 5 or 3x + 5y = 5 ...(ii)
On adding (i) and (ii), we get:
y(8 + 5) = 13
On substituting y = 13 in (i) we get
3x – 4×13 = 8
⇒ 3x = (8 + 52) = 60
⇒ x = 20
Hence the larger number is 20 and the smaller number is 13.
9. If 2 is added to each of two given numbers, their ration becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.
Solution
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 (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.
10. The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.
Solution
Let the larger numbers be x and the smaller number be y.
Then, we have:
x – y = 14 or 14 + y ...(i)
x^{2} – y^{2} = 448 ...(ii)
On substituting x = 14 + y in (ii) we get
(14 + y)^{2} – y^{2} = 448
⇒ 196 + y^{2} + 28y – y^{2} = 448
⇒ 196 + 28y = 448
⇒ 28y = (448 – 196) = 252
⇒ y = 252/28 = 9
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.
11. The sum of the digits of a twodigit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
x + y = 12 ...(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) = 18
⇒ 10y + x – 10x – y = 18
⇒ 9y – 9x = 18
⇒ y – x = 2 ...(ii)
On adding (i) and (ii), we get:
2y = 14
⇒ y = 7
On substituting y = 7 in (i) we get
x + 7 = 12
⇒ x = (12 – 7) = 5
Number = (10x + y) = 10×5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
12. A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + 7y = 7x + 7y or 3x – 6y = 0 ...(i)
Number obtained on reversing its digits = (10y + x)
(10 + x) – 27 = (10y + x)
⇒ 10x – x + y – 10y = 27
⇒ 9x – 9y = 27
⇒ 9(x – y) = 27
⇒ x – y = 3 ...(ii)
On multiplying (ii) by 6, we get:
6x – 6y = 18 ...(iii)
On subtracting (i) from (ii), we get:
3x = 18
⇒ x = 6
On substituting x = 6 in (i) we get
3×6 – 6y = 0
⇒ 18 – 6y = 0
⇒ 6y = 18
⇒ y = 3
Number = (10x + y) = 10×6 + 3 = 60 + 3 = 63
Hence, the required number is 63.
13. The sum of the digits of a twodigit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
x + y = 15 ...(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) – (10x + y) = 9
⇒ 10y + x – 10x – y = 9
⇒ 9y – 9x = 9
⇒ y – x = 1 ...(ii)
On adding (i) and (ii), we get:
2y = 16
⇒ y = 8
On substituting y = 8 in (i) we get
x + 8 = 15
⇒ x = (15 – 8) = 7
Number = (10x + y) = 10×7 + 8 = 70 + 8 = 78
Hence, the required number is 78.
14. A two – digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = 4(x + y) + 3
⇒ 10x + y = 4x + 4y + 3
⇒ 6x – 3y = 3
⇒ 2x – y = 1 ...(i)
Again, we have:
10x + y + 18 = 10y + x
⇒ 9x – 9y =  18
⇒ x – y =  2 ...(ii)
On subtracting (ii) from (i), we get
x = 3
On substituting x = 3 in (i) we get
2×3 – y = 1
⇒ y = 6 – 1= 5
Required number = (10x + y) = 10×3 + 5 = 30 + 5 = 35
Hence, the required number is 35.
15. A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number:
Solution
We know:
Dividend = Divisor × Quotient + Remainder
Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = (x + y)×6 + 0
⇒ 10x – 6x + y – 6y = 0
⇒ 4x – 5y = 0 ...(i)
Number obtained on reversing its digits = (10y + x)
∴ 10x + y – 9 = 10y + x
⇒ 9x – 9y = 9
⇒ x – y = 1 ...(ii)
On multiplying (ii) by 5, we get:
5x – 5y = 5 ...(iii)
On subtracting (i) from (iii), we get:
x = 5
On substituting (i) from (iii), we get :
5x – 5y = 5 ...(iii)
On subtracting (i) from (iii), we get:
x = 5
On subtracting x = 5 in (i) we get
4×5 – 5y = 0
⇒ 20 – 5y = 0
⇒ y = 4
∴ The number = (10x + y) = 10×5 + 4 = 50 + 4 = 54
Hence, the required number is 54.
16. A twodigit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Then we have:
xy = 35 ...(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x – 9y =  18
⇒ 9(y – x) = 18
⇒ y – x = 2 ...(ii)
We know:
(y + x)^{2} – (y – x)^{2} = 4xy
⇒ y + x = 12 ...(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
⇒ y = 7
On substituting y = 7 in (ii) we get
7 – x = 2
⇒ x = (7 – 2) = 5
∴ The number = (10x + y) = 10×5 + 7 = 50 + 7 = 57
Hence, the required number is 57.
17. A two – digits number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution
Let the tens and the units digits of the required number be x and y, respectively.
Then, we have:
xy = 18 ...(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) – 63 = 10y + x
⇒ 9x – 9y = 63
⇒ 9(x – y) = 63
⇒ x – y = 7 ...(ii)
We know:
(x + y)^{2} – (x – y)^{2} = 4xy
= ± 11
⇒ x + y = 11 ...(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
⇒ x = 9
On substituting x = 9 in (ii) we get
9 – y = 7
⇒ y = (9 – 7) = 2
∴ Number = (10x + y) = 10×9 + 2 = 90 + 2 = 92
Hence, the required number is 92.
18. The sum of a twodigit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number,
Solution
Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per question
(10y + x) + (10x + y) = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 ...(i)
Since the digits differ by 3, so
x – y = 3 ...(ii)
Adding (i) and (ii), we get
2x = 14
⇒ x = 7
Putting x = 7 in (i), we get
7 + y = 11
⇒ y = 4
Changing the role of x and y, x = 4 and y = 7
Hence, the twodigit number is 74 or 47.
19. The sum of the numerator and denominator of a fraction is 8. if 3 is added to both of the numerator and the denominator, the fraction becomes 3/4. Find the fraction.
Solution
Let the required fraction be x/y.
Then we have:
x + y = 8 ...(i)
And, (x + 3)/(y + 3) = 3/4
⇒ 4(x + 3) = 3(y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x – 3y =  3 ...(ii)
On multiplying (i) by 3, we get:
3x + 3y = 24
On adding (ii) and (iii), we get :
7x = 21
⇒ x = 3
On substituting x = 3 in (i), we get:
3 + y = 8
⇒ y = (8 – 3) = 5
∴ x = 3 and y = 5
Hence, the required fraction is 3/5.
20. If 2 is added to the numerator of a fraction, it reduces to (1/2) and if 1 is subtracted from the denominator, it reduces to (1/3). Find the fraction.
Solution
Let the required fraction be x/y.
Then, we have:
(x + 2)/y = 1/2
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x – y =  4 ...(i)
Again, x/(y – 1) = 1/3
⇒ 3x = 1(y – 1)
⇒ 3x – y =  1 ...(ii)
On substituting (i) from (ii), we get:
x = (1 + 4) = 3
On substituting x = 3 in (i), we get:
2×3 – y = 4
⇒ 6 – y =  4
⇒ y = (6 + 4) = 10
∴ x = 3 and y = 10
Hence, required fraction is 3/10.
21. The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes 3/4. Find the fraction.
Solution
Let the required fraction be x/y.
Then, we have:
y = x + 11
⇒ y – x = 11 ...(i)
Again, (x + 8)/(y + 8) = 3/4
⇒ 4(x + 8) = 3(y + 8)
⇒ 4x + 32 = 3y + 24
⇒ 4x – 3y =  8 ...(ii)
On multiplying (i) by 4, we get:
4y – 4x = 44
On adding (ii) and (iii), we get:
4y – 4x = 44
On adding (ii) and (iii), we get:
y = ( 8 + 44) = 36
On substituting y = 36 in (i), we get:
36 – x = 11
⇒ x = (36 – 11) = 25
∴ x = 25 and y = 36
Hence, the required fraction is 25/36.
22. Find a fraction which becomes (1/2) when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes (1/3) when 7 is subtracted from the numerator and 2 is subtracted from the denominator.
Solution
Let the required fraction be x/y.
Then, we have:
(x – 1)/(y + 2) = 1/2
⇒ 2(x – 1) = 1(y + 2)
⇒ 2x – 2 = (y + 2)
⇒ 2x – y = 4 ...(i)
Again, (x – 7)/(y – 2) = 1/3
⇒ 3(x – 7) = 1(y – 2)
⇒ 3x – 21 = y – 2
⇒ 3x – y = 19 ...(ii)
On subtracting (i) from (ii), we get:
x = (19 – 4) = 15
On substituting x = 15 in (i), we get:
2×15 – y = 4
⇒ 30 – y = 4
⇒ y = 26
∴ x = 15 and y = 26
Hence, the required fraction is 15/26.
23. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3. They are in the ratio of 2: 3. Determine the fraction.
Solution
Let the required fraction be x/y.
As per the question
x + y = 4 + 2x
⇒ y – x = 4 ...(i)
After changing the numerator and denominator
New numerator = x + 3
New denominator = y + 3
Therefore,
(x + 3)/(y + 3) = 2/3
⇒ 3(x + 3) = 2(y + 3)
⇒ 3x + 9 = 2y + 6
⇒ 2y – 3x = 3 ...(ii)
Multiplying (i) by 3 and subtracting (ii), we get:
3y – 2y = 12 – 3
⇒ y = 9
Now, putting y = 9 in (i), we get:
9 – x = 4
⇒ x = 9 – 4 = 5
Hence, the required fraction is 5/9.
24. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Solution
Let the larger number be x and the smaller number y.
Then, we have:
x + y = 16 ...(i)
And, 1/x + 1/y = 1/3 ...(ii)
⇒ 3(x +y) = xy
⇒ 3 × 16 = xy [Since from (i), we have: x + y = 16]
∴ xy = 48 ...(iii)
We know:
(x – y)^{2} = (x + y)^{2} – 4xy
(x – y)^{2} = (16)^{2} – 4 × 48 = 256 – 192 = 64
= ± 8
Since x is larger and y is smaller, we have:
x – y = 8 ...(iv)
On adding (i) and (iv), we get:
12 + y = 16 ⇒ y = (16 – 12) = 4
Hence, the required numbers are 12 and 4.
25. There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in each room because the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.
Solution
Let the number of students in classroom A be x.
Let the number of students are transferred from A to B, then we have:
x – 10 = y + 10
⇒ x – y = 20 ...(i)
If 20 students are transferred from B to A, then we have:
2(y – 20) = x + 20
⇒ 2y – 40 = x + 20
⇒  x + 2y = 60 ...(ii)
On adding (i) and (ii), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
x – 80 = 20
⇒ x = (20 + 80) = 100
Hence, the number of students in classroom A is 100 and the number of students in classroom B is 80.
26. Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels 80 km, he pays Rs 1330, and for travelling 90 km, he pays Rs 1490. Find the fixed charges per day and the rate per km.
Solution
Let fixed charges be Rs x. and rate per km be Rs y.
Then as per the question
x + 80y = 1330 ...(i)
x + 90y = 1490 ...(ii)
Subtracting (i) from (ii), we get
10y = 160
⇒ y = 160/10 = 16
Now, putting y = 16, we have
x + 80×16 = 1330
⇒ x = 1330 – 1280 = 50
Hence, the fixed charges be Rs 50 and the rate per km is Rs 16.
27. A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student takes food for 25 days, he has to pay 4550, as hostel charges whereas a student B, wo takes food for 30 days, pays Rs 5200 as hostel charges. Find the fixed charges. and the cost of the food per day.
Solution
Let the fixed charges be Rs x and the cost of food per day be Rs y.
Then as per the question
x + 25y = 4500 ...(i)
x + 30y = 5200 ...(ii)
Subtracting (i) from (ii), we get
5y = 700
⇒ y = 700/5 = 140
Now, putting y = 140, we have
x + 25×140 = 4500
⇒ x = 4500 – 3500 = 1000
Hence, the fixed charges be Rs 1000 and the cost of the food per day is Rs 140.
28. A man invested an amount at 10% per annum simple interest and another amount at 10% per annum simple interest. He received an annual interest of Rs 1350. But, if he had interchanged the amounts invested, he would have received Rs 45 less. What amounts did he invest at different rates?
Solution
Let the amount invested at 10% and 8% be Rs x and Rs y respectively.
Then as per the question
(x×10×1)/100 = (y×8×1)/100 = 1350
10x + 8y = 135000 ...(i)
After the amounts interchanged but the rate being the same, we have
(x×8×1)/100 = (y×10×1)/100 = 1350 – 45
8x + 10y = 130500 ...(ii)
Adding (i) and (ii) and dividing by 9, we get
2x + 2y = 29500
Subtracting (ii) from (i), we get
2x – 2y = 29500
Subtracting (ii) from (i), we get
2x – 2y = 4500
now, adding (iii) and (iv), we have
4x = 34000
x = 34000/4 = 8500
Putting x = 8500 in (iii), we get
2 × 8500 + 2y = 29500
2y = 29500 – 17000 = 12500
y = 12500/2 = 6250
Hence, the amounts invested are Rs 8,500 at 10% and Rs 6,250 at 8%.
29. The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly expenditures are in the ratio of 7 : 5. If each saves Rs 9000 per month, find the monthly income of each.
Solution
Let the monthly income of A and B are Rs x and Rs y respectively.
then as per the question
x/y = 5/4
⇒ y = 4x/5
Since each save Rs 9,000, so
Expenditure of A = Rs (x – 9000)
Expenditure of B = Rs (y – 9000)
The ratio of expenditures of A and B are in the ratio 7 : 5.
∴ (x – 9000)/(y – 9000) = 7/5
⇒ 7y – 63000 = 5x – 45000
⇒ 7y – 5x = 18000
From (i), substitute y = 4x/5 in (ii) to get
7× 4x/5 – 5x = 18000
⇒ 28x – 25x = 90000
⇒ 3x = 90000
⇒ x = 30000
Now, putting x = 30000, we get
y = (4 × 30000)/5 = 4 × 6000 = 24000
Hence, the monthly incomes of A and B are Rs 30000 and Rs 24,000.
30. A man sold a chair and a table together for Rs 1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs 1535, he would made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
Solution
Let the cost price of the chair and table be Rs. x and Rs. y respectively.
Then, as per the question
Selling price of chair + Selling price o table = 1520
(100 + 25)/100 × x + (100 + 10)/100 + 1520
⇒ 125/100.x + 110/100.y = 1520
⇒ 25x + 22y – 30400 = 0 ...(i)
When the profit on chair and table are 10% and 25% respectively, then
(100 + 10)/100 × x + (100 + 25)/100 × y = 1535
⇒ 110/100.x + 125/100.y = 1535
⇒ 22x + 25y – 30700 = 0 ...(ii)
Solving (i) and (ii) by cross multiplication, we get
x/{(22)(30700)(25)(30400)} = y/{(30400)(22)(30700)(25)} = 1/{(25)(25) – (22) – (22)}
⇒ x/(7600 – 6754) = y/(7675 – 6688) = 100/(3 × 47)
⇒ x/846 = y/987 = 100/(3×47)
⇒ x = (100×846)/(3×47, y = (100×987)/(3×47)
⇒ x = 600, y = 700
Hence, the cost of chair and table are Rs 600 and Rs 700 respectively.
31. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.
Solution
Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M.
Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM – BM) = AB
⇒ (7x – 7y) = 70
⇒ 7(x – y) = 70
⇒ (x – y) = 10 ...(i)
Case II: When the two cars move in opposite directions
In this case, let the two cars meet at point N.
Distance covered by car X in 1 hour = x km
Distance covered by car Y in 1 hour = y km
∴ AN = x km and BN = y km
⇒ AN + BN = AB
⇒ x + y = 70 ...(ii)
On adding (i) and (ii), we get:
2x = 80
⇒ x = 40
On substituting x = 40 in (i), we get:
40 – y = 10
⇒ y = (40 – 10) = 30
Hence, the speed of car X is 40 km/h and the speed of car Y is 30 km/h.
32. A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3 hours less than the scheduled time. Find the length of the journey.
Solution
Let the original speed be x kmph and let the time taken to complete the journey by y hours.
∴ Length of the whole journey = (xy) km
Case I:
When the speed is (x + 5) kmph and the time taken is (y – 3) hrs:
Total journey = (x + 5)(y – 3) km
⇒ (x + 5)(y – 3) = xy
⇒ xy + 5y – 3x – 15 = xy
⇒ 5y – 3x = 15 ...(i)
Case II:
When the speed is (x – 4) kmph and the time taken is (y + 3) hrs:
Total journey = (x – 4)(y + 3) km
⇒ (x – 4)(y + 3) = xy
⇒ xy – 4y + 3x – 12 = xy
⇒ 3x – 4y = 12 ...(ii)
On adding (i) and (ii), we get:
y = 27
On substituting y = 27 in (i), we get:
5×27 – 3x = 15
⇒ 135 – 3x = 15
⇒ 3x = 120
⇒ x = 40
∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km
33. Abdul travelled 300 km by train and 200 km by taxi taking 5 hours and 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of taxi.
SolutionLet the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question
3/x + 2/y = 11/200 …(i)
When the speeds of the train and taxi are 260 km and 240 km respectively, then
260/x + 240/y = 11/2 + 6/10
⇒ 13/x + 12/y = 28/100 ...(ii)
Multiplying (i) by 6 and subtracting (ii) from it, we get
18/x – 13/x = 66/200 – 28/100
⇒ 5/x = 10/200
⇒ x = 100
Putting x = 100 in (i), we have
3/100 + 2/y = 11/200
⇒ 2/y = 11/200 – 3/100 = 1/40
⇒ y = 80
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.
34. Places A and B are 160 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.
Solution
Let the speed of the car A and B be x km/h and y km/h respectively. Let x > y.
Case 1:
When they travel in the same direction
From the figure
AC – BC = 160
⇒ x×8 – y×8 = 160
⇒ x – y = 20
Case 2: When they travel in opposite direction
From the figure
AC + BC = 160
⇒ x×2 + y×2 = 160
⇒ x + y = 80
adding (i) and (ii), we get
2x = 100
⇒ x = 50 km/h
Putting x = 50 in (ii), we have
50 + y = 80
⇒ y = 80 – 50 = 30 km/h
Hence, the speeds of the cars are 50 km/h and 30 km/h.
35. A sailor goes 8 km downstream in 420 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.
Solution
Let the speed of the sailor in still water be x km/h and that of the current y km/h.
Speed downstream = (x + y) km/h
Speed upstream = (x – y) km/h
As per question
(x + y)× 40/60 = 8
⇒ x + y = 12 ...(i)
When the sailor goes upstream, then
(x – y)× 1 = 8
x – y = 8 ...(ii)
Adding (i) and (ii), we get
2x = 20
⇒ x = 10
Putting x = 10 in (i), we have
10 + y = 12
⇒ y = 2
Hence the speeds of the sailor in still water and the current are10 km/h and 2 km/h respectively.
36. A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Solution
Let the speed of the boat in still water be x km/h and the speed of the stream by y km/h.
Then we have
Speed upstream = (x – y) km/hr
Speed downstream = (x + y) km/hr
time taken to cover 40 km downstream = 40/(x + y) hrs
Total time taken = 8 hrs
∴ 12/(x – y) + 40/(x + y) = 8 ...(i)
Again, we have:
Time taken to cover 16 km upstream = 16/(x – y) hrs
Time taken to cover 32 km downstream = 32/(x + y) hrs
Total time taken = 8 hrs
∴ 16/(x – y) + 32/(x + y) = 8 ...(ii)
Putting 1/(x – y) = u and 1/(x + y) = v in (i) and (ii), we get:
12u + 40v = 8
3u + 10v = 2 ...(a)
And, 16u + 32v = 8
⇒ 2u + 4v = 1 ...(b)
On multiplying (a) by 4 and (b) by 10, we get:
12u + 40v = 8 ...(iii)
And, 20u + 40v = 10 ...(iv)
On subtracting (iii) from (iv), we get:
8u = 2
⇒ u = 2/8 = 1/4
On substituting u = 1/4 in (iii), we get:
40v = 5
⇒ v = 5/40 = 1/8
Now, we have:
u = 1/4
⇒ 1/(x – y) = 1/4
⇒ x – y = 4 ...(v)
v = 1/8
⇒ 1/(x + y) = 1/8
⇒ x + y = 8 ...(vi)
On adding (v) and (vi), we get:
2x = 12
⇒ x = 6
On substituting x = 6 in (v), we get:
6 – y = 4
y = (6 – 4) = 2
∴ Speed of the boat in still water = 6 km/h
And, speed of the stream = 2 km/h
37. 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.
Solution
Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man’s one day’s work = 1/x
And, one boy’s one day’s work = 1/y
2 men and 5 boys can finish the work in 4 days.
∴ (2 men’s one day’s work) + (5 boy’s one day’s work) = ¼
⇒ 2/x + 5/y = 1/4
⇒ 2u + 5v = 1/4 ...(i)
Here, 1/x = u and 1/y = v
Again, 3 men and 6 boys can finish the work in 3 days.
∴ (3 men’s one day’ work) + (6 boys one day’s work) = 1/3
⇒ 3/x + 6/y = 1/3
⇒ 3u + 6v = 1/3 ...(ii)
Here, 1/x = u and 1/y = v
On multiplying (iii) from (iv), we get:
3u = (5/3 – 6/4) = 2/12 = 1/6
⇒ u = 1/(6×3) = 1/18
⇒ 1/x = 1/18
⇒ x = 18
On substituting u = 1/18 in (i), we get:
2 × 1/18 + 5v = 1/4
⇒ 5v = (1/4 – 1/9) = 5/36
⇒ v = (5/36 × 1/5) = 1/36
⇒ 1/y = 1/36
⇒ y = 36
Hence, one man alone can finish the work is 18 days and one boy alone can finish the work in 36 days.
38. The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 meters and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.
Solution
Let the length of the room be x metres and he breadth of the room be y metres.
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
⇒ x – 2y + 3y – 6 = xy
⇒ 3y – 2x = 6
On multiplying (i) by 2, 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 metres and its breadth is 12 metres.
39. The area of a rectangle gets reduced by 8m2, when its length is reduced by 5m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2m, the area is increased by 74m^{2}. Find the length and the breadth of the rectangle.
Solution
Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq. m
Case I:
when the length is reduced by 5 m and the breadth is increased by 3m:
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)
Case II:
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 10 m.
40. The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and the breadth is decreased by 4m. If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres, Find the dimension of the rectangle.
Solution
Let the length and the breadth of the rectangle be x m and y m, respectively.
Case I: When length is increased by 3 m and the breadth is decreased by 4m:
xy – (x + 3)(y – 4) = 67
⇒ xy – xy + 4x – 3y + 12 = 67
⇒ 4x – 3y = 55 ...(i)
Case 2: When length is reduced by 1 m and breadth is increased by 4 m:
(x – 1)(y + 4) – xy = 89
⇒ xy + 4x – y – 4 – xy = 89
⇒ 4x – y = 93 ...(ii)
Subtracting (i) and (ii), we get:
2y = 38 ⇒ y = 19
On substituting y = 19 in (ii), we have
4x – 19 = 93
⇒ x = 28
Hence, the length = 28 m and breadth = 19 m.
41. A railway half ticket costs half the full fare and the reservation charge is the some on half tickets as on full ticket. One reserved first class ticket from Mumbai to Delhi costs ₹ 4150 while one full and one half reserved first class cost ₹6255. What is the basic first class full fare and what is the reservation charge?
Solution
Let the basic first class full fare be Rs x and the reservation charge be Rs y.
Case 1:
One reservation first class full ticket cost Rs 4, 150
x + y = 4150 ...(i)
Case 2:
One full and one and half reserved first class tickets cost Rs 6,255
(x + y) + (1/2.x + y) = 6255
⇒ 3x + 4y = 12510 ...(ii)
Substituting y = 4150 – x from (i) in (ii), we get
3x + 4(4150 – x) = 12510
⇒ 3x – 4x + 16600 = 12510
⇒ x = 16600 – 12510 = 4090
Now, putting x = 4090 in (i), we have
4090 + y = 4150
⇒ y = 4150 – 4090 = 60
Hence, cost of basic first full charge full fare = Rs 4,090 and reservation charge = Rs 60.
42. Five years hence, a man’s age will be three times the sum of the ages of his son. Five years ago, the man was seven times as old as his son. Find their present ages.
Solution
Let the present age of the man be x years and that of his son be years.
After 5 years man’s age = x + 5
After 5 years ago son’s age = y + 5
As per the question
x + 5 = 3(y + 5)
⇒ x – 3y = 10 ...(i)
5 years ago man’s age = x – 5
5 years ago son’s age = y – 5
As per the question
x – 5 = 7(y – 5)
⇒ x – 7y =  30 ...(ii)
Subtracting (ii) from (i), we have
4y = 40
⇒ y = 10
Putting y = 10 in (i), we get
x – 3×10 = 10
⇒ x = 10 + 30 = 40
Hence, man’s present age = 40 years and son’s present age = 10 years.
43. The present age of a man is 2 years more than five times the age of his son. Two years hence, the man’s age will be 8 years more than three times the age of his son. Find their present ages.
Solution
Let the man’s present age be x years.
Let his son’s present age be y years.
According to the question, we have:
Two years ago:
Age of the man = Five times the age of the son
⇒ (x – 2) = 5(y – 2)
⇒ x – 2 = 5y – 10
⇒ x – 5y =  8 ...(i)
Two years later:
Age of the man = Three times the age of the son + 8
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x – 3y = 12 ...(ii)
Subtracting (i) from (ii), we get:
2y = 20
⇒ y = 10
On substituting y = 10 in (i), we get:
x – 5×10 =  8
⇒ x – 50 =  8
⇒ x = (8 + 50) = 42
Hence, the present age of the man is 42 years and the present age of the son is 10 years.
44. If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95 years. Find the age of the mother and that of the son.
Solution
Let the mother’s present age be x years.
Let her son’s present age be y years.
Then, we have:
x + 2y = 70 ...(i)
And, 2x + y = 95 ...(ii)
On multiplying (ii) by 2, we get:
4x + 2y = 190 ...(iii)
On subtracting (i) from (iii), we get:
3x = 120
⇒ x = 40
On substituting x = 40 in (i), we get:
40 + 2y = 70
⇒ 2y = (70 – 40) = 30
⇒ y = 15
Hence, the mother’s present age is 40 years and her son’s present age is 15 years.
45. The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. find their present ages.
Solution
Let the woman’s present age be x years
Let her daughter’s present age be y years.
Then, we have:
x = 3y + 3
⇒ x – 3y = 3 ...(i)
After three years, we have:
(x + 3) = 2(y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x – 2y = 13 ...(ii)
Subtracting (ii) from (i), we get:
y = (3 – 13) =  10
⇒ y = 10
On substituting y = 10 in (i), we get:
x – 3×10 = 3
⇒ x – 30 = 3
⇒ x = (3 + 30) = 33
Hence, the woman’s present age is 33 years and her daughter’s present age is 10 years.
46. On selling a teaset at 5% loss and a lemonset at 15% gain, a shopkeeper gains Rs. 7. However, if he sells the teaset at 5% gain and the lemonset at 10% gain, he gains Rs 14. Find the price of the teaset and that of the lemonset paid by shopkeeper.
Solution
Let the actual price of the tea and lemon set be Rs x and rs y respectively.
When gain is Rs 7, then
y/100 × 15 – x/100 × 5 = 7
⇒ 3y – x = 140 ...(i)
when gain is Rs 14, then
y/100 × 5 + x/100 × 10 = 14
⇒ y + 2x = 280 ...(ii)
Multiplying (i) by 2 and adding with (ii), we have
7y = 280 + 280
⇒ y = 560/7 = 80
Putting y = 80 in (ii), we get
80 + 2x = 280
⇒ x = 200/2 = 100
Hence, actual price of the tea set and lemon set are Rs 100 and Rs 80 respectively.
47. A lending library has fixed charge for the first three days and an additional charge for each day thereafter. Mona paid ₹27 for a book kept for 7 days, while Tanvy paid ₹ 21 for the book she kept for 5 days find the fixed charge and the charge for each extra day.
Solution
Let the fixed charge be Rs x and the charge for each extra day be Rs y.
In case of Mona, as per the question
x + 4y = 27 ...(i)
In case of Tanvy, as per the question
x + 2y = 21 ...(ii)
Subtracting (ii) from (i), we get
2y = 6
⇒ y = 3
Now, putting y = 3 in (ii), we have
x + 2×3 = 21
⇒ x = 21 – 6 = 15
Hence, the fixed charge be Rs 15 and the charge for each extra Day is Rs 3.
48.A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution?
Solution
Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question
50% of x + 25% of y = 40% of 10
⇒ 0.50x + 0.25y = 4
⇒ 2x + y = 16 ...(i)
Since, the total volume is 10 litres, so
x + y = 10
Subtracting (ii) from (i), we get
x = 6
Now, putting x = 6 in (ii), we have
6 + y = 10
⇒ y = 4
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres
49. A jeweller has bars of 18carat gold and 12carat gold. How much of each must be milted together to obtain a bar of 16carat gold, weighing 120 gm? (Given: Pure gold is 24carat).
Solution
Let x g and y g be the weight of 18carat and 12carat gold respectively.
As per the given condition
18x/24 + 12y/24 = (120 × 16)/24
⇒ 3x + 2y = 320 ...(i)
And
x + y = 120 ...(ii)
Multiplying (ii) by 2 and subtracting from (i), we get
x = 320 – 240 = 80
Now, putting x = 80 in (ii), we have
80 + y = 120
⇒ y = 40
Hence, the required weight of 18carat and 12carat gold bars are 80 g and 40 g respectively.
50. 90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acid to be mixed to form the mixture.
Solution
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 ...(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 + 0.97y = 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.
51. The larger of the two supplementary angles exceeds the smaller by 180°. Find them.
Solution
Let x and y be the supplementary angles, where x > y.
As per the given condition
x + y = 180° ...(i)
And
x – y = 18° ...(ii)
Adding (i) and (ii), we get
2x = 198°
⇒ x = 90°
Now, substituting x = 90° in (ii), we have
99°  y = 18°
⇒ x = 99°  18° = 81°
Hence, the required angles are 99° and 81°.
52. In a ∆ABC, ∠A = x°, ∠B = (3x – 2)°, ∠C = y° and ∠C  ∠B = 9°. Find the three angles.
Solution
∵ ∠C  ∠B = 9°
∴ y°  (3x – 2)° = 9°
⇒ y°  3x° + 2° = 9°
⇒ y°  3x° = 7°
The sum of all the angles of a triangle is 180°, therefore
∠A + ∠B + ∠C = 180°
⇒ x° + (3x – 2)° + y° = 180°
⇒ 4x° + y° = 182°
Subtracting (i) from (ii), we have
7x° = 182°  7° = 175°
⇒ x° = 25°
Now, substituting x° = 25° in (i), we have
y° = 3x° + 7° = 3×25° + 7° = 82°
Thus
∠A = x° = 25°
∠B = (3x – 2)° = 75°  2° = 73°
∠C = y° = 82°
Hence, the angles are 25°, 73° and 82°.
Solution
The opposite angles of cyclic quadrilateral are supplementary, so
∠A + ∠C = 180°
⇒ (2x + 4)° + (2y + 10) ° = 180°
⇒ x + y = 83°
And
∠B + ∠D = 180°
⇒ (y + 3)° + (4x – 5)° = 180°
⇒ 4x + y = 182°
Subtracting (i) from (ii), we have
3x = 99
⇒ x = 33°
Now, substituting x = 33° in (i), we have
33° + y = 83°
⇒ y = 83°  33° = 50°
Therefore,
∠A = (2x + 4)° = (2×33 + 4)° = 70°
∠B = (y + 3)° = (50 + 3)° = 53°
∠C = (2y + 10)° = (2×50 + 10)° = 110°
∠D = (4x – 5)° = (4×33 – 5)° = 132°  5° = 127°
Hence, ∠A = 70°, ∠B = 53°, ∠C = 110° and ∠D = 127°