Problem 1 :
One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.
Solution :
Let x and y be the two numbers such that
x > y
Given : One number is greater than thrice the other number by 2.
Then, we have
x = 3y + 2 -----(1)
Given : 4 times the smaller number exceeds the greater by 5.
Then, we have
4y - x = 5 -----(2)
From (1), substitute (3y + 2) in (2).
(2)-----> 4y - (3y + 2) = 5
Simplify.
4y - 3y - 2 = 5
y - 2 = 5
Add 2 to each side.
y = 7
Substitute 7 for y in (1).
x = 3(7) + 2
x = 21 + 2
x = 23
So, the two numbers are 23 and 7.
Problem 2 :
A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
Solution :
Let xy be the two digit number.
Given : The two digit number is seven times the sum of its digits.
xy = 7(x + y) -----(1)
In the two digit number xy, x is in tens place and y is in ones place.
Then, we have
xy = 10 ⋅ x + 1 ⋅ y
xy = 10x + y
Substitute (10x + y) for xy in (1).
(1)-----> 10x + y = 7(x + y)
10x + y = 7x + 7y
Simplify.
3x - 6y = 0
Divide each side by 3.
x - 2y = 0 -----(2)
Given : The number formed by reversing the digits is 18 less than the given number.
The number formed by reversing the digits is yx.
Then, we have
xy - yx = 18
[10 ⋅ x + 1 ⋅ y] - [10 ⋅ y + 1 ⋅ x] = 18
(10x + y) - (10y + x) = 18
Simplify.
10x + y - 10y - x = 18
9x - 9y = 18
Divide each side by 9.
x - y = 2 -----(3)
Solving (2) and (3), we get
x = 4 and y = 2
So, the required two digit number is 42.
Problem 3 :
In a rectangle, if the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm . If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm . Find the area of the rectangle.
Solution :
Let x and y be the length and width of the rectangle respectively.
Then, area of the rectangle is
= x ⋅ y
Given : If the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm
Then, we have
(x + 2) ⋅ (y - 2) = x ⋅ y - 28
Simplify.
xy - 2x + 2y - 4 = xy - 28
- 2x + 2y = - 24
Divide each side by 2.
- x + y = - 12 -----(1)
Given : If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm
Then, we have
(x - 1) ⋅ (y + 2) = x ⋅ y + 33
Simplify.
xy + 2x - y - 2 = xy + 33
2x - y = 35 -----(2)
Solving (1) and (2), we get
x = 23 and y = 11
Area of the original rectangle is
= x ⋅ y
= 23 ⋅ 11
= 253 sq.cm
So, the area of the rectangle is 253 sq.cm.
Problem 4 :
Sum of incomes of A and B is $2640. If B's income is 20% more than A, find the income of A and B.
Solution :
Let x and y be the incomes of A and B respectively.
Then, we have
x + y = 2640 -----(1)
Given : B's income is 20% more than A.
Then, we have
y = 120% of x
y = 1.2x -----(2)
Substitute 1.2x for y in (1).
x + 1.2x = 2640
Simplify.
2.2x = 2640
Divide each side by 2.2.
x = 1200
Substitute 1200 for x in (2)
y = 1.2 ⋅ 1200
y = 1440
So, the income of A is $1200 and B is $1440.
Problem 5 :
Sum of the cost price of two products is $50. Sum of the selling price of the same two products is $52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.
Solution :
Let x and y be the cost prices of two products.
Given : Sum of the cost prices of the two products is $50.
Then, we have
x + y = 50 ------(1)
Given : One is sold at 20% profit and other one is sold at 20% loss.
Let us assume that x is sold at 20% profit.
Then, the selling price of x is
= 120% of x
= 1.2x
Let us assume that y is sold at 20% loss.
Then, the selling price of y is
= 80% of y
= 0.8y
Given : Sum of the selling price of the two products is $52.
Then, we have
1.2x + 0.8y = 52
Multiply both sides by 10.
12x + 8y = 520
Divide each side by 4.
3x + 2y = 130 -----(2)
Solving (1) and (2), we get
x = 30 and y = 20
So, the cost prices of two products are $30 and $20.
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