Problem 1 :
In a school, there are 880 students in total. If there is 20% more boys than girls, find the number of boys and girls in the school.
Solution :
Let "x" and "y" be the no. of boys and girls respectively.
Then, we have
x + y = 880 ------(1)
Given : There is 20% more boys than girls.
So, we have
x = 120% of y
x = 1.2y ------(2)
Substitute x = 1.2y in (1)
1.2y + y = 880
Simplify.
2.2y = 880
Divide by 2.2
y = 400
Substitute y = 400 in (2).
x = 1.2 ⋅ 400
x = 480
So, the number of boys in the school is 480 and girls is 400.
Problem 2 :
Mr. Lenin invests some amount in deposit A and some amount in deposit B. The total money invested is $2500. He gets 10% income on deposit A and 20% income on deposit B. If the total income earned be $380, find the amount invested in A and B separately
Solution :
Let "x" and "y" be the amounts invested in A and B respectively.
Then, we have
x + y = 2500
Subtract y from both sides.
x = 2500 - y ------(1)
Given : Mr. Lenin gets 10% income on deposit A, 20% income on deposit B and the total income earned be $380.
So, we have
10% of x + 20% of y = 380
0.1x + 0.2y = 380
Multiply both sides by 10.
x + 2y = 3800 -------(2)
From (1), we can plug x = 2500 - y in (2).
(2)------> 2500 - y + 2y = 3800
Simplify.
2500 + y = 3800
Subtract 2500 from both sides.
y = 1300
Substitute y = 1300 in (1).
(1)------> x = 2500 - 1300
x = 1200
So, the amount invested in the deposit A is $1200 and B is $1300.
Problem 3 :
The sum of two numbers is 209. If one number is 7 less than two times of the other, then find the two numbers.
Solution :
Let "x" and "y" be the two numbers.
Given : The sum of two numbers is 209.
x + y = 209 ------(1)
Given : One number is 7 less than two times of the other.
So, we have
x = 2y - 7 -------(2)
From (2), we can plug x = 2y - 7 in (1).
(1)------> 2y - 7 + y = 209
Simplify.
3y - 7 = 209
Add 7 to both sides.
3y = 216
Divide both sides by 3.
y = 72
Substitute y = 72 in (2).
x = 2 ⋅ 72 - 7
x = 144 - 7
x = 137
So, the two numbers are 137 and 72.
Problem 4 :
The perimeter of the rectangle is 158 cm. If the length is 7 more than 3 times the width, find the area of the rectangle.
Solution :
Let "x" and "y" be the length and width of the rectangle respectively.
Given : Perimeter of the rectangle is 158 cm.
So, we have
2x + 2y = 158
Divide both sides by 2.
x + y = 79 ------(1)
Given : The length is 7 more than 3 times the width.
So, we have
x = 3y + 7 --------(2)
From (2), we can plug x = 3y + 7 in (1).
(1)------> 3y + 7 + y = 79
Simplify.
4y + 7 = 79
Subtract 7 from both sides.
4y = 72
Divide both sides by 4.
y = 18
Substitute y = 18 in (2).
x = 3 ⋅ 18 + 7
x = 61
Therefore, the length of the rectangle is 61 cm and width is 18 cm.
Then, area of the rectangle is
= length ⋅ width
= 61 ⋅ 18
= 1098
So, the area of the rectangle is 1098 sq.cm.
Problem 5 :
A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is $43. The sum of the cost prices of two products is $150. Find the cost price of each product.
Solution :
Let "x" and "y" be the cost prices of the two products.
Then, we have
x + y = 150
Subtract y from both sides.
x = 150 - y -------(1)
Let the trader gain one third of the cost price as profit on the product whose cost price is x.
Then, profit on the product whose cost price is x :
= 1/3 ⋅ x
= x / 3
Let the trader gain one fourth of the cost price as profit on the product whose cost price is y.
Then, profit on the product whose cost price is y :
= 1/4 ⋅ y
= y / 4
Given : Total profit earned on these two products is $43.
So, we have
x/3 + y/4 = 43
L.C.M of (3, 4) is 12.
Then, we have
(4x / 12) + (3y / 12) = 43
Simplify.
(4x + 3y) / 12 = 516
Multiply both sides by 12.
4x + 3y = 516 ------(2)
From (1), we can plug x = 150 - y in (2).
4(150 - y) + 3y = 516
Simplify.
600 - y = 516
Subtract 600 from both sides.
- y = - 84
y = 84
Substitute y = 84 in (1).
(1)------> x = 150 - 84
x = 66
So, the cost prices of two products are $66 and $84.
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