Problem 1 :
Two numbers add upto 20 and they differ by 6. What are the two numbers?
Problem 2 :
Henry bought bought rulers and erasers in a total of 45. The price of each ruler is $5 and that of an eraser is $3. If he had paid a total of $185 for his purchase, what is the price of a ruler and an eraser?
Problem 3 :
A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750?
Problem 4 :
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1, it becomes ½ if we only add 1 to the denominator. What is the fraction?
Problem 5 :
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Problem 6 :
In a three digit number, the middle digit is zero and the other two digits add upto 11. If 297 be subtracted from it, the digits are reversed. Find the number.
Problem 7 :
Of two positive numbers, three times the larger number is equal to four times the smaller number. If the sum of one-half of the larger number and two-third of the smaller number is equal to 8, find the two numbers.
Problem 8 :
In a business, John has invested money in stock and bonds. He earns 5% of the money invested in stock and 7% in bonds. If his total investment is $25000, find the money invested in stock and bonds.
1. Answer :
Let x and y be the two numbers.
x + y = 4 ----(1)
x - y = 6 ----(2)
(1) + (2) :
Divide both sides by 2.
x = 13
Substitute x = 13 into (1).
13 + y = 20
Subtract 13 from both sides.
y = 7
Therefore, the numbers are 13 and 7.
2. Answer :
Let r and e be the costs of a ruler and an eraser respectively.
r + e = 45 ----(1)
5r + 3e = 185 ----(2)
3(1) - (2) :
Divide both sides by -2.
r = 25
Substitute r = 25 into (1).
25 + e = 45
Subtract 25 from both sides.
e = 20
Therefore,
number of rulers = 25
number of erasers = 20
3. Answer :
Let a and k be number of adult and kids tickets respectively.
a + k = 548 ----(1)
10a + 5k = 3750 ----(2)
5(1) - (2) :
Divide both sides by -5.
a = 202
Substitute a = 202 into (1).
202 + k = 548
Subtract 202 from both sides.
k = 346
Therefore,
number of adult tickets = 202
number of kids tickets = 346
4. Answer :
Let ˣ⁄y be the required fraction.
Given : When 1 is added to the numerator and 1 is subtracted from the denominator, the fraction reduces to 1.
⁽ˣ ⁺ ¹⁾⁄₍y – ₁₎ = 1
x + 1 = y - 1
x – y = -2 ----(1)
Given : When only 1 is added to the denominator, the fraction becomes ½.
ˣ⁄₍y ₊ ₁₎ = ½
2x = y + 1
2x – y = 1 ----(2)
(2) - (1) :
Substitute x = 3 into (1).
3 - y = -2
Subtract 3 from both sides.
-y = -5
y = 5
ˣ⁄y = ⅗
Therefore, the fraction is ⅗.
5. Answer :
Let x and y be the present ages of Nuri and Sonu respectively.
Five years ago, Nuri was thrice as old as Sonu.
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -10 ----(1)
Ten years later, Nuri will be twice as old as Sonu.
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10 ----(2)
(2) - (1) :
Substitute y = 20 into (2).
x - 2(20) = 10
x - 40 = 10
Add 40 to both sides.
x = 50
Therefore,
the present age of Nuri = 50 years
the present age of Sonu = 20 years
6. Answer :
Let x0y be the required three-digit number.
x + y = 11 ----(1)
If 297 be subtracted from it, the digits are reversed.
x0y - 297 = y0x
100(x) + 10(0) + 1(y) - 297 = 100(y) + 10(0) + 1(x)
100x + 0 + y - 297 = 100y + 0 + x
100x + y - 297 = 100y + x
99x - 99y = 297
x - y = 3 ----(2)
(1) + (2) :
Divide both sides by 2.
x = 7
Substitute x = 7 into (1).
7 + y = 11
Subtract 7 from both sides.
y = 4
x0y = 704
Therefore, the three digit number is 704.
7. Answer :
Let x and y be the two positive numbers such that x > y.
Given : Three times the larger number is equal to four times the smaller number.
3x = 4y
3x - 4y = 0 ----(1)
Given : The sum of one-half of the larger number and two-third of the smaller number is equal to 8.
(½)x + (⅔)y = 8
Least common multiple of (2, 3) = 6.
Multiply both sides by 6 to get rid of the denominators 2 and 3.
3x + 4y = 48 ----(2)
(1) + (2) :
Divide both sides by 6.
x = 8
Substitute x = 8 into (1).
3(8) - 4y = 0
24 - 4y = 0
Subtract 24 from both sides.
-4y = -24
Divide both sides by -4.
y = 6
Therefore, the two numbers are 8 and 6.
8. Answer :
John has invested money in stock and bonds. He earns 5% of the money invested in stock and 7% in bonds. If the total investment is $25000 and the total earning is $1550, find the money invested in stock and bonds.
Let x and y be the amounts of money invested stock and bonds respectively.
x + y = 25000 ----(1)
Given : John earns 5% of the money invested in stock and 7% in bonds.
5% of x + 7% of y = 1550
0.05x + 0.07y = 1550
Multiply both sides by 100.
5x + 7y = 155000 ----(2)
7(1) - (2) :
Divide both sides by 2.
x = 10000
Substitute x = 10000 into (1).
10000 + y = 25000
Subtract 10000 from both sides.
y = 15000
Therefore,
money invested in stock = $10,000
money invested in bonds = $15,000
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 01, 23 11:01 AM
Nov 30, 23 04:36 AM
Nov 30, 23 04:18 AM