Explanation
Step -1: Writing mathematical equations according to conditions given in question
Let the price of ticket for station A be Rs. x and for station B be Rs. y,
According to question,
2x + 3y = 77 → Equation 1
3x + 5y = 124 → Equation 2
Step -2: Calculating fares for station A and station B .
Solving Equation 1 and Equation 2,
Multiplying Equation 1 by 3,
6x + 9y = 231 → Equation 3
Multiplying Equation 2 by 2,
6x + 10y = 248 → Equation 4
Subtracting Equation 3 from Equation 4,
⇒ y = 248 - 231
⇒ y = 17
Substituting value of y in Equation 1,
⇒ 2x + 3(17) = 77
⇒ 2x + 51 = 77
⇒ 2x = 26
⇒ x = 13
Thus, the fare to station A is Rs. 13 and to station B is Rs.17 .
Suppose the cost of 1 table =x and cost of 1 chair =yThen according to the question⇒4x+3y=2250...........(1)⇒3x+4y=1950.............(2)
Multiply (1) by 4 and (2) by 3 and Subtract both⇒(16x+12y=9000)−(9x+12y=5850)
⇒7x=3150⇒x=450
Put x=450 in eq2
⇒450×4+3y=2250⇒3y=2250−1800⇒3y=450
⇒y=150
Cost of 1 table =450Cost of 1 chair =150∴ Cost of 1 table and 2 chair =450+150×2=750
Let no of 20p coins =xNo of 25p coins =yAccording to the questionx+y=50⇒x=50−y.....eq1And 20x+25y=1125⇒4x+5y=225....eq2Put the value of x from eq1⇒4(50−y)+5y=225......eq3⇒200−4y+5y=225⇒y=25Put y=25 in eq1⇒x+25=50⇒x=25
No of 20p coins =25
No of 25p coins =25
Let the first number be x and second number y.
According to questionx+y=8…(i)
4(x−y)=8
⇒x−y=2…(ii)
Add equations (i) and (ii),(x+y)+(x−y)=8+22x=10x=5
Substitute x=5 in (i),(5)+y=8y=8−5y=3
So, the numbers are 5 and 3.
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