59
Let the present age of Y be y years.
In 7 years, X’s age will be 55 years, and Y’s age will be y + 7 years.
According to the problem:
\( {55 \over y + 7} = {5 \over 6} \)
\(6 \times 55 = 5 \times (y + 7)\)
330 = 5y + 35
295 = 5y
y = 59
Hence, the present age of Y is 59 years.
88
Given that the average of 35 consecutive even numbers is 44, we need to find the sum of the first and last numbers.
Let the first number be x and the last number be x+68 (since the difference between each consecutive even number is 2, and there are 34 steps between the first and last numbers).
The average of these numbers is given by:
\(Average = {\text{Sum of first and last numbers} \over 2}\)
Given that the average is 44:
\(44 = {x + x + 68 \over 2}\)
Simplifying:
\(44 = {2x + 68 \over 2}\)
\(44 = x + 34\)
Solving for x:
x = 44 − 34 = 10
Thus, the first number is 10 and the last number is 10 + 68 = 78
The sum of the first and last numbers is: 10 + 78 = 88
₹ 440
Given that there are 20 students in a mountain climbing camp. When the number of students increases by 5, the mess expenditure increases by ₹35 per day, and the average expenditure per head decreases by ₹3.
Let the original total expenditure of the camp be ₹ x.
The original average expenditure per head = ₹ \({x \over 2}\).
After the increase in students, the new total expenditure = ₹ x + 35, and the number of students = 25.
The new average expenditure per head = ₹\({x + 35 \over 25}\)
According to the problem:
\({x \over 2}\) - \({x + 35 \over 25}\) = 3
Simplifying:
5x − 4(x+35) = 300
x − 140 = 300
x = 440
Hence, the original expenditure of the camp is ₹ 440.
12 km/hr, 20 km/hr
Given that Bus A goes from one city to another, taking 4 hours more than Bus B. The two cities are 120 km apart. When Bus A doubles its speed, it takes 1 hour less than Bus B.
Let the speed of Bus A be x km/hr and the speed of Bus B be y km/hr.
According to the problem:
\({120 \over x} = {120 \over y} + 4\)
When Bus A doubles its speed, the new time taken by Bus A is:
\({60 \over x} = {120 \over y} - 1\)
Simplifying the two equations:
From the first equation:
\({120 \over x} - {120 \over y} = 4\)
From the second equation:
\( {120 \over y} - {60 \over x} = 1\)
By solving these two equations, we find:
x = 12 km/hr, y = 20 km/hr
Hence, the earlier speeds of Bus A and Bus B are 12 km/hr and 20 km/hr, respectively.
Monday
To find the day of the week on 25th April 1960, we calculate the number of odd days from the reference date (1st January 1900) to 25th April 1960.
Number of years from 1900 to 1959: 60 years
Ordinary years: 45
Leap years: 15
Total number of odd days:
Odd days from 45 ordinary years: \(45 \times 1 = 45\)
Odd days from 15 leap years: \(15 \times 2 = 30\)
Total odd days = 45 + 30 = 75
Reducing the total odd days modulo 7:
75 ÷ 7 gives a remainder of 5, so there are 5 odd days.
Odd days from 1st January 1960 to 25th April 1960:
January: 31 days = 3 odd days
February (1960 is a leap year): 29 days = 1 odd day
March: 31 days = 3 odd days
April: 25 days = 4 odd days
Total odd days from 1960 = 3 + 1 + 3 + 4 = 11 odd days = 4 odd days (mod 7)
Final calculation:
Total odd days = 5 + 4 = 9 odd days
Reduced modulo 7:
9 mod 7 = 2 odd days
Day corresponding to 2 odd days = Monday