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21. The present ratio of ages of A and B is 4 : 5. 18 years ago, this ratio was 11 : 16. Find the sum total of their present ages.
18 years ago their ages
Or, 64x - 288 = 55x - 198
Or, 64x - 55x = -198 + 288
Or, 9x = 90
Or, x = = 10
Sum of the present ages = 40 + 50 = 90 years
Solution:
Let present age of A and B be 4x and 5x18 years ago their ages
Or, 64x - 288 = 55x - 198
Or, 64x - 55x = -198 + 288
Or, 9x = 90
Or, x = = 10
Sum of the present ages = 40 + 50 = 90 years
22. A dishonest milk man mixed 1 liter of water for every 3 liters of milk and thus made up 36 liters of milk. If he now adds 15 liters of milk to mixture, find the ratio of milk and water in the new mixture:
27 liter milk and 9 liter water
Now, 15 liters milk is added then milk becomes 42 liters
Now, ratio = 42 : 9 = 14 : 3
Solution:
Quantity of milk and water in the 36 liter mixture27 liter milk and 9 liter water
Now, 15 liters milk is added then milk becomes 42 liters
Now, ratio = 42 : 9 = 14 : 3
23. If the ratio of the ages of Maya and Chhaya is 6 : 5 at present, and fifteen years from now, the ratio will get changed to 9 : 8, then find Maya's present age.
And,
Or, 48x + 120 = 45x = 135
Or, 3x = 15
Or, x = 5
Present age of Maya = 6x = 30
Solution:
Let Maya's and Chhaya's present age is 6x and 5x respectivelyAnd,
Or, 48x + 120 = 45x = 135
Or, 3x = 15
Or, x = 5
Present age of Maya = 6x = 30
24. The Lucknow-Indore Express without its rake can go 24 km an hour, and the speed is diminished by a quantity that varies as the square root of the number of wagon attached. If it is known that with four wagons its speed is 20 km/h, the greatest number of wagons with which the engine can just move is
Putting the value, n = 4
we get, k = 2
Now the equation (as k = 2) become,S =
Thus, it means when n = 144, speed will be zero.
Hence, train can just move when 143 wagons are attached
Solution:
Speed = Putting the value, n = 4
we get, k = 2
Now the equation (as k = 2) become,S =
Thus, it means when n = 144, speed will be zero.
Hence, train can just move when 143 wagons are attached
25. If x varies as y then x2 + y2 varies as
x = y
Or, x - y = 0
Or, (x - y)2 = 0
Or, x2 + y2 - 2xy = 0
Or, x2 + y2 = 2xy
It means that, x2 + y2 varies as xy
Solution:
Given,x = y
Or, x - y = 0
Or, (x - y)2 = 0
Or, x2 + y2 - 2xy = 0
Or, x2 + y2 = 2xy
It means that, x2 + y2 varies as xy
26. Brindavan Express leave Chennai Central Station every day at 07.50 am and goes to Bangalore City Railway station. This train is very popular among the travelers. On 25th July 2012 number of passengers traveling by I class and II class was in the ratio 1 : 4. The fare for this travel is in the ratio 3 : 1. The total fare collected was Rs. 224000. (Rs. Two lakhs twenty four thousand only). What was the fare collected from I class passengers on that day?
Then, number of passenger traveling by second class will be 4x
But the fare is in the ratio 3 : 1
In other words, if 3y fare is collected per I class passenger, y would be collected per II class passenger
Fares of I class passengers : Fares of II class passengers
= x × 3y : 4x × y
= 3 : 4
The above ratio can be interpreted as follows
If total fare is 3 + 4 = 7, then I class passengers should pay Rs. 3
Similarly, we can calculate the fare of I class passengers when total was 224000
=
= Rs. 96000
Solution:
Let the number of passenger traveling by first class be xThen, number of passenger traveling by second class will be 4x
But the fare is in the ratio 3 : 1
In other words, if 3y fare is collected per I class passenger, y would be collected per II class passenger
Fares of I class passengers : Fares of II class passengers
= x × 3y : 4x × y
= 3 : 4
The above ratio can be interpreted as follows
If total fare is 3 + 4 = 7, then I class passengers should pay Rs. 3
Similarly, we can calculate the fare of I class passengers when total was 224000
| Total Fare | Class Fare |
| 7 | 3 |
| 224000 | ? |
=
= Rs. 96000
27. A vessel of capacity 2 litre has 25% alcohol and another vessel of capacity 6 litre had 40% alcohol. The total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water. what is the new concentration of mixture ?
= 25% of 2 litre
= 0.25 × 2 = 0.5 litre
Amount of alcohol in second vessel,
= 40% of 6 litre
= 0.4 × 6 = 2.4 litre
Total amount of alcohol out of 10 litres of mixture is
0.5 + 2.4 = 2.9 litre
Thus, Concentration of the mixture is,
= 29%
Solution:
Amount of alcohol in first vessel,= 25% of 2 litre
= 0.25 × 2 = 0.5 litre
Amount of alcohol in second vessel,
= 40% of 6 litre
= 0.4 × 6 = 2.4 litre
Total amount of alcohol out of 10 litres of mixture is
0.5 + 2.4 = 2.9 litre
Thus, Concentration of the mixture is,
= 29%
28. The number of oranges in three basket are in the ratio 3 : 4 : 5. In which ratio the no. of oranges in first two basket must be increased so that the new ratio becomes 5 : 4 : 3 ?
B1 : B2 : B3 = 5y : 4y : 3y
Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets.
Hence, 5x = 3y
x = and,
∴ 3x : 4x : 5x (putting the vale of x)
=
= 9y : 12y : 15y
And,
5y : 4y : 3y (multiple by 5) → 25y : 20y : 15y
∴Increment in first basket = 16
And, Increment in second basket = 8
Thus, Required ratio = = 2 : 1
Solution:
Let, B1 : B2 : B3 = 3x : 4x : 5x and B1 : B2 : B3 = 5y : 4y : 3y
Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets.
Hence, 5x = 3y
x = and,
∴ 3x : 4x : 5x (putting the vale of x)
=
= 9y : 12y : 15y
And,
5y : 4y : 3y (multiple by 5) → 25y : 20y : 15y
∴Increment in first basket = 16
And, Increment in second basket = 8
Thus, Required ratio = = 2 : 1
29. At a casino in Mumbai, there are 3 tables A, B and C. The payoffs at A is 10 : 1, at B is 20 : 1 and C is 30 :1. If a man bets Rs. 200 at each table and win at two of the tables, what is the maximum and minimum difference between his earnings can be ?
A → 10 : 1
B → 20 : 1
C → 30 : 1
i.e., he won B and C but lost on A
20 × 200 + 30 × 200 - 1 × 200 = 9800
Minimum earnings will be when he won on table A and B and lose on table 3
10 × 200 + 20 × 200 - 1 × 200 = 5800
Therefore, difference = 9800 - 5800 = Rs. 4000
Alternatively,
The difference,
= [(30 + 20 - 1) - (10 + 20 -1)] × 200
= Rs. 4000
Solution:
Maximum earning will be only possible when he will won on the maximum yielding tableA → 10 : 1
B → 20 : 1
C → 30 : 1
i.e., he won B and C but lost on A
20 × 200 + 30 × 200 - 1 × 200 = 9800
Minimum earnings will be when he won on table A and B and lose on table 3
10 × 200 + 20 × 200 - 1 × 200 = 5800
Therefore, difference = 9800 - 5800 = Rs. 4000
Alternatively,
The difference,
= [(30 + 20 - 1) - (10 + 20 -1)] × 200
= Rs. 4000
30. A track covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minute. The ratio of their speeds is:
Speed of bus = = = 733.33 m/minute
Ratio of their speeds = = 3 : 4
Solution:
Speed of track = 550 per minute.Speed of bus = = = 733.33 m/minute
Ratio of their speeds = = 3 : 4