Let us say x boys and x girls joined the group.
(64 + x)/(40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.
To get the solution that contains 1 part of milk and two parts of water,
they must be mixed in the ratio as
7x+6x/5y+11y = 1/2
26x = 16y
x/y = 16/26
x/y = 8/13
Here the ratio of mixtures( i.e milk , water) doesnot matter. But the important point is that whether the total amount ( either pure or mixture ) being transferred is equal or not.Since the total amount ( i.e 5 cups) being transferred from each one to another , hence A =B.
From given data,
R : M = 5 : 7
M : A = 5 : 7
R : M : A = 25 : 35 : 49
25 + 35 + 49 = 109
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.
Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.
There is a proposal to increase these seats by 40%, 50% and 75% respectively.
Then Seats for Mathematics, Physics and Biology in a school will be
1.4 X 5x, 1.5 X 7x and 1.75 X 8x and
ratio will be 7 : 10.5 : 14 => 2:3:4.
Given ratio of initial mixture of milk and water in Q = 5 : 3
Let the initial quantity of mixture in vessel Q = 8x
Let quantity of Milk = 5x and
Let quantity of water = 3x
According to the question,
=> 25x + 30 = 24x + 32
=> x = 2
Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.
The ratio of fees collected from B.Tech : MBA = 4x * 25y : 5x * 16y
= 100xy : 80xy
= 5xy : 4 xy = 5k : 4k
The amount collected only from MBA students = = Rs. 72,000
Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7
We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7
Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9
Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.
A : B
(8000x4)+(4000x8) : (9000x6)+(6000x6)
64000 : 90000
=> 32 : 45
Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k
As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)
x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.
6M + 3B = 4(1M + 1B)
6M + 3B = 4M + 4B
2M = 1B
M/B = 1/2
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