The required expression (E) = (2a+3b)/(3a+4b)
Given a/b = 4/3
So, E =[2(3/4)+3]/[3(3/4)+4]
Let the no. of 50 P, 25 P, and 20 P coins be 4X ,2X ,X Respectively
Total value of coins = 50(4X) + 25(2X) + 20(X) = 5400
270X = 5400
X = 20
So, number of 25 P coins = 2X = 2x20 = 40
Let Ajay's and Balu's marks be 4X and 5X respectively.
(4X+36)/(5X+36) = 7/8
8(4X+36) = 7(5X+36)
32X + 288 = 35X + 252
3X = 36
So, Balu's marks = 5X = 60
Let the amounts received by the old man's wife, the eldest son and youngest son be a, b and c respectively.
Then, according to the first condition
given a = 1/2(b+c)
2a = b + c
Adding 'a' on both sides,
a = (a + b + c)/3
According to the 2nd condition given,
c = (1/3)(a+b)
3c = a + b
Adding 'c' on both sides,
4c = a + b + c
c = (a + b + c)/4
So, (a+b+c)/3 - (a+b+c)/4 = 60000
a + b + c = 720000
Let the two numbers be 4X and 5X
Let K be added in both of these for the ratio to become 5:6
(4X+K)/(5X+K) = 5/6
24X+6K = 25X+5K
K = X
As X is unknown, K cannot be found.
Let, the speed of bus be x kmph and the consumption of diesel per hour be y litres.
Given, y ∝ x2,
⇒ y = kx2
Given, 1 = k(40)2
K = 1/1600
∴ y = x2/1600
Let the required speed be x kmph
Time taken to cover 400 km at c kmph = 400/x hours
∴ consumption of diesel in 400/x hours = 400/x litres
∴ cost of diesel = 400y/x(40) = ₹16000y/x
= (16000/x)(x2/1600) = 10x [∵y= x2/16000]
and other expenses for 400/x hours = 400/x (40) = ₹16000/x
Let the quantities of milk in the 1st to 5st vessels be 4x, 5x, 6x, 7x and 8x respectively.
Total quantity of milk in the vessels is 30x
Total capacity of the 5 vessels = 30x(100/75) = 40x
Capacity of each vessel is 8x
64% of 8x = 5.12x
The number of vessels of vessels which contain at least 5.12x of milk is three.
Let the truck is rented for 8 hours or less.
Then the number of hours it is rented for = 800/100 = 8 hours
Buts, if it charged ₹ 10 per km, the amount that should paid be = 120(8) = 960
But company paid only ₹800
So, The truck is rented for more than 8 hours.
So, number of hours = 800/80 = 10 hours
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