Aptitude Ratio And ProportionPage 1

1. Ratio:
The ratio of two quantities x and y in the same units, is the fractionx/yand we write it as x : y.
In the ratio x : y, we call x as the first term or antecedent and y, the second term or consequent.
Eg. The ratio 2 : 3 represents 2/3 with antecedent = 2, consequent = 3.
Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.
Eg. 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.

2. Proportion:
The equality of two ratios is called proportion.
If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.
Here a and d are called extremes, while b and c are called mean terms.
Product of means = Product of extremes.
Thus, a : b :: c : d ⇔ (b x c) = (a x d).

3. Fourth Proportional:
If a : b = c : d, then d is called the fourth proportional to a, b, c.
Third Proportional:
a : b = c : d, then c is called the third proportion to a and b.
Mean Proportional:
Mean proportional between a and b is √ab.

4. Comparison of Ratios:
We say that (a : b) > (c : d) ⇔ a/b > c/d.

Compounded Ratio:
The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).

5. Duplicate Ratios:
Duplicate ratio of (a : b) is (a2 : b2).
Sub-duplicate ratio of (a : b) is (√a : √b).
Triplicate ratio of (a : b) is (a3 : b3).
Sub-triplicate ratio of (a : b) is (a1/3 : b1/3).
If a/b = c/d , then a + b/a - b = c + d/c - d [componendo and dividendo]

6. Variations:
We say that x is directly proportional to y, if x = ky for some constant k and we write, x ∝ y.
We say that x is inversely proportional to y, if xy = k for some constant k and
we write, x ∝ 1/y.


Given a:b = 3:4, find (2a+3b)/(3a+4b)
Answer is: DThe required expression (E) = (2a+3b)/(3a+4b)
= [2(a/b)+3]/[3(a/b)+4]
Given a/b = 4/3
So, E =[2(3/4)+3]/[3(3/4)+4]
= 18/25


A Bag has coins in the denominations of 50 P, 25 P, and 20 P in the ratio 4:2:1. If the total value of the coins is Rs. 54, Find number of 25 P coins in the bag.
Answer is: CLet 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


In an exam, the ratio of Ajay's and Balu's marks is 4:5, If each them had scored 36 more marks, the ratio of their marks would be 7:8. Find Balu's marks?
Answer is: ALet 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
X =12
So, Balu's marks = 5X = 60


An old man makes a will to divide his property among his wife and two sons such that his wife gets half of the total amount received by the sons. His younger son gets a third of total amount received by his wife and elder son. If his wife gets Rs. 60000 worth of property more than the youngest son, Find the total value of the property of the man .(In lakhs of rupees)
Answer is: DLet 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


What should be added to both numbers which are in the ratio 4:5, so that their ratio becomes 5:6?
(d)cannot be determined
Answer is: DLet 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.


The consumption of diesel per hour of a bus various directly as square of its speed. When the bus is travelling at 40 kmph its consumption is 1 litre per hour. If each litre cost Rs. 40 and other expenses per hour is Rs 40. then what would be the minimum expenditure required to cover a distance of 400 km?
Answer is: CLet, 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


There are five vessels, with equal capacities, each containing some milk. The quantities of milk in the 5 vessels are in ratio 4 : 5 : 6 : 7 : 8 such that the total quantity of milk in the vessels is equal to 75% of the total capacities of the 5 vessels. How many of the vessels are at lost 64% full of milk?
Answer is: CLet 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.


A truck rental agency has the following forms. if a truck is rented for 8 hours or less the charge is Rs. 100 per hour or Rs 8 per km whichever is more. On the other hand, if the truck is rented for more than 8 hours, the charge is ₹ 80 per hour or ₹ 6 per km whichever is more. A company rented a truck from the agency, and used it for 120 km and paid ₹ 800. For how many hours did the company rent the truck?
Answer is: CLet 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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