Important Questions Test: Ratio & Proportion- 2 - GMAT MCQ

Area of semicircle = (1/2)πr²
Area of sector = (θ/360)πr²

V ∝ W², V = KW²
Where, V = Value of diamond, W = weight of diamond and K = constant
⇒ 10000 = K.(10)²
⇒ K = 100
Hence, V = 100 x 4² = 1600
also, V = 100 x 6² = 3600
Net loss = Rs.(10000 - 5200) = Rs 4800 i.e., 48%

► Let the fare of 1st, 2nd and 3rd class be 10x, 8x, and 3x respectively.
► Let the no. of passengers of 1st, 2nd and 3rd class be 3y, 4y and 10y respectively. 
Then 30xy + 32xy + 30xy = 16100
⇒ 92xy = 16100
Hence xy = 175
Required amount = 32xy = 32 x 175 = 5600 

► Price of one goat = Rs 70 ⇒ Price of 9 goats = 9 x 70 = Rs 630
► Price of 14 sheeps = Rs 630 
⇒ Price of 1 sheep = Rs 630/14 = Rs 45 ⇒ Price of 50 sheep = Rs 45 x 50 = Rs 2250
► Price of 10 oxen = 2250 ⇒ Price of 1 oxen = 225 ⇒ Price of 8 oxen  = 1800
► Price of 6 dogs = 1800 ⇒ Price of 1 dog = 300 ⇒ Price of 5 dog = 1500
► Price of two horses = 1500 ⇒ Price of one horse = 750

Given,
A ∝ B
B ∝ (1/C)
C ∝ D²
D ∝ ∛E
When A increases ⇒ B also increases ⇒ C decreases
Decrease in C results in decrease in D and decrease in D results decrease in E.

► Let the 1^stno. = A = x
2^nd no.= B , 3^rd no.= C , 4^th no. = D
► A:B = 2:3
    A/B = 2/3
    x/B = 2/3
    B = 3x/2
► B:C = 5:6
    (3x/2)/C = 5/6
    C = (6*3x)/2*5 = 9x/5
    C = 9x/5
► C:D = 8:9
    9x/5/D = 8/9
    D = (9x*9)/8*5 = 81x/40
    D = 81x/40
► A + B + C + D = 253. (Given)
    ⇒ x + 3x/2 + 9x/5 + 81x/40 = 253
    ⇒ (40x+ 60x+ 72x+81x)/40 = 253
    ⇒ 253x = 253*40
    ⇒ X = (253*40)/253 = 40
    ⇒ 1st no.(A) = X = 40
    ⇒ 2nd no.(B) = 3x/2 =( 3 × 40)/2 = 60
    ⇒ 3rd no.(C) = 9x/5 = (9×40)/5 = 72
    ⇒ 4th no.(D) = 81x/40 =( 81×40/)/40 = 81
► Average of numbers = sum of observations/ total no.of observations
► Average of 2nd no. & 3rd no.= (60+72)/2 = 132/2 = 66

► Given that a man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal.
► We are to find the total number of notes he has.
► Let x represents the number of notes of each of the three denominations.
► Then, according to the given information, we have
⇒ 1.x + 5.x + 10.x = 480
⇒ x + 5x + 10x = 480
⇒ 16x = 480
⇒ x = 30
► Therefore, the total number of notes is given by 
⇒ x + x + x = 3x = 3*30 = 90

► Let A, B, C invested Rs 4x, Rs x, Rs 15x respectively.
► According to question, Ratio of their investments for 1 year
= (4x*3 + 2x*3 + x*3 + x/2*3) : (x*3 + 2x*3 + 4x*3 + 8x*3) : (15x*12)
⇒ 45x/2 : 45x : 180x
⇒ 45 : 90 : 360
⇒ 1 : 2 : 8
Total profit = (11/2)*22000 = Rs 121000

► Let 1 black cow gives x litres milk in 1 day and 1 white cow gives y litres milk in 1 day.
► Then 5(4x + 3y) = 4(3x + 5y)
⇒ 20x + 15y = 12x + 20y 

CP = 75

► According to question, A:B:C:D = 6:5:10:14
► Now in D there are 70 chapters
Therefore, A:B:C:D = 30:25:50:70      [ Multiplying Each by 5 ]
L.C.M of 30, 25, 50, 70 is 1050
So each part should have 1050 pages.
So total pages = 4 x 1050 = 4200 ans.

► Let their present ages be 2x and 3x.
Given that

10x + 5N = 9x + 3N
X = -2N which 18 not possible.
Hence, correct option is (d).

► If P is taking 3 days to do some work, then Q takes 6 days to do the same work.
► Now, both of them will take 2 days to complete the work. So, R takes 6 days to complete the same work.
► Hence, earning should be distributed in the ratio of their efficiency, i.e., 2 : 1 : 1.

► Number of apples/Number of guavas = Number of guavas/Number of oranges
► (Number of guavas)² = Number of apples x Number of oranges.
► This gives us a hint that Number of apples x Number of oranges should be a prefect square.
Now start using the options.
Option (b) satisfies the relationship (16/20 = 20/25)

Given:

Total coin = 220

Total money = Rs. 160

There are thrice as many 1 Rupee coins as there are 25 paise coins.

Concept used:

Ratio method is used.

Calculation:

Let the 25 paise coins be 'x'

So, one rupees coins = 3x

50 paise coins = 220 – x – (3x) = 220 – (4x)

According to the questions,

3x + [(220 – 4x)/2] + x/4 =160

⇒ (12x + 440 – 8x + x)/4 = 160

⇒  5x + 440 = 640

⇒ 5x = 200

⇒ x = 40

So, 50 paise coins = 220 – (4x) = 220 – (4 × 40) = 60

∴ The number of 50 paise coin is 60.