Important Questions Test: Ratio & Proportion- 2 - CAT 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

► 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

Here’s how to work it out step by step:

1. Let the two foil prices be in the ratio 1 : 4.

   • Call the cheaper foil’s price “x” per kg, so the more expensive is “4x” per kg.

2. They’re mixed in quantity ratio 6 : 1 (cheaper : expensive).

   • In 7 kg of alloy you have 6 kg at x and 1 kg at 4x.

   • Total cost for 7 kg = 6·x + 1·4x = 10x.

   • So cost price per kg of alloy = (10x)⁄7.

3. He sells the alloy at Rs 90/kg for a 20% profit.

   • If CP is C and SP is 90, then 90 = C·(1 + 20%) = 1.2 C ⇒ C = 90⁄1.2 = 75.

   • Thus the alloy’s cost price per kg must be Rs 75.

4. Set up the equation:
   (10x)⁄7 = 75
   ⇒ 10x = 75·7 = 525
   ⇒ x = 52.5

► 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).

► 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.

Total cost = Fixed cost + Variable cost

Variable cost = k × b
where, b = number of boarders and k = variable cost per boarder.

Total cost = FC + kb

Total collection = 1600b

Case 1: Total profit = 200 × 50 = 1600 × 50 – (FC + k × 50)   …(1)
Case 2: Total profit = 250 × 75 = 1600 × 75 – (FC + k × 75)   …(2)

(2) - (1), we get
18750 – 10000 = 1600 × 25 - 25k
⇒ k = 1250 and FC = 7,500 

Now, total profit when there are 80 boarders = 1600 × 80 – (7500 + 1250 × 80) = 20,500

Alternately,
Total cost when there are 50 students = (1600 – 200) × 50 = 70,000
Total cost when there are 75 students = (1600 – 250) × 75 = 1,01,250

Due to 25 students cost increases by (101250 - 70000) = 31250
∴ Due to 5 students cost increases by = 31250/5 = 6250

∴ Total cost when there are 80 students = 1,01,250 + 6250 = 1,07,500
Total revenue for 80 students = 1600 × 80 = 1,28,000

∴ Profit = 1,28,000 – 1,07,500 = 20,500.

Hence, option (d).

Let there be 'x' men in the beginning so that after 15 days the food for them is left for 45 days.

After adding 500 men the food lasts for only 40 days.

Now (x+500) men can have the same food for 40 days.

Therefore by equating the amount of food we get,

45 * x = (x + 500) * 40

45x = (x+500) * 40

5x = 20,000

x = 4,000

Therefore there are 4,000 men in the fort.

• Let Vinod marks be 6x and Basu's is 5x. Therefore, the sum of the marks = 6x + 5x = 11x.
• But the sum of the marks is given as 275 = 11x. We get x = 25 therefore, vinod marks is 6x = 150 and Basu marks = 5x = 125.
• Therefore, the combined average of their marks = (150 + 125) / 2 = 137.5.
• If the total mark of the exam is 100 then their combined average of their percentage is 68.75

Therefore, if their combined average of their percentage is 137.5 then the total marks would be (137.5 / 68.75)*100 = 200.