Test: Ratio And Proportion- 2 - GMAT MCQ

Given:

• Let Gerald originally have g bills and James have j bills
• Number of one dollar bills that Gerald had after James gave him 16 one dollar bills = g + 16
• Number of one dollar bills that James after he gave Gerald 16 one dollar bills = j - 16
  • So, we can write  
• Number of one dollar bills that Gerald had after he returned 8 one dollar bills to James = g + 16 – 8
• Number of one dollar bills that James had after he got 8 one dollar bills from Gerald = j – 16 + 8
  • So, we can write  

To Find: J/g = ?

Approach:

1. We’re given 2 relations for j and g. Using these, we’ll:
   • Either find the values of j and g, and thus solve for the required ratio.
   • Or directly solve for the ratio J/g

Working out:

• So, the above equation simplifies to: 11x-14 = 2(9x-16)

Looking at the answer choices, we see that the correct answer is Option A

Step 1 & 2: Understand Question and Draw Inference

Given: Let revenue = R, Number of customers = C and marketing spend = M

• R = kC, where k is a proportionality constant
• C = pM, where p is a proportionality constant
• So, R = (kp)M . . . (I)

• So, to answer the question, we need the value of the product of k and p

Step 3 : Analyze Statement 1 independent

1. If M = 250,000, then R = 10⁶
   • From (I), 10⁶ = (kp)(250,000)
   • From this equation, we can get the value of kp

Thus, Statement 1 alone is sufficient to answer the question.

Step 4 : Analyze Statement 2 independent

• So, R = 4C
  • As R = kC, we have, k = 4

Since we don’t know the value of p yet, we cannot find the value of product kp Thus, Statement 2 alone is not sufficient to answer the question.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in Step 3, this step is not required

Answer: Option A

Step 1 & 2: Understand Question and Draw Inference

Let the number of shirts, trousers and suits be 4x, 5x and 6x respectively.
Total number of items = 4x +5x+6x = 15x.
 

To find: 15x = ?
To answer this question, we need to find the value of x

Step 3 : Analyze Statement 1 independent

1. The number of suits is 100 more than the number of shirts stocked in the store

6x – 4x = 100
i.e. 2x = 100
x = 50
Sufficient to answer

Step 4 : Analyze Statement 2 independent

2. The number of suits is 150 less than the number of shirts and trousers stocked in the store
(4x +5x) – 6x = 150
3x = 150
x = 50
Sufficient to answer

Step 5: Analyze Both Statements Together (if needed)

Since, we have a unique answer from steps 3 and 4, this step is not required.

Answer: D

Given:

• Let the number of literature books be l and the number of science books be s
  • l : s = 5 : 6
• Number of literature books added = 50
  • Total literature books now = l + 50
• Number of science books increased = ¼ * s
  • Total science books now = s + ¼ * s
• New Ratio of literature and science books = 4 : 3

To Find: Initial number of books in the library

• Value of l + s?

Approach:

1. To Find the value of l + s, we need to find the values of l and s.
2. We know that l : s = 5 : 6. So, we need to find another relation between l and s to find unique values of l and s.
3. We are given that the new number of literature books = l + 50 and the new number of science books is s + ¼ * s. The ratio of these literature and science books is 4 : 3
   • So, we have here another relation between l and s

Working out:

a. So, we have 3l + 150 = 5s, i.e. 5s – 3l = 150…….(2)
3. Adding equations (1) and (2), we have 3l = 150, i.e. l = 50
a. Putting the value of l in equation (2), we have 5s = 150 + 150, i.e. s = 60
4. Hence, initial books in the library = l + s = 50 + 60 = 110
Thus, the library initially had 110 books in total.

Answer : B

Given:

• Relation of scores of Alastair, Bell, Cook, Darren and Eoin in an Aptitude test
  • Alastair’s score = ½ * Darren’s Score
  • Darren’s Score = Cook’s score + 1/5* Cook’s score =3/2 * Bell’s score
  • Eoin’s score = Cook’s score + 2/5* Cook’s score
• Average score of the group = 50
  • So, total score of the group = 50*5 = 250

To Find: Range of the scores of the group

• Range of the score = Highest score – Lowest Score

Approach:

1. Range of the score = Highest score – Lowest Score
   • For finding the highest and the lowest score, we need to find the individual scores of all the 5 people
2. Let’s assume the score of any one person to be x. Let’s assume that person to be Alastair
   • As we are given the relation between the scores of all the 5 people, we will express the scores of other people in terms of x.
3. Once we get the scores of all the people in terms of x, we know that the sum of their scores is 250.
   • So, equating the sum of the scores of all the 5 people to 250 will give us an equation in x, which will give us a unique value of x.

Working out:

1. Let Alastair’s score be x……….(1)
2. Darren’s Score
   • We know that, Alastair’s score = ½ * Darren’s Score
   • So, Darren’s score = 2x………..(2)
3. Cook’s Score
   • Now, we know that Darren’s Score = Cook’s score + 1/5 * Cook’s score

4. Bell’s Score

​

5. Eoin’s Score

6. Also, sum of all the scores = 250

So, the range of the scores of the group is 40.

Answer D

Given:

• In ΔABC, let the measure of ∠BAC be a^o , ∠ABC be b^o and ∠ACB be c^o

To Find:

a^o = ?

Approach:

1. To find the value of a^o we need to draft an equation in terms of a^o only. We are given the value of c^o in terms of a^o , and using the given information, we can infer the value of b^o in terms of a^o .
2. We know another relation between a^o, b^o, c^o, - The Angle Sum Property. By using the values of b^o and c^o in terms of a in the Angle Sum relation, we can find the value of a^o

Working out:

Looking at the answer choices, we see that the correct answer is Option B

Given:

• Total students in the class = W + X + Y + Z
• Favorite subject votes were as under:
  • Mathematics = W
  • Science = X
  • History = Y
  • Literature = Z

To Find: % of students who voted for Mathematics as their favorite subject

Approach:

1.  % of students who voted for Mathematics as their favorite subject=  

​

2. So, to find the required percentage, we need to find the values of  

​​

• We’ll use the given ratios to find the other 2 required values

Working out:

• Finding Y/W

Looking at the answer choices, we see that the correct answer is Option C

Step 1 & 2: Understand Question and Draw Inference

Given:

Initial Solution

• Let the weight of
  • salt = t
  • sugar = g
  • water = w
• So, weight of whole solution = w + g + t

Final Solution

​

Step 3 : Analyze Statement 1 independent

(1) Due to the addition of water, the ratio of the weight of sugar in the solution to the weight of the solution decreases from to 

Initial Solution

Thus, Statement 1 is sufficient.

Step 4 : Analyze Statement 2 independent

(2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in each of Steps 3 and 4, this step is not required

Answer: Option D

Given:

• Number of students who did not prepare for the test =3/4 of the total student
  • Number of students who prepared for the test = 1/4 of the total student
• Number of students who prepared and passed the test = 2/3of the students who prepared for the test
• Number of students who passed the test = 25% of the total students
  • Number of students who failed the test = 75% of the total students
     

To Find: fraction of the students who had not prepared for the test and failed the test?

Approach:

1. Required Fraction = ​
   • So, we need to either find the values of both “Number of students who did not prepare and failed the test” and “Number of students who did not prepare for the test” or express both of them in terms of a common variable
   • As we are not given any information about the number of students, we will try to express both of them in terms of a common variable
2. Number of students who did not prepare for the test
   • Let’s assume the total number of students in the class be x.
     As we know that Number of students who did not prepare for the test =  3/4  of the total student, we can find the number of students who did not prepare for the test in terms of x.
   • Also that would give us the number of students who prepared for the test in terms of x.
3. Number of students who did not prepare for the test and failed the test
   • We know that number of students who failed the test = 75% of the total students
   • Also, number of students who failed the test = Number of students who prepared and failed the test + Number of students who did not prepare and failed the test
   • Now, we know that Number of students who prepared and passed the test =2/3 of the students who prepared for the test.
   • So, number of students who prepared and failed the test = (Number of students who prepared) – (number of students who prepared and passed the test)
   • Using the above relation, we can find the number of students who prepared and failed the test in terms of x.

Working out:

1. Number of students who did not prepare for the test = 3/4  of the total student = 
   • So, number of students who prepared for the test = 

2. Number of students who failed the test = 75% of x = 3/4

3. Also, number of students who failed the test = Number of students who prepared and failed the test + Number of students who did not prepare and failed the test

• 3/4 =  Number of students who prepared and failed the test + Number of students who did not prepare and failed the test.
•  Now, we are given that Number of students who prepared and passed the test =  2/3  of the students who prepared for the test =    
• So, number of students who prepared and failed the test = 

• Using (3.a), (3.b) and (3.c), we can write    Number of students who did not prepare and failed the test, which gives us the number of students who did not prepare and failed the test =  2/3

Answer  E

Given:

 where x is a constant and P and I are price of the stock and Inflation respectively

• Inflation α Local Diesel prices
  • I = bD, where b is a constant and D is the local diesel price.
• P = 100 when I = 10
• I = 12 when D = 60

To Find: By how many currency units should the local diesel price fall so that the price of the stock, which is currently at 200 currency units, increases by 25 percent?

• Current Price stock = 200 currency units.
• Increased Price stock = 200 *1.25 = 250 currency units

Approach:

1. Let the current local price of diesel be D and the local price of Diesel at which the price of the stock increases by 25 percent be D’. So, the local Diesel price should fall by (D – D’) units
   • So, we should find the value of D and D’ at the current price and the increased price of the stocks.
2. We need to find the relation between the price of a stock and the diesel prices
   • We know the relation between price of a stock and the inflation, which is given by 
   • We also know the relation between Inflation and local diesel prices, which is given by I = bD
3. Using the above relations, we will establish a relation between price of a stock and the local diesel prices.
4. Once, we establish a relation between price of a stock and the local diesel prices, we will need to find the values of constants b and x to calculate the change needed in local diesel prices when the price of a stock increases by 25 percent.
   • As we are given the price of a stock for a particular level of inflation, we can find the value of x using the relation 
   • Similarly, as we are given the value of inflation for a particular level of local diesel prices, we can find the value of b using the relation I = bD

Working out:

2. Calculating values of x and b

• Price of a stock = 100 when inflation = 10
  • Using the relation   we can write  

which gives us x = 1000

• Inflation = 12 when the local diesel prices = 60
  • Using the relation I = bD, we can write 12 =b *60, i.e. b = 1/5

4. So, the fall in local diesel price = 25 – 20 = 5
Hence, the local diesel price should fall by 5 currency units so that the price of
the stock should increase by 25% from 200.

Answer : A