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**DIRECTIONS for questions 1 to 9**: In each of these questions, two quantities are mentioned, at column A and at column B.

To mark the answers to the questions, use the following instructions.

A: If the quantity in column A is greater.

B: If the quantity in column B is greater.

C: If the two quantities are equal.

D: If the relationship cannot be determined.**Example 1:**

Column A | Column B |

The circumference of a circle with radius 2 | The sum of the circumferences of two circles, each with radius 1 |

**Ans:** C

**Sol**: Circumference of circle = 2 Ã— 3.14 Ã— r.

So, for r = 2, C = 2 Ã— 3.14 Ã— 2 = (4 Ã— 3.14).

For, r = 1, C = 2 Ã— 3.14 Ã— 1.

For two circles of this type, C = (2 Ã— 3.14 + 2 Ã— 3.14) â‡’; 4 Ã— 3.14.

We got the same result from both the columns.

Hence, the answer is (C)

**Example 2:**

Column A | Column B |

2y | 100 |

**Ans:** A

**Sol:** y = sum of first ten positive integers = (10 Ã— 11) Ã· 2 = 55

A = 2 Ã— 55 = 110.

B = 100

A is greater than B. So, the answer is option (A).**Example 3:**

Column A | Column B |

The number of ways in which 6 can be expressed as a product of two different single digit positive integers | The number of ways in which 12 can be expressed as a product of two different single digit positive integers |

**Ans:** C**Sol:**

Column A : 6 can be expressed as 6 Ã— 1 and 2 Ã— 3. (2 ways)

Column B : 12 can be expressed as 12 Ã— 1, 6 Ã— 2 and 3 Ã— 4 (3 ways). Out of these, 12 Ã— 1 contains a two-digit number. As that is not to be considered, the valid number of ways reduces to 2. Thus, in both the cases, the valid number of cases is 2. Hence, the answer is option (C).**Exmaple 4:**

Column A | Column B |

The maximum number of small doughnuts that can be made with 3 packages of mix | The maximum number of large doughnuts that can be made with 4 packages of mix |

**Ans: **A

**Sol:** Small doughnuts: 1 pack makes 12, 3 packs makes 12 x 3 = 36;

Large doughnuts: 1 pack makes 8, 4 packs makes 4 x 8 = 32. So, the answer is (A).**Example 5:** **A = {1, 2, 3} B = {4, 5, 6, 7}**

Column A | Column B |

The total number of ordered pairs (a, b) that can be formed where a is from set A and b is from set B | The total number of ordered pairs (a, b) that can be formed where a is from set AâˆªB and b from set A |

**Ans:** B

**Sol****:** n(A ) = 3, n( B) = 4, n(AâˆªB ) = 7;

Ordered pair from set A to set B is = 3 x 4 = 12,

Number of ordered pairs from set A âˆªB to set A is = 7 x 3 = 21. So, the answer is (B)**Example 6:** **Three times the sum of x and y is 18**

Column A | Column B |

Twice the sum of x and y | 12 |

**Ans:** C

**Sol:** 3 (x + y) = 18. (x + y) = 6.

2 (x + y) = 12. So, the answer is (C).**Example 7:** a + b = 0

Column A | Column B |

a.a.a | b.b.b |

**Ans:** D**Sol:** a + b = 0; a = - b.

For b = 2, a = -2,

a3 = -8, b3= 8

But for b = -2, a = 2,

a3= 8, b3= -8.

So, the answer is (D).**Example 8: The total cost of 1 apple and 1 orange is Rs. 1.70**

Column A | Column B |

The cost of one apple | The cost of one orange |

Ans :D

Sol: Let the cost of 1 apple = Rs. x, cost of 1 orange = Rs. y

Then, x + y = Rs. 1.70 but various combinations of x and y are possible. So, the answer is (D)**Example 9: When Rina was 10 years old, the price of a certain item was Rs. 100.**

Column A | Column B |

The price of the same item when Rina will be 12 years old | Rs. 100 |

**Ans:** D**Sol:** No relation between age of Rina and price of the item is mentioned. So, the answer is (D).

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