November 13, 2017

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**Answer the questions independently of each other. **

## Mathematics Test-18

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Question 1 |

**How many even integers n, where 100 ? n? 200, are divisible neither by seven nor by nine?**

A | 39 |

B | 40 |

C | 38 |

D | 37 |

Question 1 Explanation:

There are 101 integers in all, of which 51 are even. From 100 to 200, there are 14 multiples of 7, of which 7 are even. There are 11 multiples of 9, of which 6 are even. But there is one integer (i.e. 126) that is a multiple of both 7 and 9 and also even. Hence the answer is (51 – 7 – 6 + 1) = 39

Question 2 |

**Twenty-seven persons attend a party. Which one of the following statements can never be true?**

A | There is a person in the party who has an odd number of acquaintances. |

B | There is a person in the party who is acquainted with all the twenty-six others. |

C | Each person in the party has a different number of acquaintances. |

D | In the party, the is no set of three mutual acquaintances. |

Question 2 Explanation:

The number 27 has no significance here. Statement 2, will never be true for any number of people. Let us the case of 2 people. If A knows B and B only knows A, both of them have 1 acquaintance each. Thus, B should be knowing atleast one other person. Let us say he knows ‘C’ as well. So now ‘B’ has two acquaintances (A and C), but C has only acquaintance (B), which is equal to that of A. To close this loop, C will have to know A as well. In which case he will have two acquaintances, which is the same as that of C. Thus the loop will never be completed unless atleast two of them have the same number of acquaintances. Besides, statements 1, 3 and 4 can be true.

Question 3 |

**At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p>0 and q>0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?**

A | p > q |

B | p < q |

C | p = q |

D | p = q/2 |

Question 3 Explanation:

The number of goats remain the same. If the percentage that is added every time is equal to the percentage that is sold, then there should be a net decrease. The same will be the case if the percentage added is less than the percentage sold. The only way, the number of goats will remain the same is if p > q.

Question 4 |

**In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?**

A | 15 min |

B | 5 min |

C | 10 min |

D | 20 min |

Question 4 Explanation:

The ratio of the speeds of the fastest and the slowest runners is 2 : 1. Hence they should meet at only one point on the circumference i.e. the starting point (As the difference in the ratio in reduced form is 1). For the two of them to meet for the first time, the faster should have completed one complete round over the slower one. Since the two of them meet for the first time after 5 min, the faster one should have completed 2 rounds (i.e. 2000 m) and the slower one should have completed 1 round. (i.e. 1000 m) in this time. Thus, the faster one would complete the race (i.e. 4000 m) in 10 min.

Question 5 |

**A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals ...**

A | 31 |

B | 91 |

C | 75 |

D | 63 |

Question 5 Explanation:

Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91. Among these only 31 and 91 are a part of the answer choices. Among these, 10235 (31)10? (11111)2? (1011)3? (111)5 Thus, all three forms have leading digit 1. Hence the answer is 91.

Question 6 |

**Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is ...**

A | 4.0 |

B | 4.5 |

C | None of these. |

D | 1.5 |

Question 6 Explanation:

We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2 or x = 1.5. Thus smallest value of g(x) = 3.5

Question 7 |

**Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be ...**

A | 90.5 |

B | 87.5 |

C | 92.5 |

D | 85.5 |

Question 7 Explanation:

The surface area of a sphere is proportional to the square of the radius. The volume of a sphere is proportional to the cube of the radius.

Question 8 |

**When the curves y = log 10 x and y = x-1 are drawn in the x-y plane, how many times do they intersect for values x ? 1?**

A | Once |

B | Twice |

C | More than twice |

D | Never |

Question 8 Explanation:

Question 9 |

**A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than ...**

A | 3 |

B | 12 |

C | 6 |

D | 9 |

Question 9 Explanation:

Question 10 |

**The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?**

A | None of the these. |

B | 1st |

C | 12th |

D | 9th |

Question 11 |

**The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at ...**

A | x = 2.7 |

B | x = 2.5 |

C | None of these. |

D | x = 2.3 |

Question 11 Explanation:

Case 1: If x < 2, then y = 2 – x + 2.5 – x + 3.6 – x = 8.1 – 3x.
This will be least if x is highest i.e. just less than 2. In this case y will be just more than 2.1
Case 2: If ?? 2x2.5 , then y = x – 2 + 2.5 – x 3.6 – x = 4.1 – x
Again, this will be least if x is the highest case y will be just more than 1.6.
Case 3: If 2.5x3.6 ?? , then y = x – 2 + x – 2.5 + 3.6 – x = x – 0.9
This will be least if x is least i.e. X = 2.5.
Case 4: If In this case y = 1.6 X3.6 ? , then y = x – 2 + x – 2.5 + x – 3.6 = 3x – 8.1
The minimum value of this will be at x = 3.6 = 27
Hence the minimum value of y is attained at x = 2.5

Question 12 |

**Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ? 0? x + 2y – 3z = p 2x + 6y – 11z = q x – 2y + 7z = r**

A | 5p – 2q + r = 0 |

B | 5p – 2q – r = 0 |

C | 5p + 2q + r = 0 |

D | 5p + 2q – r = 0 |

Question 12 Explanation:

It is given that p?q?r?0, if we consider the first option, and multiply the first equation by 5, second by –2 and third by –1, we see that the coefficients of x, y and z all add up-to zero. Thus, 5p – 2q – r = 0 No other option satisfies this.

Question 13 |

A | Standard 60 bags, Deluxe 90 bags |

B | Standard 100 bags, Deluxe 60 bags |

C | Standard 50 bags, Deluxe 100 bags |

D | Standard 75 bags, Deluxe 80 bags |

Question 13 Explanation:

Let ‘x’ be the number of standard bags and ‘y’ be the number of deluxe bags. Thus, 4x + 5y ? 700 and 6x + 10y ? 1250.
Among the choices, Standard 50 bags, Deluxe 100 bags, and Standard 60 bags, Deluxe 90 bags do not satisfy the second equation. Standard 100 bags, Deluxe 60 bags is eliminated as, in order to maximize profits the number of deluxe bags should be higher than the number of standard bags.

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