Two barrels contain a mixture of ethabol and gasoline. The content of ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol?
Answer: B Let the volume of each barrel be 100 ml. In 1st barrel Ethanol = 60 ml, Gasoline = 40 ml. In 2nd barrel Ethanol = 30 ml, Gasoline = 70 ml. Let the ratio of mixture from 1st barrel to 2nd barrel be x : y (60x +30y)/(40x+70y) = 1/1 x/y = 2/1
Q. No. 8:
A rural child specialist has to determine the weight of five children of different ages. He knows from his past experience that each of the children would weigh less than 30 Kg and each of them would have different weights. Unfortunately, the scale available in the village can measure weight only over 30 Kg. The doctor decides to weigh the children in pairs. However, his new assistant weighed the children without noting down the names. The weights were: 35, 36, 37, 39, 40, 41, 42, 45, 46 and 47 Kg. The weight of the lightest child is:
Answer: B Let Five different ages are a, b, c, d, e, such that a < b< c < d < e The ten different pairs are: a + b, a + c, a + d, a + c, b + c, b + d, b + e, c + d, c + e, d + e The sum of all 10 combinations are:- 4(a + b + c + d + e) = 35 + 36 + 37 + 39 + 40 + 41 + 42 + 45 + 46 + 47 = 408 a + b + c + d + e = 102 The sum of two Smallest weight = a + b = 35 The sum of two Largest weight = d + e = 47 Therefore 35 + c + 47 = 102 => c = 20 Now a + c = 36 =>a + 20 = 36 =>a = 16
Q. No. 9:
Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his average by 2. 75 marks in OB further increased his average by 1. Reasoning is the next test, if he gets 51 in Reasoning, his average will be _____?
Answer: A Let the total marks of Rajiv and the number of tests he gave before giving QT be x and n respectively. 83 marks in QT increased his average by 2. Thus, x/n + 2 = (x+83)/(n+1)...........................(i) => 75 marks in OB further increased his average by 1, {(x+83)/(n+1) }+ 1 = (x+83+75)/(n+2)...........(ii) He gets 51 in his next test – Reasoning, Average = (x+158+51)/(n+3)...................(iii) Solving equations (i) and (ii), we get the value of n = 10 and x = 610 From (iii), we get Average = 63
Q. No. 10:
Consider the set S = {2, 3, 4, ……, 2n + 1}, where ‘n’ is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y?
Answer: B Sum of the odd integers in the set S => n/2* (2*3+ (n–1)2) => n(n+2) Therefore the average of the odd integers in set S = n + 2...........(i) Sum of the even integers in the set S => n/2 * (2*2+ (n-1)2) => n(n+1) Therefore the average of the even integers in the set S = n + 1.........(ii) Therefore X – Y = (n + 2) – (n + 1) = 1
Q. No. 11:
Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to
Answer: C The total age of all the eight people in the family = 231 As per the information given in the question, the total age of all the people in the family = 231 + 3 × 8 – 60 + 0 = 195 Similarly the total age of the people in the family four years ago = 195 + 3 × 8 – 60 + 0 = 159. Therefore the current average age of all the people in the family => (159+32)/8 = 24 years
Q. No. 12:
Five times P's income added to Q's income is more than Rs 51. Three times P's income is more than Rs 21 from Q's income. Find out the possible range of values of p and q if they represent P's and Q's income respectively.
Answer: C 5p+q >51..............(i) 3p-q = 21...............(ii) From (ii), p = (21+q)/3 => (21+q)/3 >9 => b>6 Combining eq(1) and b>6 we get => a>9 and b>6