1 .If mean of y and $1\over y$ is M, then what is the mean of
$y^3$and $1\over y^3$?
a. $M(M^23)\over3$ , b.$M^3$ , c.$M^3$3 , d. M( $4M^2$3)
Answer & Explanation
Answer : Option D 

Explanation : 


2 .From a series of 50 observations, an observation with value 45 is dropped but the mean remains the same. What was the mean of 50 observations?
Answer & Explanation
Answer : Option C 

Explanation : 

Let the observation mean = x
âˆ´ Sum of 50 observations = 50x According to question
âˆ´ $50x45\over49$ x
=> 50x45=49x
âˆ´ x= 45 
3 . Read the following information carefully to answer the questions that follow.
The average age of 6 persons living in a house is 23.5 years. Three of them are majors and their average age is 42 years. The difference in ages of the three minor children is same
What is the mean of the ages of minor children?
Answer & Explanation
Answer : Option C 

Explanation : 

Total age of six persons = 23.5 Ã— 6 = 141 years Total age of three major persons = 42 Ã— 3 = 126 years
âˆ´ Total age of three minor children = 141 â€“ 126 = 15 years The difference in ages of the three minor children is same. Therefore, we take ages may be: 5, 5, 5; 3, 5, 7; 2, 5, 8 and 1, 5, 9 In all the cases, median will be 5 years
Mean age of minor children = $15\over3$ = 5 years 
4 . Read the following information carefully to answer the questions that follow.
The average age of 6 persons living in a house is 23.5 years. Three of them are majors and their average age is 42 years. The difference in ages of the three minor children is same
What is the median of the ages of minor children?
Answer & Explanation
Answer : Option B 

Explanation : 

Total age of six persons = 23.5 Ã— 6 = 141 years Total age of three major persons = 42 Ã— 3 = 126 years
âˆ´ Total age of three minor children = 141 â€“ 126 = 15 years The difference in ages of the three minor children is same. Therefore, we take ages may be:
5, 5, 5; 3, 5, 7; 2, 5, 8 and 1, 5, 9
In all the cases, median will be 5 years.
Median age of minor children = 5 years 
5 .In how many ways can a committee of 4 people be chosen out of 8 people ?
Answer & Explanation
Answer : Option C 

Explanation : 

Required number of ways
= $^8$$c_4$ = $8!\over4!(8X4)!$ = $5Ã—6Ã—7Ã—8\over1Ã—2Ã—3Ã—4$ = 70 
6 .
Answer & Explanation
Answer : Option D 

Explanation : 

Male Engineers + Male Designers
40% of (18% of 10500) + 65% of (16% of 10500)
Female Engineers + Female Designers
= 60% of (18% of 10500) + 35% of (16% of 10500)
âˆ´ Required ratio = (40 Ã— 18)
+ (65 Ã— 16) : (60 Ã— 18 + 35 Ã— 16)
= (720 + 1040) : (1080 + 560)
= 1760 : 1640 = 44 : 41 
7 .
Answer & Explanation
Answer : Option C 

Explanation : 

Required % = $19\over21$ X 100 = 90% 
8 .
Answer & Explanation
Answer : Option D 

Explanation : 

% of female professionals =
= [20% of 21% + 60% of 18% + 40% of 11% + 80% of 15% + 40% of 19% + 35% of 16%]
$1\over100$ = [420 + 1080 + 440 +1200 + 760 + 560]%
= $4460\over100$% = 44.6%
âˆ´ % of male professionals
= 100% â€“ 44.6% = 55.4%
âˆ´ Required diff
= (55.4 â€“ 44.6)% of 10500
= 10.8% of 10500
= 10.8 Ã— 105 = 1134 
9 .
Answer & Explanation
Answer : Option D 

Explanation : 

Required %
= $20 of 21\over89of 15$x 100%= $20 X21\over89X15$x 100%
$420\over12$= 35% 
10 .
Answer & Explanation
Answer : Option A 

Explanation : 

Required ratio = $60x11\over20x15$=11.5 
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