Aptitude Questions (Arithmetic) for AAO and Upcoming Exams 2016

Aptitude Questions (Arithmetic) for AAO and Upcoming Exams 2016
Aptitude Questions (Arithmetic) for AAO and Upcoming Exams 2016 Set-59:
Dear Readers, Important Practice Aptitude Questions for Upcoming AAO Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

1).The average age of 7 students of a class is 16 years. When a new student replace an old student, the average age reduces by 1 years. What is the age of the new student?
a)  9 years
b)  10 years
c)  15 years
d)  cannot be determined
e)  None of these

2).Raj travels a distance of 2 km from his house to the metro station from where he boards metro train to go to his office. The speed of the metro trains 58(2/3) kmph. He takes a total of 2 hours to reach his office from his house to the metro station if the average speed for entire journey is 45 kmph.
a)  4 kmph
b)  5 kmph
c)  3 kmph
d)  7 kmph
e)  None of these

3).The average of 12, 15, 17, 19, x& 21 lies between 15 and 21. Find the value of x, given that x is an integer, greater than the average of the given numbers.
a)  15< x< 42
b)  15< x< 21
c)  16< x< 41
d)  16< x< 42
e)  None of these

4).The number of chocolates with Rani when multiplied by 12, gives the number of chocolates with Kajol. Which are as much above 130 as the number of chocolates with rani are below 130. The average of the number of chocolates with Rani and Kajol is:
a)  40
b)  90
c)  120
d)  None of these
e)  None

5).The average runs made by Dhoni decreased by 2, when he replaced the match in which he had scored 80 runs by two other matches in which he had scored 35 and 50 runs respectively. Later on, which he average 61 runs made in another match the average runs increased by 1. How many matches did he consider initially?
a)  15
b)  18
c)  16
d)  14
e)  None of these

6).The average age of 200 employees of ABC ltd was 50 years in 1992. In 1994, 40 employees having an average age of 58 years left their job. In the year 1997, the company employed 50 more employees, having an average age of 45 years. Find the average age of all the employees in the year 2000
a)  52.5 years
b)  50.2 years
c)  54.47 years
d)  55 years
e)  None of these

Directions (Q. 7-8): Seven years ago, the average age of the family of Mr.Pataudi, consisting of 4 members was 30 years. Mr.Pataudi, then got married to Sharmila and gave birth to son, called Saif. The average age of the family is still the same at present.

7).If the difference between the age of Sharmila and Saif is 24 years then what is the present age of Sharmila?
a)  26 years
b)  28 years
c)  30 years
d)  32 years
e)  None of these

8).If the present average age of Pataudi, Sharmila and Saif is 23 years, then find the age of Pataudi, 5 years from now.
a)  42 years
b)  37 years
c)  40 years
d)  38 years
e)  None of these

9).A newspaper seller distributed newspaper from house to house on his cycle. The speed of the cycle in 20 km/hr. He has to half for a fixed duration of time each hour, as a result of which his speed reduces to 15 kmph. Find the time interval for which the newspaper seller has to halt each hour.
a)  10 minutes
b)  12 minutes
c)  14 minutes
d)  None of these
e)  None

Answer:
1) d)  2) a)  3) d)  4) d)  5) b)  6) c)  7) b)  8) a)  9) d)

Solutions:

1.Cannot be determined. We know the average age is reduced by 1 year, hence the sum of the age is reduced by 7×1=7 years, but we do not know the age of the replaced student (old student), hence age of new student cannot be determined.
Answer: d)

2.We know that average speed
=(Total distance/Total time)
Average =45 and total tome =2
Hence, total distance = 45 ×2 = 90 km
Distance between house to metro station = 2km, Hence 90-2=88 km is covered by metro train at the speed 0f 58(2/3) kmph or (176/3) kmph
Time taken during metro ride = [(88×3)/176]=(3/2)hours=1.5 hours
Time taken to travel from house to station = (1/2) hour in which 2 km is covered, hence speed= 4 kmph
Answer: a)

3.Sum of the given numbers=12+15+17+19+21+x=84+x
We are given 15<[(84+x)/6]<21 i.e. sum of the numbers lies between 90< 84 + x< 126 i.e. x would lie between 6< x< 42
Also, x is greater than the average of the given numbers.
X> [(84 + x)/6]
5x> 84 or x> (84/5)=16.8
Since, x is an integer hence 16< x< 42
Answer: d)

4.Let no. of chocolates with Rani=x
No. of chocolates with Kajol = 12x, we are given that
12x-130=130-x
13x=260, x=20, no. of chocolates with Kajol
=12×20=240 and average =(20+240/2)=130
OR [simply we are h=given that no.of chocolates with Rani and Kajol are equidistant from 130, one is less and other is more by the same no, hence average would be 130 itself]
Answer: d)

5.Let the initial no. of matches br ‘x’ and initial average be ‘a’
Sum of runs made in ‘x’ matches = ax
New average =[(ax+35+50-80)/x+1]=a-2
Ax+5=(a-2)(x+1)or ax+5=ax-2x+a-2
7=a-2x    …..(1)
Also, [(ax+5+61)/x+2] = a-2+1
Ax+66=(a-1)(x+2) or ax+66=ax-x+2a-2
68=2a-x
Multiply eq. (1) by 2, 14=2a-4x[from (1) and (2)]
[(68=2a-x)/(-54=-3x)]
=> x=18& a=43
Answer: b)

6.In 1992, average age of 200 employees=50 years
In 1994 (just before leaving) average age of 200 employees would be 50+2=52 years
Average in 1994 [after 40 employees k=left]
=(52×200-40×58)/160]=[(10400-2320)/160]=50.5
Average in 1997 = 50.5+3=53.5
Average in 1997 [when new employees are recruited]
=[(53.5×160+45×50)/210]=[(8560+2250)/210]=51.476
Average in the year 200 = 51.476+3=54.476 yrs
Answer: c)

7.Seven years ago, the average age of 4 family members was 30 years.
At present the average age of 4 family members should be 30+7=37
Then, [(47×4+S+W)/6]=30
Where SàPresent age of son&
Wàpresent age of wife
So, S+W=30×6-37×4=180-148=32
Since, difference between the age of mother son is 24 years hence
W-24+W=32=>2W=56, W=28 and S=4
Answer: b)

8.The present age of sharmila and saif is 28& 4 years respectively, the average age of all the three is 23 years.
Hence saif +sharmila+pataudi=3×23
28+4+pataudi=69
Pataudi=37
5 years hence, age of pataudi = 37+5=42 years
Answer: a)

9.The distance covered is less by 5 km, which would be covered in (5/20)=(1/4) hours or (1/4) ×60=15 minutes
Answer: d)

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