Aptitude model placement papers with solution - set 11
1. In a group containing 6 cows and 4 buffalos, 4 livestock are to be selected in such a way that at least 1 cow should always be present. How many way of doing that are possible?A. 209
B. 205
C. 194
D. 163
View Answer / Hide AnswerANSWER: A. 209
Solution: Total number of ways of selecting 4 livestock from 6 cows and 4 buffalos: 10C4
There will be only one way in which no cow will be selected i.e. when all the selected animals are buffallos.
Number of ways 4 livestock are selected such that atleast 1 cow is present: 10C4 - 1
=> ((10*9*8*7)/(4*3*2*1))-1
=>209
A. 7/9
B. 5/12
C. 1/6
D. 5/9
View Answer / Hide Answer3. What is the smallest number which is a perfect square and is divisible by 16, 20 and 24?A. 2400
B. 3600
C. 7200
D. 1600
View Answer / Hide AnswerANSWER: B. 3600
Solution: 3600. The number is perfect square of 60.
It is divisible by 36 24 and 20. Only two numbers 3600 and 7200 fit these conditions.
It is clear that 3600 is the smaller number.
A. 34.25
B. 35
C. 33.25
D. 34.25
View Answer / Hide AnswerANSWER: D. 34.25
Solution = Average runs scored by Virat Kohli in 10 matches: Total runs scored/10—(1)
Average runs scored by Virat Kohli in 6 matches: (Total runs in 6 matches)/6
=> 42 = Runs/6
=>252 runs
Using (1):
=>38.9 = Runs/10
=>389 runs
Runs scored in rest 4 matches: 389 - 252 runs
=> 137 runs
Average runs scored by Virat Kohli in 4 matches: 137/4 = 34.25 runs
A. 108
B. 86
C. 74
D. 92
View Answer / Hide AnswerANSWER: A. 108
Solution: It is given that 243 has been divided into 3 numbers x y and z for instance.
Second condition mentioned in the question is:
=> x/3 = y/4 =z/2
Clearly the biggest number is y. So let us solve w.r.t. y.
=> x = 3y/4
=> x =3z/2
=> z = y/2
=> x+y+z = 243
=> y/2 + 3y/4 + y = 243
9y/4 = 243
=> 243*4/9
=> 108
A. 375
B. 1175
C. 675
D. 2125
View Answer / Hide AnswerANSWER: B. 1175
Answer: Total number of students who appeared for the examination: 2500
35% of these students failed in Engish: (35/100)*2500 = 875
42% candidates failed in Hindi: (42/100)*2500 = 1050
15% candidates failed in both English and Hindi: (15/100)*2500 = 375
Now, to find out the number of candidates who failed in either of the two but not both:
=> No. of students who failed only in English: 875 - 375 = 500
=> No. of students who failed only in Hindi: 1050 - 375 = 675
=> 500 + 675
=>1175
A. 12
B. 24
C. 6
D. 8
View Answer / Hide AnswerANSWER: A. 12
Solution: We know that 26=64
Hence, 26 = 2n/2
=> n = 12
A. 25
B. 33.333
C. 66.667
D. 40
View Answer / Hide AnswerANSWER: C. 66.667
Solution: Let listed price of the product be x
Hence at 25% discount rate the Cost Price will become: 3x/4
Now to make an overall profit of 25% over listed price:
SP = CP+(25/100)CP
=> SP = x+x/4
=>SP = 5x/4
=>Selling Price over Cost price: (5x/4)-(3x/4)
=>x/2
=>%Selling price over Cost price: (x/2)/(3x/4)*100
=>66.67%
A. Rs 2250
B. Rs 1550
C. Rs 450
D. Rs 900
View Answer / Hide AnswerANSWER: D. Rs 900
Solution = It is given that business partnership is independent of time in this case as all the investors have invested their money for same amount of time.
Let A’s capital be x, B’s capital be y and C’s capital be z.
=>x+y+z = 4650
=> 4x = 6y = 10z
=>C’s share is z so let us solve the question w.r.t. z
=>x = 5z/2
=>y = 5z/3
=>(5z/2)+(5z/3)+(z) = 4650
=>31z = 4650*6
=>z = 900
A. 12 days
B. 3 days
C. 4 days
D. 6 days
View Answer / Hide AnswerANSWER: B. 3 days
Solution:
More no of people: Less days (Inverse Relationship)
More work: More days (Direct Relationship)
Ratio is given:
Persons: 1 :: 2
Work 1 :: 1/2
Time 12 :: x
Solution of this can be explained by solving this ratio by the sort of relationships they possess with time.
1*1/2*12 = 2*1*x
=> x = 3 days
