
The logical reasoning, also known as critical thinking or analytic reasoning, involves one's ability to isolate and identify the various components of any given argument.So, it is not a surprise that the logical reasoning questions commonly appear in any placement tests, competitive exams or entrance exams.
1) 21 km road can be done by 45 workers in 300 days.100 days later only 5 km road is completed. So how many workers should more join to complete work in same time.
solution: 27
explanation :
capacity of doing work of a man before and after will be same
100*45/5=200*x/(21-5)
x=72
total increase in workers=72-45=27
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2)a,b,c,d,e are distinct numbers. If (75-a) (75-b) (75-c) (75-d) (75-e) =2299.
Then a+b+c+d=?
Hint(as given): 2299 is divisible by 11.
solution: 306
explanation:
2299 = 11×11×19×1×1 = 11× (−11)×19× (−1) × 1
but Two of the terms in the given expression should equal to 1. As all the digits are distinct, two of the terms should be negative.
One possible solution = (75 - 64)(75 - 56)(75 - 86)(75 - 74)(75 - 76)
Then a + b + c + d + e = 64 + 56 + 86 + 74 + 76 = 356
But as the sum of only 4 terms was asked, we have to subtract one term.
So given answer can be one of 292, 306, 270, 282, 280.
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3) A beaker contains 180 liters of alcohol. On 1st day, 60 l of alcohol is taken out and replaced by water. 2nd day, 60 l of mixture iss taken out and replaced by water and the process continues day after day. What will be the quantity of alcohol in beaker after 3 days ?
solution: 53.3
explanation:
Final Alcohol = Initial Alcohol ×( 1− Replacementquantity/FinalVolume) ^ n
Where n is no of days.
Here n is 3.
Final Alcohol = 180 ×( 1− 60/180) ^ 3== 180 ×( 2/3) ^ 3=53.3
Hence quantity of alcohol in beaker after 3 days is 53.3
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4) Sum of number between 200 and 300 which is multiple of 3?
solution: A) 8517
explanation:
multiple of 3 between 200 and 300
201,204,207,......297,300
hear a= 201, d=3,last term=300
Last term= a+ (n-1)d
300= 201+ (n-1)x3
Therefore n=34
sum of numbers(201+204+...+297+300)= (n(a+last term))/2 = 34(201+300)/2 = 8517
Sum of number between 200 and 300 which is multiple of 3 is 8517
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5) Mother, daughter and infant total weight is 74 kg. Mother's weight is 46 kg more than daughter and infant's weight. Infant's weight is 60% less than daughter's weight. Find daughter's weight.
solution: A) 10Kg.
explanation:
Mother, daughter and infant total weight is 74 kg.
M+D+I = 74 .....(I)
Mother's weight is 46 kg more than daughter and infant's weight
M= (D+I)+46.....(II)
Substitute value of M in eq(I)
M+D+I = 74
(D+I)+46+D+I = 74
2(D+I)=74-46
2(D+I)=28
D+I = 14.....(III)
Infant's weight is 60% less than daughter's weight.
I= D*(1-60%)=D*(40%)=D*40/100
I=D*2/5
Substitute value of I in eq(III)
D+I = 14
D+D*2/5 = 14
7D/5=14
D=10
Daughters weight is 10Kg.
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6) A man goes upstream for 10 km in 5 hour .another man goes downstream for 10 km in 3 hour .if speed of stream is 15 km/h what is the difference in speed of both men.
B) 3.85 kmph
wrong answer
C) 4.67 kmph
wrong answer
D) 6.33 kmph
wrong answer
solution: A)5.33 kmph
explanation:
A man goes upstream for 10 km in 5 hour
speed in upstream = 10/5 = 2 kmph
another man goes downstream for 10 km in 3 hour
speed in downstream = 10/3 kmph
but speed of stream is 15 km/h
upstream = 15-2 = 13 kmph
downstream = 15+10/3 = 18.33 kmph
difference in speed of both men = 18.33-13.00= 5.33 kmph
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7)Price in taai is 15rs/km. in train 21rs/km. Total distance 450 km. Total amount paid 8130 rs. Journey travel by train?
solution: 230Km
explanation:
let distance by taai is x then by train will be y.
Total distance traveled is 450 km.
x+y=450....(I)
Price in taai is 15rs/km and in train 21rs/km.Total amount paid 8130 rs.
15x+21y=8130....(II)
from (I) & (II),
x=220
y=230
Journey travel by train =y=230Km
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8)There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with a telephone lines such that every two towns are connected with three direct lines if they belong to the same zone, and with only one direct line otherwise. How many direct telephone lines are required?
solution: 90
explanation:
In each zone, there are 3 towns and so each town in any particular zone will be connected to each of the remaining two towns in the same zone with 3 lines( as given in question). So, in each such zone, there will be a total of 3 x 3 = 9 lines connecting towns of a that zone to itself and since there are four such zones, there will be a total of 9 x 4 = 36 lines.
Now, since there are four zones with 3 towns in each zone, each town of a particular zone will be connected by a single line to the remaining 9 towns with a total of 1 x 9 = 9 lines.
Similarly other two towns of first zone will connected to 9 towns.So, the towns of the first zone will be connected to the towns of the remaining three zones with a total of 3 x 9 = 27 lines.
Similarly, the towns of the second zone with now be connected to the towns of the remaining two zones with a total of 2 x 9 = 18 lines and finally the towns of the thirds zone will be connected to the towns of the fourth zone with a total of 9 lines, thus giving us a total of 27 + 18 + 9 = 54 lines.
Hence, the total number of lines = 36 + 54 = 90.
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9)A salesman buy 20 kg rice at the rate of 26 per kg and another rice item 30 kg at the rate of 40 per kg. Now he mix both and sell a part at Rs572 now on what price he sells the remaining part so that he earns total 25% profit .
solution: A) 1578
explanation:
A salesman buy 20 kg rice at the rate of 26 per kg and another rice item 30 kg at the rate of 40 per kg.
Total cost price =(20*26)+(30*40)=1720
he want to earns total 25% profit
25%profit =1720*25/100 =430
so total selling price =1720+430 =2150
he sell a part at Rs572
hence another part should be sell at 2150-572=1578 Rs
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10)A does a work in 15 days B and C does the same work in 12 and 20 days respectively. C left the work after two days and B left before completion of 3 days. Find the total no. of days in which work completed.
solution: B)7.66
explanation:
Let they complete work in "x" days.
A works for x days= x/15
B left before complimantation of 3 days.
B works for (x-3) days=(x-3)/12
C left the work after two days
C works for 2 days= 2/20
x/15+ (x-3)/12 + 2/20=1
x=7.66
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