Basic Mathematics Aptitude - Mental Ability Notes | EduRev

: Basic Mathematics Aptitude - Mental Ability Notes | EduRev

 Page 1


Question 1 
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as 
a husband, 1/4 as a good politician and 3/40 as a father. And 
the remaining 6 years as a good grand father. Then the living 
days of the man is:  
a) 72 b) 89 c) 80 d) 69 
Answer : c)80 
Solution : 
Lets say his age is X. 
So years spent as student = 5X / 16 + 2, 
as Husband = X/4 + 1, 
as Politician = X/4, 
and as Father = 3X/40. 
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + 
X/4 + 3X/40} 
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40} 
6 = X - [13X / 16 + 3 + 3X / 40] 
6 = X - [71X / 80 + 3] 
9X/80 = 9 
X = 80. 
Hence the age of the man is 80 years. 
Question 2 
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Question 1 
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as 
a husband, 1/4 as a good politician and 3/40 as a father. And 
the remaining 6 years as a good grand father. Then the living 
days of the man is:  
a) 72 b) 89 c) 80 d) 69 
Answer : c)80 
Solution : 
Lets say his age is X. 
So years spent as student = 5X / 16 + 2, 
as Husband = X/4 + 1, 
as Politician = X/4, 
and as Father = 3X/40. 
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + 
X/4 + 3X/40} 
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40} 
6 = X - [13X / 16 + 3 + 3X / 40] 
6 = X - [71X / 80 + 3] 
9X/80 = 9 
X = 80. 
Hence the age of the man is 80 years. 
Question 2 
In a village, every weekend, 1/4 th of men and 1/6 th of women 
participate in the social activity. If the total number of 
participants is 75, out of them 25 are men then the total 
number of men and women in the village is: 
a) 100 b) 200 c) 300 d) 400 
Answer : d)400 
Solution : 
Number of men who participated in social activities = 25 
Therefore, 1/4 th of the men = 25 
Or, total number of men in the village = 100 
Number of women who participated in social activities = 75 - 25 
= 50 
Therefore, 1/6 th of the women = 50 
Or, total number of women in the village = 300 
Hence the required answer is 300 + 100 = 400. 
Question 3 
A father has divided his properties in such a way that one-half 
of total properties goes to A, two-third of the remaining shared 
equally to B, C and D and the rest to E, F, G and H such they 
equally gets Rs.30,000 by their sharing property. Then the 
amount will D get is: 
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000  
Page 3


Question 1 
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as 
a husband, 1/4 as a good politician and 3/40 as a father. And 
the remaining 6 years as a good grand father. Then the living 
days of the man is:  
a) 72 b) 89 c) 80 d) 69 
Answer : c)80 
Solution : 
Lets say his age is X. 
So years spent as student = 5X / 16 + 2, 
as Husband = X/4 + 1, 
as Politician = X/4, 
and as Father = 3X/40. 
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + 
X/4 + 3X/40} 
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40} 
6 = X - [13X / 16 + 3 + 3X / 40] 
6 = X - [71X / 80 + 3] 
9X/80 = 9 
X = 80. 
Hence the age of the man is 80 years. 
Question 2 
In a village, every weekend, 1/4 th of men and 1/6 th of women 
participate in the social activity. If the total number of 
participants is 75, out of them 25 are men then the total 
number of men and women in the village is: 
a) 100 b) 200 c) 300 d) 400 
Answer : d)400 
Solution : 
Number of men who participated in social activities = 25 
Therefore, 1/4 th of the men = 25 
Or, total number of men in the village = 100 
Number of women who participated in social activities = 75 - 25 
= 50 
Therefore, 1/6 th of the women = 50 
Or, total number of women in the village = 300 
Hence the required answer is 300 + 100 = 400. 
Question 3 
A father has divided his properties in such a way that one-half 
of total properties goes to A, two-third of the remaining shared 
equally to B, C and D and the rest to E, F, G and H such they 
equally gets Rs.30,000 by their sharing property. Then the 
amount will D get is: 
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000  
Answer : a) Rs.80,000 
Solution : 
A's share = 1/2 
Then remaining = 1 - 1/2 = 1/2 ...(A) 
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D. 
Total share of B, C and D = 2/3 of 1/2 = 1/3 
Therefore, now remaining property = Total share of E,F,G and H 
= value of (A) - total share of B,C and D 
= 1/2 - 1/3 = 1/6 
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 
1/24 
Given 1/24 = Rs.30,000 
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000 
Total share of B, C and D = 1/3 x 7,20,000 
Then their individual share amount = (1/3 x 7,20,000) / 3 = 
Rs.80,000. 
Hence D's share is Rs.80000. 
Question 1 
In a factory, there are two machines A and B. A machine can 
produce 1000 watts power in 5 hours and by running both the 
machines together can produce same watts power in 3 hours. 
Find the time taken by B machine alone to produce same watts 
of power. 
Page 4


Question 1 
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as 
a husband, 1/4 as a good politician and 3/40 as a father. And 
the remaining 6 years as a good grand father. Then the living 
days of the man is:  
a) 72 b) 89 c) 80 d) 69 
Answer : c)80 
Solution : 
Lets say his age is X. 
So years spent as student = 5X / 16 + 2, 
as Husband = X/4 + 1, 
as Politician = X/4, 
and as Father = 3X/40. 
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + 
X/4 + 3X/40} 
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40} 
6 = X - [13X / 16 + 3 + 3X / 40] 
6 = X - [71X / 80 + 3] 
9X/80 = 9 
X = 80. 
Hence the age of the man is 80 years. 
Question 2 
In a village, every weekend, 1/4 th of men and 1/6 th of women 
participate in the social activity. If the total number of 
participants is 75, out of them 25 are men then the total 
number of men and women in the village is: 
a) 100 b) 200 c) 300 d) 400 
Answer : d)400 
Solution : 
Number of men who participated in social activities = 25 
Therefore, 1/4 th of the men = 25 
Or, total number of men in the village = 100 
Number of women who participated in social activities = 75 - 25 
= 50 
Therefore, 1/6 th of the women = 50 
Or, total number of women in the village = 300 
Hence the required answer is 300 + 100 = 400. 
Question 3 
A father has divided his properties in such a way that one-half 
of total properties goes to A, two-third of the remaining shared 
equally to B, C and D and the rest to E, F, G and H such they 
equally gets Rs.30,000 by their sharing property. Then the 
amount will D get is: 
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000  
Answer : a) Rs.80,000 
Solution : 
A's share = 1/2 
Then remaining = 1 - 1/2 = 1/2 ...(A) 
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D. 
Total share of B, C and D = 2/3 of 1/2 = 1/3 
Therefore, now remaining property = Total share of E,F,G and H 
= value of (A) - total share of B,C and D 
= 1/2 - 1/3 = 1/6 
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 
1/24 
Given 1/24 = Rs.30,000 
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000 
Total share of B, C and D = 1/3 x 7,20,000 
Then their individual share amount = (1/3 x 7,20,000) / 3 = 
Rs.80,000. 
Hence D's share is Rs.80000. 
Question 1 
In a factory, there are two machines A and B. A machine can 
produce 1000 watts power in 5 hours and by running both the 
machines together can produce same watts power in 3 hours. 
Find the time taken by B machine alone to produce same watts 
of power. 
a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins 
Answer : b) 7 hrs & 30 mins 
Solution : 
A type machine can produce 1000 watts power in 5 hours. 
Then A's 1 hour work = 1/5. 
Both A and B can produce in 3 hours. 
And 1 hour work of (A + B)'s = 1/3. 
B's 1 hour work = 1/3 = 1/5 = 2/15. 
Therefore, machine B can produce 1000 watts power in 15/2 
hours. 
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes. 
Question 2 
There is a building with 4 floors and each floor has equal area. 
There are three sweepers P, Q and R. P can clean 1 floor in 10 
hours, Q can clean 8/5 of the floors in 40 hours and R can clean 
4/3 of the floors in 13 hours. Then who will clean all the 4 floors 
at first? 
a) R b) R & P c) Q d) Q & P 
Answer : a) R 
Solution : 
P can sweep 1 floor in 10 hours. 
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours. 
Page 5


Question 1 
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as 
a husband, 1/4 as a good politician and 3/40 as a father. And 
the remaining 6 years as a good grand father. Then the living 
days of the man is:  
a) 72 b) 89 c) 80 d) 69 
Answer : c)80 
Solution : 
Lets say his age is X. 
So years spent as student = 5X / 16 + 2, 
as Husband = X/4 + 1, 
as Politician = X/4, 
and as Father = 3X/40. 
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + 
X/4 + 3X/40} 
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40} 
6 = X - [13X / 16 + 3 + 3X / 40] 
6 = X - [71X / 80 + 3] 
9X/80 = 9 
X = 80. 
Hence the age of the man is 80 years. 
Question 2 
In a village, every weekend, 1/4 th of men and 1/6 th of women 
participate in the social activity. If the total number of 
participants is 75, out of them 25 are men then the total 
number of men and women in the village is: 
a) 100 b) 200 c) 300 d) 400 
Answer : d)400 
Solution : 
Number of men who participated in social activities = 25 
Therefore, 1/4 th of the men = 25 
Or, total number of men in the village = 100 
Number of women who participated in social activities = 75 - 25 
= 50 
Therefore, 1/6 th of the women = 50 
Or, total number of women in the village = 300 
Hence the required answer is 300 + 100 = 400. 
Question 3 
A father has divided his properties in such a way that one-half 
of total properties goes to A, two-third of the remaining shared 
equally to B, C and D and the rest to E, F, G and H such they 
equally gets Rs.30,000 by their sharing property. Then the 
amount will D get is: 
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000  
Answer : a) Rs.80,000 
Solution : 
A's share = 1/2 
Then remaining = 1 - 1/2 = 1/2 ...(A) 
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D. 
Total share of B, C and D = 2/3 of 1/2 = 1/3 
Therefore, now remaining property = Total share of E,F,G and H 
= value of (A) - total share of B,C and D 
= 1/2 - 1/3 = 1/6 
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 
1/24 
Given 1/24 = Rs.30,000 
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000 
Total share of B, C and D = 1/3 x 7,20,000 
Then their individual share amount = (1/3 x 7,20,000) / 3 = 
Rs.80,000. 
Hence D's share is Rs.80000. 
Question 1 
In a factory, there are two machines A and B. A machine can 
produce 1000 watts power in 5 hours and by running both the 
machines together can produce same watts power in 3 hours. 
Find the time taken by B machine alone to produce same watts 
of power. 
a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins 
Answer : b) 7 hrs & 30 mins 
Solution : 
A type machine can produce 1000 watts power in 5 hours. 
Then A's 1 hour work = 1/5. 
Both A and B can produce in 3 hours. 
And 1 hour work of (A + B)'s = 1/3. 
B's 1 hour work = 1/3 = 1/5 = 2/15. 
Therefore, machine B can produce 1000 watts power in 15/2 
hours. 
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes. 
Question 2 
There is a building with 4 floors and each floor has equal area. 
There are three sweepers P, Q and R. P can clean 1 floor in 10 
hours, Q can clean 8/5 of the floors in 40 hours and R can clean 
4/3 of the floors in 13 hours. Then who will clean all the 4 floors 
at first? 
a) R b) R & P c) Q d) Q & P 
Answer : a) R 
Solution : 
P can sweep 1 floor in 10 hours. 
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours. 
Q can sweep 8/5 part in 40 hours. 
Then 1 floor will be done by Q in 40/(8/5) = 25 hours. 
And 4 floors will be done by Q in 25 x 4 = 100 hours. 
R can clean 4/3 part of the floors in 13 hours. 
Then 1 floor will be cleaned by R in 13/(4/3) = 13 x 3/4 hours. 
And 4 floors will be done by R in 13 x 3/4 x 4 = 13 x 3 = 39 
hours. 
Thus P, Q and R takes 40, 100 and 39 hours respectively. 
Hence the sweeper R will complete the work at first. 
Question 3 
A man and his son planned to paint their own house. The man 
alone can do in 20 days. He paints for 4 days and then his son 
completed in 8 days. Find how many days will they take 
working together? 
a) 12 days b) 11 days c) 7 days d) 8 days 
Answer : c) 7 days 
Solution : 
Man's 1 day work = 1/20 
Then 4 days work = 4/20 = 1/5 
Remaining work = 1 - 1/5 = 4/5 
Now, 4/5 work is done by the son in 8 days. 
Then, whole work will be done by him in (8 x 5/4) days = 10 
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