Answer :
42 people in 30 days did 3/5th of the work
Let x people are required to complete the remaining 2/5th of the work in 8 days
Applying chain rule we get x = 42 × 30/8 × (2×5)/(3×5) = 105 people
Additional no. of people required = ( 105 - 42 ) = 63.
Answer :
Tank part filled by Pipe A in 1 hour = 1/8
or
Time taken to fill 1/8 part of tank filled A = 1 hour ------------- (1)
Tank part filled by Pipe B in 1 hours = 1/6
or
Time taken to fill 1/6 part of tank filled B = 1 hour ---------(2)
=> Tank Part filled by (A+B) in 1 hour = (1/8 + 1/6)
=> Tank Part filled by (A+B) in 2 hour = 2*(1/8 + 1/6) = 2*(3+4)/24 = 2*7/24 = 7/12
After 2 hours left over portion of tank = (1-7/12) = 5/12
Time taken to fill 1/6 part of tank filled B = 1 hour
so time taken to fill 5/12 part of tank by B = 6 * 5/12 = 5/2 = 2 1/2
=> So it takes 2 1/2 hrs (2hrs 30 min) to fill 5/12 of tank.
Answer :
(C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)
Answer :
Clearly, the lady is the grandmother of man's sister's son.
i.e., the mother of the mother of man's sister's son.
=> The mother of man's sister.
So, the lady is man's mother.
Answer :
Brother of person's wife brother-in-law of the person. Hence, the son of lady's brother is brother-in-law of the person.
Therefore, the brother of the lady is the father-in-law of the person. Hence, the lady is the sister of the person's father-in-law.
Answer :
If values are proportional to 13 : 11 : 7, then the number of coins will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28.
Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.
Question 7. Simplify (0.001344 / 0.3 X 0.7) = ?
Answer :
(0.001344 / 0.3 x 0.7) = 0.0064
Question 8. The Difference Of Two Numbers Is 11 And One Fifth Of Their Sum Is 9. The Numbers Are ?
Answer :
x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28
Question 9. If A, B, C, D, E Are Five Consecutive Odd Numbers, Their Average Is ?
Answer :
b = a + 2
c = b + 2 = a + 4
d = c + 2 = a + 6
e = d + 2 = a + 8
Therefore, Required average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5
= a + 4
Answer :
If the average of remaining numbers be x, then
20 × 15 = 5 × 12 + 15x
⇒ 300 = 60 + 15x
⇒ 15x = 300 – 60 = 240
=> x = 240/15 = 16
Question 11. What Is The Next Number Of The Following Sequence 125102550(?).
Answer :
Consider the series as 1 2 5 10 25 50 ____
1*2 = 2
2*2.5 = 5
5*2 = 10
10*2.5 = 25
25*2 = 50
50*2.5 = 125
Answer :
120n/75d=1
n/d=75/120=5/8
Answer :
A investment for 12 months = 60,000*12 = 720000
B investment for 6 months = 90000 * 6 = 540000
A & B investment ratio = 72/54 => 4/3
So, B' share = 126000 * 3/7= 54000
Answer :
Lets consider c's salary = Rs. 100
B's will be 90
C's will be 100(1-25/100)=100*(75/100)=75
[(90-75)/75]*100=(15/75)*100=20%
Answer :
Let amount of sand and cement as 3x and x
3x+x = 1000kg
x = 250
Sand : cement => 750 : 250
(750+y)/250 = 6/1
=> y = 750 kg.
Answer :
Lets assume the age of all the 6 people = 20 years.
After 5 years i.e. now their ages will be 25 years.
Total age of all the 6 will be = 25*6 = 150
Including baby's age, avg is 22 years
Let x be the age of baby.
(150+x)/7 = 22
=> x = 4 years.
Answer :
114 km/h => 114*5/18 m/s => 95/3 m/s
66km/h => 66*5/18 m/s => 55/3 m/s
Time taken = (246+304)/(55/3 + 95/3) => 11 sec
Answer :
Average would be : 350 = (275 + x)/2
On solving, x = 425.
Answer :
Let d be the average daily expenditure
Original expenditure = 35 × d
New expenditure = 35 × d + 42
New average expenditure will be :
(35 × d + 42)/42 = d - 1
On solving, we get d = 12
Therefore original expenditure = 35 × 12 = 420
Answer :
Let x be the number of passengers and y be the fare taken from passengers.
3xy + 50xy = 1325 => xy = 25
Amount collected from II class passengers = 25 × 50 = Rs. 1250.
Answer :
Let d be the distance between A and B
So, d/12 + d/6 = 3 d = 12 km
Answer :
x/5 + 22 = 3x/4 ⇒ x = 40 litres
Answer :
A can give B a 100 m start and C a 150m. Start means when A runs 1000m, B runs 900m and C runs 850m.
When B runs 1000m, C will run 1000 x (850/900) m (i.e. 8500/9 m)
Thus, B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.
Answer :
Let Boys in class = B
Girls in class = 20
Now, (20B+15*20)/(B+20) = 18
⇒ B = 30
Answer :
Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock.
It looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm .
In 54 hrs , it lost 15 mins.
so 5 mins are lost in 5*54/15 = 18 hrs
so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday
Answer :
Let n be the mixture quantity of milk = 7/12 × 60 Ltr = 35 Ltr
Quantity of water = 5/12 × 60 Ltr. = 25 Ltr
12 Ltr. of mixture removed contains milk = 7/12 × 12 Ltr. = 7 Ltr and water = 5/12 × 12 Ltr = 5 Ltr
After adding 8 Ltr of pure milk,
Net milk in the mixture = 35 - 7 + 8 = 36 Ltr.
Net water in the mixture = 25 - 5 = 20 Ltr.
So the required ratio of milk and water now = 36/20 = 9/5
Answer :
Total work = 1/20 + 1/30
(3+2)/60 = 5/60 = 1/12
For both of them needed = 12 min.
But due to leakage it take = 15 min.
So we can say for filling pipe it takes 3 minutes
work for the waste pipe will be = 12 min.
3 min. of both filling pipe = 12 min. of waste pipe
12 min. of filling pipe will be = 48 min. of waste pipe i.e.
Fidelity Aptitude Related Tutorials |
|
---|---|
Accounts and Finance for Managers Tutorial | International Finance Tutorial |
Fidelity Aptitude Related Practice Tests |
|
---|---|
Financial Accounting Practice Tests | Accounts and Finance for Managers Practice Tests |
Business Finance Practice Tests | Finance Practice Tests |
All rights reserved © 2020 Wisdom IT Services India Pvt. Ltd
Wisdomjobs.com is one of the best job search sites in India.