QuantitativeAptitude Shortcuts Tricks and Methods - Formulas and Handy Notes - Download
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OK Lets Start our Tricks on Chapter TIME AND WORK
We can divided into two parts:
1.First Set/First part
of problem consists of below information
1. Nos of men
2. Nos of days
3. Nos of hours
4. Quantum of work
2.Second set/Second parts may anything missing from above
details and anyone would be common(refer problem no.1 ) which is actually the question to find out.
Formula for TIME AND WORK PROBLEMS
M1 x D1 x H1 M2 x D2 x H2
—————- = —————-
W1 W2
1. 36 men can do a piece of work in 25
days. In how many days can 30 men do it?
Here , the work
is same.. So,
(M XD) / W = (m X d) /
w
= > (M XD)
/ W = (m X d )/ w
[ Cancel W for both sides]
= > 36 X 25 = 30
X d
= > (36 X 25) /
30 = d => d = 900/3 = 30
2.If
6 men dig 100 meters in 20 days working 8 hours , how
many men are required to dig 200 meters in 10 days working 4 hours?
Sol:
6 x 20 x8 M2 x 10×4
———— = —————— (solve this equation) Answer is 48 men
100 200
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Ok Lets work with some complex Problems:
3.A man engaged 10 laborers to make 320 toys in 5 days. After 3 days he found that only 120 toys were made. How many additional men should he engage to finish the work in time?
Problem Explanation
10 laborers=1/5 days=>320 toys(engaged) but after 3days -- eq1
10 laborers=1/3 days=>120 toys (only completed from engaged)-- eq2
M laborers =1/2 days=>200 toys (sub eq1- eq2) additional men
Now put in our formula
till completed = pending
Note: above 6/60 is 6/L.C.M value(60)
Ok Lets work with some complex Problems:
3.A man engaged 10 laborers to make 320 toys in 5 days. After 3 days he found that only 120 toys were made. How many additional men should he engage to finish the work in time?
Problem Explanation
10 laborers=1/5 days=>320 toys(engaged) but after 3days -- eq1
10 laborers=1/3 days=>120 toys (only completed from engaged)-- eq2
M laborers =1/2 days=>200 toys (sub eq1- eq2) additional men
Now put in our formula
till completed = pending
(10
X 3) / 120 = (10+M) x 2 / 200
= > 10+M = 25 => X = 15
A+B+C------->6/60 = 10 answer
4.A, B and C can do a
job in 20 days, 30 days and 60 days respectively. If they work together, in how
many days will the work be finished?
A work =1/20 days , B work =1/30 days and C work = 1/60 days
A+B+C= (1/20)+(1/30)+(1/60) = (3+2+1)/60 =
6/60 = 1/10
So, the number of days is = 10
Shortcut = product/sum i.e Take L.C.M
work=Days ---> L.C.M for A,B and C is 60
A=20 ----> 3
B=30 ----> 2
C=60 ----> 1
---
(sum)6
---
Note: above 6/60 is 6/L.C.M value(60)
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Per day work of A and B = 1/20
5.Two taps A and B can
fill a tank in 10 hours and 15 hours respectively. a third tap C can empty the
full tank in 12 hours. How many hours will be required if all of them are
opened simultaneously to fill in an empty tank completely?
Here, first two are Inlets and the last one is Outlet,
So, (1/10)+(1/15)-(1/12) =
(6+4-5)/60 = 5/60 = 1/12
So, our answer is 12
6. A and B can do a
job in 12 days. B and C in 15 days and C and A in 20 days. In how many days can
they finish it if they work TOGETHER?
A+B = 12
B+C = 15
C+A = 20
So
here, A+B's One day's work = > 1/ (A+B) = 1/12
B+C's one day's work
=> 1/ (B+C) = 1/15
C+A's one day's work =
> 1/(C+A) = 1/20
Just Add them => 1/(
2A+2B+2C) = 12/60 = 1/5
=> 1/2(A+B+C) = 1/5
=>1/(A+B+C) =
1/10 [this is their one day's work TOGETHER]
So, they
can finish it in 10 days :)
7. A and B can
do a job in 12 days. B and C can do the same job in 15 days. C and A in 20
days. In how many days can A alone finish the whole task???
A+B = 1/12
B+C = 1/15
C+A = 1/20
Here we need A, so take a pair which is NOT HAVING A
and subtract it from the others,
so, A+B-(B+C)+C+A =
A+B-B-C+C+A = 2A
But according to our Problem, 2A = (1/12)-(1/15)+(1/20)
=>A = 1/30 (this is A's one day work but we need A's total work)
=> A = 30 Days
But according to our Problem, 2A = (1/12)-(1/15)+(1/20)
=>A = 1/30 (this is A's one day work but we need A's total work)
=> A = 30 Days
8. A and B
can do a piece of work in 20 days. A alone can do it in 30 days. In how many
days can B alone do it?
Per day work of A and B = 1/20
Work done by A= 1/30
So, B's one day work = 1/20 - 1/30 =
(3-2)/60 = 1/60
=>
B's work is 60 Days
Short Cut : Product/diff =
600/10 = 60 :)
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