Aptitude - Time and Work - Discussion

Discussion Forum : Time and Work - General Questions (Q.No. 6)
6.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
4 days
5 days
6 days
7 days
Answer: Option
Explanation:

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2

Solving these two equations, we get : x = 1 and y = 1 .
100 200

(15 men + 20 boy)'s 1 day's work = 15 + 20 = 1 .
100 200 4

15 men and 20 boys can do the work in 4 days.

Discussion:
193 comments Page 1 of 20.

Vinutha said:   9 months ago
6m + 8b = 1/10.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
(143)

Kumar said:   1 year ago
6m + 8b = 1/10.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
(120)

YASEEN said:   10 months ago
Here in 1st case there are 6 men and 8 boys = [6+8=14] and their work done is 10 so 10*14=140[total workdone by men and boys]
Similarly in 2nd case 26+48=74 , 74*2=140
Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35*x
from the above work done take anyone
140=35*x
x=4
(103)

Nav Raj Bhatta said:   3 years ago
Find the relation between first, second to third equation.

6M+8B=10
26M+48B=2
15M+20B=?.

Now taking first equation;
6M + 8B = 10.
60M + 80B = 1.
30M + 40B = 2.
15M + 20B = 4.
(63)

Yamuna said:   1 year ago
6m + 8b = 1/10.
2(3m + 4b) =1/10.
3m + 4b = 1/20.
5(3m + 4b) = 5(1/20).
15m + 20b = 1/4.
(54)

YASEEN said:   10 months ago
Here in 1st case, there are 6 men and 8 boys = [6+8=14] and their work done is 10.
So, 10*14 = 140[total work done by men and boys,

Similarly,
In 2nd case 26+48 = 74,
74*2 = 140.

The Total work done by 15 men and 20 boys is 15+20=35 and let their work done be x.
Now 35 * x
.
From the above work done take anyone.
140 = 35*x,
x = 4.
(52)

Arun said:   1 year ago
From given data,

6M+8B ---> 10days.
26M+48B ---> 2days.
15M+20B ---> t days.

By using Product approach
WF1*T1=WF2*T2.

(6M+8B)*10=(26M+48B)*2,
4M=8B,
M/B=2/1,
M=2;B=1.

Now substitute We get;
20 --->10 days.
100 --->2 days.
50 ---> t days.
100*2=50*t(Product approach)
t=4 i.e.,4 days.
(34)

Parkavi said:   3 years ago
(6m+8b)*10 = (26m + 48b) * 2.
(6m+8b)*5= (26m + 48b),
20m+40b =26m + 48b.
4m. = 8b,
M/b = 2/1.

Now;
(6*2+8*1) * 10 = (15 * 2 + 20 * 1) * X.
12+8 ) * 10 = (30+20) * X,
200. = 50 * X.
X = 4.
(30)

Sujan gautam said:   2 years ago
15/6=2.5, 20/8=2.5.
Then , 10(6m and 8b) /2.5=4.
(18)

Lung said:   4 months ago
(6M + 8B)× 10 = (26M+48B)×2
30M+40B = 26M+48B
M=2, B=1 {their efficiency}
Now, total work = 6×2+8×10 = 200
therefore work done by 15M + 20B = 200/15×2+20 = 4.
(17)


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