ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

A and B complete a work in = 15 days
One day's work of (A + B) = 1/15
B complete the work in = 20 days;
One day's work of B = 1/20
Then, A's one day's work = 1/15 - 1/20
= (4 - 3)/60
= 1/60
Thus, A can complete the work in = 60 days.

Work done by X in 8 days = 1/40 x 8 = 1/5
Remaining work = 1 - 1/5 = 4/5

Now, 4/5 work is done by Y in 16 days.
Whole work will be done by Y in = 16 x 5/4 = 20 days.
X's 1 day's work = 1/40
Y's 1 day's work = 1/20
(X + Y)'s 1 day's work = 1/40 + 1/20 = 3/40

Hence, X and Y will together complete the work in = 40/3 = 13(1/3) days.

(4M + 6W) × 8 = (3M + 7W) × 10
=> M / W = 11 : 1
Now , (4 × 11 + 6 × 1) × 8 = 10 × 1 × T
=> T = 40 days

P's 1 day 1 hrs work = 1/96 part.
Q's 1 day 1 hr's work = 1/80 part.
P & Q together work in 1 hr per day = 1/96 + 1/80 = 11/480 part.
P & Q together completes in 8 hr per day = 11×8/480 = 11/60 part.
So, days required = 60/11 = 5(5/11)

Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work.
Then,
10:13 :: 23:x
⇒ x = (23 × 13)/10
⇒ x = 299/10
A's 1 day's work = 1/23
B's 1 day's work = 10/299
(A + B)'s 1day's work = 1/23 + 10/299
= 23/299
= 1/13
∴ A and B together can complete the work in 13 days.

Number of pages typed by Robi in 1 hour = 32/6 = 16/3 .
Number of pages typed by Kamal in 1 hour = 40/5 = 8
Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3
Therefore, Time taken by both to type 110 pages = 110/(40/3) hours
= 110 × 3/40 hours (or) 8 hours 15 minutes.

Suppose A, B and C take x, x/2, x/3 days respectively to finish the work
Then,
1/x + 2/x + 3/x = 1/2
⇒ 6/x = 1/2
⇒ x = 12
So, B takes 12/2 = 6 days to finish the work.

Work done by X in 8 days = 1/40×8 = 1/5

Remaining work = 1−1/5 = 4/5

Now, 4/5 work is done by Y in 16 days
Whole work will be done by Y in = 16×5/4 = 20 days
∴ X's 1 day's work = 1/40

∴ Y's 1 day's work = 1/20

(X + Y)'s 1 day's work
=1/40 + 1/20
=3/40

Hence, X and Y will together complete the work in
= 40/3
= 13(1/3) days

A's 5 days work = 50%
B's 5 days work = 33.33%
C's 2 days work = 16.66% [100 - (50 + 33.33)]

Ratio of contribution of work of A, B and C = 50: 33(1/3) : 16(2/3)
= 3 : 2 : 1

A's total share = Tk. 1500
B's total share = Tk. 1000
C's total share = Tk. 500

A's one day's earning = Tk.300
B's one day's earning = Tk.200
C's one day's earning = Tk.250

(A + B + C) do 1 work in 10 days.

So (A + B + C)'s 1 day work = 1/10
And as they work together for 4 days,
so, work done by them in 4 days = 4/10 = 2/5
Remaining work = 1 - (2/5) = 3/5
(B + C) takes 10 more days to complete 3/5 work.
So ( B+C)'s 1 day work = 3/50
Now A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work
= 1/10 - 3/50
= 1/25
A does 1/25 work in in 1 day
Therefore 1 work in 25 days.

লাইভ পরীক্ষায় প্রশ্নটি অসম্পূর্ণ ছিলো। প্রশ্নের শেষ লাইনে 'A' এর কথা উল্লেখ ছিলো না। ‘A’ - এর কথা উল্লেখ থাকলে প্রশ্ন ও সমাধান ঠিক ছিলো।

Ratio of rates of working of A and B = 2:1.
So, ratio of times taken = 1:2
Therefore, A's 1 day's work = 1/9
B's 1 day's work = 1/18
(A + B)'s 1 day's work = 1/9 + 1/18 = 1/6
So, A and B together can finish the work in 6 days.

A : B : C
Ratio of efficiency - 3 : 1 : 2
Ratio of No. of days - 1/3 : 1/1 : 1/2
or 2 : 6 : 3

4 men can repair a road in 7 hours
1 men can repair a road in 7 × 4hours = 28 hours
28 hours is required by 1 men
2 hours is required by 28/2 men = 14 men

A + B = 70%
B + C = 50%
[∵ (A + B) + (B + C) − (A + B + C) = B]
=> B = 20% ; A = 50% and C = 30%
Hence A is most efficient.

Let 'B' alone can do the work in 'x' days
6/30 + 18/x = 1
=> x = 22.5
1/30 + 1/22.5 = 7/90
=> 90/7 = 12 (6/7) days

Quarter of Kg means 250 gm
Less weight, less price (Direct Proportion)
So, 250 : 200 :: 60 : x
=> x = 48
So 200 gm will cost 48 poysa.

1 2 3 4 5 6 (Shahadat) 7 (Nitu) 8 (Altaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14.

Assume first child (the youngest) get = Tk. x
According to the question ;
each son having Tk. 30 more than the younger one

Second child will get = Tk. x + 30
Third child will get = Tk. x + 30 + 30 = x + 60
Fourth child will get = Tk. x + 30 + 30 + 30 = x + 90
Fifth child will get = Tk. x + 30 + 30 + 30 + 30 = x + 120

Total amount they got = Tk. 2000

x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Tk. 340
So the youngest child will get Tk. 340.

Originally let there be x men.

Less men, More days (Indirect Proportion)

Therefore, (x - 10) : x :: 100 :110

=> (x - 10) × 110 = x × 100
=> x = 110

Let the required number of working hours per day be x.
More pumps , Less working hours per day (Indirect Proportion)
Less days, More working hours per day (Indirect Proportion)
Pumps 4 : 3 and Days 1 : 2 } :: 8:x
=> (4 × 1 × x) = (3 × 2 × 8)
=> x = 12

Let the required Price be Tk. X .
Then, Lest toys , Less cost (Direct Proportion)
Therefore 6 : 5 :: 264.37 : x
=> 6 × x = 5 × 264.37
=> x = 220.308
Therefore, Approximate price of 5 toys = Tk. 220

After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
∴ 125 : 150 :: 35 : x
⇒ 125 × x = 150 × 35
⇒ x = (150 × 35)/125
= 42

Let the required number of days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)
Spiders 1:7 and Webs 7:1 } :: 7:x
=> 1 × 7 × x = 7 × 1 × 7
=> x = 7