# Written Math Suggestion Combined 5 Bank Officer

Written Math Suggestion : Written Math Suggestion টি দেয়া হচ্ছে Combined Bank Written Exam এর জন্য। এই Written Math Suggestion  টি অনুমানমাত্র। এখান থেকে কমন পড়বেই এই আশা করে হলে গেলে আমার বলার কিছুই নাই। তবে AUST যে প্যাটার্নের প্রশ্ন করে তার আলোকে মাত্র 29 টি ম্যাথ দিলাম। আপনারা হাতে-কলমে Practice করবেন। ভাগ্য ভাল হলে কমন পড়তেও তো পারে।

## Written Math Suggestion for Combined 5 Bank Officer ## Bank Written Math Suggestion Only Questions

Math Solution গুলো নিচে দেয়া আছে। আপনারা নিজের ভাল চাইলে আগে নিজে করার চেষ্টা করুন। তারপর Solution দেখুন।

01. From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with 2/3 of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:

02. Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi’s speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi’s speed is:

03. Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

04. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

05. A train B speeding with 120 kmph crosses another train C running in the same direction, in 2 minutes. If the lengths of the trains B and C be 100m and 200m respectively, what is the speed (in kmph) of the train C?

06. What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of a m/s and (a + 1) m/s and passes them completely in b seconds and (b + 1) seconds respectively?

07. Train A passes a lamp post in 9 seconds and 700 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)

08. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

09. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

10. Two pipes A and B can fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is one – third full, a leak develops in the tank through which one – third water supplied by both the pipes gose out. The total time taken to fill the tank is?

11. If A and B together can complete a work in 18 days, A and C together in 12 days, and B and C together in 9 days, then B alone can do the work in:

12. Three pipes A, B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?

13. A group of 12 men can do a piece of work in 14 days and other group of 12 women can do the same work in 21 days. They begin together but 3 days before the completion of work, man’s group leaves off. The total number of days to complete the work is:

14. Three taps A, B and C together can fill an empty cistern in 10 minutes. The tap A alone can fill it in 30 minutes and the tap B alone in 40 minutes. How long will the tap C alone take to fill it?

15. A, B and C enter into a partnership in the ratio 7/2 : 4/3 : 6/5 . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:

16. A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew 1/4 of his capital and B withdrew 1/5 of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:

17. Rahul and Bharti are partners in a business. Rahul contributes 1/4th capital for 15 months and Bharti received 2/3 of profit. For how long Bharti money was used.

18. An article is listed at Rs. 920. A customer pays Rs. 742.90 for it after getting two successive discounts. If the rate of first discount is 15%, the rate of 2nd discount is:

19. A dealer buys an article marked at Rs. 25,000 with 20% and 5% off. He spends Rs. 1,000 for its repairs and sells it for Rs. 25,000. What is his gain or loss per cent?

20. A reduction of 10% in the price of tea enables a dealer to purchase 25 kg more tea for Rs. 22500. What is the reduced price per kg of tea?

21. Cost price of 12 oranges is equal to the selling price of 9 oranges and the discount on 10 oranges is equal to the profit on 5 oranges. What is the percentage point difference between the profit percentage and discount percentage?

22. The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost price of Rs. 500. What is the new profit percentage if instead of two successive discounts the markup price was further increased successively two times by the same percentage?

23. The profit percentage on three articles A, B and C is 10%, 20%, and 25% and the ratio of the cost price is 1: 2: 4. Also the ratio of number of articles sold of A, B and C is 2: 5: 2, then overall profit percentage is:

24. The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid.

25. In an examination, questions were asked in five sections. Out of the total students, 5% candidates cleared the cut-off in all the sections and 5% cleared none. Of the rest, 25% cleared only one section and 20% cleared four sections. If 24.5% of the entire candidates cleared two sections and 300 candidates cleared three sections. Find out how many candidates appeared at the examination?

26. Pinku, Rinku and Tinku divide an amount of Rs.4200 amongst themselves in the ratio of 7 : 8 : 6 respectively. If an amount of Rs.200 is added to each of their shares, what will be their new respective ratio of their shares of amount?

27. A, B and C together start a business, B invests 1/6 of the total capital while investments of A and C are equal. If the annual profit of this investment is Rs.33600, find the difference between the profits of B and C.

28. X and Y make a partnership. X invests Rs.8000 for 8 months and Y remains in the business for 4 months. Out of the total profit, Y claims 2 / 7 of the profit. How much money was contributed by Y?

29. Two partners invest Rs.125000 and Rs.85000, respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be treated as interest on capital. If one partner gets Rs.600 more than the other, find the total profit made in the business.

### Bank Written Math Suggestion With Solution ### AUST Bank Written Math Suggestion- 01. From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with 2/3 of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:

Solution:
Let the speed of A= x kmph and that of B = y kmph;

ATQ,
X×6 + y×6 = 60
Or, x + y = 10 ——— (i)

And,

{(2x/3) × 5} + (2y×5) = 60
Or, 10x+30y = 180;
Or, x+3y = 18; ———- (ii)
From equation (i) ×3 – (ii)
3x+3y-x-3y = 30-18
Or, 2x = 12
Hence, x = 6 kmph.

### AUST Bank Written Math Suggestion- 02. Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi’s speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi’s speed is:

Solution:
Let the speed of Ravi = x kmph;
Hence, Ajay’s speed = (x+4) kmph;
Distance covered by Ajay = 60 + 12 = 72 km;
Distance covered by Ravi = 60 – 12 = 48 km.

ATQ,
72/(x+4) = 48/x
Or, 3/(x+4) = 2/x
3x = 2x+8
Or, x = 8 kmph.

### AUST Bank Written Math Suggestion- 03. Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

Solution

Let,

The speed of the slower train = x m/sec

Then, speed of the faster train = 2x m/sec

Relative speed = (x + 2x) m/sec = 3x m/sec

So,

(100+100)/8 = 3x

Or, 24x = 200

Or, x = 25/3

So, speed of the faster train,

2x = 2×(25/3)

= 50/3 m/sec

= (50/3) ×(18/5) km/hr

= 60 km/hr

### Bank Written Math Suggestion- 04. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

Solution:

4.5 km/hr = (4.5 × 5/18) m/sec = 1.25 m/sec

And,

5.4 km/hr = (5.4 × 5/18) m/sec = 1.5 m/sec

Let,

The speed of the train = x m/sec

Then,

(x – 1.25) × 8.4 = (x – 1.5) × 8.5

Or, 8.4x  – 10.5 = 8.5x – 12.75

Or, 0.1x = 2.25

Or, x = 22.5

So, speed of the train = (22.5 × 18/5) km/hr = 81 km/hr

### Bank Written Math Suggestion- 05. A train B speeding with 120 kmph crosses another train C running in the same direction, in 2 minutes. If the lengths of the trains B and C be 100m and 200m respectively, what is the speed (in kmph) of the train C?

Solution:

Let,

Speed of C = x m/sec

Speed of B relative to C = (120-x) kmph

=(120-x) × 5/18 m/sec

= (600-5x)/18 m/sec

Distance covered = (length of B + length of C) = (100 + 200) = 300 m

Now,

300/{(600 – 5x)/18} = 120 (2 minutes = 120 sec)

Or, 300 = {120(600 – 5x)}/18

Or, 10 × 9 = 2(600 – 5x)

Or, 90 = 1200 – 10x

Or, 10x = 1110

Or, x = 111

### Bank Written Math Suggestion- 06. What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of a m/s and (a + 1) m/s and passes them completely in b seconds and (b + 1) seconds respectively?

Solution:

Let,

The length of the train = x m.

And  its speed = y m/s

Then,

x/(y – a) = b

Or, x = b(y – a) —–(1)

Again,

x/{y – (a + 1)} = (b + 1)

Or, x = (b + 1)((y – a – 1)

Or, b(y – a) = (b + 1)((y – a – 1)

Or, by – ab = by – ab – b + y – a – 1

Or, y = (a + b + 1)

### Bank Written Math Suggestion- 07. Train A passes a lamp post in 9 seconds and 700 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)

Solution:

Let the length of the train = x m

When the train crosses a lamppost in 9 second, the distance covered = length of the train

So, speed of the train = x/9

Distance covered in crossing a 700 m platform in 30 seconds,

= length of the platform + length of the train

So, speed of the train = (x + 700)/9

ATQ,

x/9 = (x + 700)/30

Or, x/3 = (x+700)/10

Or, 10x = 3x + 2100

Or, x = 2100/7 = 300

When the length of the platform be 800m, time T be taken by the train to cross the length.

So,

x/9 = (x+800)/T

Or, Tx = 9x + 7200

Or, 300T = 2700 + 7200 (x = 300)

Or, T = 9900/300 = 33

### Bank Written Math Suggestion- 08. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

Solution:

Let, 1st pipe alone takes x hrs to fill the tank

Then, 2nd and 3rd pipe will take (x-5) and (x-9) hrs respectively to fill the tank.

So,

1/x + 1/(x-5) = 1/(x-9)

Or, (x-5+x)/{x(x-5)} = 1/(x-9)

Or, (2x-5)(x-9) = x(x-5)

Or, x2 – 18x + 45 = 0

Or, (x-15)(x-3) = 0

Or, x = 15 [x=3 not acceptable]

### Bank Written Math Suggestion- 09. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

Solution:

(A+B)’s 1 hrs work = (1/12 + 1/15) = 3/20

(A+C)’s 1 hrs work = (1/12 + 1/20) = 2/15

Part filled in 2 hrs = (3/20 + 2/15) = 17/60

So, Part filled in 6 hrs = (17/60 × 6/2) = 17/20

Remaining = (1 – 17/20) = 3/20

Now, it is the turn of A & B,

3/20 part filled in 1 hr

So, total time taken to fill the tank = (6+1) = 7 hrs.

### Bank Written Math Suggestion- 10. Two pipes A and B can fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is one – third full, a leak develops in the tank through which one – third water supplied by both the pipes gose out. The total time taken to fill the tank is?

Solution:

Part filled by (A+B) in 1 hr

=(1/20 + 1/30) = 1/12

So, A and B together can fill the tank in 12 hrs.

So, 1/3 part is filled in (1/3 × 12) = 4 hrs

Since the leak empties 1/3 part

So, time taken to fill the tank = Time taken by (A+B) to fill the whole tank + Time taken by (A+B) to fill 1/3 part

= (12+4) = 16 hrs

### Bank Written Math Suggestion- 11. If A and B together can complete a work in 18 days, A and C together in 12 days, and B and C together in 9 days, then B alone can do the work in:

Solution:

1 days work of (A+B) = 1/18 ——(1)

1 days work of (A+C) = 1/12 ——(2)

1 days work of (B+C) = 1/9 ——(3)

(1)+(2)+(3),

2×(A+B+C) = 1/18 + 1/12 + 1/9 = ¼

1 days work of (A+B+C) = 1/(4×2) = 1/8

B’s 1 days work = (A+B+C) – (A+C)

= 1/8 – 1/12

= 1/24

So, B alone can do the work in 24 days.

### Bank Written Math Suggestion- 12. Three pipes A, B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?

Solution:

Let,

C alone can empty the cistern in x min

In 1 min A fills = 1/10

In 1 min B fills = 1/15

(A+B) together fill in 1 min = 1/10 + 1/15 = 1/6 part

So, (A+B) together fill in 2 mins 2 × 1/6 = 1/3 part

(A+B) together fill in 6 mins = 1 part

Now, (A+B+C) together fill in 6 min (1 – 1/3) = 2/3 part

So, in 6 min C empties = (1 – 2/3) = 1/3 part

C empties 1/3 part of the cistern in 6 mins

So, C empties the cistern in = 6 ×3 = 18 min

### Bank Written Math Suggestion- 13. A group of 12 men can do a piece of work in 14 days and other group of 12 women can do the same work in 21 days. They begin together but 3 days before the completion of work, man’s group leaves off. The total number of days to complete the work is:

Solution:

Let, x be the required number of days

Now,

12 men’s 1 day’s work = 1/14

12 women’s 1 day’s work = 1/21

12 women’s 3 day’s work = 3/21 = 1/7

Remaining work = 1 – 1/7 = 6/7

Now,

12 men’s + 12 women’s 1 day’s work = 1/14 + 1/21 = 5/42

5/42 work is done by 2 groups in 1 day

So, 6/7 work is done by 2 groups in 42/5 × 6/7 = 36/5 days

So, total time taken to complete the work = 36/5 + 3 = 51/5 days

### Bank Written Math Suggestion- 14. Three taps A, B and C together can fill an empty cistern in 10 minutes. The tap A alone can fill it in 30 minutes and the tap B alone in 40 minutes. How long will the tap C alone take to fill it?

Solution:

A, B and C together can fill 100% empty tank in 10 minutes.
Work rate of (A + B + C) = 100/10 = 10% per minute.
A alone can fill the tank in 30 minutes.
Work rate of A = 100/30 = 3.33% per minute.
B alone can fill the tank in 40 minutes.
Work rate of B = 100/40 = 2.5%.
Work rate of (A + B) = 3.33 + 2.5 = 5.83% per minute.
Work rate of C,
= Work rate of (A +B +c) – (A +B).
= 10 – 5.83 = 4.17% per minute.
So, C takes = 100/4.17 = 24 minutes to fill the tank.

### Bank Written Math Suggestion- 15. A, B and C enter into a partnership in the ratio 7/2 : 4/3 : 6/5 . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:

Solution:

Ratio of investment = (7/2 : 4/3 : 6/5 ) = 105:40:36

Let, the investment be 105x, 40x & 36x

Now,

(105x × 4 + 150/100 × 105x × 8): (40x × 12): (36x × 12)

= 1680x : 480x : 432x

= 35 : 10 : 9

Hence, B’s share,

= Rs. (21600 × 10/54)

= Rs. 4000

### Bank Written Math Suggestion- 16. A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew 1/4 of his capital and B withdrew 1/5 of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:

Solution:

A:B= [4x × 3 + (4x – 4x/4) × 7] : [5x – 5x/5) × 7]

= (12x + 21x) : (15x + 28x)

= 33x : 43x

= 33 : 43

So, A’s share = Rs. (760 × 33/70) = Rs. 330

### Bank Written Math Suggestion- 17. Rahul and Bharti are partners in a business. Rahul contributes 1/4th capital for 15 months and Bharti received 2/3 of profit. For how long Bharti money was used.

Solution:

Let, the total profit be Rs. X

Bharti’s share = x × 2/3 = 2x/3

Rahul’s share = x – 2x/3 = x/3

Rahul : Bharti = x/3 : 2x/3 = 1:2

Now,

Let, the total capital = y

And Bharti’s capital was used for n months

Then, Rahul’s capital will be = y/4

And Bharti’s capital will be = y – y/4 = 3y/4

(15 × y/4)/(n × 3y/4) = ½

Or, n = (15×2)/3 = 10

Bharti’s money was used for 10 months.

### Bank Written Math Suggestion- 18. An article is listed at Rs. 920. A customer pays Rs. 742.90 for it after getting two successive discounts. If the rate of first discount is 15%, the rate of 2nd discount is:

Solution: MP = 920.
After first discount Marked Price (MP) become,
= 920 – 15% of 920 = 782.
The Selling Price (SP) = 742.90.
Let second discount was x% on 782.
782 – x% of 782 = 742.90
Or, 782x/100 = 39.1
Or, 782x = 3910
Or, x = 5%.
Second Discount = 5%.

### Bank Written Math Suggestion- 19. A dealer buys an article marked at Rs. 25,000 with 20% and 5% off. He spends Rs. 1,000 for its repairs and sells it for Rs. 25,000. What is his gain or loss per cent?

Solution:

Marked Price = 25000.
After first discount it become,
= 25000 – 20% of 25000 = 20000.
After second discount, it becomes
= 20000 – 5% of 20000 = 19000.
So, SP = 19000.
CP for the man who bought it, as he spends 1000 on repair.
= 19000 + 1000 = 20000
Profit = 25000 – 20000 = 5000.
% Profit = (5000*100)/20000 = 25%.

### Bank Written Math Suggestion- 20. A reduction of 10% in the price of tea enables a dealer to purchase 25 kg more tea for Rs. 22500. What is the reduced price per kg of tea?

Solution:

Let, the original price of tea be x Rs. Kg

After reduction, the price becomes = x – 10% of x = 9x/10 per kg

Now,

[22500/(9x/10)] – 22500/x = 25

Or, 22500(10/9x – 1/x) = 25

Or, 25 × 9x = 22500

Or, x = 22500/(9×25) = 100

Hence, new price = 100 – 10 = 90

### Bank Written Math Suggestion- 21. Cost price of 12 oranges is equal to the selling price of 9 oranges and the discount on 10 oranges is equal to the profit on 5 oranges. What is the percentage point difference between the profit percentage and discount percentage?

Solution: Ratio of selling price and Cost Price,
SP:CP = 12:9 =4:3
Profit of 3 oranges = Re 1 (Let CP = Re 1)
Profit = 1/3 = 33.33%
and, Discount = 11.11%
Since,
CP:SP:MP = 3:4:4.5
Profit doubles that of discount.
So, % point discount = 33.33%-11.11% = 22.22% point.

### Bank Written Math Suggestion- 22. The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost price of Rs. 500. What is the new profit percentage if instead of two successive discounts the markup price was further increased successively two times by the same percentage?

Solution:

Cost Price = 500

Selling price = 576

Mark up price =500 + 80% of 500 = 900

Now,

Selling price = Mark up price × [1 – R/100]2 [ R = Rate of Discount]

576 = 900 × [1 – R/100]2

Or, R = 20%

Again,

Selling price = Mark up price × [1 + R/100]2

= 900 × [1 + 20/100]2

= 1296

New profit percentage = [(Selling price – Cost Price)/ Cost Price] × 100

= [(1296 – 500)/ 500] × 100

= 159.2%

### Bank Written Math Suggestion- 23. The profit percentage on three articles A, B and C is 10%, 20%, and 25% and the ratio of the cost price is 1: 2: 4. Also the ratio of number of articles sold of A, B and C is 2: 5: 2, then overall profit percentage is:

Solution: Ratio of CP given as 1:2:4.
Let cost of,
A = x
B = 2x
C = 4x
Ratio of Number of sell is given as 2:5:2.
Let number of items sold be,
A = 2y
B = 5y
C = 2y
Total cost (A+B+C),
= (2xy+10xy+8xy)
= 20xy
Profit of A = 0.2xy
Profit of B = 2xy
Profit of C = 2xy
Total profit = 4.2xy
% Profit = (4.2xy*100)/20xy = 21%.

### Bank Written Math Suggestion- 24. The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid.

Solution: Let each side of the cuboid be 10 unit initially.
Initial Volume of the cuboid,
= length × breadth × height

= 10 ×10×10 = 1000 cubic unit.
After increment dimensions become,
Length = (10 + 10% 0f 10) = 11 unit.
Breadth = (10 + 20% of 10) = 12 unit.
Height = (10 + 50 of 10) = 15 unit.
Now, present volume = 11 ×12 ×15 = 1980 cubic unit.
Increase in volume = 1980 – 1000 = 980 cubic unit.
% increase in volume = (980/1000)*100 = 98%.

### Bank Written Math Suggestion- 25. In an examination, questions were asked in five sections. Out of the total students, 5% candidates cleared the cut-off in all the sections and 5% cleared none. Of the rest, 25% cleared only one section and 20% cleared four sections. If 24.5% of the entire candidates cleared two sections and 300 candidates cleared three sections. Find out how many candidates appeared at the examination?

Solution: Passed in none = 5%
Passed in all = 5%
Passed in four = 20% 0f 90% = 18%
Passed in three = 24.5%
Passed in three =(100-5-5-22.5-24.5-18) = 25%.
But given 300 students passed in three.
Hence, 25% = 300.
So, 100% = 1200.
1200 students must have appeared.

### Bank Written Math Suggestion- 26. Pinku, Rinku and Tinku divide an amount of Rs.4200 amongst themselves in the ratio of 7 : 8 : 6 respectively. If an amount of Rs.200 is added to each of their shares, what will be their new respective ratio of their shares of amount?

Let Pinku’s share = 7x
Rinku’s share = 8x
Tinku’s share = 6x

According to the question,

7x + 8x + 6x = 4200

21x = 4200

X = 4200 / 21 = 200

Hence amount of Pinku = 7x
= 7 X 200 = Rs.1400

Amount of Rinku = 8x
= 8 X 200 = Rs.1600

Amout of Tinku = 6x
= 6 X 200 = Rs.1200

Now Rs.200 is added to each share
Then,
New ratio of shares = (1400 + 200 ) : (1600 + 200) : (1200 : 200)

= 1600 : 1800 : 1400
= 16 : 18 : 14 = 8 : 9 : 7

### Bank Written Math Suggestion- 27. A, B and C together start a business, B invests 1/6 of the total capital while investments of A and C are equal. If the annual profit of this investment is Rs.33600, find the difference between the profits of B and C.

Solution:

Investment of B = 1/6 of total capital

Investment of A & C each,

= ½(1 – 1/6) of total capital

= ½ × 5/6 of total capital

= 5/12 of total capital

Now, A’s share : B’s share : C’s share

= 5/12 : 1/6 : 5/12

= 5 : 2 : 5

Let, A’s share = 5x

B’s share = 2x

C’s share = 5x

ATQ,

5x + 2x + 5x = 33600

Or, 12x = 33600

Or, x = 2800

Difference in profit of B & C,

= 5x – 2x

= 3x

= 3×2800

= 8400

### Bank Written Math Suggestion- 28. X and Y make a partnership. X invests Rs.8000 for 8 months and Y remains in the business for 4 months. Out of the total profit, Y claims 2 / 7 of the profit. How much money was contributed by Y?

Solution:

Y gets 2/7 of the profit

X gets (1 – 2/7) = 5/7 of the profit

X:y = 5/7 : 2/7 = 5:2

Let, the contribution of Y be a.

Then,

8000 × 8 : a × 4 = 5 : 2

Or, 64000/4a = 5/2

Or, 20a = 128000

Or, a = 6400

### Bank Written Math Suggestion- 29. Two partners invest Rs.125000 and Rs.85000, respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be treated as interest on capital. If one partner gets Rs.600 more than the other, find the total profit made in the business.

Solution:

Let, total profit = x

(100-60)% = 40% of x is distributed in the ratio.

125000:85000 = 25 : 17

Share of 1st partner = 40% of x {25/(25+17)}

= 40% of 25x/42

= 17x/105

ATQ,

5x/21 – 17x/105 = 600

Or, x(25 – 17)/105 = 600

Or, x = (600 × 105)/8 = 7875

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Updated: November 2, 2020 — 11:37 pm

1. Please post some important topics of bangla focus writing for upcoming combined bank cash officer written exam. Thanks.

2. dada 5 no. is incomplete and something went missing

3. 5 no math ta ki vul acay dada aktu janaban…

4. 27 no math er solution correction koren. Vhul process e kora hoyeche.

1. Process ঠিক আছে।

5. Dada 21.No math ta vul ache may be..

S.p of 9= c.p of 12

So.s.p of 12 = c.p of 16

Profit =[(16-12)÷12]×100
= (100/3)℅

Again s.p of 12=c.p of 16
So.Profit of 12= c.p of 4
Profit of 5=(4×5)/12=5/3

Discount on 10 =5/3
Discount%= (5/3×10)×100=50/3 %

So difference=(100/3)% -(50/3)%=(50/3)℅

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