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Sonali Bank Written Question And Solution

Sonali Bank Written Question And Solution

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Sonali Bank Written Question And Solution:

Post: Officer (Cash)

Written Exam Date: 18 May 2018

Written Exam time: 10.00 AM to 12.00 AM

Sonali Bank Limited Written Math Solution:

Math Part Answer:

1.  720 and 1280

2.  7.2

3. 131.25 and 168.75

4. 11:8

5. 62.5%

Full Math solution:

1.  A, B and C are partners. ‘A’ whose money has been in the business for 4 months claims 1/8 of the profits. ‘B’ whose money has been in the business for 6 months claims 1/3 of the profit. If ‘C’ had Tk. 1560 in the business for 8 months, how much money did A and B contribute to the business?

Solution :

Here,

C’s share =1-(1/8+1/3)=13/24
C’s investment = 1560 × 8
If 13/24=1560×8
Then, 1/8=(1 ×1560 ×8 ×24)/(8 ×13)
= 120 × 24
A’s share is for 4 months
∴ 120 × 24 = 120 × 6 × 4
A’s share = 120 × 6 = 720
B’s share =1/3=(1 ×1560 × 8 × 24)/(3 ×13)
= 120 × 8 × 8
B’s share is for 6 months
120 × 8 × 8 = 20 × 6 × 8 × 8
B’s share = 20 × 8 × 8 = 1280

Or,

Let total profit x tk. 
Then, A’s share =x/8;B’s share =x/3
C’s share =x-(x/8+x/3)=13x/24
Ratio =x/8:x/3:13x/24=3:8:13
Again let A invested p for 4 month’s and B invested q for 6 months. So their investment ratio 4p :6q :1560*8=3:8:13
Therefore p=720 & q=1280.

Or,

Shortcut,

Solution:
A        :       B            :             C
[(960*3)/4]  :   [(960*8)/6]  :   [(1560*8)13] =720     :   1280   :     960

So, A:B= 720 And 1280

Ans: 720 and 1280

 


2. Machine A, working alone at its constant rate, produces x pounds of peanut butter in 12 minutes. Machine B, working alone at its constant rate, produces x pounds of peanut butter in 18 minutes. How many minutes will it take machines A and B, working simultaneously at their respective constant rates, to produce x pounds of peanut butter?

Solution:

Translating the given information into an algebraic expression, you see that machine A produces \frac{x}{12}

pounds of peanut butter in 1 minute, and machine B produces \frac{x}{18} pounds of peanut butter in 1
minute. Therefore, working simultaneously, machine A and machine B produce \frac{x}{12}+\frac{x}{18} pounds of peanut butter in 1 minute.Letting t be the number of minutes it takes machines A and B, working simultaneously, to produce x pounds of peanut butter, you can set up the following equation.

(\frac{x}{12}+\frac{x}{18})*t = x

Solving for t, you get

(\frac{x}{12}+\frac{x}{18})*t = 1

\frac{5x}{36}*t = 1

t = \frac{36}{5}= 7.2

 

Ans: 7.2


3. Two trains running at the rate of 75 km and 60 km an hour respectively on parallel rails in opposite directions, are observed to pass each other in 8 seconds and when they are running in the same direction at the same rate as before, a person sitting in the faster train observes that he passes the other in 31½ seconds. Find the lengths of the trains.

Solution:

Here,

Relative speed=75+60= 135*5/18=37.5 m/s.
let, length of one train be X meters
& 2nd train be Y meters.
therefore, X+Y/37.5= 8
=>X+Y=300.
relative speed = 15*5/18=75/18 m/s.
so length of one train= 75*31.5/18=131.25m
so the length of other train= 300 -131.25=168.75m

Or,

ATQ,

y = 31½(125/6 – 50/3) m = 63/2 × 25/6 = 525/4 = 131.25 m

again,
x + 525/4 = 8(125/6 + 50/3) = 300
x = 300 – 525/4 = 675/4 = 168.75m

Ans: 168.75m And  131.25m.

 


4. A gardener plants two rectangular gardens in separate regions on his property. The first garden has an area of 600 square feet and a length of 40 feet. If the second garden has a width twice that of the first garden, but only half of the area, what is the ratio of the perimeter of the first garden to that of the second garden?

Solution:

Here,

Area of first garden 600 sq m
length 40 m
so wide =600/40=15 m
Now as the wide of 2nd garden double of 1st garden
so wide of 2nd garden =15*2=30
as the area is half the area of 1st garden
so area is 600/2=300 sq m
so length of 1st garden 300/30=10
so ratio of perimeter of two garden is
2(40+15) : 2(10+30)= 11:8

Ans:11:8

 

5.   In a certain class, 1/5 of the boys are shorter than the shortest girls in the class and 1/3 of the girls are taller than the tallest boy in the class. If there are 16 students in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

Solution:

Since total boys and girls = 16
Boys must multiple of 5 and girls must multiple of 3 so
Total boys = 10
Shorter boys = 10*1/5 = 2
Total girls = 6
Taller girls = 6*1/3 = 2
ATQ,
Shortest girl = 1 and Tallest boy = 1
Shorter boys = 2 and Taller girls = 2
So
Taller than shortest girls and shorter than tallest boy =(16-2-2-1-1)=10
10*100/16 = 62.5%

Ans: 62.5%

See Sonali Bank Limited Written Math Solution in PDF below:

Sonali Bank Officer (cash) Full Math Solution-18 May 2018

See Sonali Bank Limited Written Question Solution in PDF below:

Sonali Bank Cash Question paper

 

See Sonali Bank Limited Written Question Solution in image below:

See Sonali Bank Limited Written Question Solution in image below:

 

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