SONALI BANK - OFFICER (CASH) - WRITTEN MATHS

SONALI BANK - OFFICER (CASH) - WRITTEN MATHS
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MARKS : 15 × 5 = 75
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01. 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?
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Solution:
Suppose, A contributes Tk.x and B contributes Tk. y in the business.
We know, in a partnership business, profit is distributed according to the ratio of the weighted average investments of the partners.
Here, ratio of the weighted average investments of A, B and C
= 4x/12 : 6y/12 : 1560×8/12 = 4x : 6y : 12480 = 2x : 3y : 6240

According to the question:
share of profit of A, B, and C = 1/8 : 1/3 : (1 - 1/8 - 1/3) = 1/8 : 1/3 : 13/24 = 3 : 8 : 13

So, here,
2x :6240 = 3 : 13
x = 6240×3/13×2 = 720

again,
3y : 6240 = 8 : 13
y = 6240×8/13×3 = 1280

Ans: Contribution of A = Tk. 720, B = Tk. 1280.
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02. 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?
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Solution:
Machine A can produce in 1 minute = x/12 pounds of peanut butter.
Machine B can produce in 1 minute = x/18 pounds of peanut butter.
Working simultaneously, Machine A and B can produce in 1 minute = x/12 + x/18 = 5x/36 pounds of peanut butter.
So, to produce x pounds of peanut butter, working simultaneously, it will take = x ÷ (5x/36) = 36/5 minutes = 7.2 minutes or 7 minutes 12 seconds.

Ans: 7 minutes 12 seconds.
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03. 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.
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Solution:
Suppose, the lengths of the trains are x meters and y meters respectively.
Speed of the first train = 75 kmph = 75000m/3600sec = 125/6 mps
Speed of the second train = 60 kmph = 60000m/3600sec = 50/3 mps

according to the question:
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, 131.25m.
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04. 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?
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Solution:
Width of the first garden = 600 ÷ 40 = 15 feet.
Width of the second garden = 2×15 feet = 30 feet.
Area of the second garden = 600÷2 = 300 square feet.
Length of the second garden = 300÷30 = 10 feet.

Ratio of perimeters = {2(40+15)} : {2(10+30)} = 55 : 40 = 11 : 8

Ans: 11 : 8
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05. 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?
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Solution:
As the number of boys or girls cannot be fractions, and in the problem 1/5th of the boys and 1/3rd of the girls is mentioned, we can say, the number of boys is a multiple of 5 and the number of girls is a multiple of 3.
possible number of boys|girls = 5|11, 10|6, 15|1
here, only boys|girls = 10|6 is acceptable.
Hence, the number of boys is 10 and the number of girls is 6.

Suppose, the boys are - a, b, c, d, e, f, g, h, i  and j, and among them j is the tallest boy.
Suppose, the girls are - p, q, r, s, t and u, and among them p is the shortest girl.

Number of boys shorter than p = 10/5 = 2. suppose, a and b are these two boys.
So, c, d, e, f, g, h and i are the 7 boys who are taller than p and shorter than j.

No of girls taller than j = 6/3 = 2. suppose, t and u are these two girls.
So, q, r, and s are the 3 girls who are shorter than j but taller than p.

In total there are 7+3 = 10 students who are taller than the shortest girl (p) and the tallest boy (j).

Required percentage = (10×100/16)% = 62.5%