Janata Bank Ltd. || AEO (10-01-2020) || 2020

$i f x + \frac{1}{x} = 2; f i n d t h e v a l u e o f x^{17} + \frac{1}{x^{19}} = ?$

$G i v e n t h a t, x + \frac{1}{x} = 2 = \frac{x^{2} + 1}{x} = 2 = x^{2} + 1 = 2 x = x^{2} - 2 x + 1 = 0 = x^{2} - 2 \times x \times 1 + 1^{2} = 0 = {(x - 1)}^{2} = 0 = x - 1 = \sqrt{0} = 0 ∴ x = 1 ∴ T h e v a l u e o f x^{17} + \frac{1}{x^{19}} = 1^{17} + \frac{1}{1^{19}} = 1 + \frac{1}{1} = 1 + 1 = 2 ∴ x^{17} + \frac{1}{x^{19}} = 2 (A n s w e r)$

2.

In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

.

3.

P does a work in 12 days for 8 hours and Q does a work in 8 days for 10 hours. If both work for 8 hours then how many days it needed to complete the work?

P can do the complete work in = 12x8 = 96 hours.

$∴$ AQ can do the complete work in = 8 x 10 = 80 hours.

In 1 hour P can $\frac{1}{96}$ part of the work

and In 1 hour Q can do $\frac{1}{80}$ of the work

$∴$ (P + Q) can do in 1 hour = $\frac{1}{96} + \frac{1}{80} = \frac{5 + 6}{480} = \frac{11}{480}$ part of the work

$∴$ (P + Q) can do in 8 hour = $\frac{8 \times 11}{480} = \frac{11}{60}$ part of the work

$\frac{11}{60}$ 1 of the work is done in = $\frac{60}{11} = 5 \frac{5}{11}$ day.

4.

Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for n initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

Initial annual salary of X = 12 $\times$ 300 = 3600 Tk.

Annual increment of X = 12 $\times$ 30 = 360 Tk.

Total salary of X for 10 years will be found using arithmetic progression,

with initial values 3,600 Tk. and difference 360 Tk. for 10 years.

So, here n = 10; a = 3600; d = 360

$∴$ Total salary of X = $\frac{n}{2} {2 a + (n - 1) d} = \frac{10}{2} {(2 \times 3600) + (10 - 1) 360}$

=5{(7200+(9 $\times 360)} = 5 (7200 + 3240) = 5 \times 10440 = 52200 T k .$

Initial half-yearly of Y = 6 $\times$ 200 = 1, 200Tk

and half-yearly salary of Y = 15 $\times$ 6 = 90Tk

Total salary of Y for 10 years or 20 half-yearly period will be found using arithmetic progression

with initial value Tk. 1,200 and difference Tk. 90 for 20 half-year period

So, here n = 20 a = 1, 200 d = 90

$∴$ Total salary of Y = $\frac{n}{2} {2 a + (n - 1) d} = \frac{20}{2} {2 \times 1200) + (20 - 1) 20}$

$= 10 {(2400 + (19 \times 90)} = 10 (2400 + 1710) = 10 \times 4110 = 41100 T k . ∴ T o t a l s a l a r y p a i d t o X a n d Y = 52, 200 + 41, 100 = 93, 300 T k . (A n s w e r)$

5.

A sum of taka 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Tk. 362.50 more is lent but at the rate twice the former. At the end of the year, Tk. 33.50 is earned as interest from both the loans. What was the original rate of interest?

Let, real rate of interest r%

$∴$ New rate of interest (2r)%

According to the question

$\frac{725 \times r \times 1}{100} + \frac{362.50 \times 2 r \times 1}{100 \times 3} = 33.50 = \frac{2175 r \times 725 r}{300} = 33.52 = r (2175 + 725) = 10050 = r = \frac{10050}{2175 + 725} = \frac{10050}{2900} = 3.46 ∴ r = 3.46 R e a l r a t e o f i n t e r e s t i s 3.46 %$

6.

A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is

Let, length of the garden is x meter

$∴$ Width of the garden is $\frac{100}{x}$ meter

According to question.

$x + \frac{100}{x} + x = 30 . . . . . . . . . . . . . (i) O r, \frac{100}{x} + x + \frac{100}{x} = 30 . . . . . . . . . . . . (i i) W e g e t f r o m e q u a t i o n (i) x + \frac{100}{x} + x = 30 = \frac{x^{2} + 100 + x^{2}}{x} 30 = 2 x^{2} + 100 = 30 x = 2 x^{x} - 30 x + 100 = 0 = 2 (x^{2} - 15 x + 50) = 0 = x^{2} - 15 x + 50 = 0 = (x - 10) (x - 5) = 0$

$∴$ Either x = 10

Now, if length = 10 meter

Then width = $\frac{100}{10}$ = 10 meter

Or, x = 5

Or, if length = 5 meter

Then width= $\frac{100}{5}$ = 20 meter

7.

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. How many hours will take C alone to fill the tank ?

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