Combined 7 Banks & Financial Institutions || Senior Officer (19-11-2021) || 2021

Answer the following mathematical problems:

What is the minimum number of oranges that must be added to the existing stock of 264 oranges so that the total stock can be equally distributed among 6, 7 or 8 persons?

$H e r e, L . C . M . o f 6, 7 a n d 8$

$6, 7, 8 3, 7, 4 2 ∴ L . C . M . = 2 \times 3 \times 7 \times 4 = 168$

$W e h a v e 264 o r a n g e s . ∴ W e n e e d = 168 \times 2 = 336 o r a n g e s .$

$∴ W e n e e d a d d i t i o n a l = 336 - 264 = 72$

$∴ R e q u i r e d n u m b e r o f o r a n g e s i s 72 . (a n s w e r)$

Answer the following mathematical problems:

2.

Bokul invested an amount of Tk. 15,000 in fixed deposit scheme for 3 years at interest rate 5% per annul. How much amount will she get on maturity of the fixed deposit if the interest is compounded quarterly?

$H e r e, p r i n c i p a l a m o u n t, D = 15 T k .; i n t e r e s t r a t e, r = 5 %$

$N u m b e r o f y e a r s, n = 3 n u m b e r o f m o n t h s, m = 4 [Q u a r t e r l y m e a n s 4 t i m e s a y e a r]$

$N o w, w e k n o w r e q u i r e d a m o u n t, ‍ A = p {(1 + \frac{r}{100 \times m})}^{m n} = 15000 {(1 + \frac{5}{100 \times 4})}^{4 \times 3} = 15000 {(1 + \frac{1}{80})}^{12} = 15000 {(\frac{81}{80})}^{12} = 15000 {(1.0125)}^{12} = 17411.317 T k .$

$∴ R e q u i r e d a m o u n t i s 17, 411.317 T k .$

Answer the following mathematical problems:

3.

$I f x^{4} + \frac{1}{x^{4}} = 119, P r o v e t h a t x^{3} - \frac{1}{x^{3}} = 36$

$G i v e n t h a t, x^{4} + \frac{1}{x^{4}} = 119 = {(x^{2})}^{2} + {(\frac{1}{x^{2}})}^{2} = 119 = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 \times x^{2} \times \frac{1}{x^{2}} = 119 = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 119 = {(x^{2} + \frac{1}{x^{2}})}^{2} = 121 = x^{2} + \frac{1}{x^{2}} = \sqrt{121} = \sqrt{{(11)}^{2}} = 11 = {(x - \frac{1}{x})}^{2} + 2 \times x \times \frac{1}{x} = 11 = {(x - \frac{1}{x})}^{2} + 2 = 11 = {(x - \frac{1}{x})}^{2} = 11 - 2 = 9 ∴ x - \frac{1}{x} = 3$

$N o w, L . H . S . = x^{3} - \frac{1}{x^{3}} = {(x - \frac{1}{x})}^{3} + 3 \times x \times \frac{1}{x} (x - \frac{1}{x}) = 3^{3} + 3 \times 3 = 27 + 9 = 36 ∴ L . H . S . = R . H . S . (P r o v e d)$

Answer the following mathematical problems:

4.

The hypotenuse of a right angled triangle is 2x - 6 and the other two sides are x + 2 and x. Find the value of x.

$L e t, A B C i s t h e r i g h t a n g l e t r i a n g l e$

$H e r e, H y p o t e n u s e A B = (2 x - 6)$

$O n e a r m, A C = x A n d a n o t h e r a r m, B C = (x + 2) ∴ A c c o r d i n g t o P e t h a g o r e a n T h e o r e m, {(2 x - 6)}^{2} = {(x + 2)}^{2} + x^{2}$

$= {(2 x)}^{2} - 2 \times 2 x \times 6 + 6^{2} = x^{2} + 2 \times x \times 2 + 2^{2} + x^{2}$

$= 4 x^{2} - 24 x + 36 = 2 x^{2} + 4 x + 4 = 2 x^{2} - 28 x + 32 = 0 = x^{2} - 14 x + 16 = 0$

$N o w, c o m p a r i n g, x^{2} - 14 x + 16 = 0 w i t h a x^{2} + b x + c = 0, W e g e t, x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 14) \pm \sqrt{{(- 14)}^{2} - 4 \times 1 \times 16}}{2 \times 1} = \frac{14 \pm \sqrt{196 - 64}}{2} = \frac{14 \pm \sqrt{132}}{2} = \frac{14 \pm \sqrt{4 \times 33}}{2}$

$= \frac{14 \pm 2 \sqrt{33}}{2} = \frac{2 (7 \pm \sqrt{33})}{2} = 7 \pm \sqrt{33} ∴ x = (7 + \sqrt{33}) o r (7 - \sqrt{33})$

5.

A box without lid is made of wood everything 0.5cm thick and its external dimensions are 6cm by 5cm by 4cm. Find the volume of wood used in making the box.

$G i v e n t h a t, e x t e r n a l d i m e n s i o n s a r e 6 c m, 5 c m a n d 4 c m .$

$∴ E x t e r n a l w i l l b e a s f o l l o w s, l e n g t h = 6 - (0.5 \times 2) = 6 - 1 = 5 c m . W i d t h = 5 - (0.5 \times 2) = 5 - 1 = 4 c m A n d h i g h t = 4 - (0.5 \times 1) = 4 - 0.5 = 3.5 c m . ∴ T h e r e q u i r e d v o l u m e o f w o o d = E x t e r n a l c a p a c i t y - I n t e r n a l c a p a c i t y = (6 \times 5 \times 4) - (5 \times 4 \times 3.5) = 120 - 70 = 50 c m^{3} .$

6.

A bag contains 10 white and 15 black balls. Two balls are drawn in succession. What is the probability that first is white and second is black?

$L e t, p r o b a b i l i t y o f w h i t e b a l l s b e w & B l a c k b a l l s b e b .$

$∴ p (w) = \frac{N u m b e r o f w h i t e b a l l s}{T o t a l b a l l} = \frac{10}{10 + 15} = \frac{10}{25} = \frac{2}{5} ∴ p (w) = \frac{N u m b e r o f w h i t e b a l l s}{T o t a l b a l l} = \frac{15}{10 + 15 - 1} = \frac{15}{24} = \frac{5}{8} ∴ R e q u i r e d p r o b a b i l i t y = \frac{2}{5} \times \frac{5}{8} = \frac{1}{4}$

7.

The sum of the first two digits of a three digit number is equal to the third digit. If the digits are reversed, the number so obtained is 198 more than the original number. What are the possible original numbers?

$L e t, x b e t h e 100 t h d i g i t; y b e t h e 10 t h d i g i t a n d z b e t h e u n i t d i g i t$

$∴ T h e n u m b e r i s = (100 x + 10 y + z) A c c o r d i n g q u e s t i o n t o, (x + y) = z . . . . . . . (i)$

$(100 x + 10 y + z) = x + 10 y + 100 z - 198 = x - 100 x + 10 y - 10 y + 100 z - z = 198 = 99 z - 99 x = 198 = 99 (z - x) = 198 = (z - x) = \frac{198}{99} = 2$

$= x + y - x = 2 [P u t t i n g z = x + y] ∴ y = 2 [H e r e, 10 t h d i g i t a l w a y s w i l l b e 2] W h e n x = 1 t h e n t h e p o s s i b l e n u m b e r i s = 1 + 2 = 3 : h e r e b y t h e n u m b e r i s 123 . ∴ x = 2 t h e n t h e p o s s i b l e n u m b e r i s 2 + 2 = 4; h e r e b y t h e n u m b e r i s 224 . ∴ x = 3 t h e n t h e p o s s i b l e n u m b e r i s = 3 + 2 = 5; h e r e b y t h e n u m b e r i s 325 . ∴ x = 4 t h e n t h e p o s s i b l e n u m b e r i s = 4 + 2 = 6 : h e r e b y t h e n u m b e r i s 426 . ∴ x = 5 t h e n t h e p o s s i b l e n u m b e r i s = 5 + 2 = 7; h e r e b y t h e n u m b e r i s 527 . ∴ x = 6 t h e n t h e p o s s i b l e n u m b e r i s = 6 + 2 = 8; h e r e b y t h e n u m b e r i s 62 . ∴ x = 7 t h e n t h e p o s s i b l e n u m b e r i s = 7 + 2 = 9 : h e r e b y t h e n u m b e r i s 729 . T h e n u m b e r c a n b e 123, 224, 325, 426, 527, 628, 729 .$

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