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Combined 5 Banks || Officer (Cash) (23-04-2022) || 2022

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Mathematics

What is the value of $\frac{2^{x + 4} - {4.2}^{x + 1}}{2^{x + 2} + 2} = ?$

$G i v e n t h a t, \frac{2^{x + 4} - 4 . 2^{x + 1}}{2^{x + 2 - 1}} = \frac{2^{x} \times 2^{4} - 4 \times 2^{x} \times 2^{1}}{\frac{2^{x + 2}}{2^{1}}} = \frac{16 \times 2^{x} - 8 \times 2^{x}}{2^{x + 2 - 1}} = \frac{8 \times 2^{x}}{2^{x + 1}} = \frac{8 \times 2^{x}}{2^{x} \times 2^{1}} = \frac{8 \times 2^{x}}{2^{x \times 2}} = 4$

Mathematics

A pipe can fill up an empty tank in 15 minutes. Another pipe flows out 20 liter of water per minute. If the two pipes are opened together and the empty tank is filled in 135 minutes, how much water does the tank contain?

$L e t, t h e c a p a c i t y o f t h e \tan k b e x l i t e r s . B y 1 s t p i p e I n 15 m i n u t e s p i p e f i l l = x p a r t o f t h e \tan k ∴ I n 1 m i n u t e p i p e f i l l = \frac{x}{15} p a r t o f t h e \tan k$

$A c c o r d i n g t o q u e s t i o n, 135 (\frac{x}{15} - 20) = x = 135 (\frac{x - 300}{15}) = x = 9 (x - 300) = x = 9 x - 2700 = x = 8 x = 2700 ∴ x = \frac{2700}{8} = 337.5 l i t e r s ∴ T h e c a p a c i t y o f t h e \tan k i s 337.5 l i t e r s .$

Mathematics

Ms. Wazaiha invested a total amount of BDT 50,000 in two honds namely BEXIMCO and DBBL that pay annual interest at the rate of 6.5% and 9%, respectively. If she likes to receive BDT 4,000 as interest income, how much should she invest in each bond?

$L e t, s h e s h o u l d i n v e s t x T k . i n B E X I M C O ∴ S h e s h o u l d i n v e s t = (50000 - x) T k i n D B B L A c c o r d i n g t o q u e s t i o n 6.5 % o f x + 9 % o f (50000 - x) = 4000 = \frac{6.5 c}{100} + \frac{9 (50000 - x)}{100} = 4000 = 6.5 x + 450000 - 9 x = 4000000 = 6.5 x - 9 x = 450000 - 4000000 = 2.5 x = 50000 ∴ x = \frac{50000}{2.5} = 20000$

$∴ T h e a m o u n t o f B E X I M C O b o n d b e 20, 000 T k . ∴ D B B L a m o u n t w o u l d b e = 50000 - 20000 = 30000 T k .$

Mathematics

Faiyaz and Ehan are standing 90 yards apart on a north-south axis, Faiyaz walks 70 yards to west. Ehan walks 50 yards to the east, and then both stop. What is the straight-line distance in yards between them now?

$L e t, u s d e p i c t t h e i m a g e a c c o r d i n g t o q u e s t i o n : N o w, F a i y a z s \tan d s a t p o i n t A A n d E h a n s \tan d s a t p o i n t C H e r e, d i r e c t d i s \tan c e b e t w e e n t h e m i s A C △ A B C i s a r i g h t a n g l e t r i a n g l e w h e r e A C = H y p o t e n u s e$

$∴ A p p l y i n g P y t h a g o r e a n t h e o r e m, w e g e t {(A C)}^{2} = {(A B)}^{2} + {(B C)}^{2} = {(A C)}^{2} = {(50 + 70)}^{2} + {(90)}^{2} = {(120)}^{2} + {(90)}^{2} = {(A C)}^{2} = 14400 + 8100 = 22500 = {(150)}^{2} ∴ A C = 150 ∴ T h e s t r a i g h t d i s \tan c e b e t w e e n F a i y a z a n d E h a n i s 150 y a r d s .$

Mathematics

Suppose the number of people in Cumilla city is 1 million. If the number of males increases by 10% and the number of females decreases by 6%, then the number of people in city remains unchanged. What are the numbers of male and female in that city?

$L e t, n u m b e r o f m a l e b e x ∴ N u m b e r o f f e m a l e b e = (1000000 - x) A c c o r d i n g t o q u e s t i o n 110 % o f x + 94 % o f (1000000 - x) = 1000000 = \frac{110 X}{100} + \frac{94 (1000000 - X)}{100} = 1000000 = \frac{11 X}{10} + \frac{9.4 (1000000 - X)}{10} = 1000000 = 11 X + 9400000 - 9.4 X = 10000000 = 1.6 X = 10000000 - 9400000 = 600000 ∴ X = \frac{600000}{1.6} = 375000 ∴ N u m b e r o f m a l e = 3, 75, 000 ∴ N u m b e r o f f e m a l e = 10, 00, 000 - 3, 75, 000 = 6, 25, 000$

Job Solution | Combined Bank (সমন্বিত ব্যাংক)

Combined 5 Banks || Officer (Cash) (23-04-2022) || 2022

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