Given that, 2x+4-4.2x+12x+2-1=2x×24-4×2x×212x+221 =16×2x-8×2x2x+2-1=8×2x2x+1=8×2x2x×21=8×2x2x×2=4 Let, the capacity of the tank be x liters. By 1st pipe In 15 minutes pipe fill=x part of the

Mathematics

Created: 3 years ago | Updated: 10 months ago

Updated: 10 months ago

Combined Bank Combined Bank Officer (Cash) - 2022 গণিত 2022

306

(a)

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

উত্তরঃ

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

PRONAY TIRKI

2 years ago

(b)

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?

উত্তরঃ

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

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

PRONAY TIRKI

2 years ago

(c)

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?

উত্তরঃ

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

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

PRONAY TIRKI

2 years ago

(d)

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?

উত্তরঃ

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

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

PRONAY TIRKI

2 years ago

(e)

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?

উত্তরঃ

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