Janata Bank Ltd. || Officer (Cash) (13-11-2020) || 2020

Solve the followings:

1.

Simplify: $[(3 - x) (x + 2) + (- x + 4) (7 x + 2) - (x - y) (2 x - y)] - 3 x^{2} - 7 x + 5$

$G i v e n t h a t, [(3 - x) (x + 2) + (- x + 4) (7 x + 2) - (x - y) (2 x - y)] - 3 x^{2} - 7 x + 5$

$= [(3 x + 6 - x - 2 x) + (- x^{2} - 2 x + 28 x + 8) - (2 x^{2} - x y - 2 x y + y^{2})] - 3 x ² - 7 x + 5$

$= [(x - x^{2} + 6) + (26 x + 8 - 7 x^{2}) - (2 x^{2} - 3 x y + y^{2})] - 3 x^{2} - 7 x + 5$

$= x - x^{2} + 6 + 26 x + 8 - 7 x^{2} - 2 x^{2} + 3 x y - y^{2} - 3 x^{2} - 7 x + 5$

$= (- 13 x^{2} + 3 x y - y^{2} + 20 x + 19) a n s w e r$

Solve the followings:

2.

A salesman sells two shirts at BDT 594 each. On one he gains 10% and on the other, he loses 10%. Find his gain or loss percent on the whole.

$A t 10 % p r o f i t,$

$T h e \cos t p r i c e o f f i r s t s h i r t = \frac{100}{100 + 10} \times 594 = \frac{100}{110} \times 594 = 540 T k .$

$A g a i n, a t 10 % l o s s$

$T h e \cos t p r i c e o f s e c o n d s h i r t = \frac{100}{100 - 10} \times 594 = \frac{100}{90} \times 594 = 660 T k .$

$N o w, t h e t o t a l c o s t p r i c e o f t w o s h i r t s = 540 + 660 = 1, 200 T k .$

$A n d t h e t o t a l s e l l p r i c e o f t w o s h i r t s = 594 x 2 = 1, 188 T k .$

$∴ L o s s 1.200 - 1, 188 = 12 T k .$

$∴ P e r c e n t a g e l o s s = (\frac{12}{1200} \times 100) % = 1 % (a n s w e r)$

Solve the followings:

3.

A solution made up of 40% alcohol by volume is mixed with 4 liters of solution that is 10% alcohol by volume. How much, in liters of the 40% alcohol solution is needed to make a mixture that is 25% alcohol by volume?

$L e t, x l i t e r s o f 40 % a l c o h o l n e e d s t o b e a d d e d t o m a k e a m i x t u r e t h a t i s 25 % a l c o h o l .$

$A c c o r d i n g t o q u e s t i o n,$

$40 % o f x + 10 % o f 4 = 25 % o f (x + 4)$

$= \frac{40 x}{100} + \frac{10 \times 4}{100} = \frac{25 (x + 4)}{100}$

= $40 x + 40 = 25 x + 100 [M u l t i p l y i n g b o t h s i d e s b y 100]$

= $40 x - 25 x = 100 - 40$

= $15 x = 60$

$∴ x = \frac{60}{15} = 4 (a n s w e r)$

Solve the followings:

4.

If 24 machines can make 5 devices in 30 minutes, how many hours will it take 4 machines to make 15 devices?

$H e r e, 30 m i n u t e s = \frac{1}{2} h o u r$

$24 m a c h i n e s c a n m a k e 5 d e v i c e s i n = \frac{1}{2} h o u r$

$∴ I m a c h i n e c a n m a k e 1 d e v i c e s n = \frac{1 \times 24}{2 \times 5} h o u r$

$∴ 4 m a c h i n e s c a n m a k e 15 d e v i c e s n = \frac{1 \times 24 \times 15}{2 \times 5 \times 4} = 9 h o u r s (a n s w e r)$

Solve the followings:

5.

An equilateral triangle is inscribed into a circle with radius 10cm. What is the area of the triangle?

$A s s h o w n i n t h e f i g u r e,$

$△ A B C i s a n e q u i l a t e r a l t r i a n g l e i n s c r i b e d i n a c i r c l e c e n t e r e d a t O .$

$H e r e, t h e r a d i u s o f t h e c i r c l e i s 10 c m .$

$A s t h e t r i a n g l e i s a n e q u i l a t e r a l, s o t h e v a l u e o f e a c h a n g l e i s 60 ° a n d h a l f o f i t i s 30 ° .$

$H e r e, O D ⊥ B C a n d O D l i n e d i v i d e s B C e q u a l l y .$

$N o w, f r o m △ B O D,$

$C o s 30 ° = \frac{B a s e}{H y p o t e n u s e} = \frac{B D}{O B} = \frac{B D}{10}$

$= \frac{\sqrt{3}}{2} = \frac{B D}{10}$

$∴ B D = 5 \sqrt{3}$

$∴ O n e A r m o f t h e t r i a n g l e i s = B D + C D = 5 \sqrt{3} + 5 \sqrt{3} = 10 \sqrt{3} c m$

$N o w, w e k n o w t h a t,$

$A r e a o f a e q u i l a t e r a l t r i a n g l e i s \frac{\sqrt{3}}{4} {(A r m)}^{2} c m^{2}$

$= \frac{\sqrt{3}}{4} {(10 \sqrt{3})}^{2} c m^{2} = \frac{\sqrt{3}}{4} (100 \times 3) c m^{2} = 75 \sqrt{3} c m^{2} (a n s w e r) .$

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