Rupali Bank Limited ‘Officer (Cash)’ Recruitment Test-2018

33. If \sin A + {\sin ^2}A = 1, then the value of the expression {\cos ^2}A + {\cos ^4}A  is-
(a) 1           (b) 1/2             (c) 2                 (d) 3
Ans. (a) 1

Explanation:
Given that,
\sin A + {\sin ^2}A = 1
\Rightarrow \sin A = 1 - {\sin ^2}A
\Rightarrow \sin A = {\cos ^2}A\,\,\,\,\,\,\,\,[\because {\cos ^2}A = 1 - {\sin ^2}A]
\Rightarrow {\sin ^2}A = {\cos ^4}A\,\,\,     [Squaring both sides]
\Rightarrow 1 - {\cos ^2}A = {\cos ^4}A\,\,\,\,[\because {\sin ^2}A = 1 - {\cos ^2}A]
\therefore {\cos ^2}A + {\cos ^4}A\, = 1

34. A pole 6m high casts a shadow 2\sqrt 3m long on the ground, then the sun’s elevation is-
(a) {60^0}      (b) {45^0}            (c) {30^0}      (d) {90^0}
Ans. (a) {60^0}
Explanation:
চিত্র
\tan A = \frac{6}{{2\sqrt 3 }}
\Rightarrow\tan A = \frac{3}{{\sqrt 3 }}
\Rightarrow \tan A = \sqrt 3
\Rightarrow \tan A = \tan {60^0}
\therefore A = {60^0}

35. If a+1, 2a+1, 4a-1 are in Arithmetic progression, then the value of ‘a’ is:
(a) 1           (b) 2              (c) 3              (d) 4
Ans. (b) 2 
Explanation:
we can test option,
If a=1 then, progression is 2,3,3 [it is not equally successive]
If a=2 then, progression is 3,5,7 [it is equally successive]
If a=3 then, progression is 4,7,11 [it is not equally successive]
If a=4 then, progression is 5,9,15 [it is not equally successive]
So, a=2 is right.

36. Suppose today is Friday. What day of the week will it be 65 days from now?
(a) saturday               (b) Monday              (c) Tuesday          (d) Friday
Ans. (a) saturday
Explanation:
We know that 1 week=7 days
If 65\div 9=7 and Remaninder will be 2
So day will be 2 days from today
i.e. day will be saturday.

37. A median of a triangle divides it into two-
(a) congruent triangles  (b) triangles of equal area  (c) isosceles triangles (d) right triangles
Ans. (b) triangles of equal area
Explanation:
Let ABC a triangle
Here, BD=CD
AE \bot BC
Area of \Delta ABD
=\frac{1}{2} \times BC \times AE
\frac{1}{2} \times BC \times AE  [\because BD=CD]
= Area of \Delta ADC
\therefore A median of a triangle divides it into two triangles of equal area.

38. Which of the following angle can be constructed with the help of a ruler and a pair of compasses? 
(a) {35^0}   (b) {40^0}     (c) {37.5^0}   (d) {47.5^0}
Ans. (c) {37.5^0}

39. If a, b and c are the lengths of the three sides of a triangle, then which of the following is true? 
(a) a+b<c       (b) a-b<c        (c) a+b=c          (d) a+b\geqslant c
Ans. (b)  a-b<c 
Explanation: The difference between any two sides of triangle is less than the third side.
40. Which line is parallel to y=x-2 ?
(a) y=2x+1       (b) 2y=2x-6        (c) 2y=x+7       (d) y=3x+1
Ans. (b) 2y=2x-6
Explanation: 
Given line of equation, y=x-2
Slop of this equation is 1
From, option (b)
2y=2x-6
or, y=x-3
\therefor Slop of this equation is 1
Since slop is equal. so both equation are parallel.
41. The area of a triangle with sides 3 cm, 5 cm and 6 cm  is-
(a) 2\sqrt 3 c{m^2}  (b) 4\sqrt 14 c{m^2}  (c) 2\sqrt 14 c{m^2}  (d) 2\sqrt 5 c{m^2}
Ans.  (c) 2\sqrt 14 c{m^2}
Explanation:
Given that, Sides of the triangle 3cm, 5cm and 6cm
Perimerter=2s=3+5+6
Half perimeter=s=\frac{{14}}{2}=7
Area of the triangle,
=\sqrt {s(s - a)(s - b)(s - c)}
=\sqrt {7(7 - 3)(7 - 5)(7 - 6)}
=\sqrt {7 \times 4 \times 2 \times 1}
=2\sqrt {14}

43. The value of k, if (x-1) is a factor of 4{x^3} + 3{x^2} - 4x + k is-
(a) 1                (b) 2                      (c) -3               (d) 3
Ans. (c) -3 
Explanation:
f(x)=4{x^3} + 3{x^2} - 4x + k
\therefore f(1)={4.1^3} + {3.1^2} - 4.1 + k    [\because x-1 is a factor]
=4+3-4+k
=K+3
Now, K+3=0
\therefore k=-3

44. There are 5 red and 3 black balls in a bag. Probability of drawing a black ball is-
(a) 5/8                   (b) 1/2                    (c) 3/8                      (d) 1/4
Ans. (c) 3/8
Explanation: 
Number of red ball=5
Number of black ball=3
Total ball=5+3=8
\therefore Probability of a black ball=\frac{3}{8}

45. The total surface area of a hemisphere of radius r is-
(a) 4\pi {r^2}   (b)   \pi {r^2} (c)  2\pi {r^2}  (d)3\pi {r^2}
Ans.   (d)3\pi {r^2}
Explanation:
Total Surface area of a hemisphere
=curved surface arc+area of the base
=2\pi {r^2}+ \pi {r^2}=3\pi {r^2}

46. The roots of the equation 9{x^2} - bx + 81 = 0 will be equal, if the value of b is-
(a) \pm 9         (b) \pm 18             (c) \pm 27       (d) \pm 54
Ans.  (d) \pm 54
Explanation: 
a{x^2} + bx + c =0 is a quadratic equation.
If {b^2} - 4ac = 0 then roots of the equation will be equal.
Given quadratic equation  9{x^2} - bx + 81 = 0
here, a=9, c=81
so, {b^2} - 4.9.81 = 0
\Rightarrow {b^2} - 2916 = 0
\Rightarrow b = \sqrt {2916}
\Rightarrow b = \pm 54

47. If \sec \theta + \tan \theta = x, then \tan \theta is-
(a) ({x^2} + 1)/x   (b) ({x^2} - 1)/x  (c)  ({x^2} + 1)/2x  (d)({x^2} - 1)/2x
Ans. (d)({x^2} - 1)/2x
Explanation: We know that,
{\sec ^2}\theta - {\tan ^2}\theta = 1
\Rightarrow \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) = 1
\Rightarrow x\left( {\sec \theta - \tan \theta } \right) = 1
\Rightarrow \sec \theta - \tan \theta = \frac{1}{x} .. ……. (i)
Given, \sec \theta + \tan \theta = x ……………. (ii)
from (i)-(ii), we get,
-2tan \theta=\frac{1}{x} - x
\Rightarrow - 2\tan \theta = \frac{{1 - {x^2}}}{x}
\Rightarrow \tan \theta = \frac{{{x^2} - 1}}{{2x}}
\therefore \Rightarrow \tan \theta = \frac{{{x^2} - 1}}{{2x}}

48. Consider that w+x=-4, x+y=25 and y+w=15. Then the average of w, x, y is-
(a) 3                 (b) 4                (c) 5                 (d) 6
Ans. (d) 6
Explanation: 
w+x=-4
x+y=25
y+w=15
—————-
2(x+y+w)=36       [adding]
\therefore  x+y+w=\frac{{36}}{2}=18
Now, Average of x,y &w =\frac{{x + y + w}}{3}=\frac{{18}}{3} = 6

49. What is the original price of a T-shirt. if the sale price after 15% discount is 272?
(a) 300                 (b) 280                      (c) 320                   (d) 314
Ans.  (c) 320
Explanation:
After 15% discount sale price=100%-15%=85%
85% of the price=272 tk.
\therefore 1% of the price = \frac{{272}}{{85}}
\therefore 100% of the price=\frac{{272 \times 100}}{{85}}=320 tk.

50. Tk. 500 is deposited in a savings account which pays 7% annual interest compounded semi-annually. To the nearest taka, how much is in the account at the end of the year?
(a) 542              (b) 536                (c) 512                (d) 524
Ans. (b) 536
Explanation:

51. If {\log _x}\frac{1}{4} = - 2, then x =?
(a) - \frac{1}{2}           (b) \frac{1}{2}        (c) 2          (d) 3
Ans. (c) 2
Explanation: 
since,  {\log _x}\frac{1}{4} = - 2
\therefore {x^{ - 2}} = \frac{1}{4}
\Rightarrow  {x^{ - 2}} = \frac{1}{{{2^2}}}
\Rightarrow {x^{ - 2}} = {2^{ - 2}}
\therefore x=2

52. If 5% is gained by selling an article for BDT 350 than selling it for BDT 340, the cost of the article is-
(a) BDT 180            (b) BDT 150          (c) BDT 200             (d) BDT 250
Ans. (c) BDT 200
Explanation: 
5% = 10 tk.
\therefore 1%= \frac{{10}}{5}
\therefore 100% = \frac{{10 \times 100}}{5}= 200 tk.

53. If x = {y^a}y = {z^b} and z = {x^c} then the value of abc is-
(a) 1               (b) 0                   (c) 0.5             (d) Infinity
Ans. (a) 1
Explanation: Given that,
x = {y^a}
\Rightarrow x = {z^{ba}}     [\because y = {z^b}]
\Rightarrow x={x^{abc}}    [\because z = {x^c}]
\therefore abc=1

54. If  x is 30% greater than y, what percent of y is x?
(a) 70          (b) 77             (c) 120                (d) 130
Ans. (d) 130
Explantion: 
Let, y=100 then, x=130
so, x is 130% of y

55. The lengths of two sides of a right-angled triangle are 13cm and 5cm respectively. The length of the third side is-
(a) 13                 (b) 17              (c) 11               (d) 12
Ans. (d) 12
Explanation:
Let, length of third side = x cm
since right-angled triangle,
so, {x^2} + {5^2} = {13^2}
\Rightarrow {x^2} = {13^2} - {5^2}
\Rightarrow {x^2} = 169 - 25
\Rightarrow {x^2} = 144
\Rightarrow x = \sqrt {144}
\Rightarrow x = 12

56. The present age of Habib and shikha are in the ratio of 6:4. Five years ago their ages were in the ratio of 5:3. How old is Habib now?
(a) 24              (b) 30                       (c) 36                (d) 42
Ans. (b) 30
Explanation:
Given that, Ratio of present age of Habib and shikha=6:4
Let, Age of Habib = 6x year and Age of Shikha = 4x year
So, Five years ago age of Habib= 6x-5
and age of Shikha = 4x-5
regarding question,
\frac{{6x - 5}}{{4x - 5}} = \frac{5}{3}
\Rightarrow 5\left( {4x - 5} \right) = 3\left( {6x - 5} \right)
\Rightarrow 20x - 25 = 18x - 15
\Rightarrow 20x - 18x = - 15 + 25
\Rightarrow 2x = 10
\Rightarrow x = \frac{{10}}{2}
\therefore x = 5