Bank Math

Question: What is the H.C.F. of 20/48, 28/72, and 36/90?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M of denominators)

H.C.F of numerators: H.C.F. of 20, 28 and 36 = 4
& L.C.M of denominators: L.C.M. of 48, 72 and 90 = 720

∴ Required H.C.F. = 4/720 = 1/180

Question: If the difference between the circumference and diameter of a circle is 90 cm, then the diameter of the circle is-

Solution:
Let,
Radius of the circle = r
Diameter of the circle = 2r
Circumference of the circle = 2πr

According to the question,
2πr - 2r = 90
⇒ 2r(π - 1) = 90
⇒ r(π - 1) = 45
⇒ r = 45/(π - 1)
⇒ r = 45/(22/7 - 1) ; [π = 22/7]
⇒ r = 45/{(22 - 7)/7}
⇒ r = 45/(15/7)
⇒ r = 45 × (7/15)
∴ r = 21 cm

∴ Diameter of the circle = 2r = 2 × 21 = 42 cm

So the diameter of the circle is 42 cm.

Total C.P. = Cost + Overhead Expenses
= Tk. (150 + 12% of 150)
= Tk. (150 + 18)
= Tk. 168
∴S.P.=Tk.(110/100×168)=Tk. 184.80

Question: ∠P and ∠Q are complementary to each other. If ∠P = 3x + 9° and ∠Q = 2x - 4°, find the value of ∠Q.

Solution:
Here,
∠P = 3x + 9° and ∠Q = 2x - 4°

For complementary angles,
∠P + ∠Q = 90°
⇒ (3x + 9°) + (2x - 4°) = 90°
⇒ 3x + 9° + 2x - 4° = 90°
⇒ 5x + 5° = 90°
⇒ 5x = 85°
∴ x = 17°

So, ∠Q = 2x - 4°
= (2 × 17) - 4°
= 30°

ধরি,
মোট member = 100
∴ মহিলা member = 70 এবং পুরুষ member = 30 জন।
30 জন পুরুষের মধ্যে অবিবাহিত = 30 × 2/3 = 20 জন।
সুতরাং বিবাহিত = 10 জন।
মোট বিবাহিত = 60 জন।
∴ মহিলাদের মধ্যে বিবাহিত = 60 - 10 = 50 জন।
∴ নির্ণেয় fraction = 50/70
= 5/7

Question: The cost price of 8 apples equals the selling price of 7 apples. The profit or loss percent in the transaction is?
 
Solution:
দেওয়া আছে,
8টি আপেলের ক্রয়মূল্য = 7টি আপেলের বিক্রয়মূল্য

ধরি, 1টি আপেলের ক্রয়মূল্য = x টাকা
তাহলে,
8টি আপেলের ক্রয়মূল্য = 8x টাকা
7টি আপেলের বিক্রয়মূল্য = 8x টাকা

∴ 1টি আপেলের বিক্রয়মূল্য = 8x / 7 টাকা

∴ লাভ = বিক্রয়মূল্য − ক্রয়মূল্য
= (8x/7) − x
= (8x − 7x)/7
= x/7

∴ x টাকায় লাভ হয় x/7 টাকা
1 টাকায় লাভ হয় = (x/7)/x = 1/7
100 টাকায় লাভ হয় = (1/7) × 100 ≈ 14.29 টাকা

∴ লাভ হয় ≈ 14.29%

S.I. in 3 years = 6500 - 5600 = 900
Interest every year = 900/3 = 300
So, principal = 5600 - 2 × 300 = 5000

We know, I = pnr
Or, Interest rate, r = I/pn
= 600/(5000×2) × 100
= 6%

Question: If tanA = 3/4, then sec A =

Solution:
দেওয়া আছে,
tanA = 3/4
∴ অতিভুজ = √(3² + 4²) = √(9 + 16) = 5

আমরা জানি,
secA = অতিভুজ/ভূমি
= 5/4

Question: The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Solution: 
et the breadth of the rectangle be B cm.
Perimeter of a rectangle = 2(Length + Breadth) = 2(L + B)
Given the Ratio of perimeter to breadth = 5 : 1
Now,
2(L + B)/B = 5/1
⇒ 2L + 2B = 5B
⇒ 3B = 2L
∴ B = (2/3)L

Area of the rectangle = L . B = 216 sq. cm.
⇒ L . (2/3)l = 216
⇒ L² = (216 × 3)/2
⇒ L² = 324 = 18²
∴ l = 18 cm

So the length of the rectangle is 18 cm.

Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

Question: The simple interest on a certain sum at 10% per annum for 2 years is Tk. 2,000. Find the compound interest on the same sum at the same rate for 2 years.

Solution:
We know,
I = Pnr

Given,
I = 2,000 Tk.
r = 10%
n = 2 years

∴ P = I/nr
= (2,000 × 100)/(2 × 10)
= 10000

Now,
Compound Principal = 10,000 × (1 + 10/100)²
= 10,000 × (110/100)²
= 10,000 × (11/10)²
= 10,000 × (121/100)
= 12,100

So, Compound interest = 12,100 - 10,000
= 2,100 Tk.

Question: , then what is the value of m?

Solution:

We have 4^m > 1 
Now, if m = -1, then 4^m = 4 ^-1 = 1/4 = 0.25 < 1, so incorrect.
if m = 1, then 4^m = 4¹ = 4 > 1, correct.

∴ m = 1

Let, the carpet's length = x
Given, x² = 169 cm²
∴ carpet's length = √169 = 13 cm
Width of the room = (13 - 2) cm = 11 cm
So, area of the rectangular room = (13 × 11) = 143 cm²

Question: If the ratios A : B = 1 : 2, B : C = 4 : 3, hold true, and their total sum is 630, determine C.

Solution: 
Given,
A : B = 1 : 2 = 2 : 4
B : C = 4 : 3

∴ A : B : C = 2 : 4 : 3

∴ Value of C = 630 × 3/9
= 210

Question: Three lights blink at intervals of 18 sec, 24 sec, and 32 sec. If they blink together now, how many times will they blink together in 6 hours?

Solution:
The lights will blink together at LCM of their intervals.
Intervals: 18, 24, 32 seconds

Factorize:
18 = 2 × 3²
24 = 2³ × 3
32 = 2⁵

LCM = product of highest powers of all primes:
LCM = 2⁵ × 3² =3² × 9 = 288 seconds
Convert 6 hours to seconds:
6 hours = 6 × 60 × 60 = 21600 seconds

Number of times they blink together:
Times = 21600/288 + 1 (+1 because they blink together now, at time 0)
= 75 + 1
= 76

Note: If you only count after the first blink at time 0, it is 75 times; including the initial moment, it is 76 times.

Question: A salesperson earns a 7.5% commission on sales. If he earned Tk. 1125 in commission, what was the total value of the goods he sold?

Solution:
Given that,
Commission rate = 7.5%
Commission earned = Tk. 1125

Let the total sales amount be x.
Then,
(7.5/100) ​× x = 1125
⇒ x = (1125 ​× 100)/7.5
⇒ x = 15000

So the total value of the goods he sold was Tk. 15000

Distance covered downstream = 64 km
Time is taken in downstream = 12 hours.
Rate of downstream = distance/time = a = 64 km/12 hours
= 16/3 km/hr.
Distance covered in upstream = 32km
Time is taken upstream = 8 hours.
Rate of upstream = distance/time = b = 32km/8 hours
= 4 km/hr.
Speed in still water = (a + b)/2 = (1/2){(16/3) + 4} km/hr
= (1/2)(28/3)
= 14/3 km/hr.

Let the interest rate be r%
We know that,
S.I = PTR/100
=> (1540 x 5 x r)/100  +  (1800 x 4 x r)/100 = 1788
=> r = 178800/14900 = 12%

The Distance covered by the motorcycle with speed 60km/hr in 3 hours = (60 x 3)
= 180 km
Now,
Speed = Distance/Time
Since the train covers the same 180 km in 3/2 hours (we can write 1 and half hours as 3/2hours)
Then the speed of the train = 180/ (3/2)
= 180 × (2/3)
= 120 km/hr.
Hence, the train travelled at the speed of 120km/hr to cross 180km in 1 and half hour.

Question: A square garden is surrounded by a path of uniform width 2 meters. If the area of the path is 48 square meters, find the side length of the garden.

Solution:
Let the side of the garden = x meters.
Then, the side of the garden including the path = x + (2 × 2)
= x + 4 meters.

Area of path = Area of garden with path - Area of garden
⇒ 48 = (x + 4)² - x²
​⇒ 48 = x² + 8x + 16 - x²
​⇒ 8x + 16 = 48
⇒ 8x = 48 - 16
⇒ x = 32/8
∴ x = 4 meters

∴ Therefore, side length of the garden is 4 meters.

Question: If A = 30°, then what is the value of (1 - tan²A)/(1 + tan²A)?

Solution:
Here, A = 30°

Now,
(1 - tan²A)/(1 + tan²A)
= {1 - (tan30°)²}/{1 + (tan30°)²}
= {1 - (1/√3)²}/{1 + (1/√3)²}
= (1 - 1/3)/(1 + 1/3)
= (2/3)/(4/3)
= 1/2

Number of rounds to be completed = 4 / 1/4 = 16 times
Since the ratio of speeds of A and B is 5 : 4
⇒ When A covers 5 rounds, B will complete 4 rounds and they will meet.
⇒ A will pass B after 5^th, 10^th & 15^th round.
∴ The winner passes the other thrice.

Question: Two students appeared at an examination. One of them secured 7 marks more than the other and his marks was 52% of the sum of their marks. The marks obtained by them are-

Solution:

ধরি, 
তাদের প্রাপ্ত নাম্বার যথাক্রমে x এবং x + 7

প্রশ্নমতে,
x + 7 = (52/100) × (x + x + 7)
⇒ x + 7 = (13/25) × (2x + 7)
⇒ 25x + 175 = 26x + 91
⇒ 26x - 25x = 175 - 91
⇒ x = 84

∴ অপর জনের প্রাপ্ত নাম্বার = 84 + 7
= 91

Question: An observer who is 1.6 meters tall is standing 25 meters away from a tower. If the angle of elevation from his eye to the top of the tower is 45°, what is the height of the tower?

Solution:


পর্যবেক্ষকের উচ্চতা, CD = 1.6 মিটার
এখানে, CD = EB
টাওয়ারের উচ্চতা = AB
এখন,
tan∠C = AE/CE
⇒ tan45° = AE/25
⇒ 1 = AE/25
∴ AE = 25

∴ AB = AE + BE
= 25 + 1.6 
= 26.6
∴ টাওয়ারটির উচ্চতা 26.6 মিটার।

Question: The simple interest on a certain sum at the rate of 5% per annum for 3 years and 7 years differs by Tk. 400. What is the principal amount?

Solution:
Let the principal amount be Tk. P
Then, ATQ
400 = {(P × 5 × 7)​/100} - {(P × 5 × 3​)/100}
⇒ 400 = (35P - 15P)/100
⇒ 40000 = 20P
⇒ P = 40000/20
∴ P = 2000

So the principal amount is Tk. 2000

Question: Which of the following is a leap year?

Solution:
অধিবর্ষ বা লিপ ইয়ার নির্ণয়ের দুটি প্রধান নিয়ম রয়েছে:
১. সাধারণ বছর: বছরটি 4 দ্বারা নিঃশেষে বিভাজ্য হতে হবে।
২. শতাব্দী বছর (100 দ্বারা বিভাজ্য): বছরটি 400 দ্বারা নিঃশেষে বিভাজ্য হতে হবে।

এখন,
ক) 1900 সাল (শতাব্দী বছর): 1900 ÷ 400 দ্বারা বিভাজ্য নয় (ভাগশেষ 300)। ⇒ অধিবর্ষ নয়।

খ) 2000 সাল (শতাব্দী বছর): 2000 ÷ 400 = 5 (ভাগশেষ 0)। ⇒ অধিবর্ষ।

গ) 2022 সাল: এটি 4 দ্বারা বিভাজ্য নয়। (2022 ÷ 4 ⇒ ভাগশেষ 2)। ⇒ অধিবর্ষ নয়।

ঘ) 2010 সাল: এটি 4 দ্বারা বিভাজ্য নয়। (2010 ÷ 4 ⇒ ভাগশেষ 2)। ⇒ অধিবর্ষ নয়।

অতএব, 2000 সালটি অধিবর্ষ।

Let the numbers be 7x and 4x
ATQ,
(7x + 8) : (4x + 8) = 13 : 8
Or, 56x + 64 = 52x + 104
Or, 4x = 40

The smaller number is 40

log₁₀1000 
= log₁₀10³
= 3log₁₀10
= 3×1=3

Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 - 15 = 39)

Let x number of people come after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days

More men, Less days (Indirect Proportion)

(men) 2000 : (2000 + x) :: 20 : 39
⇒ 2000 × 39 = (2000 + x) 20
⇒ 100 × 39 = (2000 + x)
⇒ 3900 = 2000 + x
⇒ x = 3900 − 2000
= 1900

A hostel had provisions for 250 men for 40 days

If 50 men leaves the hostel, remaining men = 250 - 50 = 200
We need to find out how long the food will last for these 200 men.

Let the required number of days = x days.

More men, Less days (Indirect Proportion)
(men) 250 : 200 :: x : 40
⇒ 250 × 40 = 200x
⇒ 5 × 40 = 4x
⇒ x = 5 × 10
= 50

Question: If ⁷P_r = 210, what is the value of r?

Solution:
We know,
ⁿP_r = n!/(n - r)!

Given that,
⁷P_r = 210
⇒ 7!/(7 - r)! = 210
⇒ 5040/(7 - r)! = 210 (7! = 5040)
⇒ (7 - r)! = 5040/210
⇒ (7 - r)! = 24
⇒ (7 - r)! = 4!   ;(4! = 4 × 3 × 2 × 1 = 24)
⇒ 7 - r = 4
⇒ r = 7 - 4
∴ r = 3

Question: Find the area of a triangle with side lengths of 5 meters, 6 meters, and 7 meters.

Solution:
দেওয়া আছে,
ত্রিভুজের তিন বাহুর দৈর্ঘ্য, a = 5 মি., b = 6 মি. এবং c = 7 মি.

আমরা জানি,
ত্রিভুজের অর্ধপরিসীমা, s = (a + b + c)/2
= (5 + 6 + 7)/2
= 18/2
= 9 মি.

∴ ত্রিভুজের ক্ষেত্রফল = √{s(s - a)(s - b)(s - c)}
= √{9(9 - 5)(9 - 6)(9 - 7)}
= √(9 × 4 × 3 × 2)
= √(36 × 6)
= 6√6 বর্গ মি.

অতএব, ত্রিভুজটির ক্ষেত্রফল 6√6 বর্গ মি.

Let the number A, B and C be 12a , 15a and 25a respectively
Then from the question 12a + 15a + 25a = 312
⇒ 52a = 312
⇒ a = 312/52
⇒ a = 6
Required ratio = {(15 × 6) - (12 × 6)}/{(25 × 6) - (15 × 6)}
= 3/10
= 3 : 10.

Question: Two friends invested TK.1500 and TK. 3500 in a business. They earned a profit of TK. 1000. One-half of the profit was divided equally between them and the other half was divided in proportion to their capitals. How much did each of them receive ?

Solution:
Ratio of shares
= 1500 : 3500
= 3 : 7

Share of first friend
= TK. [(500/2) + {500 × (3/10)}]
= TK.[250 + 150]
= TK. 400

Share of second friend
= TK. [(500/2) + {500 × (7/10)}]
= TK.[250 + 350]
=  TK. 600

Let the weight of the grocery bag be 1000 gm.
Now, the shopkeeper sells his grocery using weights 10% less than true weights.
Hence, actual weight of bag = 90% of 1000 gm = 900 gm
If each gram = Tk.1, C.P. of each bag containing 900 gm = Tk. 900
The shopkeeper sells with a gain of 30% on true C.P.

Calculate the S.P.
Selling price = 100 + gain% × C.P.
Therefore,
Selling price = (130/100) × Tk. 1000
= Tk. 1300
Gain = S.P. - C.P. = 1300 - 900 = Tk. 400
Gain% = (400/900) × 100
= 44.44%