PrepBank · বিষয়ভিত্তিক প্রশ্ন
PrepBank · পাতা ১০০ / ১৬১ · ৯,৯০১–১০,০০০ / ১৬,১২৪
Let BD be the lighthouse and A and C be the two points on the ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°,∠BCD = 60°
tan45° = BD/BA
1 = 60/BA
BA = 60 m .......(i)
tan60° = BD/BC
√3= 60/BC
BC = 60/√3
= 60 × √3)/(√3 × √3)
= 20√3
= 20 × 1.73
= 34.6 m .......(ii)
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (1) and (2)]
= 94.6 m
Speed of the stream = 1
Motor boat speed in still water be = x kmph
Downstream = x + 1 kmph
Up Stream = x - 1 kmph
[35/(x + 1)] + [35/(x - 1)] = 12
x = 6 kmph
salt = 16 × 1/4 = 4 Kg
water = 16 - 4
= 12 kg.
এখন water/salt = 4
∴ water = 4 × 4 = 16 Kg
∴ (16 - 12) = 4 Kg water add করতে হবে।
Question: If tan (5x - 10°) = cot (5y + 20°), then the value of (x + y) is
Solution:
tan (90° - θ) = cotθ
∴ tan (5x - 10°) = cot (5y + 20°)
⇒ tan (5x - 10°) = tan {90° - (5y + 20°)}
⇒ 5x - 10° = 90° - (5y + 20°)
⇒ 5x - 10° = 90° - 5y - 20°
⇒ 5x + 5y = 70° + 10°
⇒ 5 (x + y) = 80°
∴ x + y = 16°
Question: If a pole 12 m high casts a shadow 4√3 m long on the ground, then the elevation of the sun is -
Solution:
ধরি,
AB = 12, BC = 4√3
ABC সমকোণী ত্রিভুজ হতে পাই,
tanθ = AB/BC
⇒ tanθ = 12/4√3
⇒ tanθ = 3/√3
⇒ tanθ = (√3 × √3)/√3
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°
So the elevation of the sun is 60°.
Question: If 35% of a certain number is 84, then find the number-
Solution:
Let the number be x.
Then,
35% of x = 84
⇒ (35/100) × x = 84
⇒ 7x/20 = 84
⇒ x = (20 × 84)/7
⇒ x = 1680/7
∴ x = 240
Question: A reduction of 25% in the price of sugar enables a housewife to purchase 5 kg more for 600 Taka. What is the original price per kg of sugar?
Solution:
Let original price per kg = p Taka
Reduced price = p - p of 25% = 0.75p
Original quantity for 600 Taka
Q1 = 600/p kg
And
New quantity for 600 Taka
Q2 = 600/0.75p = 800/p kg
Now, quantity difference,
Q2 - Q1 = 5
(800/p) - (600/p) = 5
200/p = 5
p = 200/5
p = 40
∴ Original price per kg = Tk. 40
Question: If x2 - 5x + 1 = 0, and x > 1, then what is the value of x - (1/x)?
Solution:
We are given:
x2 - 5x + 1 = 0
⇒ x - 5 + 1/x = 0
∴ x + 1/x = 5
Now,
(a - b)2 = (a + b)2 - 4ab
⇒ (x - 1/x)2 = (x + 1/x)2 - 4 × x × (1/x) [Here, a = x, b = 1/x]
Substitute the values:
(x - 1/x)2 = 25 - 4
⇒ (x - 1/x)2 = 21
⇒ (x - 1/x)2 = 21
∴ x - 1/x = √21
(d/80 + d/40) - (d/60 + d/60) = 2
(3d + 6d - 4d - 4d)/240 = 2
d = 480 km
Let,
The amount of work P does in 1 day = x
Amount of work Q does in 1 day = 2x
Amount of work R does in 1 day = 3x
Amount of work P, Q,R together do in 1 day = x + 2x + 3x = 6x
they can together complete the work in 1 day = (1/6x) days
Given,
1/6x = 2
⇒ 12x = 1
⇒ x = 1/12
Therefore, amount of work Q does in 1 day = 2 × (1/12) = 1/6
That is, Q needs 6 days to complete the work.
Question: By selling a product for 3500, a seller makes 15% profit. If the profit is decreased to 9%, then selling price will be:
Solution:
১৫% লাভে,
একটি পণ্যের ক্রয়মূল্য ১০০ টাকা হলে বিক্রয়মূল্য = (১০০ + ১৫) = ১১৫ টাকা
বিক্রয়মূল্য ১১৫ টাকা হলে ক্রয়মূল্য ১০০ টাকা
বিক্রয়মূল্য ১ টাকা হলে ক্রয়মূল্য (১০০/১১৫) টাকা
বিক্রয়মূল্য ৩৫০০ টাকা হলে ক্রয়মূল্য (১০০/১১৫) × ৩৫০০ টাকা
= ৩০৪৩.৪৮ টাকা
আবার,
৯% লাভে,
একটি পণ্যের ক্রয়মূল্য ১০০ টাকা হলে বিক্রয়মূল্য = (১০০ + ৯) = ১০৯ টাকা
ক্রয়মূল্য ১০০ টাকা হলে বিক্রয়মূল্য ১০৯ টাকা
ক্রয়মূল্য ১ টাকা হলে বিক্রয়মূল্য (১০৯/১০০) টাকা
ক্রয়মূল্য ৩০৪৩.৪৮ টাকা হলে বিক্রয়মূল্য (১০৯/১০০) × ৩০৪৩.৪৮ টাকা
= ৩৩১৭.৩৯ টাকা
∴ বিক্রয়মূল্য হবে ৩৩১৭.৩৯ টাকা।
Question: A tap can fill a tank in 12 hours. After one-third of the tank has been filled, two more similar taps are opened. What is the total time taken to fill the tank completely?
Solution:
A tap can fill a tank in 12 hours.
∴ A tap can fill 1/12 part in one hour.
Tap can fill 1 part in 12 hours
⇒ Tap can fill 1/3 part in 12 / 3 = 4 hours
⇒ The rest part = 1 - 1/3 = 2/3.
After one-third of the tank has been filled, two more identical taps are opened.
3 similar tap can fill 3/12 = 1/4 part in one hour.
3 similar tap can fill 1 part in = 4 hour.
3 similar tap can fill 2/3 part in = 4 × 2/3 = 8/3 hour.
8/3 hours means 2 hours 40 minutes.
∴Total time taken: 6 hours 40 minutes.
Speed upstream = 2/2 = 1 km/hr
Speed downstream = 1/(20/60) = 3 km/hr
Speed in still water = (1/2)(3+1) = 2 km/hr
Time taken to travel 5 km in still water = 5/2
= 2(1/2)
= 2 hr 30 min
S.I. on Tk. 1800 = T.D. on Tk. 1872
P.W. of Tk. 1872 is Tk. 1800
Tk. 72 is S.I. on Tk. 1800 at 12%
Time = (100×72)/(12×1800)
= 1/3 years = 4months
Question: Bipul ate 2/3 of a cake. His friend Rafi ate 2/3 of what was left. Then Rafi's sister ate 2/3 of what was still left. What fraction of the cake remains uneaten?
Solution:
বিপুল খেয়েছে = 2/3
∴ অবশিষ্ট = 1 - 2/3 = 1/3
রাফি খেয়েছে = 1/3 × 2/3 = 2/9
∴ অবশিষ্ট = 1/3 - 2/9 = (3 - 2)/9 = 1/9
রাফির বোন খেয়েছে = 1/9 × 2/3 = 2/27
∴ অবশিষ্ট = 1/9 - 2/27 = (3 - 2)/27 = 1/27
4 x 162 = 648. Sum of decimal places = 6.
So,
0.04 x 0.0162
= 0.000648
= 6.48 x 10-4
Question: What is the value of (255 - 55) ÷ 4 × 15 - 504 ÷ 3 = ?
Solution:
(255 - 55) ÷ 4 × 15 - 504 ÷ 3
= 200 ÷ 4 × 15 - 504 ÷ 3
= 50 × 15 - 168
= 750 - 168
= 582
30 % of 50 = 15
and 75% of 80 = 60.
∴ The fraction is = 15/60 = 1/4.
Question: What is the length of a chord that is 6 cm away from the center of a circle with a radius of 10 cm?
Solution:
Given that,
Radius of the circle, r = 10 cm
Distance from the center of the circle to the chord, d = 6 cm
Then, the length of the chord = 2 × √(radius2 - distance from center2)
= 2 × √(102 - 62) cm
= 2 × √(100 - 36)
= 2 × √64
= 2 × 8
= 16 cm
So the length of the chord is 16 cm.
Question: Three sides of a triangle measure 6 cm, 10 cm and p cm. The minimum integral value of p is:
Solution:
আমরা জানি,
- ত্রিভুজের যে কোন দুই বাহুর সমষ্টি তার তার তৃতীয় বাহু অপেক্ষা বৃহত্তর।
- ত্রিভুজের যে কোন দুই বাহুর অন্তর বা ব্যবধান তৃতীয় বাহু অপেক্ষা ক্ষুদ্রতর।
অপশন
ক) 1 + 6 = 7 < 10
খ) 2 + 6 = 8 < 10
গ) 3 + 6 = 9 < 10
ঘ) 5 + 6 = 11 > 10
Question: A rectangular garden has an area of 800 square feet. It will be fenced on three sides, leaving one side of 40 feet uncovered. How many feet of fencing is required?
Solution:
দেওয়া আছে, আয়তকার বাগানের ক্ষেত্রফল = 800 square feet
এবং যে বাহুটি খোলা থাকবে তার দৈর্ঘ্য = 40 feet
ধরি, আয়তকার ক্ষেত্রের অন্য বাহুর দৈর্ঘ্য = x feet
প্রশ্নমতে,
40 × x = 800
⇒ x = 800/40
⇒ x = 20 feet
যেহেতু বাগানটির তিন দিকে বেড়া দেওয়া হবে এবং 40 feet দৈর্ঘ্যের একটি বাহু খোলা থাকবে,
∴ প্রয়োজনীয় বেড়ার দৈর্ঘ্য = x + x + 40
= 20 + 20 + 40
= 80
∴ বাগানটি ঘেরাও করতে মোট 80 feet বেড়া লাগবে।
From the question, you know that R = 6%, T = 4 years, S.I. = Tk.1600
If you apply the above values in the simple interest formula S.I. = PRT/100, you will get
1600 = (P x 4 x 6)/100
⇒ P = (1600 x 100)/6 × 4
⇒ P = 6333.33
Using the above value of P, you have to now calculate C.I. as shown below:
CI = [P(1 + R/100)n] – P
= [6333.33(1 + 6/100)4] - 6333.33
= [6333.33 (106/100)4] - 6333.33
= [6333.33× 53/50 × 53/50 × 53/50 × 53/50] - 6333.33
= 7995.68 - 6333.33
= Tk.1662.35
We know,
n(B U E U F) = n(B) + n(E) + n(F) - n(B ∩ E) - n(B ∩ F) - n(E ∩ F) + n(B ∩ E ∩ F)
Or, 24 = 6 + 12 + 15 + 1 - 2 - 2 - n(E ∩ F)
Or, n(E ∩ F) = 6
Question: The average of the first five multiples of 3 is:
Solution:
The first five multiples of 3 are:
3, 6, 9, 12, 15.
∴ Average = (3 + 6 + 9 + 12 + 15)/5
= 45/5
= 9
Question: For a rhombus with diagonals 8 m and 16 m, determine the diagonal of a square that covers the same area as the rhombus.
Solution:
area of rhombus = (1/2) × 8 × 16
= 64 m2
area of square = 64 m2
side of square = √64 m
= 8 m
∴ diagonal of the square = 8√2 m