উত্তর
ব্যাখ্যা
Solution:
Total outcome = 16 + 32 = 48
Favorable outcome = 16
P (getting a prize) =16/48
= 1/3
PrepBank · বিষয়ভিত্তিক প্রশ্ন
PrepBank · পাতা ৯১ / ১৬১ · ৯,০০১–৯,১০০ / ১৬,১২৪
Given, Captain's age 26 years and Wicketkeeper's age 29 years.
Let, Average age be X years
ATQ,
9(x - 1) + 26 + 29 = 11x
⇒ 9x - 9 + 26 + 29 = 11x
⇒ 2x = 46
⇒ x = 23
tan (x - 30°) = 1/√3 = tan 30°
Or, x - 30° = 30°
Or, x = 60°
∴ cos 60° = 1/2
Question: How many different registration numbers can be formed using two distinct letters followed by two distinct digits?
Solution:
Here,
Number of ways to choose and arrange two distinct letters out of 26 alphabets = 26P2
= 26 × 25
= 650
Number of ways to choose and arrange two distinct digits out of 10 digits (0 - 9) = 10P2
= 10 × 9
= 90
Total number of registration numbers = 650 × 90
= 58,500
Question: A bag contains 6 red balls, 9 black balls, and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is neither red nor white?
Solution:
Total balls = 6 + 9 + 5 = 20
Favorable outcomes = balls that are neither red nor white, that is, black balls = 9
∴ P(black) = Favorable outcomes/Total outcomes
= 9/20
Therefore, the probability is 9/20.
First daughter, in 1 hour, did 1/3 part
Second daughter, in 1 hour, did 1/6 part
Mother, in 1 hour, did (1/3 + 1/6) part
= 3/6 part
= 1/2 part
so, mother did it in 2 days.
Question: If the average of 'p' numbers is 3q2 and the average of 'q' numbers is 3p2, what is the average of the combined (p + q) numbers?
Solution:
দেওয়া আছে,
'p' সংখ্যার গড় = 3q2
∴ p সংখ্যার সমষ্টি = p × 3q2
'q' সংখ্যার গড় = 3p2
∴ 'q' সংখ্যার সমষ্টি = q × 3p2
∴ মোট সমষ্টি = (p × 3q2) + (q × 3p2)
= 3pq2 + 3qp2
= 3pq(q + p)
∴ তাদের গড় = মোট সমষ্টি/(p + q)
= 3pq(p + q)/(p + q)
= 3pq
Let the required number of rounds be x
More radius, less rounds(Indirect proportion)
Hence we can write as
(radius) 14 : 20 :: x : 70
⇒ 14 × 70 = 20x
⇒ 14 × 7 = 2x
⇒ x = 7 × 7
= 49 days.
Question: A wheel of an engine of 450 cm in circumference makes 20 revolutions in 6 seconds. What is the speed of the wheel in km/h?
Solution:
Total distance = (450 × 20) cm
= 9000 cm
= 9000/100 m
= 90 m
We know,
Speed = (Total distance ÷ Time)
= (90 ÷ 6) m/sec
= 15 m/sec
= (15 × 18/5) km/h
= 54 km/h
Question: A truck can carry 24 motorcycles or 36 scooters at a time. If there are 10 motorcycles on the truck, how many scooters can be loaded onto it?
Solution:
Here,
24 motorcycles = 36 scooters
∴ 1 motorcycle = 36/24 scooters = 3/2 scooters
∴ 10 motorcycles = (36 × 10)/24 scooters = 15 scooters
∴ Maximum number of scooters that can still be loaded = 36 - 15 = 21 scooters
Question: In a 500-meter race, Q starts 50 meters ahead of P, yet P defeats Q by a margin of 25 meters. What distance did Q cover when P reached the finish line?
Solution:
Total distance Q needed to cover = 500 - 50 = 450 meters
Distance covered by P = 500 meters
But P defeats Q by 25 meters
∴ Distance covered by Q when P reaches the finish line = 450 - 25 = 425 meters
∴ Q covered a distance of 425 meters when P reached the finish line.
Question: A shop offers a ‘Buy 2, Get 1 Free’ deal. What is the equivalent percentage discount?
Solution:
Assume the price of each item = 1 Taka.
Then,
Cost of 2 items = 2 Taka
Since 1 item is received for free, total items = 2 + 1 = 3
After the discount,
Cost of 3 items = 2 Taka
∴ Cost per item = 2/3 Taka
Actual price per item = 1 Taka
∴ Discount per item = 1 − (2/3) = 1/3 Taka
So, discount on 1 Taka = 1/3 Taka
∴ Discount on 100 Taka = 100 × (1/3) = 33.33 Taka
Therefore, in a “Buy 2, Get 1 Free” offer, the equivalent discount is 33.33%.
অতিক্রান্ত সময়, t = (d1 + d2)/(v1 + v2)
= (105 + 90)/{(45 + 72) × 5}/18
= 195/(117 × 5)/18
= 195 × 18/(117 × 5)
= 6 seconds
3log2+2log3
=log23+log32
=log8+log9
=log(8×9)
=log72
2log6+log2
=log62+log2
=log36+log2
=log72
3log2+2log3 / 2log6+log2
=log72 / log72
= 1
Question: Find the value of n, if 9n - 1 = 243.
Solution:
9n - 1 = 243
⇒ (32)n - 1 = 35
⇒ 32(n - 1) = 35
⇒ 32n - 2 = 35
⇒ 2n - 2 = 5
⇒ 2n = 5 + 2
⇒ 2n = 7
⇒ n = 7/2
⇒ n = 3.5
Question: A jar contains only marbles of three colour: red, green and yellow. The red and green marbles are in the ratio of 2 : 5 and the yellow and red marbles are in ration of 5 : 6. Which of the following could be the total number of marbles?
Solution:
Red : Green = 2 : 5 = 6 : 15
Yellow : Red = 5 : 6 = 5 : 6
Red : Green : Yellow = 6 : 15 : 5
অনুপাতের রাশিগুলোর সমষ্টি = 6 + 15 + 5 = 26
মার্বেল সংখ্যা হবে 26 এর গুণিতক।
মার্বেল সংখ্যা হতে পারে 26, 52, 78, 104,..............
Question: What is the H.C.F. of the following fractions? 6/7, 12/21, 18/35.
Solution:
আমরা জানি,
ভগ্নাংশের গসাগু = (লবের গসাগু)/(হরের লসাগু)
এখানে লবসমূহ = 6, 12 এবং 18
6 = 2 × 3
12 = 22 × 3
18 = 2 × 32
∴ লবসমূহের গসাগু (H.C.F.) = 2 × 3 = 6
এখানে হরসমূহ = 7, 21 এবং 35
7 = 7 × 1
21 = 3 × 7
35 = 5 × 7
∴ হরসমূহের লসাগু (L.C.M.) = 3 × 5 × 7 = 105
ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 6/105
= 2/35
Question: If 2 jackets and 3 sweaters cost Tk. 4,400, and 3 jackets and 2 sweaters cost Tk. 4,700, find the cost of a single jacket.
Solution:
ধরি, একটি জ্যাকেটের মূল্য x টাকা এবং একটি সোয়েটারের মূল্য y টাকা।
প্রশ্নমতে,
2x + 3y = 4400 ............... (i)
3x + 2y = 4700 .............. (ii)
(ii) × 3 - (i) × 2 ⇒
(9x + 6y) - (4x + 6y) = 14100 - 8800
⇒ 9x - 4x = 5300
⇒ 5x = 5300
⇒ x = 5300/5
⇒ x = 1060
সুতরাং, একটি জ্যাকেটের মূল্য 1060 টাকা।
Question: A rectangle and a square have the same area. The square has a perimeter of 32 meters and the length of the rectangle is 4 meters. What is the width of the rectangle (in meters)?
Solution:
Given that,
Perimeter of square = 32 m
Area of rectangle = Area of square
And length of rectangle = 4 m
Now,
Perimeter of square,
4s = 32
⇒ s = 32/4 = 8
∴ s = 8 m
∴ Area of square = s2 = 82 = 64 m2
According to the Question,
Area of rectangle = Area of square
∴ Area of rectangle = 64 m2
∴ Area of rectangle = length × width
64 = 4 × w
⇒ w = 64/4
∴ w = 16 m
So the width of the rectangle = 16 meters
Question: A boat goes 10 km upstream in 50 minutes, and the speed of the stream is 3 kmph. Find the speed of the boat in still water (in km/h).
Solution:
Let the speed of the boat in still water = x km/h
Speed of the stream = 3 km/h (given)
Upstream speed = x - 3 km/h
Distance upstream = 10 km
Time upstream = 50 minutes = 50/60 hours = 5/6 hours
We know,
Speed = Distance/Time
Upstream speed = 10 /(5/6) = 10 × (6/5) = 12 km/h
So,
⇒ x - 3 = 12
⇒ x = 12 + 3
∴ x = 15 km/h
The speed of the boat in still water is 15 km/h.
Question: If a + 1/a = √3, then what is the value of a30 + a24 + a6 + 1?
Solution:
Given, a + 1/a = √3
Now,
a3 + 1/a3 = (a + 1/a)3 - 3 . a . (1/a)(a + 1/a)
⇒ a3 + 1/a3 = (√3)3 - 3(√3) [∵ a + 1/a = √3]
⇒ a3 + 1/a3 = 3(√3) - 3(√3)
⇒ a3 + 1/a3 = 0
⇒ a6 + 1 = 0 [Multiplying both sides by a3]
Then,
a30 + a24 + a6 + 1
= a24 (a6 + 1) + (a6 + 1)
= (a24 × 0) + 0
= 0
ধরি, y বেতন পায় 100 টাকা
x পায় (133.33% of 100) টাকা = 133.33 টাকা
এবং, z পায় (100 × 100 /75) = 400/3 = 133.33 টাকা
x : y : z = 133.33 : 100 : 133.33 = 400 : 300 : 400 = 4 : 3 : 4
∴ z পায় = 5500 এর 4/(4 + 3 + 4) টাকা = 2000 টাকা
Let the numbers be x and y.
Then, 2x + 3y = 39 .......(i) and
3x + 2y = 30 .........(ii)
On solving (i) and (ii), we get : x = 6 and y = 9.
larger number = 9
Question: A solid cylindrical block has a radius of 7 meters and a height of 10 meters. If the material of the cylinder costs Tk. 20 per cubic meter, find the total cost of the material required.
Solution:
Given,
Radius of the cylinder, r = 7 m
Height of the cylinder, h = 10 m
Cost per cubic meter = Tk. 20
The volume of the cylinder:
V = πr2h
= (22/7) × (7)2 × 10
= (22/7) × 49 × 10
= 22 × 7 × 10
= 1540 cubic metres
Total cost = Volume × Cost per cubic metre
= 1540 × 20
= Tk. 30,800
∴ The total cost of the material is Tk. 30,800.
Question: The ratio of milk and water in a solution is 7 : 4. After adding 8 liters of water, the ratio of milk and water becomes 3 : 2. Find the final amount of water in the solution.
Solution:
Let the initial amount of milk = 7x liters
Let the initial amount of water = 4x liters
According to the question,
7x/(4x + 8) = 3/2
⇒ 2 × 7x = 3 × (4x + 8)
⇒ 14x = 12x + 24
⇒ 14x - 12x = 24
⇒ 2x = 24
⇒ x = 12
∴ Final amount of water = 4x + 8
= 4 × 12 + 8
= 48 + 8 = 56 liters
Question: If logx 1/216 = - 3, then x = ?
Solution:
Given that,
logx 1/216 = - 3
⇒ x- 3 = 1/216
⇒ 1/x3 = 1/216
⇒ x3 = 216
⇒ x3 = 63
∴ x = 6
Question: The sum of seventh and eleventh term of an arithmetic progression is 18. What is the sum of the first seventeen terms of that progression?
Solution:
In an arithmetic progression.
Let first term = a
Common difference = d
We know,
a = a + (n - 1)d
∴ a7 = a + 6d and a11 = a + 10d
Given that,
Seventh term + eleventh term = 18
⇒ a7 + a11 = 18
⇒ a + 6d + a + 10d = 18
⇒ 2a + 16d = 18
⇒ 2(a + 8d) = 18
∴ a + 8d = 9 ........(1)
We need the sum of the first 17 terms.
S17 = (n/2) × [2a + (n - 1)d]
= (17/2) × [2a + 16d]
= 17/2 × 2(a + 8d)
= 17 × (a + 8d)
= 17 × 9
= 153
So the sum of the first seventeen terms is 153.
Milk : water = 17:3 = 17x : 3x
∴ 17x + 3x = 200
⇒ x =10litre
so, Milk = 170 litre and water = 30 litre in initial mixture.
Let 'y' litre of milk added in mixture
i.e. (170+y) : 30 = 7:1
⇒ (170 + y)/ 30= 7/1
∴ y = 210 - 170 = 40 litres
Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3
= (13 x 12 x 11)/(1 x 2 x 3)
= 66 x 13
= 858.
Let E be the event of taking 3 bottles of the same variety
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= (6 x 5 x 4 )/ (1 x 2 x 3) + 1 + (4 x 3 x 2) / (1 x 2 x 3)
= 20 + 1 + 4
= 25
The probability of taking 3 bottles of the same variety = n(E)/n(S)
= 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858
= 833/858.
Question: A man completes a journey in 8 hours. He travels the first half of the journey at the rate of 40 km/hr and the second half at the rate of 60 km/hr. Find the total journey in km.
Solution:
ধরা যাক, মোট যাত্রার দূরত্ব হলো D কিমি।
তাহলে, যাত্রার প্রথম অর্ধেকের দূরত্ব হবে D/2 কিমি
এবং দ্বিতীয় অর্ধেকের দূরত্বও হবে D/2 কিমি।
প্রথম অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (D/2)/40 ঘন্টা
= D/80 ঘন্টা
দ্বিতীয় অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (D/2)/60 ঘন্টা
= D/120 ঘন্টা
প্রশ্নমতে,
D/80 + D/120 = 8
⇒ (3D + 2D)/240 = 8
⇒ 5D/240 = 8
⇒ 5D = 8 × 240
⇒ 5D = 1920
⇒ D = 1920/5
⇒ D = 384 কিমি
∴ মোট যাত্রার দূরত্ব 384 কিলোমিটার।
Question: Sonar Bangla express, normally reaches its destination at 72 km/h in 6 hours. Find the speed at which it must travel to reduce the travel time by 2 hours?
Solution:
Distance to be covered = Speed × Time
= 72 × 6
= 432 km
Reduced time = 6 - 2 = 4 hours
∴ Required speed = 432/4
= 108 km/h
∴ The train must travel at 108 km/h to reduce the time by 2 hours.
Remaining books = 1 - 2/3 = 1/3
1/3 of the books = 41 books
So, 2/3 of the books = 41×(2/3) / (1/3) = 82
∴ Total amount received for the sold books = 82 × 3.20 = 262.4 TK
There are 12 people, so this is our n value.
So, 12C2= 66
Question: On simplification √{(0.65)2 - (0.16)2} reduces to-
Solution:
Given that,
√{(0.65)2 - (0.16)2}
Since, a2 - b2 = (a - b)( a + b)
= √{(0.65 + 0.16)(0.65 - 0.16)}
= √{(0.81)(0.49)}
= √{(0.9)(0.9)×(0.7)(0.7)}
= 0.9 × 0.7
= 0.63
Question: A batsman scored 45, 68, 52, 71, and 59 runs in five innings. How many runs must he score in the sixth innings to have an average of exactly 60 runs?
Solution:
ধরি, 6-তম ইনিংসে রান = x
প্রথম 5টি ইনিংসে মোট রান = (45 + 68 + 52 + 71 + 59) = 295
প্রশ্নমতে,
(295 + x)/6 = 60
⇒ 295 + x = 60 × 6
⇒ 295 + x = 360
⇒ x = 360 - 295
⇒ x = 65
∴ 6-তম ইনিংসে 65 রান করতে হবে।
Let,
The required number of days be x.
Less Man, More Days [Indirect proportion]
∴ 42 : 56 :: 24 : x
⇒ 42x = 56 × 24
⇒ x = (56 × 24)/42
⇒ x = 32.
The seller manages to make a 10% profit.
Value of this 10% profit = Tk. 77
% Profit = Profit/Cost price × 100
∴ 10 = 77/C.P. × 100
∴ CP = Tk. 770
And S.P. = 770+77 = Tk. 847
Price after Commission = Selling Price
= (100-23)% of L.P.
= 77% of L.P.
∴ 847 = 77/100 × L.P.
∴ L.P. = List Price = Tk. 1100