উত্তর
ব্যাখ্যা
Solution:
Let
a = 1
b = 3
ab = 1 × 3 = 3, which is odd.
a + 2b + 2 = 1 + 2 × 3 + 2 = 1 + 6 + 2 = 9, which is odd.
a + b + 1 = 1 + 3 + 1 = 5, which is odd.
2a + 4b = 2 × 1 + 4 × 3 = 2 + 12 = 14, which is even.
PrepBank · বিষয়ভিত্তিক প্রশ্ন
PrepBank · পাতা ১১৯ / ১৬১ · ১১,৮০১–১১,৯০০ / ১৬,১২৪
Question: If a + 1/a = √3, then what is the value of a36 + a30 + a6 + 2?
Solution:
Given, a + 1/a = √3
Now,
a3 + 1/a3 = (a + 1/a)3 - 3a × (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,
a36 + a30 + a6 + 2
= a36 (a6 + 1) + (a6 + 1) + 1
= (a36 × 0) + 0 + 1
= 0 + 1
= 1
Question: If tanθ = 1, then the the value of
Solution:
Given,
tanθ = 1
⇒ tanθ = tan45°
∴ θ = 45°
Now,
3√32 = 2x
⇒ 25/3 = 2x
⇒ x = 5/3
Question: A car travels at 50 km/h. If it had traveled 10 km/h faster, it would have reached its destination 30 minutes earlier. What is the distance of the journey?
Solution:
Let the distance be x km.
Time at 50 km/h = x/50 hours
Time at 60 km/h = x/60 hours
According to the question, the difference in time = 0.5 hours
⇒ x/50 − x/60 = 0.5
⇒ (6x − 5x) / 300 = 0.5
⇒ x / 300 = 0.5
⇒ x = 0.5 × 300
⇒ x = 150 km
∴ Distance of the journey = 150 km
Question: A square field is surrounded by a path of uniform width 5 meters. If the area of the path is 220 square meters, find the side length of the field.
Solution:
Let the side of the field = x meters.
Then, the side of the field including the path = x + (2 × 5)
= x + 10 meters.
Area of path = Area of field with path - Area of field
⇒ 220 = (x + 10)2 - x2
⇒ 220 = x2 + 20x + 100 - x2
⇒ 220 = 20x + 100
⇒ 20x = 220 - 100
⇒ 20x = 120
⇒ x = 120/20
⇒ x = 6 meters
∴ Therefore, the side length of the field is 6 meters.
Question: Karim completes one-fourth of a piece of work in 4 hours. Rahim works at three times Karim’s speed and finishes the remaining portion of the work. How many hours does Rahim take to complete it?
Solution:
Let, Whole work = 1
Karim completes one-fourth (1/4) of a piece of work in 4 hours
∴ Karim’s rate = (1/4)/4 = 1/16 per hour.
∴ Rahim works at three times Karim’s speed
∴ Rahim’s rate = 3 × 1/16 = 3/16 per hour.
∴ Remaining work = (1 - 1/4) = 3/4
∴ Rahim complets 3/4 part work in = (3/4) × (16/3)
= 4 hours
Question: At what angle the hands of a clock are inclined at 15 minutes past 5?
Solution:
Hours hand moves in 15 past.
5 from 12 p.m = (5 + 15/60) hours = 21/4 hours
Angle of hours hand = (360/12) × (21/4)
= 157.5°
Minutes hands makes angle of = (360/60) × 15
= 90°
Angle between hours and minutes hands = (157.5° - 90°)
= 67.5°
In every odd jump (1st, 3rd, 5th, 7th etc) it will reach either to B or D. So, it is not possible for the frog to reach point C in 7th Jump.
The logic behind asking this question is to check if the student gets the point that if the frog takes odd number of jumps he can never reach vertex C.
Since the question asks the number of ways the frog can get to point C with 7 jumps which is an odd number, the answer to this question then becomes 0 as the frog can never reach point C with 7 jumps.
Question: The value of tan60° is:
(Officer General 2022 অনুযায়ী)
Solution:
cos60° = 1/2
sin60° = √3/2
Hence, tan60°
= sin60°/cos60°
= (√3/2)/(1/2)
= √3
Question: A mixture contains two liquids 'A' and 'B' are in the ratio 4 : 1. If 10 litres of mixture is withdrawn and replaced with 10 litres of 'B', then the ratio becomes 2 : 3. What was the initial quantity of A?
Solution:
ধরি, মিশ্রণের প্রাথমিক পরিমাণ = 5x লিটার
A এর পরিমাণ = 4x লিটার
B এর পরিমাণ = x লিটার
∴ 10 লিটার মিশ্রণ তুলে নেওয়ার পর,
A এর পরিমাণ = 4x - (4/5) × 10 = 4x - 8 লিটার
B এর পরিমাণ = x - (1/5) × 10 = x - 2 লিটার
আবার,
B তে 10 লিটার যোগ করার পর,
B এর পরিমাণ = x - 2 + 10 = x + 8 লিটার
∴ প্রদত্ত অনুপাত,
⇒ (4x - 8)/(x + 8) = 2/3
⇒ 12x - 24 = 2x + 16
⇒ 10x = 16 + 24
⇒ x = 40/10
⇒ x = 4
∴ A এর পরিমাণ = 4 × 4 = 16 লিটার
Question: 40% of 650 + 80% 720 - 600 = ?
Solution:
40% of 650 + 80% 720 - 600
= (40/100) × 650 + (80/100) × 720 - 600
= (26000/100) + (57600/100) - 600
= 260 + 576 - 600
= 236
Question: If WORD = DROW, then PALE = ?
Solution:
এখানে,
WORD = DROW এর W আর D এবং O আর R নিজেদের মধ্যে স্থান পরিবর্তন করেছে।
একইভাবে,
PALE শব্দের P আর E এবং A আর L নিজেদের মধ্যে স্থান পরিবর্তন করবে।
∴ PALE = ELAP
Question: What values of x satisfy the inequality 2 - 3x > 1?
Solution:
দেওয়া আছে,
2 - 3x > 1
⇒ 2 - 3x - 2 > 1 - 2 ; [অসমতার উভয় পক্ষ থেকে ২ বিয়োগ করে পাই]
⇒ - 3x > - 1
⇒ - 3x/- 3 < - 1/- 3 ; [একটি ঋণাত্মক সংখ্যা দ্বারা অসমতাকে ভাগ করলে, তখন অসমতার চিহ্নটি উল্টে যায়।]
∴ x < 1/3
Let the numbers be 2x, 3x, 5x and 7x respectively.
Then, their L.C.M = (2 × 3 × 5 × 7)x = 210x
[∵ 2, 3, 5, 7 are prime numbers ]
So, 20x = 630
or x = 3
∵ The numbers are 6, 9, 15 and 21.
Required difference = 21 - 6 = 15.
Answer : 15
Question:
Solution:
Question: A can do a work in 15 days, and B in 25 days. They work together for 5 days. How much of the work is left?
Solution:
A, 15 দিনে করতে পারে কাজটির 1 অংশ
A, 1 দিনে করতে পারে কাজটির 1/15 অংশ
B, 25 দিনে করতে পারে কাজটির 1 অংশ
B, 1 দিনে করতে পারে কাজটির 1/25 অংশ
A ও B 1 দিনে করতে পারে কাজটির = (1/15) + (1/25) অংশ
= (5 + 3)/75 অংশ
= 8/75 অংশ
A ও B 5 দিনে করতে পারে কাজটির (5 × 8)/75 অংশ
= 8/15 অংশ
∴ কাজ বাকি থাকে = 1 - 8/15 অংশ
= (15 - 8)/15 অংশ
= 7/15 অংশ
Question: : On a 12% discount sale, an article costs Tk. 704. What was the original price of the article?
Solution:
Let the original price be Tk. x.
According to the question,
x - 12% of x = 704
⇒ x - (12x/100) = 704
⇒ (100x - 12x)/100 = 704
⇒ 88x = 70400
∴ x = 800
So, the original price of the article was Tk. 800.
Question: A number reduced by 25% becomes 225. What percent should it be increased so that it becomes 390?
Solution:
Let
the number be x
Now
x - 25% of x = 225
⇒ x - 25x/100 = 225
⇒ (100x - 25x)/100 = 225
⇒ 75x/100 = 225
⇒ 75x = (225 × 100)
⇒ x = (225 × 100)/75
∴ x = 300
∴ Required increase
= (390 - 300)
= 90
∴ Increase percentage :
= (90/300) × 100%
= 30%
Question: The average of A and B is 45 and the sum of B & C is 78. What is the value of A - C?
Solution:
Given that,
Average of A and B = 45
Sum of B and C = 78
Now,
Average of A and B = 40, so-
⇒ (A + B)/2 = 45
∴ A + B = 90 ........(1)
And, B + C = 78 .........(2)
Subtract (2) from (1) than we get,
⇒ A + B - (B + C) = 90 - 78
⇒ A + B - B - C = 12
⇒ A - C = 12
So the value of A - C is 12.
Question: Find the remainder when 711 + 7111 + 71111 is divided by 8.
Solution:
When, 7 is divided by 8 then remainder = 7
When, 72 is divided by 8 then remainder = 1
When, 73 is divided by 8 then remainder = 7
When, 74 is divided by 8 then remainder = 1
Odd exponents give remainder = 7
Even exponents give remainder = 1
In 1st term, exponent is 11, which is odd so remainder = 7
In 2nd term, exponent is 111, which is also odd so remainder = 7
In 3rd term, exponent is 1111, which is odd so remainder = 7
So, (711 + 7111 + 71111) mod 8 = (7 + 7 + 7) mod 8 = 21 mod 8 = 5
Let, base = x
Then, height = 2x
Area = base × height
= x × 2x
= 2x2
Area is given as 72 cm2
2x2 = 72 cm2
⇒ x2 = 36 cm2
⇒ x = 6 cm
Hence, the length of the base is 6 cm.
Since, AB = BC
∠A = ∠C
let ∠A = x
so, ∠C is also equal to x
Now, ∠A + ∠B + ∠C = 180°
Or, 2x + 90 = 180
Or, x = (180-90)/2
Or, x = 45°
Let's father's age x and son's age is y
ATQ, x + y = 42 ....(i)
x - y = 22 .... (ii)
(i) - (ii) = 2y = 20
∴ y = 10
∴ son's age is 10 years
Question: Today is Friday. After 58 days, it will be:
Solution:
আমরা জানি যে সপ্তাহের প্রতিটি দিন 7 দিন পর পুনরাবৃত্তি হয়।
58 ÷ 7 = 8 (ভাগশেষ 2)
অর্থাৎ, (7 × 8) = 56 দিন পর আবার শুক্রবার হবে।
∴ 58 দিন পর হবে (শুক্রবার + 2 দিন) = রবিবার।
Question: ∠X and ∠Y are complementary to each other. If ∠X = 25° + 2y and ∠Y = 3y, find the value of ∠Y.
Solution:
Here,
∠X = 25° + 2y and ∠Y = 3y
For complementary angles,
∠X + ∠Y = 90°
⇒ (25° + 2y) + 3y = 90°
⇒ 25° + 5y = 90°
⇒ 5y = 65°
∴ y = 13°
So, ∠Y = 3 × 13° = 39°
Question: If θ is a positive angle and 9sin2θ - 9 = 0, then the value of tan(θ - 30°) is equal to?
Solution:
Given,
9sin2θ - 9 = 0
⇒ 9sin2θ = 9
⇒ sin2θ = 1
⇒ sinθ = 1
⇒ sinθ = sin90°
∴ θ = 90°
Now,
tan(θ - 30°) = tan(90° - 30°)
= tan60°
= √3
এখানে,
দূরত্বকে একটি ধারা মনে করে পাই,
ধারাঃ 100m 108m 114m 118m
পার্থক্যঃ 8m 6m 4m
সুতরাং সঠিক উত্তর 118m
Question: A train passes a stationary pole in 8 seconds. The train also passes a 200 m long bridge in 28 seconds. What is the length of the train?
Solution:
Given that,
Time to pass a pole = 8 s
Time to pass a 200 m bridge = 28 s
Let the length of the train = L meters
When passing a pole, the train covers distance = L in 8 s
And when passing a bridge, distance = (L + 200) in 28 s
Now,
From pole,
Speed = Distance/Time = L/8 m/s
And,
From bridge,
Speed = Distance/Time = (L + 200)/28 m/s
ATQ,
L/8 = (L + 200)/28
⇒ 28L = 8L + 1600
⇒ 28L - 8L = 1600
⇒ 20L = 1600
⇒ L = 1600/20
∴ L = 80 m
So the length of the train is 80 meters.
C.P. = (100/122.5)x 392
= (1000/1225)x 392
= 320
Profit = (392 - 320) = 72.
Given,
a/b + b/a = 2
⇒ (a2 + b2)/ab = 2
⇒ (a2 + b2) = 2ab
⇒ a2 + b2 - 2ab = 0
⇒ (a - b)2 = 0
⇒ a - b = 0.
Question: Eight years ago, Karim was five times as old as his son. Now, Karim is 28 years older than his son. What is the present age of the son?
Solution:
ধরি,
বর্তমানে Karim-এর বয়স = x বছর এবং তার ছেলের বয়স = y বছর
৮ বছর আগে, Karim তার ছেলের বয়সের পাঁচ গুণ ছিল। অর্থাৎ
⇒ x - 8 = 5(y - 8).......... (১)
আবার,
বর্তমানে Karim তার ছেলের চেয়ে ২৮ বছর বড়।
x = y + 28
x এর মান (১) এ বসিয়ে পাই,
⇒ (y + 28) - 8 = 5(y - 8)
⇒ y + 20 = 5y - 40
⇒ 20 + 40 = 5y - y
⇒ 60 = 4y
∴ y = 15
∴ ছেলের বর্তমান বয়স = 15 বছর।
Question: In a class, the number of girls is 5/6 of the number of boys. The total students of the class is 121. If 5 more girls join the class, what will be the new ratio of girls to boys?
Solution:
Let the number of boys be x
Then, number of girls = 5/6 of x = 5x/6
ATC,
x + (5x/6) = 121
⇒ (6x + 5x)/6 = 121
⇒ 6x + 5x = 6 × 121
⇒ 11x = 6 × 121
⇒ x = (6 × 121)/11
∴ x = 66
So, number of boys = x = 66
number of girls = 5/6 of x = 5/6 of 66 = 55
Now, 5 more girls join,
∴ New number of girls = 55 + 5 = 60
∴ ratio (girls : boys) = 60/66
= 10/11
= 10 : 11
Question: A 180 meter long train running at the speed of 72 km/h crosses another train running in the opposite direction at the speed of 108 km/h in 10 seconds. What is the length of the other train?
Solution:
Relative speed = (72 + 108) km/h
= 180 × (5/18) m/sec
= 50 m/sec
Let,
Length of the other train = x metres
Then,
(x + 180)/10 = 50
⇒ x + 180 = 500
⇒ x = 500 - 180
∴ x = 320
∴ The length of the other train is 320 meters.