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PrepBank · পাতা ৯৯ / ১৬১ · ৯,৮০১–৯,৯০০ / ১৬,১২৪

৯,৮০১.

If log₃{log₂(x² - 4x - 37)} = 1, where ‘x’ is a natural number, find the value of x.

ক
10
খ
4
গ
7
ঘ
6
ঙ
9

ব্যাখ্যা

Question: If log₃{log₂(x² - 4x - 37)} = 1, where ‘x’ is a natural number, find the value of x.

Solution:
We have log₃{log₂ (x² - 4x - 37)} = 1
⇒ log₂(x² - 4x - 37) = 3
⇒ x² - 4x - 37 = 8
⇒ x² - 4x - 45 = 0
⇒ (x - 9) (x + 5) = 0
⇒ x = 9, - 5
Since x is a natural number, so x = 9.

৯,৮০২.

A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the problem is solved?

ক
32/441
খ
25/147
গ
122/147
ঘ
1/2
ঙ
None of these

ব্যাখ্যা

Question: A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the problem is solved?

Solution:
P এর সমাধান করার সম্ভাবনা = 2/7
∴ P এর সমাধান না করার সম্ভাবনা = 1 - 2/7 = 5/7

Q এর সমাধান করার সম্ভাবনা = 4/7
∴ Q এর সমাধান না করার সম্ভাবনা = 1 - 4/7 = 3/7

R এর সমাধান করার সম্ভাবনা = 4/9
∴ R এর সমাধান না করার সম্ভাবনা = 1 - 4/9 = 5/9

∴ তিনজন এর সমাধান না করার সম্ভাবনা = (5/7) × (3/7) × (5/9)
= 75/441
= 25/147

∴ তিনজনের সমাধান করার সম্ভাবনা = 1 - 25/147
= 122/147

৯,৮০৩.

A complete 1/8 of a work in one day. B with double efficiency can do the full work in -

ক
ক) 6 days
খ
খ) 7 days
গ
গ) 4 days
ঘ
ঘ) 2 days

ব্যাখ্যা

Question: A complete 1/8 of a work in one day. B with double efficiency can do the full work in -

Solution:
যেহেতু B এর দক্ষতা A এর দ্বিগুণ তাই B একদিনে A এর চেয়ে দ্বিগুণ কাজ করবে।
∴ B একদিনে করবে = ২ × ১/৮ = ১/৪ অংশ

সম্পূর্ণ কাজ করবে = ৪ দিনে

৯,৮০৪.

Which of the following is a leap year?

ক
1900
খ
2000
গ
2022
ঘ
2010

ব্যাখ্যা

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।

৯,৮০৫.

Find the average of the first 24 consecutive natural numbers.

ক
11.5
খ
14.5
গ
12.5
ঘ
13.5

ব্যাখ্যা

Question: Find the average of the first 24 consecutive natural numbers.

Solution:
The average of the first n consecutive natural numbers is
=(n + 1)/2

So, average = (24 + 1)/2 [Here, n=24]
= 25/2
= 12.5

৯,৮০৬.

A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?

ক
ক) 6 years
খ
খ) 10 years
গ
গ) 16 years
ঘ
ঘ) 19 years

ব্যাখ্যা

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
∴ (2x + 2) + 2x + x = 27
⇒ 5x = 25
⇒ x = 5.
Hence,
B's age = 2x = 10 years.

৯,৮০৭.

The compound interest on 300 tk. at 7% per annum is 21 taka. The period is:

ক
ক) 1 year
খ
খ) 2 year
গ
গ) 3 year
ঘ
ঘ) 4 year

ব্যাখ্যা

Question: The compound interest on 300 tk. at 7% per annum is 21 taka. The period is:

Solution:
চক্রবৃদ্ধি সুদাসল = P (1 + r)ⁿ
= 300 + 21
= 321 tk.

321 = 300 × (1 + 7/100)ⁿ
⇒ (1 + 7/100)ⁿ = 321/300
⇒ (107/100)ⁿ = 107/100
⇒ (107/100)ⁿ = (107/100)¹
∴ n = 1 year

৯,৮০৮.

A dress on sale in a shop is marked at Tk. Z. During the discount sale its price is reduced by 15%. Staffs are allowed a further 10% reduction on the discounted price. If a staff member buys the dress what will she have to pay in terms of Z?

ক
0.75Z
খ
0.76Z
গ
0.765Z
ঘ
None

ব্যাখ্যা

Question: A dress on sale in a shop is marked at Tk. Z. During the discount sale its price is reduced by 15%. Staffs are allowed a further 10% reduction on the discounted price. If a staff member buys the dress what will she have to pay in terms of Z?

Solution:
A dress on sale in a shop is marked at Tk. Z
After 15% discount the price is = Z - 15% of Z = Z - 0.15Z = 0.85Z

For staff member, further 10% reduction the final price will be = 0.85Z - 10% of 0.85Z = 0.85Z - 0.085Z = 0.765Z

৯,৮০৯.

The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.

ক
35
খ
42
গ
20
ঘ
18

ব্যাখ্যা

Question: The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.

Solution:
Let the three numbers be x, y and z.
Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131
Therefore,
x² + y² + z² = 138
xy + yz + zx = 131

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
(x + y + z )² = 138 + 2(131)
(x + y + z )² = 400
Hence, (x + y + z) = 20

৯,৮১০.

If the ratio of present age of Ashik and Vishal is 9 : 4 and their sum of present age is 52 years, find the present age of Vishal.

ক
16
খ
17
গ
18
ঘ
14

ব্যাখ্যা

Question: If the ratio of present age of Ashik and Vishal is 9 : 4 and their sum of present age is 52 years, find the present age of Vishal.

Solution:
Let, age of Ashik be 9x and age of Vishal be 4x
Then, sum of both ages = 9x + 4x = 13x
⇒ 13x = 52
⇒ x = 4 years

∴ Age of Vishal = 4x = 4 × 4 = 16 years

৯,৮১১.

A and B invest in a business in the ratio 6 : 4. If 10% of the total profit goes to charity and A's share is Tk. 486, what is the total profit?

ক
ক) Tk. 800
খ
খ) Tk. 700
গ
গ) Tk. 900
ঘ
ঘ) Tk. 950

ব্যাখ্যা

Question: A and B invest in a business in the ratio 6 : 4. If 10% of the total profit goes to charity and A's share is Tk. 486, what is the total profit?

Solution:
Let,
The total profit be Tk. 100.
After paying 10% to charity,
∴ A's share = 90 × (6/10) = Tk. 54

If A's share is Tk. 54 then total profit = Tk. 100.
If A's share is Tk. 1 then total profit = Tk. 100/54
∴ If A's share is Tk. 486 then total profit = Tk. (100 × 486)/54
= Tk. 900

৯,৮১২.

If (√11 - 2)/(√11 + 2) = a√11 + b, then the value of a is-

ক
3/7
খ
4/7
গ
7/4
ঘ
- (4/7)

ব্যাখ্যা

Question: If (√11 - 2)/(√11 + 2) = a√11 + b, then the value of a is-

Solution:
L.H.S = (√11 - 2)/(√11 + 2)
= {(√11 - 2)/(√11 + 2)} × (√11 - 2)/(√11 - 2)
= (√11 - 2)²/{(√11)²- 2²}
= (11 + 4 - 2 × 2 × √11)/(11 - 4)
= (15 - 4√11)/7
= (15/7) - (4/7) × √11
= - (4/7) × √11 + (15/7)
= a√11 + b (R.H.S)
(Compare the coefficients of √11 and constant term)
a = - (4/7)
b = (15/7)

∴ the value of a = - (4/7)

৯,৮১৩.

If x ≥ 7 and y ≤ 4 which of the following must be true?

ক
x + y ≥ 3
খ
x - y ≥ 3
গ
x + y ≤ 3
ঘ
x - y ≤ 3

ব্যাখ্যা

Question: If x ≥ 7 and y ≤ 4 which of the following must be true?

Solution:
Given that,
x ≥ 7
and y ≤ 4
⇒ - y ≥ - 4

Now,
x - y ≥ 7 - 4
∴ x - y ≥ 3

৯,৮১৪.

A train moving at speed of 90 km/hr crosses a pole in 7 seconds. Find the length of the train.

ক
150 m
খ
165 m
গ
175 m
ঘ
170 m

ব্যাখ্যা

Question: A train moving at speed of 90 km/hr crosses a pole in 7 seconds. Find the length of the train.

Solution:
Length of the train is equal to the distance covered by train to cross the pole.
So, we will find the distance travelled by the train in 7 seconds by applying the following formula:
Distance= Speed × Time

Speed is given in Km/hr so we will convert it into m/s as answers are given in meters.
Speed = 90 × (5/18) = 25 m/s

Time = 7 seconds
Distance = 25 × 7 = 175 meters

৯,৮১৫.

(1/2)(logx + logy) will equal to log{(x + y)/2} if -

ক
y = 0
খ
x = y
গ
x = y/2
ঘ
x = √y

ব্যাখ্যা

Question: (1/2)(logx + logy) will equal to log{(x + y)/2} if -

Solution:
(1/2)(logx + logy) = log{(x + y)/2}
⇒ (1/2)log(xy) = log{(x + y)/2}
⇒ log(xy)^1/2 = log{(x + y)/2}
⇒ (xy)^1/2 = (x + y)/2
⇒ xy = {(x + y)/2}²
⇒ 4xy = x² + y² + 2xy
⇒ x² + y² - 2xy = 0
⇒ (x - y)² = 0
⇒ x - y = 0
∴ x = y

৯,৮১৬.

A grocer has a sale of Tk. 6435, Tk. 6927, Tk. 6855, Tk. 7230 and Tk. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Tk. 6500?

ক
Tk. 4991
খ
Tk. 5991
গ
Tk. 6001
ঘ
Tk. 6991

ব্যাখ্যা

Question: A grocer has a sale of Tk. 6435, Tk. 6927, Tk. 6855, Tk. 7230 and Tk. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Tk. 6500?

Solution:

Total sale for 5 months = Tk. (6435 + 6927 + 6855 + 7230 + 6562) = Tk. 34009.

∴ Required sale = Tk. [ (6500 × 6) - 34009 ]

= Tk. (39000 - 34009)

= Tk. 4991

৯,৮১৭.

a is greater than b by 2 and b is greater than c by 10.
If (a + b + c) = 130, then (b + c) - a = ?

ক
42
খ
34
গ
38
ঘ
44

ব্যাখ্যা

b = c + 10
a = b + 2 = c + 10 + 2 = c + 12
a + b + c = 130
c + 12 + c + 10 + c = 130
3c + 22 = 130
3c = 130 - 22
3c = 108
c = 36
Now, (b + c) - a = c + 10 + c - c - 12 = c - 2 = 36 - 2 = 34

৯,৮১৮.

From a point within an equilateral triangle, perpendiculars drawn to the three sides are 6 cm, 7 cm and 8 cm respectively. The length of the side of the triangle is

ক
ক) 7√3 cm
খ
খ) 14√3 cm
গ
গ) 9√3 cm
ঘ
ঘ) 12√3 cm

ব্যাখ্যা

Question: From a point within an equilateral triangle, perpendiculars drawn to the three sides are 6 cm, 7 cm and 8 cm respectively. The length of the side of the triangle is

Solution:
Let each side of the triangle be a cm.

Area of ΔABC = (√3/4) a²
Area of ΔAOB = (1/2) × (a × 6)
Area of ΔBOC = (1/2) × (a × 7)
Area of ΔAOC = (1/2) × (a × 8)

Then, ΔAOB + ΔBOC + ΔAOC = ΔABC
(1/2) × (a × 6) + (1/2) × (a × 7) + (1/2) × (a × 8) = (√3/4) a²
⇒ a/2 (6 + 7 + 8) = (√3/4) a²
⇒ 21a/2 = (√3/4) a²
⇒ (√3/4) a² = 21a/2
⇒ a = (21/2) × (4/√3)
∴ a = 14√3 cm

∴ Each side of the triangle is 14√3 cm.

৯,৮১৯.

In the first 15 overs of a cricket game, the run rate was only 3.4. What should be the rate in the remaining 35 overs to reach the target of 256 runs?

ক
5.47
খ
5.86
গ
6.14
ঘ
4.98

ব্যাখ্যা

Question: In the first 15 overs of a cricket game, the run rate was only 3.4. What should be the rate in the remaining 35 overs to reach the target of 256 runs?

Solution:
First 15 overs total run was = (3.4 × 15) = 51

Required run rate = (256 - 51)/35
= 205/35
= 5.86

৯,৮২০.

A boat running upstream takes 12 hours to cover a certain distance, while it takes 8 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

ক
4 : 1
খ
5 : 1
গ
3 : 5
ঘ
5 : 2

ব্যাখ্যা

Question: A boat running upstream takes 12 hours to cover a certain distance, while it takes 8 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

Solution:
ধরি,
স্থির পানিতে নৌকার গতি = b কিমি/ঘণ্টা
স্রোতের গতি = c কিমি/ঘণ্টা
দূরত্ব = d কিমি

স্রোতের প্রতিকূলে নৌকার গতিবেগ = (b - c) কিমি/ঘণ্টা
∴ d/(b - c) = 12
⇒ d = 12(b - c)

স্রোতের অনুকূলে নৌকার গতিবেগ = (b + c) কিমি/ঘণ্টা
∴ d/(b + c) = 8
⇒ d = 8(b + c)

এখন, উভয় ক্ষেত্রে দূরত্ব সমান হওয়ায়,
12(b - c) = 8(b + c)
⇒ 3(b - c) = 2(b + c)
⇒ 3b - 3c = 2b + 2c
⇒ b = 5c
⇒ b/c = 5/1

∴ b : c = 5 : 1

৯,৮২১.

A train of 180 m long is moving at 72 km/h. The time taken by the train to cross a tunnel of 420 m long is-

ক
25 seconds
খ
30 seconds
গ
35 seconds
ঘ
40 seconds

ব্যাখ্যা

Question: A train of 180 m long is moving at 72 km/h. The time taken by the train to cross a tunnel of 420 m long is-

Solution:
মোট অতিক্রান্ত দূরত্ব = ট্রেনের দৈর্ঘ্য + টানেলের দৈর্ঘ্য
= (180 + 420) মিটার
= 600 মিটার

ট্রেনের গতিবেগ = 72 কিমি/ঘন্টা
= (72 × 1000) মিটার/3600 সেকেন্ড
= 20 মিটার/সেকেন্ড

সময় = দূরত্ব ÷ গতিবেগ
= 600 মিটার ÷ 20 মিটার/সেকেন্ড
= 30 সেকেন্ড

সুতরাং, টানেলটি অতিক্রম করতে ট্রেনটির 30 সেকেন্ড সময় লাগবে।

৯,৮২২.

The profit earned by selling an article for Tk 90 is double the loss incurred when the same article is sold for Tk 45. At what price should the article be sold to make 25% profit?

ক
Tk. 75
খ
Tk. 80
গ
Tk. 60
ঘ
Tk. 85

ব্যাখ্যা

Question: The profit earned by selling an article for Tk 90 is double the loss incurred when the same article is sold for Tk 45. At what price should the article be sold to make 25% profit?

Solution:
Let,
The cost price be Tk. x.

ATQ,
90 - x = 2(x - 45)
⇒ 90 - x = 2x - 90
⇒ 2x + x = 90 + 90
⇒ 3x = 180
⇒ x = 180/3
∴ x = 60

∴ Cost price = Tk. 60,
Profit = 25%

∴ The selling price of the article = Tk. (60 × 125)/100
= Tk. 75

৯,৮২৩.

What is the maximum value of cosθ?

ক
1
খ
- 1
গ
0
ঘ
90

ব্যাখ্যা

Question: What is the maximum value of cosθ?

Solution:
cosθ এর সর্বনিম্ন মান - 1 এবং সর্বোচ্চ মান 1
sinθ এর সর্বনিম্ন মান - 1 এবং সর্বোচ্চ মান 1

৯,৮২৪.

The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the oldest child?

ক
12 years
খ
4 years
গ
16 years
ঘ
20 years
ঙ
None of the above

ব্যাখ্যা

Question: The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the oldest child?

Solution:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
or, 5x = 20
or, x = 4.
∴ Age of the oldest child = x + 12 = 4 + 12 = 16 years.

৯,৮২৫.

How many terms of the arithmetic should be progression 3, 7, 11, ... taken to make its sum equals to 820?

ক
10
খ
15
গ
20
ঘ
25

ব্যাখ্যা

Question: How many terms of the arithmetic should be progression 3, 7, 11, ... taken to make its sum equals to 820?

Solution:
এটি একটি সমান্তর ধারা,
যার ১ম পদ, a = 3
সাধারণ অন্তর, d = 4

আমরা জানি,
n- তম পদের সমষ্টি,
Sn = (n/2)[2a + (n - 1)d]

প্রশ্নমতে,
(n/2)[2a + (n - 1)d] = 820
⇒ (n/2)[6 + (n - 1)4] = 820
⇒ (n/2)[6 + 4n - 4] = 820
⇒ (n/2)[2(1 + 2n)] = 820
⇒ n(2n + 1) = 820
⇒ 2n² + n - 820 = 0
⇒ 2n² - 40n + 41n - 820 = 0
⇒ n(2n - 40) + 41(n - 40) = 0
⇒ (2n - 40)(n + 41) = 0
হয়,
⇒ 2n - 40 = 0
⇒ 2n = 40
n = 20

অথবা,
n + 41 = 0
n = - 41 ;[যা গ্রহণযোগ্য নয়]

সুতরাং, প্রদত্ত ধারাটিতে পদ আছে 20 টি।

৯,৮২৬.

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

ক
ক) 50%
খ
খ) 54%
গ
গ) 60%
ঘ
ঘ) 64%

ব্যাখ্যা

Let number is 100.
Actual Multiplication = (100 × 5)/3 = 500/3.

Student multiplied = (100 × 3)/5 = 300/5.
Error,
= (500/3) - (300/5)
= (2500 - 900)/15
= 1600/15

%Error = (1600 ×100 ×3)/(15×500)% = 64%.

৯,৮২৭.

If n(U) = 50, n(A) = 28, n(B) = 26 and n(A ∩ B) = 12 then n(A ∪ B)′ = ?

ক
8
খ
14
গ
26
ঘ
42

ব্যাখ্যা

Question: If n(U) = 50, n(A) = 28, n(B) = 26 and n(A ∩ B) = 12 then n(A ∪ B)′ = ?

Solution:
আমরা জানি,
n(A ∪ B)= n(A) + n(B) - (A ∩ B)
= 28 + 26 - 12
= 42

এখন,
n(A ∪ B)′= n(U) - n(A ∪ B)
= 50 - 42
= 8

সুতরাং, n(A ∪ B)′ = 8

৯,৮২৮.

What will come at the place of question mark?
9, 17, 33, 65, ?

ক
115
খ
119
গ
131
ঘ
129
ঙ
130

ব্যাখ্যা

Question: What will come at the place of question mark ?
9, 17, 33, 65, ?

Solution:
Here,
First term = 9
Second term = (9 × 2 -1) = 17
Third term = (17 × 2 -1) = 33
Fourth term = (33 × 2 -1) = 65
∴ Fifth term = (65 × 2 -1) = 129

৯,৮২৯.

If a/b = 5/4, then (4a + 3b)/(4a - 3b) =?

ক
4
খ
8
গ
2
ঘ
12

ব্যাখ্যা

Question: If a/b = 5/4, then (4a + 3b)/(4a - 3b) =?

Solution:

৯,৮৩০.

If 6 × P's capital = 8 × Q's capital = 10 × R's capital, find the ratio of their capitals?

ক
15 : 18 : 23
খ
19 : 27 : 32
গ
20 : 15 : 12
ঘ
12 : 15 : 25

ব্যাখ্যা

Question: If 6 × P's capital = 8 × Q's capital = 10 × R's capital, find the ratio of their capitals?

Solution:
Let,
6 × P's capital = 8 × Q's capital = 10 × R's capital = x
Now,
P's capital = x/6
Q's capital = x/8
R's capital = x/10

∴ Ratio of capitals of P, Q, R = x/6 : x/8 : x/10
= 20 : 15 : 12

৯,৮৩১.

A boat can travel from point A to point B and return back to point A in 9 hours. Speed of the boat in still water is 8 km/h and the speed of the stream is 4 km/h. Find the distance between A and B.

ক
29 km
খ
27 km
গ
25 km
ঘ
24 km
ঙ
20 km

ব্যাখ্যা

Speed of boat along the stream = 8 + 4 = 12 km/h
Speed of boat against stream = 8 - 4 = 4 km/h
ATQ,
x/12 + x/4 = 9
⇒ (x+3x)/12 = 9
⇒ 4x = 108
⇒ x = 27 km

৯,৮৩২.

Two equal glasses respectively contain one-fourth and two-fifth of milk. What is the ratio of milk and water if the rest of the glasses are filled with water and then mixed in a tumbler?

ক
3 : 7
খ
13 : 17
গ
13 : 27
ঘ
10 : 20

ব্যাখ্যা

Question: Two equal glasses respectively contain one-fourth and two-fifth of milk. What is the ratio of milk and water if the rest of the glasses are filled with water and then mixed in a tumbler?

Solution:
in the first glass,
milk = 1/4
water = 1 - (1/4)
= 3/4

in the second glass,
milk = 2/5
water = 1 - (2/5)
= 3/5

so,
milk : water
(1/4 + 2/5) : (3/4 + 3/5)
(5+8)/20 : (15+12)/20
13 : 27

৯,৮৩৩.

A certain distance is covered at a certain speed. If half the distance is covered in double the time, the ratio of the two speeds is:

ক
1 : 3
খ
4 : 3
গ
4 : 1
ঘ
2 : 3
ঙ
None of the above

ব্যাখ্যা

Question: A certain distance is covered at a certain speed. If half the distance is covered in double the time, the ratio of the two speeds is:

Solution:
Let x km be covered in y hours.

Then, speed = (x/y) km/hr

In the second case, x/2 km is covered in 2y hours.

∴ New speed = (x/2) × (1/2y ) km/hr = (x/4y) km/hr

∴ Ratio of speeds = (x/y) : (x/4y)
= 1 : (1/4)
= 4 : 1

৯,৮৩৪.

A train covers the distance x between two cities in y hours, arriving 2 hours late. What rate would permit the train to arrive on schedule?

ক
(x/y) - 2
খ
x/(y - 2)
গ
xy - 2
ঘ
x/(y + 2)
ঙ
None of these

ব্যাখ্যা

Since the train is late, the allotted time = Y - 2.
D = T × S
X = (Y - 2) × S
or, S = X/(Y - 2)

৯,৮৩৫.

What is the 3^rd term of the sequence: sin⁡(nπ/6)

ক
1/2
খ
√3/2
গ
1
ঘ
1/√2

ব্যাখ্যা

Question: What is the 3^rd term of the sequence: sin⁡(nπ/6)

Solution:
এখানে,
sin(nπ/6) এর তৃতীয় পদ = {sin(3 × π)/6}
= {sin(3 × 180°)/6}
= sin90°
= 1

৯,৮৩৬.

A sum of money at simple interest amounts to Tk. 815 in 3 years and to Tk. 854 in 4 years. The sum is-

ক
Tk. 628
খ
Tk. 608
গ
Tk. 789
ঘ
Tk. 698
ঙ
None of these

ব্যাখ্যা

Question: A sum of money at simple interest amounts to Tk. 815 in 3 years and to Tk. 854 in 4 years. The sum is-

Solution:
Simple interest for 1 year = Tk. (854 - 815)
= Tk. 39

∴ Simple interest for 3 years = Tk.(39 × 3)
= Tk. 117

∴ Sum = (815 - 117)
= Tk. 698

৯,৮৩৭.

If the volume of a cube is 512 cm³, then the surface area of the cube will be-

ক
ক) 348 cm²
খ
খ) 438 cm²
গ
গ) 834 cm²
ঘ
ঘ) 384 cm²

ব্যাখ্যা

Volume of a cube with side a is a³
Given that
Volume of a cube is 512 cm³
a³=512
a³ = 8³
a = 8
Surface area of cube is 6a²= 6 × 8²= 384 cm²

৯,৮৩৮.

On a certain sum of money, the simple interest for 2 years is Tk. 450 at the rate of 4% per annum. If it was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?

ক
9 Tk
খ
11 Tk
গ
8 Tk
ঘ
7 Tk

ব্যাখ্যা

Question: On a certain sum of money, the simple interest for 2 years is Tk. 450 at the rate of 4% per annum. If it was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?

Solution:
Simple Interest (I) = (P × r × n)/100
⇒ 450 = (P × 2 × 4)/100
∴ P = 5625

∴ Compound interest = P(1 + r)ⁿ - p
= 5625(1 + 0.04)² - 5625
= 6084 - 5625
= 459

∴ Difference = 459 - 450 = 9 Tk

৯,৮৩৯.

A train runs at the speed of 72 kmph and crosses a 250 meter long platform in 26 seconds. What is the length of the train?

ক
ক) 270 meter
খ
খ) 210 meter
গ
গ) 340 meter
ঘ
ঘ) 130 meter

ব্যাখ্যা

Distance covered in 26 seconds.
= 26 × 72 × (5/18)
= 520 meter.
Length of the train = (520 - 250) meter
= 270 meter.

৯,৮৪০.

x + y = 3, xy = 2 হলে, x³ + y³ এর মান কত?

ক
9
খ
18
গ
19
ঘ
27

ব্যাখ্যা

প্রশ্ন: x + y = 3, xy = 2 হলে, x³ + y³ এর মান কত?

সমাধান:
দেওয়া আছে,
x + y = 3
এবং xy = 2

∴ x³ + y³ = (x + y)³ - 3 · x · y (x + y)
= 3³ - 3 × 2 × 3
= 27 - 18
= 9

৯,৮৪১.

The difference of two numbers is 12. If 1 is added to the greater number, it becomes twice the smaller number. Calculate the two numbers?

ক
ক) 35, 23
খ
খ) 20, 8
গ
গ) 30, 18
ঘ
ঘ) 25, 13

ব্যাখ্যা

Question: The difference of two numbers is 12. If 1 is added to the greater number, it becomes twice the smaller number. Calculate the two numbers?

Solution:
ধরি,
বড় সংখ্যাটি = x
ছোট সংখ্যাটি = y

১ম শর্তমতে,
x - y = 12 ............................. (1)

২য় শর্তমতে,
x + 1 = 2y ............................. (2)

(1) নং হতে (2) নং সমীকরণ বিয়োগ করে পাই,
x - y - x - 1 = 12 - 2y
বা, - y = - 13
∴ y = 13

(1) নং হতে পাই,
x - 13 = 12
বা, x = 12 + 13
∴ x = 25

∴ সংখ্যা দুটি 25 ও 13

৯,৮৪২.

5^{3q - 2} = 625, find the value of q.

ক
3
খ
0
গ
1
ঘ
2

ব্যাখ্যা

Question: 5^{3q - 2} = 625, find the value of q.

Solution:
5^{3q - 2} = 625
⇒ 5^{3q - 2} = 5⁴
⇒ 3^{q - 2} = 4
⇒ 3q = 6
∴ q = 2

৯,৮৪৩.

If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

ক
ক) 3
খ
খ) 4
গ
গ) 5
ঘ
ঘ) 6

ব্যাখ্যা

Given that, (10a + b) - (10b + a) = 27
⇒ 9a - 9b = 27
⇒ a - b = 3

৯,৮৪৪.

The distance from the point P to two vertices A and B of an equilateral triangle are ।PA। = 2 and ।PB। = 3. What is the greatest possible value of ।PC।?

ক
5
খ
4
গ
3
ঘ
2

ব্যাখ্যা

Greatest possible value
= s(s - a)(s - b)(s - c) ≥ 0 Where s = (a + b + c)/2 and a = 2, b = 3
= {(5 + c)/2}{5 + c)/2 - 2}{5 + c)/2 - 3}{5 + c)/2 - c} ≥ 0
= (25 - c²)(c² - 1)/16 ≥ 0
Therefore, (25 - c²) ≥ 0; possible values of c are ±1, ±2, ±3, ±4, ±5
Hence, the greatest possible value of ।PC। = 5
-----------------------------------------------------------------------

video link - https://www.youtube.com/watch?v=yy6JP_Co06o
----------------------------------------------------------------------
সংক্ষেপে,
।PA। = 2 এবং ।PB। = 3
।PA। ও ।PB। এর মানের যোগফল হচ্ছে ।PC। এর সবচেয়ে বড় মান।
।PC। এর সবচেয়ে বড় মান = 2 + 3 = 5

৯,৮৪৫.

A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both, he must invest the money in the ratio-

ক
ক) 3 : 4
খ
খ) 16 : 15
গ
গ) 3 : 5
ঘ
ঘ) 4 : 5

ব্যাখ্যা

Question: A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both, he must invest the money in the ratio-

Solution:
For an income of Tk. 1 in 9% stock at 96,
investment = Tk. (96/9) = Tk. 32/3

For an income of Tk. 1 in 12% stock at 120,
investment = Tk. 120/12 = Tk. 10

∴ Ratio of investments =32/3 : 10 = 32 : 30 = 16 : 15

৯,৮৪৬.

The difference between two positive numbers is 3 and the sum of their squares is 369. Then the sum of the numbers is

ক
ক) 27
খ
খ) 33
গ
গ) 20
ঘ
ঘ) 81

ব্যাখ্যা

Question: The difference between two positive numbers is 3 and the sum of their squares is 369. Then the sum of the numbers is-

Solution:
Let the numbers be x and (x + 3)
Then,
x² + (x + 3)²=369
⇒ x² + x²+ 9 + 6x=369
⇒ 2x²+ 6x - 360=0
⇒ x² + 3x - 180=0
⇒ (x + 15)(x - 12)=0
⇒ x = 12

So, the numbers are 12 and 15

∴ Required sum = (12 + 15) = 27

৯,৮৪৭.

ক
16
খ
18
গ
24
ঘ
36

ব্যাখ্যা

Question:

Solution:

৯,৮৪৮.

What is the total number of integers between 100 and 200 that are divisible by 3?

ক
31
খ
32
গ
35
ঘ
33
ঙ
30

ব্যাখ্যা

First, identify the number that is a multiple of 3 more than 100.
That type of number is 102.
So, 102//3 = 34.

Second, we have to identify the number that is multiple of 3 but nearest less than 200.
That type of number is 198.
Now, 198/3 = 66.

Answer is (66 - 34) + 1 = 33

৯,৮৪৯.

If (2x + y)/(x + 4y) = 3, then find the value of (x + y)/(x + 2y) = ?

ক
ক) 7/11
খ
খ) 10/9
গ
গ) 11/7
ঘ
ঘ) 9/10

ব্যাখ্যা

Question: If (2x + y)/(x + 4y) = 3, then find the value of (x + y)/(x + 2y) = ?

Solution:
(2x + y)/(x + 4y) = 3
2x + y = 3(x + 4y)
2x + y = 3x + 12y
3x - 2x = y - 12y
x = - 11y

(x + y)/(x + 2y) =(- 11y + y)/(- 11y + 2y)
- 10y/ - 9y
= 10/9

৯,৮৫০.

A tank is filled in 9 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

ক
60 hours
খ
20 hours
গ
63 hours
ঘ
none of the above

ব্যাখ্যা

Question: A tank is filled in 9 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

Solution:
Suppose,
Pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.

Now,
⇒ (1/x) + (2/x) + (4/x) = 1/9
⇒ 7/x = 1/9
∴ x = 63
∴ Pipe A alone takes 63 hours to fill the tank.

৯,৮৫১.

Rakib is walking at 5m/s. Tanvir was 200m behind Rakib and started walking at 7m/s towards Rakib. After how much time will Tanvir catch Rakib?

ক
100 seconds
খ
110 seconds
গ
90 seconds
ঘ
200 seconds

ব্যাখ্যা

Question: Rakib is walking at 5m/s. Tanvir was 200m behind Rakib and started walking at 7m/s towards Rakib. After how much time will Tanvir catch Rakib?

Solution:
let after t seconds Tanvir will catch Rakib.
in t seconds,
Rakib will cover = 5t m
Tanvir will cover = 7t m

∴ 7t = 5t + 200
2t = 200
t = 100 seconds

৯,৮৫২.

The present age of a mother is the thrice that of her daughter. After 13 years, the age of the mother will be twice that of her daughter. The present age of the daughter is-

ক
ক) 10 years
খ
খ) 11 years
গ
গ) 12 years
ঘ
ঘ) 13 years

ব্যাখ্যা

Question: The present age of a mother is the thrice that of her daughter. After 13 years, the age of the mother will be twice that of her daughter. The present age of the daughter is-

Solution:
Let
the daughter's present age be x years.
Then, mother present age = 3x years

According to the question
3x + 13 = 2 (x + 13)
⇒ 3x + 13 = 2x + 26
⇒ x = 13
Present age of daughter = 13 years

৯,৮৫৩.

A tank is 1/3 parts full with water. If 8 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?

ক
16
খ
24
গ
48
ঘ
None

ব্যাখ্যা

Question: A tank is 1/3 parts full with water. If 8 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?

Solution:
ধরি,
ট্যাংকের ধারণক্ষমতা = x লিটার

প্রশ্নমতে,
(x/3) + 8 = 5x/6
⇒ (5x/6) - (x/3) = 8
⇒ (5x - 2x)/6 = 8
⇒ 3x = 48
⇒ x = 48/3
⇒ x = 16

অর্থাৎ ট্যাংকের ধারণক্ষমতা = 16 লিটার

৯,৮৫৪.

5^{- 3} + 5^{- 3} + 5^{- 3} + 5 ^{- 3} + 5 ^{- 3} =

ক
5^{- 2}
খ
25^{- 3}
গ
25^{- 13}
ঘ
5^{- 15}

ব্যাখ্যা

Question: 5^{- 3} + 5^{- 3} + 5^{- 3} + 5 ^{- 3} + 5 ^{- 3} =

Solution:
5^{- 3} + 5^{- 3} + 5^{- 3} + 5 ^{- 3} + 5^{- 3}
= 5. 5 ^-3
= 5 ^{1 - 3}
= 5 ^-2

৯,৮৫৫.

A number is 2/5 times of another number. If the sum of two numbers is 98, what is the two numbers?

ক
ক) 60, 38
খ
খ) 70, 28
গ
গ) 80, 18
ঘ
ঘ) 68, 30

ব্যাখ্যা

Question: A number is 2/5 times of another number. If the sum of two numbers is 98, what is the two numbers?

Solution:
Let, a number is x
another number is 2x/5

ATQ,
x + 2x/5 = 98
⇒ (5x + 2x)/5 = 98
⇒ 7x/5 = 98
⇒ 7x = 98 × 5
⇒ 7x = 490
∴ x = 70

So, a number is 70
another number is 2x/5 = (2 × 70)/5 = 28

৯,৮৫৬.

If the side of a square is increased by 8 cm, its area is increased by 120 cm². Find the side of the square:

ক
ক) 3 cm
খ
খ) 2 cm
গ
গ) 3.5 cm
ঘ
ঘ) 5 cm

ব্যাখ্যা

Let
the original length of the side of the square be x cm
Area = x²
If the length increases by 8 cm, area increased by 120 cm²
(x + 8)² - x² = 120
x² + 16x + 64 - x² = 120
16x = 120 - 64
16x = 56
∴ x = 3.5 cm

৯,৮৫৭.

P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal amounts of both alloys are melted to form a third alloy R, then the ratio of gold and copper in R will be -

ক
19 : 5
খ
11 : 6
গ
14 : 13
ঘ
7 : 3
ঙ
7 : 5

ব্যাখ্যা

Question: P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal amounts of both alloys are melted to form a third alloy R, then the ratio of gold and copper in R will be -

Solution:
Alloy P (Gold : Copper) = 7:2
Gold fraction in P = 7/9
Copper fraction in P = 2/9

Alloy Q (Gold : Copper) = 7:11
Gold fraction in Q = 7/18
Copper fraction in Q = 11/18

Let, 1 kg of each alloy is mixed;
Gold in Alloy R = 7/9 + 7/18 = 21/18

Copper in Alloy R = 2/9 + 11/18 = 15/18

∴ The ratio of Gold:Copper in R = (21/18) : (15/18)
= 21 : 15
= 7 : 5

৯,৮৫৮.

A works twice as fast as B. If B can complete a work in 12 days immediately, the number of days in which A and B can together finish the work-

ক
3 days
খ
4 days
গ
5 days
ঘ
6 days

ব্যাখ্যা

Question: A works twice as fast as B. If B can complete a work in 12 days immediately, the number of days in which A and B can together finish the work-

Solution:
B can complete a work in 12 days
So, A can complete the work in 6 days

(A + B)'s 1 days work = (1/6 + 1/12) = 1/4 part
So, (A + B) can complete the work in 4 days

৯,৮৫৯.

Express the following inequality using absolute value notation:
- 18 < x < - 6

ক
|x - 12| < 6
খ
|x + 6| < 12
গ
|x + 12| < 6
ঘ
|x - 6| < 12

ব্যাখ্যা

Question: Express the following inequality using absolute value notation:
- 18 < x < - 6

Solution:
Given: - 18 < x < - 6
The midpoint (average) of - 18 and - 6 is,
Midpoint = {- 18 + (- 6)}/2
= - 24/2
= - 12

Now add 12 to all parts of the inequality to center it at zero.
- 18 + 12 < x + 12 < - 6 + 12
⇒ - 6 < x + 12 < 6
This is equivalent to |x + 12| < 6

৯,৮৬০.

Which of the following is a complete and accurate listing of all crops that will grow alone in field 2 if the only pesticide or fertilizer used is Y?

ক
ক) F
খ
খ) F and H
গ
গ) G and H
ঘ
ঘ) G, H, and J

ব্যাখ্যা

Question: Which of the following is a complete and accurate listing of all crops that will grow alone in field 2 if the only pesticide or fertilizer used is Y?

Solution:
G, H, I, J ফিল্ড ২ তে উৎপাদন করা যেতে পারে।
J, H একই ফিল্ডে উৎপাদন করা যায় না।
সার Z ছাড়া I উৎপাদন করা যায় না।

অতএব, সার Y ব্যবহার করলে, GJ বা GH ফিল্ড ২ এ উৎপাদন করা যাবে।

৯,৮৬১.

A square and an equilateral triangle have equal perimeter. if the diagonal of the square is 12√2 cm then area of triangle is-

ক
64√5
খ
64√3
গ
62√3
ঘ
60√3
ঙ
65√5

ব্যাখ্যা

Let the side of the square be a cm.
Then, its diagonal = √2 a cm.
Now, √2 a = 12√2 ⇒ a = 12 cm.
Perimeter of square = 4a = 48 cm.
Perimeter of equilateral triangle = 48 cm.
Each side of the triangle = 16 cm.
Area of the triangle = ((√3/4)×16×16) cm² = 64√3 cm²

৯,৮৬২.

A pump removes water at a rate of 6 gallons per minute. How many hours will it take to remove 1800 gallons?

ক
3 hours
খ
4 hours
গ
5 hours
ঘ
6 hours

ব্যাখ্যা

Question: A pump removes water at a rate of 6 gallons per minute. How many hours will it take to remove 1800 gallons?

Solution:
Required time = 1800/6 mins
= 300/60 hours
= 5 hours

৯,৮৬৩.

In a race of 1000m, A can beat B by 100m. In a 400m, B beats C by 40m. In a race of 500m. A will beat C by-

ক
75m
খ
85m
গ
90m
ঘ
95m

ব্যাখ্যা

Question: In a race of 1000m, A can beat B by 100m. In a 400m, B beats C by 40m. In a race of 500m. A will beat C by-

Solution:
While A covers 1000 B covers 900
while A covers 500 B covers 450m

While B covers 400, C covers 360m
While B covers 450, C covers (360 × 450)/400 = 405m

∴ in a 500m race A will beat C by = (500 - 405) = 95m

৯,৮৬৪.

If the simple interest on Tk. M at M% per annum for 4 years is Tk. M, what is the value of M?

ক
25
খ
30
গ
20
ঘ
15

ব্যাখ্যা

Question: If the simple interest on Tk. M at M% per annum for 4 years is Tk. M, what is the value of M?

Solution:
Given,
P = M
r = M% = M/100
n = 4 years
I = M

We know,
I = Prn
⇒ M = M × (M/100) × 4
⇒ M = M × M/25
∴ M = 25

৯,৮৬৫.

9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps are needed to empty the same reservoir working 6 hours a day in 16 days?

ক
21
খ
16
গ
15
ঘ
24

ব্যাখ্যা

Question: 9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps are needed to empty the same reservoir working 6 hours a day in 16 days?

Solution:
The total work required to empty the reservoir is equal to the number of pumps multiplied by the hours they work per day and the number of days.
So, total work = 9 pumps × 8 hours/day × 20 days = 1440 pump-hours.

Let x be the number of pumps needed. These pumps will work 6 hours per day for 16 days. So, total work done by these pumps = x pumps × 6 hours/day × 16 days = 96 x pump-hours.

Since the total work is the same, 1440 pump-hours = 96 × x pump-hours.

Divide 1440 by 96
We get, x = 15.

৯,৮৬৬.

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

ক
ক) 40 days
খ
খ) 36 days
গ
গ) 32 days
ঘ
ঘ) 34 days

ব্যাখ্যা

Question: 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Solution:
let, work done by one man in one day = x
work done by one woman in one day = y

so,
in one day 4 men and 6 women can do = 1/8
∴ 4x + 6y = 1/8 . . . . (i)

and 3x + 7y = 1/10 . . . (ii)

multiplying (i) by 3 and (ii) by 4 and then subtracting we get,
12x + 18y - 12x - 28y = 3/8 - 4/10
- 10y = - 1/40
10y = 1/40

hence, 10 women can finish the work in 40 days

৯,৮৬৭.

In a survey about gym membership, 80 percent of people said they would join a new fitness center. Of those who said they would join, 75 percent actually joined, and of those who said they would not join, 10 percent ended up joining. What percent of the group joined the fitness center?

ক
56%
খ
60%
গ
58%
ঘ
62%
ঙ
None

ব্যাখ্যা

Question: In a survey about gym membership, 80 percent of people said they would join a new fitness center. Of those who said they would join, 75 percent actually joined, and of those who said they would not join, 10 percent ended up joining. What percent of the group joined the fitness center?

Solution:
Let's assume,
Total group = 100 people

Now,
75% of 80 = 80 × (75/100) = 60 people who said YES and actually joined
and, 10% of 20 = 20 × (10/100) = 2 people who said NO but actually joined

Total joined = Those who said YES and joined + Those who said NO but joined
= 60 + 2
= 62 people or 62%

৯,৮৬৮.

If x percent of 40 is y, then 10x equals?

ক
ক) 4y
খ
খ) 10y
গ
গ) 25y
ঘ
ঘ) 100y

ব্যাখ্যা

Given, x% of 40 = y
⇒ 40x/100 = y
⇒ x = 100y/40
⇒ x = 5y/2
∴ 10x = 25y

৯,৮৬৯.

Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is

ক
48
খ
45.5
গ
54
ঘ
38

ব্যাখ্যা

Question: Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is-

Solution:
Let, the second number be = x
First number = 2x
∴ Third number = 2x/3

∴ 2x + x + (2x/3) = 49.5 × 3
⇒ 6x + 3x + 2x = 49.5 × 9
⇒ 11x = 445.5
⇒ x = 445.5/11
∴ x = 40.5

∴ Required difference = 2x - (2x/3)
= 4x/3
= (4 × 40.5)/3
= 54

৯,৮৭০.

Soma was asked to multiply a certain number by 36. She multiplied it by 63 instead and got an answer 3834 more than the correct one. What was the number to be multiplied?

ক
126
খ
142
গ
148
ঘ
152

ব্যাখ্যা

Let,
the number be x
then, 63x - 36x = 3834
⇒ 27x = 3834
⇒ x = 3834/27
⇒ x = 142.

৯,৮৭১.

A farmer had 17 hens. All but 9 died. How many live hens were left?

ক
ক) 0
খ
খ) 9
গ
গ) 8
ঘ
ঘ) 16

ব্যাখ্যা

Answer is given in the question. All but 9 died means - except 9 all other hens are died.
বাক্যটির বাংলা অর্থ - এক কৃষকের ১৭টি মুরগী ছিলো। নয়টি বাদে বাকি সবগুলো মারা গিয়েছিলো।
So there are 9 alive hens.

৯,৮৭২.

The value of (489.1375 x 0.0483 x 1.956)/ (0.0873 x 92.581 x 99.749) is closest to:

ক
ক) 0.006
খ
খ) 0.06
গ
গ) 0.60
ঘ
ঘ) 6

ব্যাখ্যা

(489.1375 x 0.0483 x 1.956) / (0.0873 x 92.581 x 99.749)
≈ 489 x 0.05 x 2 / 0.09 x 93 x 100
= 489/ (9 x 93 x 10)
= (163/279)x (1/10)
= 0.58/ 10
= .058
≈ .06

৯,৮৭৩.

A circle and a rectangle have the same perimeter. The sides of the rectangle are 20 cm and 46 cm. What is the area of the circle?

ক
920 sq. cm.
খ
1248 sq. cm.
গ
1080 sq. cm.
ঘ
1386 sq. cm.

ব্যাখ্যা

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 20 cm and 46 cm. What is the area of the circle?

Solution:
আয়তক্ষেত্রটির পরিসীমা = 2 × (20 + 46) সেমি
= 132 সেমি।

যেহেতু বৃত্তের পরিধি ও আয়তক্ষেত্রের পরিসীমা সমান, তাই বৃত্তের পরিধিও 132 সেমি।

বৃত্তের পরিধি, C = 2πr
⇒ 2πr = 132
⇒ r = 132/(2 × 22/7)
∴ r = 21 সেমি।

বৃত্তের ক্ষেত্রফল, A = πr²
= (22/7) × (21)²
= 1386 বর্গ সেমি।

সুতরাং, বৃত্তটির ক্ষেত্রফল হলো 1386 বর্গ সেমি।

৯,৮৭৪.

A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is half the radius of the rod. Find the number of balls.

ক
ক) 3
খ
খ) 4
গ
গ) 5
ঘ
ঘ) 6

ব্যাখ্যা

Volume of cylinder = πr²h
As height = radius
So, πr²h = πr²×r = πr³
Volume of sphere = 4/3 πr³
As the radius is Halfen, Here r = r/2
So, volume of sphere = (4/3)π × (r/2)³ = (πr³)/6
∴ Number of balls = πr³/(πr³/6) = 6

৯,৮৭৫.

An observer 1.6 m tall stands 20 meters away from a tree. The angle of elevation from his eye to the top of the tree is 45°. What is the height of the tree?

ক
25 m
খ
21.6 m
গ
18.5 m
ঘ
27 m

ব্যাখ্যা

Question: An observer 1.6 m tall stands 20 meters away from a tree. The angle of elevation from his eye to the top of the tree is 45°. What is the height of the tree?

Solution:

মনে করি,
গাছটির উচ্চতা AB। পর্যবেক্ষকের চোখ C বিন্দুতে আছে এবং তার উচ্চতা CD = 1.6 m
পর্যবেক্ষক থেকে গাছটির দূরত্ব BD = 20 m
এখানে, A, C এবং E বিন্দু দ্বারা গঠিত ACE হলো একটি সমকোণী ত্রিভুজ, যার ∠C = 45°।

আমরা জানি,
tan θ = লম্ব/ভূমি
এখানে, লম্ব = AE এবং ভূমি = CE
∴ tan 45° = AE/20
∴ 1 = AE/20
∴ AE = 20 মিটার

গাছটির মোট উচ্চতা, AB = AE + EB
= 20 + 1.6
= 21.6 মিটার

সুতরাং, গাছটির উচ্চতা হলো 21.6 মিটার।

৯,৮৭৬.

Today is Friday. After 61 days, it will be:

ক
Saturday
খ
Sunday
গ
Tuesday
ঘ
Wednesday
ঙ
None of these

ব্যাখ্যা

Question: Today is Friday. After 61 days, it will be:

Solution:
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Friday.
After 62 days, it will be Thursday.
∴ After 61 days, it will be Wednesday.

৯,৮৭৭.

The area of a triangle with sides 3 cm, 5 cm and 6 cm is-

ক
ক) 2√3cm²
খ
খ) 2√14cm²
গ
গ) 5√12cm²
ঘ
ঘ) 4√14cm²

ব্যাখ্যা

প্রশ্ন: The area of a triangle with sides 3 cm, 5 cm and 6 cm is-

সমাধান:
ত্রিভুজটির অর্ধপরিসীমা, s = (৩ + ৫ + ৬)/২ সে.মি.
= ১৪/২ সে.মি.
= ৭ সে.মি.

ত্রিভুজটির ক্ষেত্রফল = √(৭× ৪ × ২ × ১) বর্গসে.মি.
= √৫৬ বর্গসে.মি.
= √(৪ × ১৪) বর্গসে.মি.
= ২√১৪ বর্গসে.মি.

৯,৮৭৮.

In a college, the ratio of foreign to local students is 3 : 7. If three-fourths of the local students are female and one-quarter of the foreign students is female. What fraction of the combined students is female?

ক
50%
খ
60%
গ
64%
ঘ
65%

ব্যাখ্যা

Question: In a college, the ratio of foreign to local students is 3 : 7. If three-fourths of the local students are female and one-quarter of the foreign students is female. What fraction of the combined students is female?

Solution:
Since the ratio of foreign : local is 3 : 7,
let's use a total of 10 parts
So if we have a total of 100 students, that means
Foreign students = 30 students
Local students = 70 students

Foreign female students = (1/4) × 30 = 15/2 students
Local female students = (3/4) × 70 = 105/2 students

Total female students = (15/2) + (105/2)
= (15 + 105)/2
= 60 students or 60% of the combined students are female.

৯,৮৭৯.

In how may different ways can the letters of the word EXTRA be arranged so that the vowels are never come together?

ক
ক) 72
খ
খ) 48
গ
গ) 120
ঘ
ঘ) 36

ব্যাখ্যা

The word 'EXTRA' contains 5 different letters.
The word EXTRA can be arranged in 5! ways = 120 ways

When the vowels EA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters XTR(EA).
Now, 4 letters can be arranged in 4! ways
= 24 ways.
The vowels (EA) can be arranged among themselves in 2! = 2 ways.
∴ The word EXTRA can be arranged in such a way that the vowels will be together = (24 x 2) =48

The letters of the words EXTRA be arranged so that the vowels are never together = (120 - 48) = 72 ways.

৯,৮৮০.

28√a + 1426 = three-fourths of 2984, then a =?

ক
859
খ
841
গ
659
ঘ
694

ব্যাখ্যা

Question: 28√a + 1426 = three-fourths of 2984, then a =?

Solution:
28√a + 1426 = (3/4) × 2984
⇒ 28√a + 1426 = 2238
⇒ 28√a = 2238 - 1426
⇒ 28√a = 812
⇒ √a = 812/28
⇒ √a = 29
⇒ (√a)² = (29)²
∴ a = 841

৯,৮৮১.

If the total surface area of the hemisphere be 36π sq. cm, then its radius is -

ক
3√2
খ
2√2
গ
2√3
ঘ
3√3

ব্যাখ্যা

Question: If the total surface area of the hemisphere be 36π sq. cm, then its radius is -

Solution:
We know surface area of hemisphere = 3πr²

ATQ,
3πr² = 36π
⇒ r² = 12 = 4×3
⇒ r = 2√3

৯,৮৮২.

Sahil and Nitish rent a stable for 9 months. Sahil puts in 84 horses for 5 months. How many horses can Nitish put in remaining month, if Nitish pay twice of Sahil?

ক
120
খ
190
গ
210
ঘ
230

ব্যাখ্যা

Question: Sahil and Nitish rent a stable for 9 months. Sahil puts in 84 horses for 5 months. How many horses can Nitish put in remaining month, if Nitish pay twice of Sahil?

Solution:
Let X be the number of horses put by the Nitish for 4 months;
ATQ,
(84 × 5)/(X × 4)= 1/2
⇒ 420/4x = 1/2
⇒ 4x = 840
∴ x = 210

৯,৮৮৩.

If a man was to sell his table for Tk. 750, he would lose 25%. To gain 25% he should sell it for:

ক
Tk. 1100
খ
Tk. 1225
গ
Tk. 1250
ঘ
Tk. 1200

ব্যাখ্যা

Question: If a man was to sell his table for Tk. 750, he would lose 25%. To gain 25% he should sell it for:

Solution:
Let the Cost price of the table be x.

Selling price = x - 25% of x
⇒ 750 = x - (25x/100)
⇒ 750 = 75x/100
⇒ 75x = 75000
∴ x = 1000

To gain 25% = 1000 + 25% of 1000
= 1000 + 250
= Tk. 1250

৯,৮৮৪.

If p and q are the roots of the equation 3x² − 7x + 2 = 0, then what is the value of (1/p) + (1/q)?

ক
1/5
খ
3
গ
7/2
ঘ
5/2

ব্যাখ্যা

Question: If p and q are the roots of the equation 3x²− 7x + 2 = 0, then what is the value of (1/p) + (1/q)?

Solution:

3x²− 7x + 2 = 0
⇒ 3x²- 6x - x + 2 = 0
⇒ 3x (x - 2) - 1 (x - 2) = 0
⇒ (3x - 1)(x - 2) = 0
⇒ x = 1/3 = p
∴ x = 2 = q

Now,
1/p + 1/q
= 1/(1/3) + 1/2
= 3 + 1/2
= 7/2

৯,৮৮৫.

Find the length of the arc when a circle has a radius of 14 cm and a central angle of 90°.

ক
30 cm
খ
17 cm
গ
22 cm
ঘ
11 cm

ব্যাখ্যা

Question: Find the length of the arc when a circle has a radius of 14 cm and a central angle of 90°.

Solution:
দেওয়া আছে,
কোণ, θ = ৯০° এবং ব্যাসার্ধ, r = ১৪ সে. মি.

আমরা জানি,
চাপের দৈর্ঘ্যের, L = (θ/৩৬০°) × ২πr
= (৯০°/৩৬০°) × ২ × (২২/৭) × ১৪
= (১/৪) × ৪ × ২২
= ২২ সে. মি.

৯,৮৮৬.

What is the minimum number of mangos that must be added to the existing stock of 1000 mangos so that the total stock can be equally distributed among 6, 15, 20 and 24 persons?

ক
46
খ
50
গ
80
ঘ
72

ব্যাখ্যা

Question: What is the minimum number of mangos that must be added to the existing stock of 1000 mangos so that the total stock can be equally distributed among 6, 15, 20 and 24 persons?

Solution:
L.C.M of 6, 15, 20 and 24 is = 120

Here,
Divisor = 120
Divisible = 1000
Quotient = 8
Remainder = 40

∴ To make the total number of mangoes a multiple of 120, we need to add = 120 - 40 = 80

৯,৮৮৭.

Three unbiased coins are tossed. What is the probability of getting at most two heads?

ক
ক) 7/8
খ
খ) 5/8
গ
গ) 1/3
ঘ
ঘ) 4/5

ব্যাখ্যা

Getting at most Two heads means 0 to 2 but not more than 2
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
∴P(E) = n(E)/n(S) = 7/8

৯,৮৮৮.

If (n - 1) is an odd number, what are the two other odd numbers nearest to it?

ক
n - 3, n + 5
খ
n, n - 2
গ
n - 3, n + 1
ঘ
n, n - 1

ব্যাখ্যা

Question: If (n - 1) is an odd number, what are the two other odd numbers nearest to it?

Solution:
n - 1 is an odd number
previous odd number = n - 1 - 2 = n - 3
next odd number = n - 1 + 2 = n + 1

৯,৮৮৯.

A manufacturer sells an article to a wholesale dealer at a profit of 20% and the wholesale dealer sells it to retail merchant at a loss of 5%. Find the resultant loss or profit.

ক
ক) 10% loss
খ
খ) 14% profit
গ
গ) 18% loss
ঘ
ঘ) None of the above

ব্যাখ্যা

Question: A manufacturer sells an article to a wholesale dealer at a profit of 20% and the wholesale dealer sells it to retail merchant at a loss of 5%. Find the resultant loss or profit.

Solution:
Let, production cost = 100
Then first selling price after 20% profit = 120
Now, second selling price at 5% loss = 120 - (5% of 120)
⇒ 120 - 6 = 114
So, overall profit = 114 -100 = 14%

৯,৮৯০.

A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was -

ক
64
খ
72
গ
80
ঘ
85

ব্যাখ্যা

Question: A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was -

Solution:
Let the cricketer takes x wickets before last match

Total run = 12.4x + 26
New average = (12.4x + 26) / (x + 5)

ATQ,
(12.4x + 26) / (x + 5) = 12.4 - 0.4
⇒ (12.4x + 26) = 12 (x + 5)
⇒ 12.4x - 12x = 60 - 26
⇒ 0.4x = 34
⇒ x = 34/0.4
= 85

The number of wickets taken by him till the last match was = 85 wickets

৯,৮৯১.

If 4^{n - 2} = 128, find the value of n.

ক
3
খ
4.5
গ
5.5
ঘ
6
ঙ
8

ব্যাখ্যা

Question: If 4^{n - 2} = 128, find the value of n.

Solution:
4^{n - 2} = 128
⇒ (2²)^{n - 2} = 2⁷
⇒ 2^{2(n - 2)} = 2⁷
⇒ 2^{2n - 4} = 2⁷
⇒ 2n - 4 = 7
⇒ 2n = 7 + 4
⇒ 2n = 11
⇒ n = 11/2
∴ n = 5.5

৯,৮৯২.

If A = 45° , then what is the value of (1 - tan²A)/(1 + tan²A)?

ক
1
খ
2
গ
1/2
ঘ
0

ব্যাখ্যা

Question: If A = 45° , then what is the value of (1 - tan²A)/(1 + tan²A)?

Solution:
Here, A = 45°

Now,
(1 - tan²A)/(1 + tan²A)
= {1 - (tan45°)²}/{1 + (tan45°)²}
= (1 - 1²)/(1 + 1²)
= 0/2
= 0

৯,৮৯৩.

A certain sum of money yields Tk. 1261 as compound interest for three years at 5% per annum. The sum is ?

ক
8000 Tk.
খ
6000 Tk.
গ
5000 Tk.
ঘ
10000 Tk.

ব্যাখ্যা

Question: A certain sum of money yields Tk. 1261 as compound interest for three years at 5% per annum. The sum is ?

Solution:
Let the principal be Tk. P

Now,
1261 = P(1 + r)n - P
1261 = P(1 + 5/100)3 - P
1261 = P(105/100)3 - P
1261 = P(21/20)3 - P
1261 = P(9261/8000 - 1)
1261 = P (1261/8000)
1 = P/8000
P = 8000

৯,৮৯৪.

Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} having 4 elements.

ক
ক) 340
খ
খ) 370
গ
গ) 320
ঘ
ঘ) 330
ঙ
ঙ) None of these

ব্যাখ্যা

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in ¹¹C₄ ways = 330 ways

৯,৮৯৫.

The cost price of 60 articles is the same as the selling price of 'q' articles. If the profit is 20%, then the value of 'q' is -

ক
40
খ
45
গ
50
ঘ
54

ব্যাখ্যা

Question: The cost price of 60 articles is the same as the selling price of 'q' articles. If the profit is 20%, then the value of 'q' is -

Solution:
Let,
The cost price of each article be Tk. 1
The cost price of 60 articles = Tk. 60
The selling price of 'q' articles = Tk. 60
∴ Profit = Tk. (60 - q)

ATQ,
(60 - q)/q = 20%
⇒ 100(60 - q)/q = 20
⇒ 6000 - 100q = 20q
⇒ 120q = 6000
∴ q = 50
Therefore, the value of 'q' is 50.

৯,৮৯৬.

A rectangular plot measuring 90 metres by 50 metre needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?

ক
30
খ
60
গ
56
ঘ
44

ব্যাখ্যা

Length of the wire fencing = perimeter
= 2(90 + 50)
= 2 × 140
= 280. m
Two poles are kept 5 meters apart. Note that the poles are placed along the perimeter of the rectangular plot, not in a single straight line.
Hence, the number of poles required
= 280/5
= 56.

৯,৮৯৭.

Find the area of a rhombus whose side is 25cm and one of the diagonals is 30cm?

ক
ক) 225 sq.cm
খ
খ) 360 sq.cm
গ
গ) 600 sq.cm
ঘ
ঘ) 480 sq.cm

ব্যাখ্যা

Question: Find the area of a rhombus whose side is 25cm and one of the diagonals is 30cm?

Solution:
ধরি
অপর কর্ণের দৈর্ঘ্য = 2x সে.মি.
দেয়া আছে,
রম্বসের একটি বাহুর দৈর্ঘ্য = 25 সে.মি.
রম্বসের একটি কর্ণের দৈর্ঘ্য = 30 সে.মি.

আমরা জানি,
রম্বসের কর্ণদ্বয় পরস্পর সমকোণে সমদ্বিখন্ডিত করে।
আমরা জানি,
25² = (30/2)² + (2x/2)²
⇒ 625 = 225 + x²
⇒ x² = 400
⇒ x = 20

অপর কর্ণের দৈর্ঘ্য = 2x = 2 × 20 = 40 সে.মি.
রম্বসের ক্ষেত্রফল = (1/2) × 30 × 40 = 600 বর্গ সে.মি.

৯,৮৯৮.

Solve: log(2x² + 17) = log(x - 3)²

ক
- 4, - 2
খ
4, - 2
গ
- 4, 2
ঘ
4, 2
ঙ
None of these

ব্যাখ্যা

Question: Solve: log(2x² + 17) = log(x - 3)²

Solution:
log(2x² + 17) = log(x - 3)²
⇒ log(2x² + 17)= log(x² - 6x + 9)
⇒ 2x² + 17 = x² - 6x + 9
⇒ x² + 6x + 8 = 0
⇒ x² + 4x + 2x + 8 = 0
⇒ x(x + 4) + 2(x + 4) = 0
⇒ (x + 4)(x + 2)=0
∴ x = - 4, - 2

৯,৮৯৯.

A sum of money at simple interest amounts to Tk.815 in 3years and to Tk.854 in 4 years. What is the sum?

ক
ক) Tk. 650
খ
খ) Tk. 700
গ
গ) Tk. 698
ঘ
ঘ) Tk. 690

ব্যাখ্যা

S.I. for 1 year = 854 - 815
= 39
S.I. for 3 years = 39 × 3
= 117
∴ Required Sum = 815 - 117
= Tk. 698.

৯,৯০০.

একটি ত্রিভুজাকৃতি জমির তিন দিকের দৈর্ঘ্য যথাক্রমে ২৫ মি., ২০ মি., ১৫ মি.। প্রতি বর্গমিটার ২.৫০ টাকা হিসেবে ঐ জমিতে ঘাস লাগাতে কত টাকা লাগবে?

ক
২৯০ টাকা
খ
৩৩৮ টাকা
গ
৩৭৫ টাকা
ঘ
৩৯৯ টাকা
ঙ
৪১০ টাকা

ব্যাখ্যা

প্রশ্ন: একটি ত্রিভুজাকৃতি জমির তিন দিকের দৈর্ঘ্য যথাক্রমে ২৫ মি., ২০ মি., ১৫ মি.। প্রতি বর্গমিটার ২.৫০ টাকা হিসেবে ঐ জমিতে ঘাস লাগাতে কত টাকা লাগবে?

সমাধান:
ত্রিভুজের পরিসীমা ২S হলে,
আমরা জানি,
ত্রিভুজের পরিসীমা ২S = (a + b + c)
⇒ S = (a + b + c)/২
⇒ S = (২৫ + ২০ + ১৫)/২
⇒ S = ৬০/২
∴ S = ৩০

আমরা জানি,
ত্রিভুজক্ষেত্রের ক্ষেত্রফল = √S(S - a)(S - b)(S - c)
= √৩০(৩০ - ২৫)(৩০ - ২০)(৩০ - ১৫) বর্গ মি.
= √৩০ × ৫ × ১০ × ১৫ বর্গ মি.
= √২২,৫০০ বর্গ মি.
= ১৫০ বর্গ মি.

১ বর্গ মি. ঘাস লাগাতে খরচ হয় = ২.৫০ টাকা
∴ ১৫০ বর্গ মি. ঘাস লাগাতে খরচ হয় = (১৫০ × ২.৫০) টাকা
= ৩৭৫ টাকা

∴ ঐ জমিতে ঘাস লাগাতে খরচ হবে ৩৭৫ টাকা।