Bank Math

PrepBank · পাতা ২৭ / ১৬১ · ২,৬০১–২,৭০০ / ১৬,১২৪

২,৬০১.

If b < 2 and 2x - 3b = 0, which of the following must be true?

ক
ক) x > - 3
খ
খ) x < 2
গ
গ) x = 3
ঘ
ঘ) x < 3
ঙ
ঙ) x > 3

ব্যাখ্যা

Question: If b < 2 and 2x - 3b = 0, which of the following must be true?

Solution:
Given that,
2x - 3b = 0
⇒ 2x = 3b
⇒ 2x < (3×2) [ i.e b < 2]
∴ x < 3

২,৬০২.

A student has three iron rods of lengths 44 cm, 22 cm, and 55 cm. He needs to cut them into rods of the largest possible length such that no iron is wasted. What is the longest rod length he can achieve?

ক
17 cm
খ
15 cm
গ
13 cm
ঘ
11 cm

ব্যাখ্যা

Question: A student has three iron rods of lengths 44 cm, 22 cm, and 55 cm. He needs to cut them into rods of the largest possible length such that no iron is wasted. What is the longest rod length he can achieve?
(একজন ছাত্রকে ৩টি বিভিন্ন দৈর্ঘ্যের লোহার টুকরা দেওয়া হয়েছে – যথাক্রমে ৪৪ সেমি, ২২ সেমি ও ৫৫ সেমি। তাকে এমন একটি সর্বোচ্চ দৈর্ঘ্যের রড তৈরি করতে হবে যাতে কোনো লোহার অপচয় না হয়। রডটির সর্বাধিক দৈর্ঘ্য নির্ণয় করো।)

Solution:
H.C.F. = গ.সা.গু
এ ধরনের রডের সর্বাধিক সম্ভাব্য দৈর্ঘ্য = (44, 22, 55 এর গ.সা.গু) cm = 11 cm.

২,৬০৩.

In a race of 200 meters, B can give a start of 10 meters to A, and C can give a start of 20 meters to B. The starts that C can give to A, in the same race is:

ক
27 meters
খ
24 meters
গ
29 meters
ঘ
18 meters
ঙ
None

ব্যাখ্যা

Question: In a race of 200 meters, B can give a start of 10 meters to A, and C can give a start of 20 meters to B. The starts that C can give to A, in the same race is:

Solution:
ATQ,
When B runs 200 m meters, A runs 190 meters;
Hence, when B runs 180 meters,
A runs = 190 × (180/200) meters
= 171 meters

When C runs 200m, B runs 180 meters.
Hence,
C will give a start to A by = (200 - 171) meters
= 29 meters

২,৬০৪.

A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is-

ক
ক) 7/19
খ
খ) 4/19
গ
গ) 12/19
ঘ
ঘ) 21/95

ব্যাখ্যা

Question: A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is-

Solution:
Total bulbs 20
Number of defective bulbs 4
∴ Number of non-defective bulbs (20 - 4) = 16.

The probability of non-defective bulbs is ¹⁶C₂/²⁰C₂= 120/190 = 12/19

∴ The probability of at least 1 bulb is defective = 1 - (12/19) = 7/19

২,৬০৫.

In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 28, how many people were in the meeting?

ক
7
খ
8
গ
9
ঘ
10
ঙ
None of the above

ব্যাখ্যা

Question: In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 28, how many people were in the meeting?

Solution:
Let,
the number of people be n.

ATQ,
number of total handshakes,
n(n - 1)/2 = 28
⇒ n(n - 1) = 28 × 2
⇒ n² - n = 56
⇒ n² - n - 56 = 0
⇒ n² - 8n + 7n - 56 = 0
⇒ n(n - 8) + 7(n - 8) = 0
⇒ (n - 8)(n + 7) = 0

∴ n = 8, - 7

So, the number of people be 8.

২,৬০৬.

One-fifth of half of a number is 20. Then 20% of that number is:

ক
ক) 20
খ
খ) 25
গ
গ) 30
ঘ
ঘ) 40

ব্যাখ্যা

Question: One-fifth of half of a number is 20. Then 20% of that number is:

Solution:
Let the number is = x

ATQ,
1/5 of 1/2 of x = 20
⇒ 1/5 × 1/2 × x = 20
⇒ x/10 = 20
∴ x = 200

∴ 20% of 200 = (20/100) × 200
= 40

২,৬০৭.

A sum of money amounts to Tk 9800 after 5 years and Tk. 12005. after 8 years at the same rate of simple interest. The rate of interest per annum is -

ক
ক) 5%
খ
খ) 8%
গ
গ) 12%
ঘ
ঘ) 15%

ব্যাখ্যা

Simple interest for 3 years
= 12005 - 9800 = 2205
Simple interest for 5 years
= (2205/3) × 5 = 3675

Some of money = 9800 - 3675 = 6125

∴ Rate of interest = (100 × 2205)/(6125 × 3)
= 12%

২,৬০৮.

If a³ - b³ = 56 and a - b = 2, then the value of a² + b² =?

ক
15
খ
20
গ
24
ঘ
26

ব্যাখ্যা

Question: If a³ - b³ = 56 and a - b = 2, then the value of a² + b² =?

Solution:
a³ - b³ = 56
⇒ (a - b) (a² + ab + b²) = 56
⇒ a² + ab + b² = 56/2 = 28
⇒ (a - b)² + 3ab = 28
⇒ 3ab = 28 - 4 = 24
⇒ ab = 24/3 = 8

a² + ab + b² = 28
⇒ a² + b² = 28 - 8 = 20

২,৬০৯.

On the same side of the tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is -

ক
ক) 63.5 m
খ
খ) 76.9 m
গ
গ) 86.7 m
ঘ
ঘ) 90 m

ব্যাখ্যা

Let AB be the tower and C and D be the objects.
Then, AB = 150 m,
∠ACB = 45° and
∠ADB = 60°

AB/AD = tan 60° = √3
AD = AB/√3
= 150/√3 m.

AB/AC = tan 45° = 1
AC = AB = 150 m.

∴CD = (AC - AD)
= {150 - (150/√3)} m
= [{150(√3 - 1)/√3} × {(√3)/(√3)}] m
= 50(3 - √3) m
= (50 × 1.27) m
= 63.5 m.

২,৬১০.

A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in-

ক
ক) 30 years
খ
খ) 38 years
গ
গ) 45 years
ঘ
ঘ) 49 years

ব্যাখ্যা

Question: A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in-

Solution:
let the sum P

2P = P (1 + r)¹⁵
⇒ (1 + r)¹⁵ = 2

let, sum will 8 times in n years
8P = P(1 + r)ⁿ
⇒ 8 = (1 + r)ⁿ
⇒ 2³ = (1 + r)ⁿ
⇒ ((1 + r)¹⁵)³ = (1 + r)ⁿ
⇒ (1 + r)⁴⁵ = (1 + r)ⁿ
∴ n = 45 years

২,৬১১.

A man read (2/5)^th of a book on the first day. He read (1/3)^rd more on the second day than he read on the first day. 15 pages were left for the third day. The number of pages in the book is -

ক
155
খ
182
গ
215
ঘ
225
ঙ
None

ব্যাখ্যা

Question: A man read (2/5)^th of a book on the first day. He read (1/3)^rd more on the second day than he read on the first day. 15 pages were left for the third day. The number of pages in the book is -

Solution:
Let,
the number of pages in the book = x

∴ On the first day, the man read (2/5 of x) = 2x/5 pages

∴ On the second day, he read 1/3 more than he read on the first day i.e.:
Second day reading = 2x/5 + {1/3 of (2x/5)} pages
= (2x/5 + 2x/15) pages
= 8x/15 pages

ATQ,
x - (2x/5 + 8x/15) = 15
⇒ x - {(6x + 8x)/15} = 15
⇒ x - (14x/15) = 15
⇒ (15x - 14x)/15 = 15
∴ x = 225

২,৬১২.

Q. 31 - 60: Read the following questions carefully and choose the right answer.
৩১) A man can walk uphill at the rate of 2.5km/hr and downhill at the rate of 3.25km/hr. If the total time required walking a certain distance up the hill and return to the starting position is 4 hr 36 min, what is the distance he walked up the hill?

ক
ক) 3.5km
খ
খ) 4.5km
গ
গ) 5.5 km
ঘ
ঘ) 6.5 km

ব্যাখ্যা

Let the distance he walked up the hill be x km.
So, the distance he walked down the hill is x km.
Now,
According to the question
x/2.5 + x/3.25 = 4 + 36/60
⇒ 10x/25 + 100x/325 = 23/5
⇒ (130x + 100x)/325 = 23/5
⇒ 230x = (23 × 325)/5
⇒ 230x = 23 × 65
x = 6.5
∴ the distance he walked up the hill is 6.5 km.

২,৬১৩.

A, B, C subscribe Tk. 50,000 for a business. A subscribes Tk. 4000 more than B and B Tk. 5000 more than C. Out of a total profit of Tk. 30,000, A receives

ক
ক) Tk. 10800
খ
খ) Tk. 7200
গ
গ) Tk. 10200
ঘ
ঘ) Tk. 12600

ব্যাখ্যা

Question: A, B, C subscribe Tk. 50,000 for a business. A subscribes Tk. 4000 more than B and B Tk. 5000 more than C. Out of a total profit of Tk. 30,000, A receives

Solution:
Let
C = x.
B = x + 5000
A = x + 5000 + 4000 = x + 9000
Now
So, x + x + 5000 + x + 9000 = 50000
⇒ 3x = 36000
⇒ x = 12000
A : B : C = 21000 : 17000 : 12000
= 21 : 17 : 12

So A's Share = Tk. 30000 × (21/50) =Tk. 12600

২,৬১৪.

A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

ক
352
খ
370
গ
253
ঘ
145

ব্যাখ্যা

Question: A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

Solution:
let, the digits are x, y, z

x + y + z = 10
y = x + z

y + y = 10
⇒ y = 5

x + z = 5

(100z + 10y + x) - (100x + 10y + z) = 99
⇒ 99z - 99x = 99
⇒ z - x = 1

x + z + z - x = 5 + 1
⇒ 2z = 6
∴ z = 3

x = 5 - 3 = 2

The number is = 253

২,৬১৫.

Complete the following series:
350, 520, 738, ?, 1342

ক
1124
খ
984
গ
1186
ঘ
1010

ব্যাখ্যা

Question: Complete the following series:
350, 520, 738, ?, 1342

Solution:
350 = 7³ + 7
520 = 8³ + 8,
738 = 9³ + 9,
?
1342 = 11³ + 11,
So the pattern is n³ + n

So missing number is,
10³ + 10 = 1000 + 10 = 1010

২,৬১৬.

একজন মোটর সাইকেল চালক একটি নির্দিষ্ট দূরত্ব ৫ ঘণ্টায় অতিক্রম করতে পারে। সে এক-তৃতীয়াংশ দূরত্ব ৬০ কি.মি./ঘণ্টা গতিতে এবং বাকি অংশ ৮০ কি.মি./ ঘণ্টা গতিতে সম্পন্ন করে। মোট দূরত্ব কত?

ক
ক) ৩৬০ কি.মি.
খ
খ) ৩৪০ কি.মি.
গ
গ) ৩২০ কি.মি.
ঘ
ঘ) ৩০০ কি.মি.

ব্যাখ্যা

মনেকরি
মোট দূরত্ব = ক কি.মি.

{(ক/৩)/৬০} + {(২ক/৩)/৮০} = ৫
(ক /১৮০) + (ক/১২০) = ৫
(২ক + ৩ক)/৩৬০ = ৫
৫ক/৩৬০ = ৫
ক /৩৬০ = ১
ক = ৩৬০

মোট দূরত্ব = ৩৬০ কি.মি.

২,৬১৭.

Find the compound interest on Tk 8000 at 15% interest per annum for 2 years, compounded annually-

ক
ক) 9800 Tk
খ
খ) 10580 Tk
গ
গ) 11200 Tk
ঘ
ঘ) 11560 Tk

ব্যাখ্যা

Question: Find the compound interest on Tk 8000 at 15% interest per annum for 2 years, compounded annually-

Solution:
Compound Principal for 2 years = 8000(1 + 15/100)²
= 8000(115/100)²
= (8000 × 115 × 115)/(100 × 100)
= 10580

২,৬১৮.

The speed of A and B are in the ratio 3 : 4. A takes 30 minutes more than B to reach a destination. Time in which A reach the destination?

ক
100 minutes
খ
105 minutes
গ
110 minutes
ঘ
115 minutes
ঙ
None of the above

ব্যাখ্যা

Question: The speed of A and B are in the ratio 3 : 4. A takes 30 minutes more than B to reach a destination. Time in which A reach the destination?

Solution:
Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ 1/Time, When distance remains constant)

Let the time taken by A and B be 4x and 3x hours respectively.

Then,
4x - 3x = 30/60
Or, x = 1/2

Hence, time taken by A = 4x = 4 × (1/2) = 2 hours = 120 minutes

২,৬১৯.

The 2nd term of a geometric sequence is 27/8, and the 5th term is 1. What is the common ratio?

ক
2/3
খ
5/8
গ
1/2
ঘ
2/5
ঙ
- 3/8

ব্যাখ্যা

Question: The 2nd term of a geometric sequence is 27/8, and the 5th term is 1. What is the common ratio?

Solution:
আমরা জানি, একটি গুণোত্তর ধারার n-তম পদ, a_n = a.r^{n - 1}

দেয়া আছে,
2য় পদ, a₂ = 27/8
⇒ ar = 27/8 …...(1)

5ম পদ, a₅ = 1
⇒ a.r⁴ = 1 …....(2)

এখন, সমীকরণ (2) ÷ সমীকরণ (1) ⇒
(ar⁴)/(ar) = 1/(27/8)
⇒ r³ = 8/27
⇒ r³ = (2/3)³
⇒ r = 2/3

∴ সাধারণ অনুপাত (common ratio) হলো 2/3

২,৬২০.

If a ⊕ b = (a × b) + b then 7 ⊕ 6 equals to -

ক
ক) 48
খ
খ) 42
গ
গ) 38
ঘ
ঘ) 32

ব্যাখ্যা

Question: If a ⊕ b = (a × b) + b then 7 ⊕ 6 equals to -

Solution:
⇒7 ⊕ 6 =(7 × 6) + 6
= 42 + 6
= 48

২,৬২১.

500 shares, of par value TK. 20 each, are purchased from Rasel by Himel at a price of Tk. 25 each. Find the amount required to purchase the shares.

ক
ক) 10000 tk.
খ
খ) 12000 tk.
গ
গ) 12500 tk.
ঘ
ঘ) 20000 tk.

ব্যাখ্যা

Question: 500 shares, of par value TK. 20 each, are purchased from Rasel by Himel at a price of Tk. 25 each. Find the amount required to purchase the shares.

Solution:
Face value of each share = Tk. 20
The market value of each share = Tk. 25
Number of shares = 500
Amount required to purchase the shares = 500 × 25
= 12500 tk.

২,৬২২.

The ratio of the total amount distributed in all the males and females as salary is 6 : 5. The ratio of the salary of each male and female is 2 : 3. Find the ratio of the number of males and females.

ক
5 : 9
খ
5 : 7
গ
7 : 5
ঘ
9 : 5

ব্যাখ্যা

Question: The ratio of the total amount distributed in all the males and females as salary is 6 : 5. The ratio of the salary of each male and female is 2 : 3. Find the ratio of the number of males and females.

Solution:
Let,
The number of males be x
The number of females be y

Each male's salary 2a
Each female's salary 3a

Total salary of males 2a × x = 2ax
Total salary of females 3a × y = 3ay

ATQ,
2ax/3ay = 6/5
⇒ 2x/3y = 6/5
⇒ x/y = 18/10
⇒ x/y = 9/5
∴ x : y = 9 : 5

২,৬২৩.

A sold a watch to B at a gain of 20% and B sold it to C at loss of 10%. If C bought the watch for TK. 432 at what price did A purchase it?

ক
Tk. 300
খ
Tk. 200
গ
Tk. 400
ঘ
Tk. 250

ব্যাখ্যা

Question: A sold a watch to B at a gain of 20% and B sold it to C at loss of 10%. If C bought the watch for TK. 432 at what price did A purchase it?

Solution:
A purchase at x taka
Selling price of A = purchasing price of B = x + 20% of x = 1.2x

Buying price of C = 1.2x - 10% of 1.2x = 1.2x - 0.12x = 1.08x

1.08x = 432
⇒ x = 432/1.08 = Tk. 400

২,৬২৪.

Which of the numbers below is not equivalent to 4%?

ক
ক) 1/25
খ
খ) 4/100
গ
গ) 0.40
ঘ
ঘ) 0.04

ব্যাখ্যা

প্রশ্ন: Which of the numbers below is not equivalent to 4%?

সমাধান:
ক) (১/২৫) × ১০০% = ৪%

খ) (৪/১০০) × ১০০% = ৪%

গ) ০.৪০ = (৪০/১০০) × ১০০% = ৪০%

ঘ) ০.০৪ = (৪/১০০) × ১০০% = ৪%

এখানে, ৪% এর সমান নয় ০.৪০

২,৬২৫.

A person sold a product at a gain of 15%. Had he bought it for 25% less and sold it for Tk. 1200 less, he would have made a profit of 32%. The cost price of the product was?

ক
Tk. 6600
খ
Tk. 6800
গ
Tk. 7280
ঘ
Tk. 7500

ব্যাখ্যা

Question: A person sold a product at a gain of 15%. Had he bought it for 25% less and sold it for Tk. 1200 less, he would have made a profit of 32%. The cost price of the product was?

Solution:
Let the original cost price = Tk. x

Now, Selling price = x + 15% of x
= 115x/100
= 23x/20

And Cost price = x - 25% of x
= x - (25x/100)
= 75x/100
= 3x/4

∴ Selling price = (3x/4) + 32% of (3x/4)
= (3x/4) + (32/100) × (3x/4)
= 99x/100

ATQ,
(23x/20) - (99x/100) = 1200
⇒ (115x - 99x)/100 = 1200
⇒ 16x = 1200 × 100
⇒ x = (1200 × 100)/16
∴ x = 7,500

২,৬২৬.

The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at

ক
8 : 21 : 12 hrs
খ
8 : 27 : 48 hrs
গ
8 : 28 : 22 hrs
ঘ
8 : 27 : 36 hrs
ঙ
None of these

ব্যাখ্যা

Question: The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at

Solution:
Interval of change = (L.C.M. of 24, 36, 72) sec. = 72 sec. So, the lights will change after every 72 seconds, i.e. 1 min. 12 sec.
So, the next simultaneous change will take place at 8 : 21 : 12 hrs.

২,৬২৭.

The sum of the ages of the father and son is 80 years. If ten years ago the age of father was three times that of his son, find the present age of the father.

ক
60 years
খ
45 years
গ
50 years
ঘ
55 years

ব্যাখ্যা

Question: The sum of the ages of the father and son is 80 years. If ten years ago the age of father was three times that of his son, find the present age of the father.

Solution:
Let the age of son 10 years ago = x
The age of father 10 years ago = 3x

⇒ x + 10 + 3x + 10 = 80
⇒ 4x + 20 = 80
⇒ 4x = 60
⇒ x = 15

Son's present age = 15 + 10 = 25 years
∴ Present age of father = 3 × 15 + 10 = 45 + 10 = 55 years

২,৬২৮.

A relief truck carries enough food to feed 180 women or 270 children. If 90 children have already been fed, how many women can be fed with the remaining food?

ক
90 women
খ
100 women
গ
120 women
ঘ
130 women

ব্যাখ্যা

Question: A relief truck carries enough food to feed 180 women or 270 children. If 90 children have already been fed, how many women can be fed with the remaining food?

Solution:
Total food = for 270 children
Already taken = 90 children

Remaining = (270 − 90) = 180 children

ATQ,
270 children = 180 women
1 children = 180/270 women
180 children = (180 × 180)/270 women
= 120 women

২,৬২৯.

How many times in a day, are the hands of a clock in straight line but opposite in direction?

ক
ক) 22
খ
খ) 24
গ
গ) 48
ঘ
ঘ) 20

ব্যাখ্যা

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. Because between 5 and 7 they point in opposite directions at 6 o'clock only.
So, in a day, the hands point in the opposite directions 22 times.

২,৬৩০.

A gentleman has two sons. The present age of the elder son is twice the present age of the younger son. Three years ago, the elder son’s age was three times the age of the younger son. Find the current age of the elder son.

ক
16 years
খ
14 years
গ
12 years
ঘ
6 years

ব্যাখ্যা

Question: A gentleman has two sons. The present age of the elder son is twice the present age of the younger son. Three years ago, the elder son’s age was three times the age of the younger son. Find the current age of the elder son.

Solution:
ধরি,
ছোট ছেলের বয়স = x বছর
বড় ছেলের বয়স = 2x বছর

3 বছর পূর্বে,
ছোট ছেলের বয়স ছিলো = x - 3 বছর
বড় ছেলের বয়স ছিলো = 2x - 3 বছর

প্রশ্নমতে,
(2x - 3) = 3(x - 3)
বা, 2x - 3 = 3x - 9
বা, 3x - x = 9 - 3
বা, x = 6

অতএব বড় ছেলের বয়স = (2 × 6) = 12 বছর

২,৬৩১.

A and B are two alloys in which ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?

ক
12 : 19
খ
15 : 17
গ
20 : 27
ঘ
25 : 34

ব্যাখ্যা

Question: A and B are two alloys in which ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?

Solution:
Ratio of Gold and Copper in Alloy A = 5 : 3
Ratio of Gold and Copper in Alloy B = 5 : 11

Amount of Gold in Alloy A = 5/8
Amount of Gold In Alloy B = 5/16

Amount of Copper in A = 3/8
Amount of Copper in B = 11/16

Amount of Gold In C,
= (Amount of gold in A + Amount of gold in B) = (5/8) + (5/16)
= (10 + 5)/16
= 15/16
Amount of Copper in C,
= Amount of Copper in A + Amount of Copper in B = (3/8) + (11/16)
= 17/16

So, Ratio of Gold and Copper in C,
= (15/16) : (17/16) = 15 : 17

২,৬৩২.

5 machines, working 6 hours a day, can produce 300 units in 3 days. How many hours a day must 10 machines work to produce 300 units in 1 day?

ক
11 hours
খ
8 hours
গ
9 hours
ঘ
10 hours

ব্যাখ্যা

Question: 5 machines, working 6 hours a day, can produce 300 units in 3 days. How many hours a day must 10 machines work to produce 300 units in 1 day?

Solution:
5 machines, working 6 hours a day, can produce 300 units in 3 days
Formula used: M1 × T1 = M2 × T2
Where M1 and M2 is machines and T1 and T2 is time

Let H hours be the number of hours required
Applying the above formula
⇒ 5 × 6 × 3 = 10 × 1 × H
⇒ 90 = 10H
⇒ H = 90/10
⇒ H = 9 hours
∴ 10 machines need to work 9 hours to complete the work in 1 day.

২,৬৩৩.

A bag contains 8 red and 5 green balls. Two balls are drawn randomly. Find the probability that one ball is red and other is green.

ক
ক) 5/13
খ
খ) 8/13
গ
গ) 2/13
ঘ
ঘ) 20/39

ব্যাখ্যা

Question: A bag contains 8 red and 5 green balls. Two balls are drawn randomly. Find the probability that one ball is red and other is green.

Solution:
A bag contains 8 red and 5 green balls.
Total balls = 8 + 5 = 13
Two balls drawn ¹³C₂ = 78
One ball is red and other is green = ⁸C₁ × (⁵C₁) = 8 × 5 = 40

Required probability = 40/78 = 20/39

২,৬৩৪.

Let, m and n be natural numbers. If m² + mn + m = 14 and n² + nm + n = 28, then (2m + n) =?

ক
7
খ
8
গ
10
ঘ
12

ব্যাখ্যা

Question: Let, m and n be natural numbers. If m² + mn + m = 14 and n² + nm + n = 28, then (2m + n) =?

Solution:
n² + nm + n + m² + mn + m = 28 + 14
⇒ m² + 2mn + n² + m + n = 42
⇒ (m + n)² + m + n = 42
⇒ (m + n) (m + n + 1) = 42

m + n = 6, m + n + 1 = 7

Given,
m² + mn + m = 14
⇒ m(m + n + 1) = 14
⇒ m × 7 = 14
⇒ m = 2

n = 6 - m = 6 - 2 = 4

∴ 2m + n = 2 × 2 + 4 = 4 + 4 = 8

২,৬৩৫.

A boat makes a return journey from point A to point B and back in 5 hours 36 minutes. One way it travels with the stream and on the return it travels against the stream. If the speed of the stream increases by 2 km/hr, the return journey takes 9 hours 20 minutes. What is the speed of the boat in still water? (The distance between A and B is 16 km.)

ক
7 km/hr
খ
8.5 km/hr
গ
9 km/hr
ঘ
6 km/hr

ব্যাখ্যা

Question: A boat makes a return journey from point A to point B and back in 5 hours 36 minutes. One way it travels with the stream and on the return it travels against the stream. If the speed of the stream increases by 2 km/hr, the return journey takes 9 hours 20 minutes. What is the speed of the boat in still water? (The distance between A and B is 16 km.)

Solution:
Let x be speed of u/s
and y be the speed of d/s.

∴ (16/x) + (16/y) = (28/5)
and 16/(y+2) + 16/(x-2) = 28/3

Solving these 2 equations,
we get x = 4km/hr
and y = 10km/hr

∴ speed of boat in still water = (4 + 10)/2 = 7 km/hr.

২,৬৩৬.

If 2x - 1 ≥ - 3, then

ক
x ≤ - 2
খ
x ≥ - 2
গ
x ≤ - 1
ঘ
x ≥ - 1

ব্যাখ্যা

Question: If 2x - 1 ≥ - 3, then

Solution:
2x - 1 ≥ - 3
2x - 1 + 1 ≥ - 3 + 1
2x ≥ - 2
2x/2 ≥ - 2/2
x ≥ - 1

২,৬৩৭.

A problem is assigned to three students. The chances of solving it for the first, second, and third students are 1/3, 1/4, and 1/6, respectively. What is the probability that the problem will be solved by at least one of them?

ক
5/12
খ
7/12
গ
7/10
ঘ
4/9
ঙ
None

ব্যাখ্যা

Question: A problem is assigned to three students. The chances of solving it for the first, second, and third students are 1/3, 1/4, and 1/6, respectively. What is the probability that the problem will be solved by at least one of them?

Solution:
Probability of 1st student solving the problem = 1/3
Probability of 1st student not solving the problem = 1 - 1/3 = 2/3

Probability of 2nd student solving the problem = 1/4
Probability of 2nd student not solving the problem = 1 - 1/4 = 3/4

Probability of 3rd student solving the problem = 1/6
Probability of 3rd student not solving the problem = 1 - 1/6 = 5/6

Probability that none of the students solve the problem = (2/3) × (3/4) × (5/6)
= 5/12

∴ Probability that the problem will be solved = 1 - 5/12
= 7/12

∴ The probability that the problem will be solved is 7/12.

২,৬৩৮.

In a graph there are two curves, y₁ = 2x₁ - 5 and y₂ = - x₂ + 10. y₂ will be greater than y₁ when -

ক
ক) x > 5
খ
খ) x < 5
গ
গ) - 1 < x
ঘ
ঘ) x < 9

ব্যাখ্যা

Given,
y₁ = 2x - 5 and y₂ = - x + 10
As, y₂ > y₁
Or, - x + 10 > 2x - 5
Or, 15 > 3x
Or, 3x < 15
Or, x<5

২,৬৩৯.

The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?

ক
66, 77
খ
70, 84
গ
94, 108
ঘ
66, 106
ঙ
84, 96

ব্যাখ্যা

Question: The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?

Solution:
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fifth option satisfies.

84, 96
96 - 84 = 12
HCF of 84 and 96 is 12

২,৬৪০.

If sin x = 3/4, then cos x = ?

ক
ক) 2/3
খ
খ) √2/3
গ
গ) √7/4
ঘ
ঘ) 1/2

ব্যাখ্যা

cosx
= √(1 - sin²x)
= √{1 - (3/4)²}
= √(7/16)
= √7/4

২,৬৪১.

An aeroplane covers a certain distance at a speed of 400 kmph in 3 hours. To cover the same distance in 3/2 hours, it must travel at a speed of:

ক
650 kmph
খ
720 kmph
গ
800 kmph
ঘ
1000 kmph
ঙ
none of these

ব্যাখ্যা

Question: An aeroplane covers a certain distance at a speed of 400 kmph in 3 hours. To cover the same distance in 3/2 hours, it must travel at a speed of:

Solution:
দেওয়া আছে, প্রথম ক্ষেত্রে গতিবেগ = 400 kmph এবং সময় = 3 hours

আমরা জানি, দূরত্ব = গতিবেগ × সময়
∴ দূরত্ব = 400 × 3 = 1200 km

আবার, দ্বিতীয় ক্ষেত্রে অতিক্রান্ত দূরত্ব একই থাকবে।
দূরত্ব = 1200 km এবং সময় = 3/2 hours

আমরা জানি, গতিবেগ = দূরত্ব/সময়
= 1200/(3/2)
= (1200 × 2)/3
= 400 × 2
= 800 kmph

∴ উড়োজাহাজটিকে 800 kmph গতিবেগে চলতে হবে।

২,৬৪২.

If 2x + 3y = 24 and y = 2x, then find y - x.

ক
3
খ
4
গ
5
ঘ
6

ব্যাখ্যা

Question: If 2x + 3y = 24 and y = 2x, then find y - x.

Solution:
দেওয়া আছে, 2x + 3y = 24 এবং y = 2x
⇒ 2x + 3(2x) = 24
⇒ 2x + 6x = 24
⇒ 8x = 24
⇒ x = 3

∴ y = 2×3 = 6
তাহলে, y - x = 6 - 3 = 3

২,৬৪৩.

The sum of two integers is 55, and the difference is 9. What is the bigger of the two numbers?

ক
ক) 21
খ
খ) 23
গ
গ) 29
ঘ
ঘ) 32

ব্যাখ্যা

let, the smaller number be x and bigger number be y
So, y + x = 55 ..... (i) and y - x = 9 ...... (ii)
by solving equation (i) and (ii) we get y = 32 and x = 23

২,৬৪৪.

The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time do they meet?

ক
10 a.m
খ
10:30 a.m
গ
11 a.m
ঘ
11:30 a.m

ব্যাখ্যা

Question: The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time do they meet?

Solution:
Suppose they meet x hrs after 8 a.m
then,
[Distance moved by first in x hrs] + [Distance moved by second in (x - 1) hrs] = 330.
Therefore, 60x + 75(x - 1) = 330
⇒ 60x + 75x - 75 = 330
⇒ 135x = 405
∴ x = 3

So,they meet at (8 + 3) = 11 a.m.

২,৬৪৫.

The area of a square inscribed in a circle is 140 cm². What is the area of the semi-circle?

ক
220 cm²
খ
100 cm²
গ
210 cm²
ঘ
110 cm²

ব্যাখ্যা

Question: The area of a square inscribed in a circle is 140 cm². What is the area of the semi-circle?

Solution:
The area of a square inscribed in a circle is 140 cm²
side of square = √140 cm
= 2√35 cm

diagonal of the square = √2 × 2√35
= 2√70 cm

diameter of circle = 2√70 cm
radius of the circle = √70 cm
∴ area of the circle = π (√70)2 cm²
= (22/7) × 70 cm²
= 220 cm²

area of semi-circle = 220/2
= 110 cm²

২,৬৪৬.

secA - tanA = 3/5 , then secA + tanA - (5/3) =?

ক
5/3
খ
3
গ
1/2
ঘ
0

ব্যাখ্যা

প্রশ্ন: secA - tanA = 3/5 , then secA + tanA - (5/3) =?

সমাধান:
আমরা জানি,
sec²A - tan²A = 1
⇒ (secA - tanA ) (secA + tanA) = 1
⇒ (3/5) (secA + tanA) = 1
⇒ secA + tanA = 5/3

∴ secA + tanA - (5/3)
= (5/3) - (5/3)
= 0

২,৬৪৭.

What is the probability that a randomly chosen 4-digit number has all distinct digits?

ক
63/125
খ
423/500
গ
18/25
ঘ
89/100

ব্যাখ্যা

Question: What is the probability that a randomly chosen 4-digit number has all distinct digits?

Solution:
Total 4-digit numbers = 9999 − 1000 + 1 = 9000

A 4-digit number has digits in the form: ABCD, where:

A (thousands place): Can be 1-9 (9 options, cannot be 0).

B (hundreds place): Can be 0-9, except A (9 options).

C (tens place): Can be 0-9, except A and B (8 options).

D (units place): Can be 0-9, except A, B, and C (7 options).

Total numbers with distinct digits = 9 × 9 × 8 × 7 = 4536

∴ Probability = 4536/9000 = 63/125

২,৬৪৮.

A cistern can be filled by a tap in 6 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time the cistern will get filled?

ক
18 hours
খ
20 hours
গ
16 hours
ঘ
12 hours

ব্যাখ্যা

Question: A cistern can be filled by a tap in 6 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time the cistern will get filled?

Solution:
The cistern fill in 1 hour = ( 1/6 ) - ( 1/9 ) part = 1/18 part
The cistern fill 1/18 part = 1 hour
The cistern fill full = ( 1 × 18 ) /1 hour = 18 hours

২,৬৪৯.

By selling a book for Tk 525, a seller incurs a loss equal to 2/5 of his cost price. If he sells it for Tk 950, what is his gain or loss percentage?

ক
8.57% gain
খ
10% loss
গ
12.67% gain
ঘ
13.45% loss
ঙ
No gain no loss

ব্যাখ্যা

Question: By selling a book for Tk 525, a seller incurs a loss equal to 2/5 of his cost price. If he sells it for Tk 950, what is his gain or loss percentage?

Solution:
Let cost price = x Tk
If the book’s selling price is 525 Tk, then loss = (x - 525) Tk

∴ x - 525 = x × (2/5)
⇒ 5(x - 525) = 2x
⇒ 5x - 2625 = 2x
⇒ 3x = 2625
⇒ x = 875

If the book’s selling price is 950 Tk, then gain = (950 - 875) = 75 Tk

∴ Gain percentage = (75 × 100)/875 = 8.57%

২,৬৫০.

∠P এবং ∠Q পরস্পর পূরক কোণ। যদি ∠P = ২০° + ৪ক এবং ∠Q = ৬ক হয়, তবে ∠Q এর মান কত?

ক
৪০°
খ
৫০°
গ
৫৫°
ঘ
৪২°
ঙ
কোনোটিই নয়

ব্যাখ্যা

প্রশ্ন: ∠P এবং ∠Q পরস্পর পূরক কোণ। যদি ∠P = ২০° + ৪ক এবং ∠Q = ৬ক হয়, তবে ∠Q এর মান কত?

সমাধান:
এখানে,
∠P = ২০° + ৪ক এবং ∠Q = ৬ক

পূরক কোণের ক্ষেত্রে,
∠P + ∠Q = ৯০°
⇒ (২০° + ৪ক) + ৬ক = ৯০°
⇒ ২০° + ৪ক + ৬ক = ৯০°
⇒ ২০° + ১০ক = ৯০°
⇒ ১০ক = ৯০° − ২০°
⇒ ১০ক = ৭০°
∴ ক = ৭°

∴ ∠Q = ৬ × ৭° = ৪২°

২,৬৫১.

In a river flowing at 2 km/hr, a boat travels 32 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr, find the total journey time.

ক
16 hours
খ
14 hours
গ
10 hours
ঘ
12 hours

ব্যাখ্যা

speed of the boat = 6 km/hr
Speed downstream = (6+2) = 8 km/hr
Speed upstream = (6-2) = 4 km/hr
Distance travelled downstream = Distance travelled upstream = 32 km
Total time taken
= Time taken downstream + Time taken upstream
= 32/8 + 32/4
= 4 + 8
= 12 hours.

২,৬৫২.

Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2 men. In how many different ways can it be done?

ক
10
খ
20
গ
30
ঘ
23
ঙ
None of these

ব্যাখ্যা

Question: Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2 men. In how many different ways can it be done?

Solution:
Number of selections = Number of ways of selecting 2 men out of 5 men × Number of ways of selecting 1 woman out of 3 women
= ⁵C₂× ³C₁
= 10 × 3
= 30

২,৬৫৩.

ক
ক) 9
খ
খ) 16
গ
গ) 25
ঘ
ঘ) 36

ব্যাখ্যা

Question:

Solution:
√{1 + (x/144)} = 13/12
⇒ [√{1 + (x/144)}]² = (13/12)²
⇒ 1 + (x/144) = 169/144
⇒ x/144 = (169/144) - 1
⇒ x/144 = 25/144
∴ x = 25

২,৬৫৪.

The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

ক
30, 40
খ
25, 30
গ
30, 35
ঘ
35, 40

ব্যাখ্যা

Question: The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

Solution:
As the ratio of their present ages is 3 : 4 ,
let their present ages be 3X and 4X.
So, 5 years ago, as the ratio of their ages was 5 : 7,
we can write,
(3x - 5) : (4x - 5) = 5 : 7
⇒ (3x - 5)/(4x - 5) = 5/7
⇒ 21x - 35 = 20x - 25
⇒ x = 10

Hence, their present ages are 3X = 30 and 4X = 40

২,৬৫৫.

The ages of Nishi and Rimi are in the ratio 6 : 5 respectively. After 9 years, the ratio of their ages will be 9 : 8. What is the difference in their ages now ?

ক
3 years
খ
5 years
গ
7 years
ঘ
9 years

ব্যাখ্যা

Question: The ages of Nishi and Rimi are in the ratio 6 : 5 respectively. After 9 years, the ratio of their ages will be 9 : 8. What is the difference in their ages now?

Solution:
Let,
Nishi’s age be 6x years.
Then, Rimi’s age = 5x years.

Now
∴ (6x + 9)/(5x + 9) = 9/8
⇒ 8(6x + 9) = 9(5x + 9)
⇒ 48x - 45x = 81 - 72
⇒ 3x = 9
⇒ x = 3

∴ Difference in their ages = (6x - 5x) years
= x years
= 3 years.

২,৬৫৬.

A man completes a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

ক
220 km
খ
224 km
গ
230 km
ঘ
234 km
ঙ
236km

ব্যাখ্যা

(1/2)x/21 + (1/2)x/24 = 10
⇒ x/21 + x/24 = 20
⇒ 15x = 168 x 20
∴ x = 224 km

২,৬৫৭.

A coin is tossed four times. What is the probability of getting head on all tosses?

ক
ক) 1/4
খ
খ) 1/8
গ
গ) 1/16
ঘ
ঘ) 3/16

ব্যাখ্যা

Probability of getting head on all tosses = 1/2 × 1/2 × 1/2 × 1/2 = 1/16

২,৬৫৮.

In an election 10% of the voters on the voters' list did not cast their votes and 60 voters cast their ballot papers blank. There were only two candidates. The winner was supported by 47% of all the voters in the list and he got 308 votes more than his rival. The number of voters on the list was?

ক
5530
খ
5800
গ
6200
ঘ
6445

ব্যাখ্যা

Question: In an election 10% of the voters on the voters' list did not cast their votes and 60 voters cast their ballot papers blank. There were only two candidates. The winner was supported by 47% of all the voters in the list and he got 308 votes more than his rival. The number of voters on the list was?

Solution:
Let, total voters in the list = x
Votes got by the winner = 47x/100

Votes got by the loser,
x - (x/10) - 60 - (47x/100)
= (9x/10) - (47x/100) - 60
= (90x - 47x)/100 - 60
= (43x/100) - 60

ATQ,
(47x/100) - (43x/100) + 60 = 308
⇒ 4x/100 = 308 - 60
⇒ 4x/100 = 248
⇒ 4x = 248 × 100
⇒ x = (248 × 100)/4
∴ x = 6200

২,৬৫৯.

The HCF and LCM of two numbers are 24 and 168 and the numbers are in the ratio 1 ∶ 7. Find the greater of the two numbers.

ক
ক) 168
খ
খ) 250
গ
গ) 124
ঘ
ঘ) 136

ব্যাখ্যা

Given that
HCF = 24
LCM = 168
Ratio of numbers = 1 ∶ 7.
Let numbers be x and 7x.

We know
Product of numbers = LCM × HCF
x × 7x = 24 × 168
⇒ x² = 24 × 24
⇒ x = 24
∴ Larger number = 7x
= 24 × 7
= 168

২,৬৬০.

The 4th term of a geometric sequence is 81 and the 2nd term is 9. What is the common ratio?

ক
3
খ
5
গ
9
ঘ
6
ঙ
None of these

ব্যাখ্যা

Question: The 4th term of a geometric sequence is 81, and the 2nd term is 9. What is the common ratio?

Solution:
Given,
The 2nd term of a geometric sequence, a₂ = 9
The 4th term of the same sequence, a₄ = 81

In a geometric sequence we know,
a_n = ar^{n - 1}

So,
a₂= ar^{2 - 1}= ar .......(1)
And
a₄ = ar^{4 - 1}= ar³........ (2)

Now (2) ÷ (1),
ar³/ar = 81/9
⇒ r² = 9
⇒ r = √9
∴ r = 3

Since all terms are positive (2nd term is 9 and 4th is 81), the common ratio is 3.

২,৬৬১.

The present ages of Rahim and Sakib are in the ratio 4 : 5. After 6 years, the ratio of their ages will be 5 : 6. What is the difference in their present ages?

ক
10 years
খ
8 years
গ
6 years
ঘ
12 years

ব্যাখ্যা

Question: The present ages of Rahim and Sakib are in the ratio 4 : 5. After 6 years, the ratio of their ages will be 5 : 6. What is the difference in their present ages?

Solution:
Let, their present ages be 4x and 5x.

After 6 years,
Rahim's age = 4x + 6
Sakib's age = 5x + 6

According to the question,
(4x + 6)/(5x + 6) = 5/6
⇒ 6(4x + 6) = 5(5x + 6)
⇒ 24x + 36 = 25x + 30
⇒ 25x - 24x = 36 - 30
⇒ x = 6

Rahim's present age = 4 × 6 = 24 years
Sakib's present age = 5 × 6 = 30 years

∴ Difference = 30 - 24 = 6 years

২,৬৬২.

The mode and mean is given by 8 and 9, respectively. Then the median is-

ক
22/9
খ
26/3
গ
17/3
ঘ
72/17

ব্যাখ্যা

প্রশ্ন: The mode and mean is given by 8 and 9, respectively. Then the median is-

সমাধান:

We know from Empirical formula,
3Median = 2Mean + Mode
⇒ Median = (2Mean + Mode)/3
= {(2 × 9) + 8}/3
= 26/3

২,৬৬৩.

30% of a number when subtracted from 91, gives the number itself. Find the number?

ক
60
খ
65
গ
75
ঘ
70

ব্যাখ্যা

Let the number be x
According to the question,

91 - (30/100) = x
⇒ 9100 - 30x = 100x
⇒ 9100 = 130x
⇒ x = 9100/130
Hence, x = 70.

২,৬৬৪.

By how much percent is four-fifths of 70 lesser than five-sevenths of 112?

ক
12%
খ
24%
গ
36%
ঘ
40%
ঙ
None

ব্যাখ্যা

Question: By how much percent is four-fifths of 70 lesser than five-sevenths of 112?

Solution:

Here,
four-fifth of 70

= 4/5 × 70

= 56

Again,

five-seventh of 112

= 5/7 × 112

= 80

% lesser

= [(80 - 56)/80] × 100%

= (24/80) × 100%

= 30%

২,৬৬৫.

Find the ratio in which rice at Tk. 7.20 a kg be mixed with rice at Tk. 5.70 a kg to produce a mixture worth Tk. 6.30 a kg.

ক
1 : 3
খ
2 : 3
গ
3 : 4
ঘ
4 : 5

ব্যাখ্যা

Question: Find the ratio in which rice at Tk. 7.20 a kg be mixed with rice at Tk. 5.70 a kg to produce a mixture worth Tk. 6.30 a kg.

Solution:
Quantity of rice at Tk. 7.20 per kg = x
Quantity of rice at Tk. 5.70 per kg = y
By the rule of alligation:

⇒ x : y = (7.20 - 6.30) : (6.30 - 5.70) = 0.9 : 0.6 = 9 : 6 = 3 : 2

২,৬৬৬.

A merchant has 500 kg of sugar, part of which he sells at 8% and the remaining at 18% profit. He gains 14% on the whole. Find the quantity of sugar that he sold at 8% profit.

ক
300 kg
খ
250 kg
গ
200 kg
ঘ
150 kg

ব্যাখ্যা

Question: A merchant has 500 kg of sugar, part of which he sells at 8% and the remaining at 18% profit. He gains 14% on the whole. Find the quantity of sugar that he sold at 8% profit.

Solution:
Let,
The cost price of sugar be tk. x per kg
∴ Total cost price = Tk. 500x
∴ Total sell price = 500x + {500x × (14/100)}
= 500x + 70x
= 570x

The sugar sold at 8% gain by y kg
The sugar sold at 18% gain by (500 - y) kg

Now
{(108xy)/100} + [{118x(500 - y)}/100] = 570x
⇒ (108xy)/100 + (59000x - 118xy)/100 = 570x
⇒ 108xy + 59000x - 118xy =57000x
⇒ 10xy = 2000x
⇒ y = 2000/10
y = 200

∴ The sugar sold at 8% gain by 200 kg

২,৬৬৭.

A student attempts x number of questions. He answers 15 correctly out of the first 20 questions and of the remaining questions, he answers 1/3 correctly. If all questions have the same credit and the student gets 50% marks, then find the value of x.

ক
ক) 30
খ
খ) 35
গ
গ) 45
ঘ
ঘ) 50

ব্যাখ্যা

1) Student attempts x questions.
2) Out of 20 questions he answered 15 correctly and of (x – 20) questions he answered 1/3 correctly.
3) The student gets 50% marks.
Therefore,
15 + 1/3(x – 20)= 50% of x
⇒ 15 + 1/3(x – 20) = (50/100) × x
⇒ 15 + 1/3(x – 20) = x/2
⇒ 90 + 2(x - 20) = 3x
⇒ x = 50
Hence, the number of questions attempted by the students = 50.

২,৬৬৮.

The distance between two stations is 240 km. When it strikes 5 pm in the clock, a train starts from each of these stations and travels towards the other one. They meet at a junction after 12 hrs. One of the trains is slower to the other one by 14km/hr. Find the speed of the slower train.

ক
ক) 3 km/hr
খ
খ) 5 km/hr
গ
গ) 7 km/hr
ঘ
ঘ) 13 km/hr

ব্যাখ্যা

Let the speed of slower train = S km/hr
Speed of faster = (S + 14) km/hr
Trains meet after 12 hours.

Distance travelled by slower train in 12 hrs. = 12S
Distance travelled by faster train in 12 hrs. = 12(S + 14)

The total distance to be travelled between the two stations is given.
So, 12S + 12(S + 14) = 240
2S + 14 = 20
S = 3 km/hr.

Hence, The speed of the slower train is 3 km/hr.

২,৬৬৯.

If (a/b) + (b/a) = 4, find the value of (a³/b³) + (b³/a³).

ক
52
খ
64
গ
76
ঘ
112

ব্যাখ্যা

Question: If (a/b) + (b/a) = 4, find the value of (a³/b³) + (b³/a³).

Solution:
দেওয়া আছে, (a/b) + (b/a) = 4

প্রদত্ত রাশি = (a³/b³) + (b³/a³)
= (a/b)³ + (b/a)³= {(a/b) + (b/a)}³ - 3 (a/b)(b/a) . {(a/b) + (b/a)}
= 4³ - (3 × 1 × 4)
= 64 - 12
= 52

∴ প্রদত্ত রাশির মান 52

২,৬৭০.

A rectangular tank with a length of 4m and a width of 2m can store 20000 liters. What is the height of the tank?

ক
2 m
খ
1.5 m
গ
3.5 m
ঘ
2.5 m

ব্যাখ্যা

প্রশ্ন: A rectangular tank with a length of 4m and a width of 2m can store 20000 liters. What is the height of the tank?

Solution:
দেয়া আছে,
ট্যাংকের দৈর্ঘ্য (l) = 4 m, প্রস্থ (b) = 2 m, এবং আয়তন (V) = 20000 লিটার।

ধরি, ট্যাংকটির উচ্চতা হল h মিটার।

আমরা জানি,
আয়তাকার ঘনবস্তুর আয়তন, V = l × b × h ঘন একক
= (4 × 2 × h) m³
= 8h m³

এখন,
1 m³ = 1000 লিটার।

প্রশ্নমতে,
(8h × 1000) = 20000
বা, 8h = 20000/1000
বা, 8h = 20
∴ h = 2.5

সুতরাং, ট্যাংকটির উচ্চতা হল 2.5 মিটার।

২,৬৭১.

The area of a rectangle R with width 4 ft is equal to the area of a square S, which has a perimeter of 24 ft, the perimeter of the rectangle R, in feet, is

ক
ক) 9
খ
খ) 16
গ
গ) 24
ঘ
ঘ) 26

ব্যাখ্যা

Given,
Perimeter of the square = 24 ft
Length of the side of the square = 24/4 =6 ft
So, its area = 6² = 36 ft²

ATQ,
Area of the rectangle is, length × width = 36 ft²
⇒ length = 36/4 = 9 [As the rectangle's width is 4 ft]

∴ Perimeter of the rectangle = 2(9 + 4) = 26 ft

২,৬৭২.

log 2 + log 4 + log 8 + ............ Find the sum of the first 19th term-

ক
55 log 2
খ
120 log 2
গ
190 log 2
ঘ
210 log 2

ব্যাখ্যা

Question: log 2 + log 4 + log 8 + ............ Find the sum of the first 19th term-

Solution:
given that,
log2 + log4 +log8 + ............
= log 2¹+ log 2² + log 2³ + ............
= log 2 + 2 log 2 + 3 log 2 + ............
= (1 + 2 + 3 + .........) × log 2

The sum of the first 19 natural numbers is given by the formula:
Sum = n(n+1)/2
where n = 19
∴ Sum = 19(19 + 1)/2
= 19 × 10
= 190

So, the sum of the first 19 terms = 190 log 2

২,৬৭৩.

A cuboid has dimensions in the ratio 1:2:3 and a total surface area of 88 cm². What is its volume?

ক
24 cm³
খ
36 cm³
গ
48 cm³
ঘ
60 cm³

ব্যাখ্যা

Question: A cuboid has dimensions in the ratio 1:2:3 and a total surface area of 88 cm². What is its volume?

Solution:
Let the dimensions be x, 2x, and 3x.

Total surface area of the cube = 2(x.2x + 2x.3x + 3x.x)
= 22x²

According to the question,
22x² = 88
x² = 4
∴ x = 2

So, volume of the cuboid = 2 × 4 × 6
= 48 cm³

২,৬৭৪.

A merchant buys 30 kg of rice at Tk. 40/kg, and another 20 kg of rice at Tk. 30/kg. He mixes them and sells half of the mixture at Tk. 36/kg. At what price should he sell the remaining mixture to get an overall profit of 30%?

ক
Tk. 1460
খ
Tk. 1440
গ
Tk. 1420
ঘ
Tk. 1400

ব্যাখ্যা

Question: A merchant buys 30 kg of rice at Tk. 40/kg, and another 20 kg of rice at Tk. 30/kg. He mixes them and sells half of the mixture at Tk. 36/kg. At what price should he sell the remaining mixture to get an overall profit of 30%?

Solution:
Total cost for the entire quantity of rice = (30 × 40) + (20 × 30) = 1200 + 600 = 1800
If his profit is 30%, then the sales realization = (130/100) × 1800 = 2340.
Total rice = 30 + 20 = 50 kg
He sells 25 kg at Tk. 36/kg = 900.

Therefore to make the said amount of profit, he should sell the remaining 25 kg of rice at 2340 - 900 = 1440

২,৬৭৫.

Solution set of the inequality: 3y + 4 ≥ 2y - 5 is:

ক
[- 9, ∞)
খ
(- 9, ∞)
গ
(- 9, ∞]
ঘ
(9, - ∞)

ব্যাখ্যা

Question: Solution set of the inequality: 3y + 4 ≥ 2y - 5 is-

Solution:
Given that,
3y + 4 ≥ 2y - 5
⇒ 3y + 4 - 2y ≥ 2y - 5 - 2y
⇒ y + 4 ≥ - 5
⇒ y ≥ - 5 - 4
⇒ y ≥ - 9

∴ Solution set of the inequality is [- 9, ∞)

২,৬৭৬.

Two trains of equal length are running on parallel lines in the same direction at 50 km/hr and 60 km/hr. The faster train passes the slower train in 72 seconds. The length of each train is:

ক
150 m
খ
120 m
গ
80 m
ঘ
100 m

ব্যাখ্যা

Question: Two trains of equal length are running on parallel lines in the same direction at 50 km/hr and 60 km/hr. The faster train passes the slower train in 72 seconds. The length of each train is:

Solution:
Let the length of each train be x metres.
Then, distance covered = 2x metres.

Relative speed = (60 - 50) km/hr
= 10 km/hr
= (10 × 1000)/3600 m/s
= 25/9 m/s

Now,
2x/72 = 25/9
⇒ 18x = 1800
∴ x = 100

২,৬৭৭.

The difference between three times and seven times of a number comes to 36. What is the number?

ক
12
খ
11
গ
8
ঘ
7
ঙ
None of the above

ব্যাখ্যা

Question: The difference between three times and seven times of a number comes to 36. What is the number?

Solution:
Let the number be p

According to the question,
7p - 3p = 36
⇒ 4p = 36
⇒ p = 9

Hence, the required Number is 9

২,৬৭৮.

The ratio of A : B is 2 : 3, and the ratio of B : C is 4 : 5. If A 16, what is the value of C?

ক
15
খ
20
গ
24
ঘ
30

ব্যাখ্যা

Question: The ratio of A : B is 2 : 3, and the ratio of B : C is 4 : 5. If A 16, what is the value of C?

Solution:
Given that,
A : B = 2 : 3
B : C = 4 : 5
And A = 16

Now,
A : B = 2 : 3
⇒ A/B = 2/3
⇒ 2B = 3A
⇒ B = (16 × 3)/2 ; [A = 16]
∴ B = 24

And,
B : C = 4 : 5
B/C = 4/5
⇒ 24/C = 4/5
⇒ C = (24 × 5) / 4
∴ C = 30

So the value of C is 30.

২,৬৭৯.

A square that has a side length of 4m is reduced by 50% of its area. The new side length will be -

ক
ক) 2m
খ
খ) 3m
গ
গ) 2√3m
ঘ
ঘ) 2√2m

ব্যাখ্যা

Question: A square that has a side length of 4m is reduced by 50% of its area. The new side length will be -

Solution:
here, the side length is = 4m
the area of the square = 4² =16m²

after reducing the area by 50% the remaining area is = 16 - (50% of 16)
= 16 - 8 = 8m²

the new side length is = √8 = 2√2 m

২,৬৮০.

The value of - 5 - (- 10) is how much greater than the value of - 10 - (- 5)?

ক
14
খ
12
গ
10
ঘ
8

ব্যাখ্যা

Question: The value of - 5 - (- 10) is how much greater than the value of - 10 - (- 5)?

Solution:
- 5 -(- 10)
= - 5 + 10
= 5

And,
- 10 - (- 5)
= - 10 + 5
= - 5

Now,
5 - (- 5)
= 5 + 5
= 10

∴ The value of - 5 - (- 5) is 10 greater than the value of - 10 -(- 5)

২,৬৮১.

What is the value of cosA + secA is -

ক
ক) 2/5
খ
খ) 5/2
গ
গ) 1/2
ঘ
ঘ) 5/3

ব্যাখ্যা

Question:
What is the value of cosA + secA is -

Solution:
প্রদত্ত চিত্র হতে,
ΔABC সমকোণী ত্রিভুজ হতে,
অতিভূজ AC = 2 এবং ভূমি, BC = √3

আমরা জানি,
(লম্ব)² = (অতিভূজ)² - (লম্ব)²
বা, AC² = (2)² - (√3)²
বা, AC² = 4 - 3
∴ AC = 1

cosA = AB/AC = 1/2
secA = AC/AB = 2/1 = 2
∴ cosA + secA = 1/2 + 2 = 5/2

২,৬৮২.

A retailer bought a glass at wholesale and marked it up 80% to its initial retail price of Tk. 45. By how many more taka does he need to increase the price to achieve a 100% markup?

ক
Tk. 3
খ
Tk. 5
গ
Tk. 10
ঘ
None of these

ব্যাখ্যা

Question: A retailer bought a glass at wholesale and marked it up 80% to its initial retail price of Tk. 45. By how many more taka does he need to increase the price to achieve a 100% markup?

Solution:
let, wholesale price x

1.8x = 45
⇒ x = 45/1.8
∴ x = 25 taka

to achieve a 100% markup retail price = 25 + 25 = 50 taka

he needs to increase = 50 - 45 taka
= Tk. 5

২,৬৮৩.

secA + tanA = 4/3, then find, secA - tanA = ?

ক
4/5
খ
3/4
গ
5/12
ঘ
3/2

ব্যাখ্যা

Question: secA + tanA = 4/3, then find, secA - tanA = ?

Solution:
Given that,
secA + tanA = 4/3

We know,
sec²A - tan²A = 1
⇒ (secA + tanA)(secA - tanA) = 1
⇒ (4/3)(secA - tanA) = 1
⇒ (secA - tanA) = 1/(4/3)
∴ secA - tanA = 3/4

২,৬৮৪.

The average temperature for the first 4-days of a week is 40.2º C and that of the last 4-days is 41.3° C. If the average temperature for the whole week is 40.6° C, then temperature on the fourth day is-

ক
38.5° C
খ
41.8° C
গ
41.3° C
ঘ
40.8° C

ব্যাখ্যা

Question: The average temperature for the first 4-days of a week is 40.2º C and that of the last 4-days is 41.3° C. If the average temperature for the whole week is 40.6° C, then temperature on the fourth day is-

Solution:
দেওয়া আছে,
সপ্তাহের গড় তাপমাত্রা = 40.6° সেলসিয়াস
∴ সপ্তাহের মোট তাপমাত্রা = (40.6 × 7)° সেলসিয়াস
= 284.2° সেলসিয়াস।

আবার,
প্রথম 4 দিনের গড় তাপমাত্রা = 40.2º সেলসিয়াস
∴ প্রথম 4 দিনের মোট তাপমাত্রা = (40.2 × 4)° সেলসিয়াস
= 160.8° সেলসিয়াস।

শেষ 4 দিনের গড় তাপমাত্রা = 41.3° সেলসিয়াস
∴ শেষ 4 দিনের মোট তাপমাত্রা = (41.3 × 4)° সেলসিয়াস
= 165.2° সেলসিয়াস।

∴8 দিনের মোট তাপমাত্রা = 160.8° + 165.2° = 326° সেলসিয়াস

চতুর্থ দিনের তাপমাত্রা = (326° - 284.2°) = 41.8° সেলসিয়াস

২,৬৮৫.

Find the length of the altitude of an equilateral triangle of side 3√3 cm.

ক
6.5 cm
খ
5.5 cm
গ
4.5 cm
ঘ
7 cm

ব্যাখ্যা

Question: Find the length of the altitude of an equilateral triangle of side 3√3 cm.

Solution:
Given that,
Side of equilateral triangle = 3√3 cm

We know,
Altitude (height) of an equilateral triangle is,
h = (√3/2) × side
h = (√3/2) × 3√3
= (3√3 × √3)/2
= (3 × 3)/2
= 9/2
∴ h = 4.5 cm

So the length of the altitude is 4.5 cm or 9/2 cm.

২,৬৮৬.

An item is sold at 25% discount making a profit of 20%. If the discount is reduced to 15%, what will be the new profit percentage?

ক
32%
খ
46%
গ
24%
ঘ
36%

ব্যাখ্যা

Question: An item is sold at 25% discount making a profit of 20%. If the discount is reduced to 15%, what will be the new profit percentage?

Solution:
Let,
the cost price is 100
So, the selling price is 120

Giving 25% discount, then
⇒ x - 25% of x = 120
⇒ x - (x/4) = 120
⇒ 3x/4 = 120
∴ x = 160

Now,
giving 15% discount,
160 - 15% of 160 = 160 - 24 = 136
So the new selling price is 136 taka.

So profit = 136 - 100 = 36

When the discount is reduced to 15%, the profit becomes 36%.

২,৬৮৭.

0.95 expressed as a percent of 1.9 is-

ক
6%
খ
10%
গ
50%
ঘ
90%
ঙ
None

ব্যাখ্যা

Question: 0.95 expressed as a percent of 1.9 is-

Solution:
Required Percentage = (0.95 ×100)/1.9
= (0.95 × 100 × 10)/(19 × 100)
= 50%

২,৬৮৮.

A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. The number of days for which the remaining food will last, is:

ক
ক) 30 days
খ
খ) 42 days
গ
গ) 45 days
ঘ
ঘ) 52 days

ব্যাখ্যা

After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
125 : 150 :: 35 : x
125x = 150 × 35
x = 42

২,৬৮৯.

If x = 3 + 22, then find the value of √x - 1/√x.

ক
5/24
খ
5
গ
24/5
ঘ
25/4

ব্যাখ্যা

Question: If x = 3 + 22, then find the value of √x - 1/√x.

Solution:
Given,
x = 3 + 22
⇒ x = 25
⇒ 5² = x
⇒ 5 = √x
∴ √x = 5

Now,
√x - 1/√x
= 5 - 1/5
= (25 - 1)/5
= 24/5

২,৬৯০.

If b < 2 and 2x - 3b = 0, which of the following must be true?

ক
x > - 3
খ
x < 2
গ
x = 3
ঘ
x < 3
ঙ
x > 3

ব্যাখ্যা

Question: If b < 2 and 2x - 3b = 0, which of the following must be true?

Solution:
2x - 3b = 0
⇒ 3b = 2x
∴ b = (2x)/3

b < 2
∴ (2x)/3 < 2
⇒ 2x < 6
⇒ x < 3

২,৬৯১.

A man wanted to sell his bat at 5% discount. His brother who was a cricketer wanted to buy that bat so the man sells it at 8% discount. In this deal, the man makes Tk. 60/- less in profit. How much was the marked price of the bat?

ক
ক) Tk. 1800
খ
খ) Tk. 2000
গ
গ) Tk. 6000
ঘ
ঘ) Tk. 9000

ব্যাখ্যা

Difference in discount = 8% - 5% = 3%
Due to this 3% man makes Tk. 60 less in profit
That means 3% of Marked Price = 60
(3/100) × M.P. = 60
∴ M.P. = Tk. 2000

২,৬৯২.

At what angle are the hands of a clock inclined at 10 minutes past 5?

ক
85°
খ
105°
গ
90°
ঘ
100°
ঙ
95°

ব্যাখ্যা

Question: At what angle are the hands of a clock inclined at 10 minutes past 5?

Solution:
The standard formula for the smaller angle θ between the hour and minute hands is.
θ = {|60H - 11M|}/2 ; [where H = hour, M = minutes]
= {|(60 × 5) - (11 × 10)|}/2
= |300 - 110|/2
= 190/2
= 95°

So the hands of the clock are inclined at 95° at 10 minutes past 5.

২,৬৯৩.

The speed of a boat in still water is 10 km/h. The time it takes to travel downstream is one-third the time it takes to travel upstream. What is the speed of the stream?

ক
3 km/h
খ
5 km/h
গ
6 km/h
ঘ
4 km/h

ব্যাখ্যা

Question: The speed of a boat in still water is 10 km/h. The time it takes to travel downstream is one-third the time it takes to travel upstream. What is the speed of the stream?

Solution:
Let the speed of the current be = x km/h

Then,
Downstream speed = (10 + x) km/h
Upstream speed = (10 − x) km/h

We know, time = distance/speed

According to the question:
distance/(10 + x) = distance/{3 × (10 - x)}
⇒ (10 + x) = 3(10 - x)
⇒ 4x = 20
⇒ x = 5
∴ The speed of the current = 5 km/h

২,৬৯৪.

If m is an integer such that (- 2)^2m = 2^{9 - m}, then what is the value of m?

ক
3
খ
- 3
গ
1/2
ঘ
4

ব্যাখ্যা

Question: If m is an integer such that (- 2)^2m = 2^{9 - m}, then what is the value of m?

Solution:
দেওয়া আছে,
(- 2)^2m = 2^{9 - m}
⇒ 2^2m = 2^{9 - m}
কোনো ঋণাত্মক সংখ্যার Power যদি জোড় সংখ্যা হয় তবে সংখ্যাটি ধনাত্মক হবে। আর যদি Power বিজোড় হয় তবে সংখ্যাটি ঋণাত্মক হবে। যেহেতু m একটি পূর্ণ সংখ্যা সেহেতু 2m একটি জোড় সংখ্যা। তাই (- 2)^2m সংখ্যাটি ধনাত্মক সংখ্যা হবে।

∴ 2m = 9 - m
⇒ 3m = 9
⇒ m = 9/3
⇒ m = 3

২,৬৯৫.

Find the LCM of the fractions 5/8, 3/4, 1/2.

ক
15
খ
40
গ
15/2
ঘ
15/8

ব্যাখ্যা

Question: Find the LCM of the fractions 5/8, 3/4, 1/2.

Solution:
The fractions are 5/8, 3/4 and 1/2
LCM of the fraction = LCM of numerator/HCF of denominator

LCM of numerators = LCM of (5, 3 , 1 = 5 × 3 × 1 = 15
HCF of denominators = HCF of ( 8, 4, 2) = 2

∴ LCM of fraction = 15/2

২,৬৯৬.

In a right triangle, the length of one of the legs is 8 and the length of the hypotenuse is 17. What is the length of the other leg?

ক
10
খ
12
গ
15
ঘ
18

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 8 and the length of the hypotenuse is 17. What is the length of the other leg?

Solution:
এখানে,
সমকোণী ত্রিভুজের (right triangle) অতিভুজ (hypotenuse)= 17 একক
সমকোণ সংলগ্ন এক বাহু = 8 একক
সমকোণ সংলগ্ন অপর বাহু = a একক

প্রশ্নমতে,
a² + 8² = 17²
⇒ a² + 64 = 289
⇒ a² = 289 - 64
⇒ a² = 225
⇒ a = √225
∴ a = 15

২,৬৯৭.

The average of 6 numbers is 9. The average of three numbers of them is 6 . What will be the average of remaining numbers?

ক
ক) 9
খ
খ) 10
গ
গ) 11
ঘ
ঘ) 12

ব্যাখ্যা

Average of 6 numbers = 9
Sum of 6 numbers = 6 × 9= 54
Average of three numbers = 6
Sum of three numbers = 6 × 3 = 18
∴ Sum of the remaining three numbers = 54 - 18 = 36
∴ Required average
= 36/3
= 12

২,৬৯৮.

57 is 75 percent of which number given below?

ক
ক) 65
খ
খ) 87
গ
গ) 76
ঘ
ঘ) 98

ব্যাখ্যা

ধরি ,
x এর 75% = 57
x এর 75/100 = 57
3x/4 = 57
x = (57 × 4)/3 = 76

২,৬৯৯.

If x + y : y + z : z + x = 6 : 7 : 8 and x + y + z = 14, find the value of z.

ক
5
খ
4
গ
6
ঘ
7

ব্যাখ্যা

Question: If x + y : y + z : z + x = 6 : 7 : 8 and x + y + z = 14, find the value of z.

Solution:
Given,
x + y : y + z : z + x = 6 : 7 : 8
x + y + z = 14...........(i)

Let,
x + y = 6k, y + z = 7k, z + x = 8k

adding them we get,
2( x + y + z) = 21k
or, 2 × 14 = 21k
or, k = (2 × 14)/21
∴ k = 4/3

as, x + y = 6k
or, x + y = 6(4/3)
∴ x + y = 8

from (i) we get,
x + y + z = 14
8 + z = 14
z = 14 - 8
z = 6

২,৭০০.

An employee's annual salary was increased Tk.30,000. If her new annual salary now equals Tk. 105,000 what was the percent increase?

ক
ক) 20%
খ
খ) 40%
গ
গ) 30%
ঘ
ঘ) None of the above

ব্যাখ্যা

প্রশ্ন : An employee's annual salary was increased Tk.30,000. If her new annual salary now equals Tk. 105,000 what was the percent increase?

সমাধান :

New annual salary = 105,000
Salary increase = 30,000.
Original salary = 105,000 - 30,000 = 75,000

% Increase = (30,000/ 75,000 )×100 = 40%