১০,৩০১.
In a railway compartment 6 seats are vacant on a bench. In how many ways can 3 passengers sit on them?
ব্যাখ্যা
6 সিটে 3 জন যাত্রীকে বসানো যাবে
6P3 = 6 × 5 × 4 = 120 উপায়ে।
6P3 = 6 × 5 × 4 = 120 উপায়ে।
PrepBank · বিষয়ভিত্তিক প্রশ্ন
Bank Math
Bank Math
PrepBank · পাতা ১০৪ / ১৬১ · ১০,৩০১–১০,৪০০ / ১৬,১২৪
১০,৩০২.
A man deposits certain amount in his bank account. After a few days. he withdraws half of the money deposited and deposits Tk. 500 more. If he has a balance of Tk. 2000 in his bank account, find the amount deposited initially.
ব্যাখ্যা
Question: A man deposits certain amount in his bank account. After a few days. he withdraws half of the money deposited and deposits Tk. 500 more. If he has a balance of Tk. 2000 in his bank account, find the amount deposited initially.
Solution:
ধরি,
প্রথমে সে বিনিয়োগ করে ক টাকা
শর্তমতে,
ক - (ক/২) + ৫০০ = ২০০০
⇒ ক - (ক/২) = ১৫০০
⇒ ক/২ = ১৫০০
∴ ক = ৩০০০
Solution:
ধরি,
প্রথমে সে বিনিয়োগ করে ক টাকা
শর্তমতে,
ক - (ক/২) + ৫০০ = ২০০০
⇒ ক - (ক/২) = ১৫০০
⇒ ক/২ = ১৫০০
∴ ক = ৩০০০
১০,৩০৩.
A mixture of 20 kg of spirit and water contains 10% water. How much water must be added to this mixture to raise the percentage of water to 25%?
ব্যাখ্যা
In 1st mixture, water = 10/100 × 20 = 2 kg
So, Spirit = 20-2 = 18 kg
In 2nd mixture where the water is 25%,
75 kg of spirit is contained in 100 kg mixture
So, 18 kg spirit is contained in = (100×18)/75 = 24 kg
So, water to be added = 24-20 = 4 kg
১০,৩০৪.
Aman and Ajay started a business by investing Tk. 85000 and Tk. 15000 each. What be will the ratio of profit earned after 2 years between Aman and Ajay respectively?
ব্যাখ্যা
Question: Aman and Ajay started a business by investing Tk. 85000 and Tk. 15000 each. What be will the ratio of profit earned after 2 years between Aman and Ajay respectively?
Solution:
the ratio of profit earned after 2 years between Aman and Ajay respectively = (85000 × 2) : (15000 × 2)
= 170000 : 30000
= 17 : 3
Solution:
the ratio of profit earned after 2 years between Aman and Ajay respectively = (85000 × 2) : (15000 × 2)
= 170000 : 30000
= 17 : 3
১০,৩০৫.
A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7?
ব্যাখ্যা
Question: A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7?
Solution:
Let X be the event that the number selected would be divisible by 3
and Y be the event that the selected number would be divisible by 7.
Then X ∪ Y denotes the event that the number would be divisible by 3 or 7.
Now,
X = {3, 6, 9, 12, 15, 18}
and Y = {7, 14}
∴ X ∪ Y = {3, 6, 7, 9, 12, 14, 15, 18}
∴ n(X ∪ Y) = 8
∴ The probability that it would be divisible by 3 or 7 = 8/20
= 2/5
Solution:
Let X be the event that the number selected would be divisible by 3
and Y be the event that the selected number would be divisible by 7.
Then X ∪ Y denotes the event that the number would be divisible by 3 or 7.
Now,
X = {3, 6, 9, 12, 15, 18}
and Y = {7, 14}
∴ X ∪ Y = {3, 6, 7, 9, 12, 14, 15, 18}
∴ n(X ∪ Y) = 8
∴ The probability that it would be divisible by 3 or 7 = 8/20
= 2/5
১০,৩০৬.
800 grams of sugar solution has 30% sugar in it. How much sugar should be added to make 50% in the solution?
ব্যাখ্যা
Question: 800 grams of sugar solution has 30% sugar in it. How much sugar should be added to make 50% in the solution?
Solution:
Amount of sugar = 800 × 30/100
= 240 grams
Let, x gm sugar to be added
ATQ,
(240 + x)/(800 + x) = 50%
⇒ (240 + x)/(800 + x) = = 1/2
⇒ 480 + 2x = 800 + x
⇒ 2x - x = 800 - 480
∴ x = 320 gm
Solution:
Amount of sugar = 800 × 30/100
= 240 grams
Let, x gm sugar to be added
ATQ,
(240 + x)/(800 + x) = 50%
⇒ (240 + x)/(800 + x) = = 1/2
⇒ 480 + 2x = 800 + x
⇒ 2x - x = 800 - 480
∴ x = 320 gm
১০,৩০৭.
If 2 < x < 5 and 3 < y < 5, which of the following best describes x - y?
ব্যাখ্যা
Question: If 2 < x < 5 and 3 < y < 5, which of the following best describes x - y?
Solution:
দেয়া আছে,
2 < x < 5
3 < y < 5
এখন, আমরা x - y এর সীমা বের করতে চাই। এর জন্য, y এর অসমতাকে - y এর অসমতায় রূপান্তর করতে হবে।
3 < y < 5
⇒ - 3 > - y > - 5 [- 1 দ্বারা গুণ করে]
⇒ - 5 < - y < - 3
এইবার x এবং - y এর অসমতা দুটি যোগ করি,
⇒ (2 < x < 5) + (- 5 < - y < - 3)
⇒ −3 < x - y < 2
১০,৩০৮.
Jahid ran a 2 mile race at an average speed of 8 miles per hour. If Alam ran the same race at an average speed of 6 miles per hour, how many minutes longer than Jahid did Alam take to complete the race?
ব্যাখ্যা
Question: Jahid ran a 2 mile race at an average speed of 8 miles per hour. If Alam ran the same race at an average speed of 6 miles per hour, how many minutes longer than Jahid did Alam take to complete the race?
Solution:
We know,
Time = Distance ÷ Speed
Jahid takes,
2 = Time × 8
⇒ Time = 1/4 hours
∴ Time = 15 minutes
Alam Takes,
2 = Time × 6
⇒ Time = 1/3 hours
∴ Time = 20 minutes
Difference = (20 - 15) = 5 minutes
Solution:
We know,
Time = Distance ÷ Speed
Jahid takes,
2 = Time × 8
⇒ Time = 1/4 hours
∴ Time = 15 minutes
Alam Takes,
2 = Time × 6
⇒ Time = 1/3 hours
∴ Time = 20 minutes
Difference = (20 - 15) = 5 minutes
১০,৩০৯.
If 1/y = 7/2 then 1/(y + 2) = ?
ব্যাখ্যা
দেয়া আছে,
1/y = 7/2
y = 2/7
1/(y + 2) =1/{(2/7) + 2}
= 1/{(2 + 14)/7}
= 1/(16/7)
= 7/16
1/y = 7/2
y = 2/7
1/(y + 2) =1/{(2/7) + 2}
= 1/{(2 + 14)/7}
= 1/(16/7)
= 7/16
১০,৩১০.
The angle between the minute hand and the hour hand of a clock when the time is 4:20, is -
ব্যাখ্যা
Question: The angle between the minute hand and the hour hand of a clock when the time is 4:20, is -
Solution:
কোণ = ।11M - 60H।/2
= ।(11 × 20) - (60 × 4)।/2
= ।220 - 240। /2
= 10°
Solution:
কোণ = ।11M - 60H।/2
= ।(11 × 20) - (60 × 4)।/2
= ।220 - 240। /2
= 10°
১০,৩১১.
a2 - a - 12 = 0, then a = ?
ব্যাখ্যা
a2 - a - 12 = 0
a2 - 4a + 3a - 12 = 0
a(a - 4) + 3(a - 4) = 0
(a - 4)(a + 3) = 0
a = 4 , - 3
a2 - 4a + 3a - 12 = 0
a(a - 4) + 3(a - 4) = 0
(a - 4)(a + 3) = 0
a = 4 , - 3
১০,৩১২.
Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is-
ব্যাখ্যা
Question: Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is-
Solution:
Let the third number be x
Then, first number = 120% of x = 120x/100 = 6x/5
Second number = 150% of x = 150x/100 = 3x/2
∴ Ratio of first two numbers = 6x/5 : 3x/2
= 12x : 15x
= 4 : 5
Solution:
Let the third number be x
Then, first number = 120% of x = 120x/100 = 6x/5
Second number = 150% of x = 150x/100 = 3x/2
∴ Ratio of first two numbers = 6x/5 : 3x/2
= 12x : 15x
= 4 : 5
১০,৩১৩.
For which value of p will the square root of 4x2 - px + 9 be an integer?
ব্যাখ্যা
Question: For which value of p will the square root of 4x2 - px + 9 be an integer?
Solution:
4x2 - px + 9
= (2x)2 - 2.2x.3 + 32 - px + 2.2x.3
= (2x - 3)2 + 12x - px
রাশিটি পূর্ণবর্গ হলে,
12x - px = 0
বা, px = 12x
∴ p = 12
Solution:
4x2 - px + 9
= (2x)2 - 2.2x.3 + 32 - px + 2.2x.3
= (2x - 3)2 + 12x - px
রাশিটি পূর্ণবর্গ হলে,
12x - px = 0
বা, px = 12x
∴ p = 12
১০,৩১৪.
A man rows to a place 30 km distant and comes back in 8 hours. He finds that he can row 5 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
ব্যাখ্যা
Question: A man rows to a place 30 km distant and comes back in 8 hours. He finds that he can row 5 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
সমাধান:
ধরি,
লোকটি স্রোতের অনুকূলে 5 কিমি এবং স্রোতের প্রতিকূলে 3 কিমি যেতে x ঘন্টা সময় নেয়।
∴ স্রোতের অনুকূলে গতিবেগ = 5/x কিমি/ঘন্টা।
স্রোতের প্রতিকূলে গতিবেগ = 3/x কিমি/ঘন্টা।
এখন,
মোট সময় = স্রোতের অনুকূলে যাওয়ার সময় + স্রোতের প্রতিকূলে আসার সময়
⇒ 8 = {30/(5/x)} + {30/(3/x)} [সময় = দূরত্ব/বেগ]
⇒ 8 = (30x/5) + (30x/3)
⇒ 8 = 6x + 10x
⇒ 8 = 16x
⇒ x = 8/16
⇒ x = 1/2
তাহলে, স্রোতের অনুকূলে গতিবেগ = 5/(1/2) = 10 কিমি/ঘন্টা।
স্রোতের প্রতিকূলে গতিবেগ = 3/(1/2) = 6 কিমি/ঘন্টা।
স্রোতের গতিবেগ = (স্রোতের অনুকূলে গতিবেগ - স্রোতের প্রতিকূলে গতিবেগ)/2
= (10 - 6)/2 কিমি/ঘন্টা
= 4/2 কিমি/ঘন্টা
= 2 কিমি/ঘন্টা
১০,৩১৫.
The first number is 5% greater and the second number is 25% greater than a third number. What is the ratio of first two numbers?
ব্যাখ্যা
Question: The first number is 5% greater and the second number is 25% greater than a third number. What is the ratio of first two numbers?
Solution:
Let
the third number be x
Then, first number = 105% of x
= 105x/100
= 21x/20
Second number = 125% of x
= 125x/100
= 5x/4
∴ Ratio of first two numbers = 21x/20 : 5x/4
= (21x × 4) : (5x × 20)
= 21x : 25x
= 21 : 25
Solution:
Let
the third number be x
Then, first number = 105% of x
= 105x/100
= 21x/20
Second number = 125% of x
= 125x/100
= 5x/4
∴ Ratio of first two numbers = 21x/20 : 5x/4
= (21x × 4) : (5x × 20)
= 21x : 25x
= 21 : 25
১০,৩১৬.
A fair die is thrown once. What is the probability of getting a prime number?
ব্যাখ্যা
Question: A fair die is thrown once. What is the probability of getting a prime number?
Solution:
A standard fair die has 6 faces numbered: 1, 2, 3, 4, 5, 6
The prime numbers from 1 to 6 are: 2, 3, 5
Probability = (Number of favorable outcomes) / (Total number of outcomes)
= 3/6
= 1/2
Solution:
A standard fair die has 6 faces numbered: 1, 2, 3, 4, 5, 6
The prime numbers from 1 to 6 are: 2, 3, 5
Probability = (Number of favorable outcomes) / (Total number of outcomes)
= 3/6
= 1/2
১০,৩১৭.
log10{(x + y)/4} = (1/2)(log10x + log10y) হলে (x/y) + (y/x) এর মান কত?
ব্যাখ্যা
প্রশ্ন: log10{(x + y)/4} = (1/2)(log10x + log10y) হলে (x/y) + (y/x) এর মান কত?
সমাধান:
log10{(x + y)/4} = (1/2)(log10x + log10y)
⇒ log10{(x + y)/4} = (1/2)log10(xy)
⇒ log10{(x + y)/4} = log10(xy)(1/2)
⇒ (x + y)/4 = (xy)(1/2)
⇒ {(x + y)/4}2 = {(xy)(1/2)}2
⇒(x + y)2/16 = xy
⇒ x2 + 2xy + y2 = 16xy
⇒ x2 + y2 = 16xy - 2xy
⇒ x2 + y2 = 14xy
⇒ (x2/xy) + (y2/xy) = 14
∴ (x/y) + (y/x) = 14
সমাধান:
log10{(x + y)/4} = (1/2)(log10x + log10y)
⇒ log10{(x + y)/4} = (1/2)log10(xy)
⇒ log10{(x + y)/4} = log10(xy)(1/2)
⇒ (x + y)/4 = (xy)(1/2)
⇒ {(x + y)/4}2 = {(xy)(1/2)}2
⇒(x + y)2/16 = xy
⇒ x2 + 2xy + y2 = 16xy
⇒ x2 + y2 = 16xy - 2xy
⇒ x2 + y2 = 14xy
⇒ (x2/xy) + (y2/xy) = 14
∴ (x/y) + (y/x) = 14
১০,৩১৮.
Two similar triangles have areas in the ratio 4 : 9. If the perimeter of the smaller triangle is 20, what is the perimeter of the larger triangle?
ব্যাখ্যা
Question: Two similar triangles have areas in the ratio 4 : 9. If the perimeter of the smaller triangle is 20, what is the perimeter of the larger triangle?
Solution:
Given that,
Two similar triangles
Areas ratio = 4 : 9
Perimeter of smaller triangle = 20
For similar triangles, the ratio of areas = square of the ratio of corresponding sides.
(Area of larger/Area of smaller) =(side of larger/side of smaller)2
⇒ (9/4) = (k/1)2 ; [Let the ratio of sides = k : 1 (larger : smaller)]
⇒ k2 = 9/4
∴ k = 3/2
∴ Perimeter of larger : Perimeter of smaller = 3 : 2
∴ Perimeter of larger = 20 × (3/2) = 30 ; [Perimeter of smaller = 20]
So the perimeter of the larger triangle is 30.
১০,৩১৯.
A car covers four successive 7km distance at speeds of 10km/hr, 20 km/hr, 30 km/hr and 60km/hr respectively. Its average speed over this distance is:
ব্যাখ্যা
মোট অতিক্রম করে = 4 × 7 = 28 km
মোট সময় = (7/10) + (7/20) + (7/30) + (7/60)
= 7{(6 + 3 + 2 + 1)/60}
= 7/5
গড় গতিবেগ = 28 × (5/7) = 20 kmph
মোট সময় = (7/10) + (7/20) + (7/30) + (7/60)
= 7{(6 + 3 + 2 + 1)/60}
= 7/5
গড় গতিবেগ = 28 × (5/7) = 20 kmph
১০,৩২০.
The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is -
ব্যাখ্যা
Question: The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is -
Solution:
Let,
Cost price of each article be Tk. 1
Cost price of x articles = Tk. x
Selling price of x articles = Tk. 20
∴ Profit = Tk. (20 - x)
ATQ,
(20 - x)/x = 25%
Or, 100(20 - x)/x = 25
Or, 2000 - 100x = 25x
Or, 125x = 2000
∴ x = 16
Solution:
Let,
Cost price of each article be Tk. 1
Cost price of x articles = Tk. x
Selling price of x articles = Tk. 20
∴ Profit = Tk. (20 - x)
ATQ,
(20 - x)/x = 25%
Or, 100(20 - x)/x = 25
Or, 2000 - 100x = 25x
Or, 125x = 2000
∴ x = 16
১০,৩২১.
In a 120m race, Kamal defeats Jamal by 6 seconds. If the speed of Kamal is 18 kmph, then the speed of Jamal is:
ব্যাখ্যা
Question: In a 120m race, Kamal defeats Jamal by 6 seconds. If the speed of Kamal is 18 kmph, then the speed of Jamal is:
Solution:
Time taken by Kamal = 120/{18 × (5/18)}
= 24 seconds
Time taken by Jamal = 24 + 6 = 30 second
Jamal's speed=120/30
= 4 m/s
= (4 × 18)/5 kmph
=14.4 kmph
Solution:
Time taken by Kamal = 120/{18 × (5/18)}
= 24 seconds
Time taken by Jamal = 24 + 6 = 30 second
Jamal's speed=120/30
= 4 m/s
= (4 × 18)/5 kmph
=14.4 kmph
১০,৩২২.
What is the mean of the range, mode and median of the data given below?
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
ব্যাখ্যা
Question: What is the mean of the range, mode and median of the data given below?
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
Solution:
Arranging the given data in ascending order 2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here,
Most frequent data is 4
so Mode = 4
Total terms in the given data, (n) = 15 (It is odd)
Median = {(n + 1)/2}th term
= {(15 + 1)/2}th term
= (8)th term
= 6
Now,
Range = Maximum value - Minimum value = 19 - 2 = 17
Mean of Range, Mode and median = (Range + Mode + Median)/3 = (17 + 4 + 6)/3 = 27/3 = 9
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
Solution:
Arranging the given data in ascending order 2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here,
Most frequent data is 4
so Mode = 4
Total terms in the given data, (n) = 15 (It is odd)
Median = {(n + 1)/2}th term
= {(15 + 1)/2}th term
= (8)th term
= 6
Now,
Range = Maximum value - Minimum value = 19 - 2 = 17
Mean of Range, Mode and median = (Range + Mode + Median)/3 = (17 + 4 + 6)/3 = 27/3 = 9
১০,৩২৩.
ব্যাখ্যা
Question:
Solution:
Solution:
১০,৩২৪.
The difference between a number consisting of two digits and a number formed by interchanging the digits is always divisible by -
ব্যাখ্যা
Question: The difference between a number consisting of two digits and a number formed by interchanging the digits is always divisible by -
Solution:
ধরি,
একক স্থানীয় অংকটি y
দশক স্থানীয় অংকটি x
তাহলে সংখ্যাটি = 10x + y
ডিজিটগুলো জায়গা পরিবর্তন করলে নতুন সংখ্যাটি = 10y + x
পার্থক্য = 10x + y - 10y - x
= 9x - 9y
= 9(x - y)
এখানে 9(x - y) সংখ্যাটি অবশ্যই 9 দ্বারা বিভাজ্য হবে।
Solution:
ধরি,
একক স্থানীয় অংকটি y
দশক স্থানীয় অংকটি x
তাহলে সংখ্যাটি = 10x + y
ডিজিটগুলো জায়গা পরিবর্তন করলে নতুন সংখ্যাটি = 10y + x
পার্থক্য = 10x + y - 10y - x
= 9x - 9y
= 9(x - y)
এখানে 9(x - y) সংখ্যাটি অবশ্যই 9 দ্বারা বিভাজ্য হবে।
১০,৩২৫.
Today is Friday. What day will it be after 352 days?
ব্যাখ্যা
Question: Today is Friday. What day will it be after 352 days?
Solution:
এখানে,
৩৫২ কে ৭ দিয়ে ভাগ করলে ভাগশেষ থাকে ২।
৩৫২ দিন থেকে ২ দিন বাদ দিলে হয় ৩৫০ দিন।
আমরা জানি,
যে কোনো তারিখ হতে ৭ দিন পর পর (৮ম দিনে) একই বার পাওয়া যায়।
অর্থ্যাৎ, শুক্রবারের ৭ দিন পর বা ৮ম দিনে গিয়ে আবার শুক্রবার পাওয়া যাবে।
অর্থাৎ,
৩৫১ তম দিন শুক্রবার।
৩৫২ তম দিন শনিবার
৩৫৩ তম দিন রবিবার
১০,৩২৬.
12 examiners (men) work 16 hours a day to check 24000 answer sheets in 18 days. Now, 24 examiners would work how many hours per day to check 36000 answer sheets in 36 days?
ব্যাখ্যা
Question: 12 examiners (men) work 16 hours a day to check 24000 answer sheets in 18 days. Now, 24 examiners would work how many hours per day to check 36000 answer sheets in 36 days?
Solution:
১২ জন ১৮ দিনে ২৪০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে ১৬ ঘণ্টা
১২ জন ১৮ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে ১৬/২৪০০০ ঘণ্টা
১২ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮)/২৪০০০ ঘণ্টা
১ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২)/২৪০০০ ঘণ্টা
২৪ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২)/(২৪০০০ × ২৪) ঘণ্টা
২৪ জন ১ দিনে ৩৬০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২ × ৩৬০০০)/(২৪০০০ × ২৪) ঘণ্টা
২৪ জন ৩৬ দিনে ৩৬০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২ × ৩৬০০০)/(২৪০০০ × ২৪ × ৩৬) ঘণ্টা
= ৬ ঘণ্টা
Solution:
১২ জন ১৮ দিনে ২৪০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে ১৬ ঘণ্টা
১২ জন ১৮ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে ১৬/২৪০০০ ঘণ্টা
১২ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮)/২৪০০০ ঘণ্টা
১ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২)/২৪০০০ ঘণ্টা
২৪ জন ১ দিনে ১টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২)/(২৪০০০ × ২৪) ঘণ্টা
২৪ জন ১ দিনে ৩৬০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২ × ৩৬০০০)/(২৪০০০ × ২৪) ঘণ্টা
২৪ জন ৩৬ দিনে ৩৬০০০টি উত্তর পত্র যাচাই করতে দৈনিক কাজ করে (১৬ × ১৮ × ১২ × ৩৬০০০)/(২৪০০০ × ২৪ × ৩৬) ঘণ্টা
= ৬ ঘণ্টা
১০,৩২৭.
A number reduced by 30% becomes 280. What percent should it be increased so that it becomes 500?
ব্যাখ্যা
Question: A number reduced by 30% becomes 280. What percent should it be increased so that it becomes 500?
Solution:
Let, the number be x
Now, x - 30% of x = 280
⇒ x - (30x/100) = 280
⇒ (100x - 30x)/100 = 280
⇒ 70x/100 = 280
⇒ 70x = (280 × 100)
⇒ x = (280 × 100)/70
⇒ x = 400
Required increase = 500 - 400 = 100
∴ Increase % = (100/400) × 100%
= 25%
Solution:
Let, the number be x
Now, x - 30% of x = 280
⇒ x - (30x/100) = 280
⇒ (100x - 30x)/100 = 280
⇒ 70x/100 = 280
⇒ 70x = (280 × 100)
⇒ x = (280 × 100)/70
⇒ x = 400
Required increase = 500 - 400 = 100
∴ Increase % = (100/400) × 100%
= 25%
১০,৩২৮.
8 men entered a lounge simultaneously. If each person shook hands with the other, then find the total no. of hand shakes?
ব্যাখ্যা
Question: 8 men entered a lounge simultaneously. If each person shook hands with the other, then find the total no. of hand shakes?
Solution:
Required no. of hand shakes = 8C2
= 8!/2!(8 - 2)!
= 8!/2! · 6!
= 28
Solution:
Required no. of hand shakes = 8C2
= 8!/2!(8 - 2)!
= 8!/2! · 6!
= 28
১০,৩২৯.
If a gas tank that is one-fifth full needs 32 additional gallons to reach three-sevenths of its capacity, find the total capacity of the tank.
ব্যাখ্যা
Question: If a gas tank that is one-fifth full needs 32 additional gallons to reach three-sevenths of its capacity, find the total capacity of the tank.
Solution:
Solution:
১০,৩৩০.
A large field of 700 hectares is divided into two parts. The difference between the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
ব্যাখ্যা
Let, the area of larger part = x hector
∴ area of smaller part = (700 - x) hectors
The difference between the areas of the two parts
= x - (700 - x)
= 2x - 700
One-fifth of the average of the two areas
(1/5) × (700/2) [total area = 700]
= 70
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
2x−700 = 70
⇒ 2x = 770
⇒ x = 385.
∴ Smaller part of the land is = (700 - 385) = 315 hectares.
১০,৩৩১.
A fruit seller sells 20 apples for Tk. 900 and suffers a loss equal to the cost price of 5 apples. Find the cost price of one apple.
ব্যাখ্যা
Question: A fruit seller sells 20 apples for Tk. 900 and suffers a loss equal to the cost price of 5 apples. Find the cost price of one apple.
Solution:
Let,
cost price of 1 apple is = Tk. x
∴ cost price of 20 apples is = Tk. 20x
∴ cost price of 5 apples is = Tk. 5x
We know,
Cost price - Selling price = Loss
20x - 900 = 5x
⇒ 20x - 5x = 900
⇒ 15x = 900
⇒ x = 900/15
∴ x = 60
∴ Cost price of 1 apple is Tk. 60
Solution:
Let,
cost price of 1 apple is = Tk. x
∴ cost price of 20 apples is = Tk. 20x
∴ cost price of 5 apples is = Tk. 5x
We know,
Cost price - Selling price = Loss
20x - 900 = 5x
⇒ 20x - 5x = 900
⇒ 15x = 900
⇒ x = 900/15
∴ x = 60
∴ Cost price of 1 apple is Tk. 60
১০,৩৩২.
The sum of 3 consecutive odd numbers is 57. The middle one is
ব্যাখ্যা
Question: The sum of 3 consecutive odd numbers is 57. The middle one is-
Solution:
Let,
the 3 consecutive odd numbers be x, x + 2, x + 4
ATQ,
x + x + 2 + x + 4 = 57
⇒ 3x = 51
∴ x = 17
So the middle number is 17 + 2 = 19
Solution:
Let,
the 3 consecutive odd numbers be x, x + 2, x + 4
ATQ,
x + x + 2 + x + 4 = 57
⇒ 3x = 51
∴ x = 17
So the middle number is 17 + 2 = 19
১০,৩৩৩.
What will be the least number that when doubled will be exactly divisible by 12, 18, 21 and 30?
ব্যাখ্যা
Question: What will be the least number that when doubled will be exactly divisible by 12, 18, 21 and 30?
Solution:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5
L.C.M. of 12, 18, 21 and 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260
Hence, the required number = 1260/2 = 630
Solution:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5
L.C.M. of 12, 18, 21 and 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260
Hence, the required number = 1260/2 = 630
১০,৩৩৪.
In a box, there are 7 yellow, 8 black, and 5 white balls. One ball is picked randomly. What is the probability that it is neither yellow nor white?
ব্যাখ্যা
Question: In a box, there are 7 yellow, 8 black, and 5 white balls. One ball is picked randomly. What is the probability that it is neither yellow nor white?
Solution:
মোট বলের সংখ্যা = 7 + 8 + 5 = 20 টি।
ধরি, E হলো এমন ঘটনা যেখানে বলটি হলুদ বা সাদা কোনোটিই নয়, অর্থাৎ বলটি কালো।
∴ অনুকূল ফলাফলের সংখ্যা, n(E) = 8
সম্ভাব্যতা = (অনুকূল ফলাফলের সংখ্যা)/(মোট ফলাফলের সংখ্যা)
= 8/20
= 2/5
অতএব, বলটি হলুদ বা সাদা না হওয়ার সম্ভাব্যতা হলো 2/5।
১০,৩৩৫.
96 + 96 + 96 = ?
ব্যাখ্যা
Question: 96 + 96 + 96 = ?
Solution:
Given,
96 + 96 + 96
= (32)6 +(32)6 +(32)6
= (3)12 +(3)12 +(3)12
= 3 × 312
= 31+12
= 313
Solution:
Given,
96 + 96 + 96
= (32)6 +(32)6 +(32)6
= (3)12 +(3)12 +(3)12
= 3 × 312
= 31+12
= 313
১০,৩৩৬.
Which of the following fractions is the smallest?
ব্যাখ্যা
Question: Which of the following fractions is the smallest?
Solution:
33/40 = 0.825
13/18 = 0.722
5/6 = 0.833
52/67 = 0.776
এখানে দেখা যায় যে 13/18 ভগ্নাংশটি সবচেয়ে ছোট।
Solution:
33/40 = 0.825
13/18 = 0.722
5/6 = 0.833
52/67 = 0.776
এখানে দেখা যায় যে 13/18 ভগ্নাংশটি সবচেয়ে ছোট।
১০,৩৩৭.
A rectangular garden has length 40 m and breadth 20 m. If fencing costs Tk. 25 per meter, find the total cost of fencing the garden.
ব্যাখ্যা
Question: A rectangular garden has length 40 m and breadth 20 m. If fencing costs Tk. 25 per meter, find the total cost of fencing the garden.
Solution:
Given that,
Length 40 m and Breadth 20 m
We know,
Perimeter = 2 × (Length + Breadth) = 2 × (40 + 20) = 2 × 60 = 120 meters
∴ Total cost = 120 × 25 = Tk. 3000
Solution:
Given that,
Length 40 m and Breadth 20 m
We know,
Perimeter = 2 × (Length + Breadth) = 2 × (40 + 20) = 2 × 60 = 120 meters
∴ Total cost = 120 × 25 = Tk. 3000
১০,৩৩৮.
If Set A = {1, 2, 3} and Set B = {1}, which of the following is true?
ব্যাখ্যা
Question: If Set A = {1, 2, 3} and Set B = {1}, which of the following is true?
Solution:
A = {1, 2, 3}, B = {1}
∴ A - B = {1, 2, 3} - {1}
= {2, 3} ; true
খ)
A ∩ B ; common elements
A ∩ B = {1}
not equal to {1, 2, 3} ; false
গ)
A × B ; set of all ordered pairs (a, b) where a ∈ A and b ∈ B
A × B = {(1, 1), (2, 1), (3, 1)}
This is not equal to {1, 2, 3} ; false
ঘ)
A ∪ B ; all elements from A or B
A ∪ B = {1, 2, 3}
not equal to {1} ; false
Final answer ক) A - B = {2, 3}
১০,৩৩৯.
In an election between two candidates, 20% of the total votes cast were invalid. One candidate received 55% of the valid votes. If the total number of votes cast was 9,000, how many valid votes did the other candidate receive?
ব্যাখ্যা
Question: In an election between two candidates, 20% of the total votes cast were invalid. One candidate received 55% of the valid votes. If the total number of votes cast was 9,000, how many valid votes did the other candidate receive?
Solution:
Number of valid votes = 80% of 9000 = 7200
Valid votes polled by other candidate = 45% of 7200
= (7200 × 45)/100
= 3240
১০,৩৪০.
A bookstore purchased 120 textbooks for Taka 250 each and sold them at a profit of 15%. Calculate the total profit.
ব্যাখ্যা
Question: A bookstore purchased 120 textbooks for Taka 250 each and sold them at a profit of 15%. Calculate the total profit.
Solution:
Cost Price of 120 textbooks = 120 × Taka 250 = Taka 3,0000
Profit Percentage = 15%
Profit Amount = Profit Percentage × Cost Price = 15% × Taka 3,0000 = Taka 4500
Total Profit = Profit Amount = 4500
Solution:
Cost Price of 120 textbooks = 120 × Taka 250 = Taka 3,0000
Profit Percentage = 15%
Profit Amount = Profit Percentage × Cost Price = 15% × Taka 3,0000 = Taka 4500
Total Profit = Profit Amount = 4500
১০,৩৪১.
In a class, the number of girls is 20% more than that of boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is -
ব্যাখ্যা
ধরি,
বালক আছে x জন
বালিকা আছে = x এর 120% = 120x/100
= 1.2x জন।
প্রশ্নমতে, x + 1.2x = 66
⇒ 2.2x = 66
⇒ x = 66/2.2
⇒ x = 30
অতএব বালিকা আছে = 1.2x = 1.2 × 30 = 36
4 জন বালিকা ভর্তি হলে = 36 + 4 = 40 জন
∴ বালকঃ বালিকা = 30 : 40
= 3 : 4
১০,৩৪২.
A pole of 66 metre long breaks into two parts without complete separation and makes an angle 30° with the ground. Find the length of the broken part of the pole.
ব্যাখ্যা
Question: A pole of 66 metre long breaks into two parts without complete separation and makes an angle 30° with the ground. Find the length of the broken part of the pole.
Solution
sin30 = x/(66 - x)
⇒ 1/2 = x/(66 - x)
⇒ 66 - x = 2x
⇒ 3x = 66
⇒ x = 66/3 = 22
the length of the broken part of the pole = 66 - 22 = 44 m
Solution
sin30 = x/(66 - x)
⇒ 1/2 = x/(66 - x)
⇒ 66 - x = 2x
⇒ 3x = 66
⇒ x = 66/3 = 22
the length of the broken part of the pole = 66 - 22 = 44 m
১০,৩৪৩.
One number exceeds another number by 5. Two times the smaller number diminished by the greater number equals 9. What is the larger number?
ব্যাখ্যা
Let the numbers be x and 5+x.
2x - (5+x) = 9,
or, 2x - x = 9+5,
or, x = 14.
The larger number is 19.
2x - (5+x) = 9,
or, 2x - x = 9+5,
or, x = 14.
The larger number is 19.
১০,৩৪৪.
If, cosec A + cot A = 5/2, find the value of, cosec A - cot A.
ব্যাখ্যা
Question: If, cosec A + cot A = 5/2, find the value of, cosec A - cot A.
Solution:
Given that,
cosec A + cot A = 5/2
We know,
cosec2A - cot2A = 1
⇒ (cosec A + cot A)(cosec A - cot A) = 1
⇒ (5/2)(cosec A - cot A) = 1
⇒ cosec A - cot A = 1 ÷ (5/2)
∴ cosec A - cot A = 2/5
১০,৩৪৫.
A is faster than B. A and B each walk 24km. The sum of their speeds is 7km/hr and the sum of times taken by them is 14 hours. Then, A's speed is equal to -
ব্যাখ্যা
Question: A is faster than B. A and B each walk 24km. The sum of their speeds is 7km/hr and the sum of times taken by them is 14 hours. Then, A's speed is equal to -
Solution:
Let, A's speed be = x km/hr
B's speed is = (7 - x) km/hr
ATQ,
24/x + 24/(7 - x) = 14
or, {24(7 - x) + 24x}/{x(7 - x) = 14
or, 24(7 - x) + 24x = 14x(7 - x)
or, 14x2 - 98x + 168 = 0
or, x2 - 7x + 12 = 0
or, (x - 3)(x - 4) = 0
∴ x = 3 or, x = 4
as A's speed is greater than B.
so, x = 4
Solution:
Let, A's speed be = x km/hr
B's speed is = (7 - x) km/hr
ATQ,
24/x + 24/(7 - x) = 14
or, {24(7 - x) + 24x}/{x(7 - x) = 14
or, 24(7 - x) + 24x = 14x(7 - x)
or, 14x2 - 98x + 168 = 0
or, x2 - 7x + 12 = 0
or, (x - 3)(x - 4) = 0
∴ x = 3 or, x = 4
as A's speed is greater than B.
so, x = 4
১০,৩৪৬.
A student obtained 78, 82, 69, 91 marks in four subjects. What should be the 5th subject's mark to get an average of 80?
ব্যাখ্যা
Question: A student obtained 78, 82, 69, 91 marks in four subjects. What should be the 5th subject's mark to get an average of 80?
Solution:
Desired average = 80
Number of subjects = 5
Total marks needed = 80 × 5 = 400
Sum of the first four subjects obtained = 78 + 82 + 69 + 91 = 320
∴ Required marks in the fifth subject = 400 - 320 = 80
Therefore, the student must obtain 80 in the fifth subject's to achieve an average of 80.
১০,৩৪৭.
The monthly income of a person was Tk. 13500 and his monthly expenditure was Tk. 9000. Next year, his income increased by 14% and his expenditure by 7%. Find the percentage increase in his savings.
ব্যাখ্যা
Question: The monthly income of a person was Tk. 13500 and his monthly expenditure was Tk. 9000. Next year, his income increased by 14% and his expenditure by 7%. Find the percentage increase in his savings.
Solution:
Savings before increased income = Tk. (13500 - 9000)
= Tk. 4500
Increased income = 114% of 13500 = Tk. (114/100) × 13500
= Tk. 15390
Increased expenditure = 107% of 9000 = Tk. (107/100) × 9000
= Tk. 9630
Saving after Increased income = Tk. (15390 - 9630) = Tk. 5760
Increased Savings = Tk. (5760 - 4500)
= Tk. 1260
∴ Increase % in savings = (1260/4500) × 100 %
= 28%
Solution:
Savings before increased income = Tk. (13500 - 9000)
= Tk. 4500
Increased income = 114% of 13500 = Tk. (114/100) × 13500
= Tk. 15390
Increased expenditure = 107% of 9000 = Tk. (107/100) × 9000
= Tk. 9630
Saving after Increased income = Tk. (15390 - 9630) = Tk. 5760
Increased Savings = Tk. (5760 - 4500)
= Tk. 1260
∴ Increase % in savings = (1260/4500) × 100 %
= 28%
১০,৩৪৮.
If 3(a - b) = 27 and 3(a + b) = 243, then what is the value of a?
ব্যাখ্যা
Question: If 3(a - b) = 27 and 3(a + b) = 243, then what is the value of a?
Solution:
3(a - b) = 27
⇒ 3(a - b) = 33
⇒ a - b = 3 ............(1)
3(a + b) = 243
⇒ 3(a + b) = 35
⇒ a + b = 5 ...........(2)
(1) + (2) ⇒
a - b + a + b = 3 + 5
⇒ 2a = 8
∴ a = 4
Solution:
3(a - b) = 27
⇒ 3(a - b) = 33
⇒ a - b = 3 ............(1)
3(a + b) = 243
⇒ 3(a + b) = 35
⇒ a + b = 5 ...........(2)
(1) + (2) ⇒
a - b + a + b = 3 + 5
⇒ 2a = 8
∴ a = 4
১০,৩৪৯.
How will the word "ANALOGY" appear in a mirror?
ব্যাখ্যা
Question: How will the word "ANALOGY" appear in a mirror?
Solution:
The mirror reflection of the word "ANALOGY" is:
Solution:
The mirror reflection of the word "ANALOGY" is:
১০,৩৫০.
If 5% is gained by selling an article for BDT 350 than selling it for BDT 340, the cost of the article is-
ব্যাখ্যা
Let, buying price is x Taka
ATQ, 5% of x = 350 - 340 = 10
⇒ x = (100 × 10) / 5
∴ x = 200
১০,৩৫১.
There is 75% increase in an amount in 5 years at simple interest. What will be the compound interest of TK. 16,000 after 2 years at the same rate?
ব্যাখ্যা
Question: There is 75% increase in an amount in 5 years at simple interest. What will be the compound interest of TK. 16,000 after 2 years at the same rate?
Solution:
Let,
P = Tk. 100
Then, S.I = Tk. 75
and n = 5 years
So the rate r = (100 × 75)/(100 × 5)
= 15%
Now, P = Tk. 16,000
n = 2 years
r = 15% p.a.
= 3/20 p.a
C.I = 16,000 × {(1 + 3/20)2 - 1}
= 16,000 × {(23/20)2 - 1}
= 16,000 × (529/400 - 1)
= 16000 × 129/400
= 5,160
Solution:
Let,
P = Tk. 100
Then, S.I = Tk. 75
and n = 5 years
So the rate r = (100 × 75)/(100 × 5)
= 15%
Now, P = Tk. 16,000
n = 2 years
r = 15% p.a.
= 3/20 p.a
C.I = 16,000 × {(1 + 3/20)2 - 1}
= 16,000 × {(23/20)2 - 1}
= 16,000 × (529/400 - 1)
= 16000 × 129/400
= 5,160
১০,৩৫২.
A person walks 7 meters towards east, 4 meters towards north, then 4 meters towards west. What is the direct distance of the destination from the starting point?
ব্যাখ্যা
Question: A person walks 7 meters towards east, 4 meters towards north, then 4 meters towards west. What is the direct distance of the destination from the starting point?
Solution:
ধরি, যাত্রা শুরু করার স্থান A এবং গন্তব্যের স্থান B।
সরাসরি দূরত্ব নির্ণয় করতে, আমরা পিথাগোরাসের উপপাদ্য ব্যবহার করব।
এখানে, অতিক্রান্ত মোট উল্লম্ব দূরত্ব (উত্তর দিকে) হলো 4 মিটার।
এবং অতিক্রান্ত মোট অনুভূমিক দূরত্ব (পূর্ব ও পশ্চিমের পার্থক্য) হলো (7 - 4) = 3 মিটার।
সুতরাং,
AB2 = (মোট অনুভূমিক দূরত্ব)2 + (মোট উল্লম্ব দূরত্ব)2 [অতিভুজ2 = ভূমি2 + লম্ব2]
⇒ AB2 = (3)2 + (4)2
⇒ AB2 = 9 + 16
⇒ AB2 = 25
⇒ AB = 5 মিটার
∴ সরাসরি দূরত্ব 5 মিটার।
১০,৩৫৩.
In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for noun of the three subjects?
ব্যাখ্যা
Question: In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for noun of the three subjects?
Solution:
P = {2,4,6,…,118,120}
C = {5,10,15,…,115,120}
M = {7,14,21,…,112,119}
n(P) = 60, n(C) = 24, n(M) = 17
(Here, P= Physics, C= Chemistry, M= Maths)
The elements in P∩C will be the multiples of LCM (2,5).
So, n(P∩C) = 12
Similarly, n(C∩M) = 3,
n(P∩M) = 8
and n(P∩C∩M)=1
∴n(P∪C∪M) = n(P) + n(C) + n(M) − n(P∩C) − n(C∩M) − n(P∩M) + n(P∩C∩M)
= 60 + 24 + 17 − (12 + 3 +8) + 1
= 79
∴ Number of students who opt for none of the subjects =120 − 79 = 41
Solution:
P = {2,4,6,…,118,120}
C = {5,10,15,…,115,120}
M = {7,14,21,…,112,119}
n(P) = 60, n(C) = 24, n(M) = 17
(Here, P= Physics, C= Chemistry, M= Maths)
The elements in P∩C will be the multiples of LCM (2,5).
So, n(P∩C) = 12
Similarly, n(C∩M) = 3,
n(P∩M) = 8
and n(P∩C∩M)=1
∴n(P∪C∪M) = n(P) + n(C) + n(M) − n(P∩C) − n(C∩M) − n(P∩M) + n(P∩C∩M)
= 60 + 24 + 17 − (12 + 3 +8) + 1
= 79
∴ Number of students who opt for none of the subjects =120 − 79 = 41
১০,৩৫৪.
Find the HCF of 56, 72 and 90.
ব্যাখ্যা
প্রশ্ন: 56, 72, 90 এর গ.সা.গু কত?
সমাধান:
56 = 7 × 23
72 = 23 × 33
90 = 2 × 32 × 5
∴ গ.সা.গু = 2
সমাধান:
56 = 7 × 23
72 = 23 × 33
90 = 2 × 32 × 5
∴ গ.সা.গু = 2
১০,৩৫৫.
A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 woman. In how many different ways can it be done?
ব্যাখ্যা
Required number of ways = 6C1× 8C3 + 6C2× 8C2 + 6C3× 8C1+ 6C4× 8C0
= {6 × (8 × 7 × 6)/3!} + {(6 × 5)/(2 × 1) × (8 × 7)/(2 × 1)} + {(6 × 5 × 4)/3! × 8} + {(6 × 5)/(2 × 1) × 1}
= (336 + 420 + 160 + 15)
= 931.
১০,৩৫৬.
As the price of chocolate has decreased by 15%, 2 more chocolates are now available for tk 10 . What is the current price of 36 chocolates?
ব্যাখ্যা
প্রশ্ন: As the price of chocolate has decreased by 15%, 2 more chocolates are now available for tk 10 . What is the current price of 36 chocolates?
সমাধান:
১৫% কমায় ১০ টাকায় কমে = ১০ × (১৫/১০০) = ১.৫ টাকা
∴ ১.৫ টাকা কমায় ২ টি চকলেট বেশি পাওয়া যায়।
অর্থাৎ,
২টি চকলেটের বর্তমান মূল্য = ১.৫ টাকা
∴ ১টি চকলেটের বর্তমান মূল্য = ১.৫/২ টাকা
∴ ৩৬টি চকলেটের বর্তমান মূল্য = (১.৫/২) × ৩৬ টাকা
= ২৭ টাকা
১০,৩৫৭.
A pole 6m high casts a shadow 2√3m long on the ground, then Sun's elevation is -
ব্যাখ্যা
We know,
⇒ tanΘ = perpendicular/base
⇒ tanΘ = AB/BC
⇒ tanΘ = 6/2√3
⇒ tanΘ = √3
⇒ tanΘ = tan60°
∴ Θ = 60°
১০,৩৫৮.
What should be added to 9p2 + 14p so that the sum is a perfect square?
ব্যাখ্যা
Question: What should be added to 9p2 + 14p so that the sum is a perfect square?
Solution:
(3p)2 + 2.3p.(7/3) + (7/3)2
= (3p + 7/3)2
so, (7/3)2 or, 49/9 should be added to make it a perfect square.
Solution:
(3p)2 + 2.3p.(7/3) + (7/3)2
= (3p + 7/3)2
so, (7/3)2 or, 49/9 should be added to make it a perfect square.
১০,৩৫৯.
How many terms are there in the geometric progression (GP) 5 + 20 + 80 + 320 +........... + 20480?
ব্যাখ্যা
Question: How many terms are there in the geometric progression (GP) is 5 + 20 + 80 + 320 +........... + 20480?
Solution:
First term, a = 5
Common ratio, r = 20/5 = 4
And
Last term ,l = 20480
We know,
an = a⋅rn - 1
⇒ 20480 = 5 × (4n - 1)
⇒ 4n - 1 = 20480/5 = 4096
⇒ 4n - 1 = 46
⇒ n - 1 = 6
∴ n = 7
So the number of terms is 7
১০,৩৬০.
A man spent 1/2 of his money and then lost 1/4 of the remainder. He was left with TK. 3,600. How much did he start with?
ব্যাখ্যা
Let,
he has TK. x
ATQ, x - x/2 - x/2×1/4 = 3600
or, x/2 - x/8 = 3600
or, (4x - x)/8 = 3600
∴ x = (3600×8)/3 = 9600
১০,৩৬১.
What will come at the place of question mark ?
6, 13, 28, 59, ?, 249.
6, 13, 28, 59, ?, 249.
ব্যাখ্যা
Question: What will come at the place of question mark ?
6, 13, 28, 59, ?, 249.
Solution:
First term ⇒ 6
Second term ⇒ (6 × 2 + 1) = 13
Third term ⇒ (13 × 2 + 2) = 28
Fourth term ⇒ (28 × 2 + 3) = 59
Fifth term ⇒ (59 × 2 + 4) = 122
Sixth term ⇒ (122 × 2 + 5) = 249
So, the required term = 122.
১০,৩৬২.
What is the measure of each interior angle in a regular Pentagon?
ব্যাখ্যা
Question: What is the measure of each interior angle in a regular Pentagon?
Solution:
সুষম বহুভুজের বাহুর সংখ্যা n হলে তার কোণগুলোর সমষ্টি (2n - 4) সমকোণ।
সুতরাং সুষম পঞ্চভুজের পাঁচ কোণের সমষ্টি = (2 × 5 - 4) সমকোণ
= (10 - 4) × 90°
= 6 × 90°
= 540°
সুতরাং সুষম পঞ্চভুজের একটি শীর্ষ কোণ = 540°/5
= 108°
Solution:
সুষম বহুভুজের বাহুর সংখ্যা n হলে তার কোণগুলোর সমষ্টি (2n - 4) সমকোণ।
সুতরাং সুষম পঞ্চভুজের পাঁচ কোণের সমষ্টি = (2 × 5 - 4) সমকোণ
= (10 - 4) × 90°
= 6 × 90°
= 540°
সুতরাং সুষম পঞ্চভুজের একটি শীর্ষ কোণ = 540°/5
= 108°
১০,৩৬৩.
Four milkmen rented a pasture. A grazed 24 cows for 3 months; B grazed 10 cows for 5 months; C grazed 35 cows for 4 months and D grazed 21 cows for 3 months. If A’s share of rent is Tk. 720, find the total rent of the field?
ব্যাখ্যা
Ratio of the shares of A,B,C and D = (24 × 3) : (10 × 5) : (35 × 4) : (21 × 3)
= 72:50:140:63
Total rent be X then A’s share = x × (72/325)
= 72x/325
A/Q,
72x/325 = 720
Or, X = (720 × 325)/72
= 3250
১০,৩৬৪.
What is the profit of Tk. 650 in 6 years at the rate of profit Tk. 7.5 percent per annum?
ব্যাখ্যা
Required profit
= 6 × 650 × 7.5/100
= Tk. 292.50
= 6 × 650 × 7.5/100
= Tk. 292.50
১০,৩৬৫.
A man travels the distance of his journey 3/4 by bus, 1/6 by rickshaw and remaining 2 km on foot. The total distance travelled by the man is:
ব্যাখ্যা
Question: A man travels the distance of his journey 3/4 by bus, 1/6 by rickshaw and remaining 2 km on foot. The total distance travelled by the man is:
Solution:
Let the man travels 1 unit distance
So, remaining distance
= 1 - (1/6 + 3/4)
= 1 - 22/24
=1/12
∵ 1/12 unit = 2 km
So, 1 unit = 24 km.
Solution:
Let the man travels 1 unit distance
So, remaining distance
= 1 - (1/6 + 3/4)
= 1 - 22/24
=1/12
∵ 1/12 unit = 2 km
So, 1 unit = 24 km.
১০,৩৬৬.
Marked price of a product is Tk. 240 and 25% discount is provided on it. Find the selling price.
ব্যাখ্যা
Question: Marked price of a product is Tk. 240 and 25% discount is provided on it. Find the selling price.
Solution:
Discount = SP × 25% = 240 × (25/100) = Tk. 60
Discount = SP × 25% = 240 × (25/100) = Tk. 60
Selling price = MP - Discount = 240 - 60 = Tk. 180.
১০,৩৬৭.
A certain business printer can print 40 characters per second, which is 4 times as fast as an average printer. If an average printer can print 5 times as fast as an electric typewriter, how many characters per minute can an electric typewriter print?
ব্যাখ্যা
Question: A certain business printer can print 40 characters per second, which is 4 times as fast as an average printer. If an average printer can print 5 times as fast as an electric typewriter, how many characters per minute can an electric typewriter print?
Solution:
Rate at which business printer can print = 40 char per second
Rate at which an average printer can print = (40/4) char per second = 10 char per second
Average printer's rate = 5 × Electric typewriter's rate
⇒ Electric typewriter's rate = 10/5 = 2 char per second
⇒ 2 × 60 char per min = 120 char per min
Solution:
Rate at which business printer can print = 40 char per second
Rate at which an average printer can print = (40/4) char per second = 10 char per second
Average printer's rate = 5 × Electric typewriter's rate
⇒ Electric typewriter's rate = 10/5 = 2 char per second
⇒ 2 × 60 char per min = 120 char per min
১০,৩৬৮.
At what rate percentage per annum will the simple interest on sum of money be 3/5 of the amount in 10 years?
ব্যাখ্যা
Let
sum = x
S. I = 3x/5
Time = 10 years
Rate = {(3x × 100)/(x × 5 × 10)}% = 6%
sum = x
S. I = 3x/5
Time = 10 years
Rate = {(3x × 100)/(x × 5 × 10)}% = 6%
১০,৩৬৯.
If the cost price is 25% of selling price, then what is the percentage of profit?
ব্যাখ্যা
ধরি,
বিক্রয়মূল্য = ১০০ টাকা
ক্রয়মূল্য = ১০০ এর ২৫%
= ২৫ টাকা
লাভ = ১০০ - ২৫ = ৭৫ টাকা
শতকরা লাভ = {(৭৫/২৫) × ১০০}% = ৩০০%
বিক্রয়মূল্য = ১০০ টাকা
ক্রয়মূল্য = ১০০ এর ২৫%
= ২৫ টাকা
লাভ = ১০০ - ২৫ = ৭৫ টাকা
শতকরা লাভ = {(৭৫/২৫) × ১০০}% = ৩০০%
১০,৩৭০.
A sum of 20,000 Taka is invested at 8% per annum. If the interest is compounded quarterly, what is the amount after 9 months?
ব্যাখ্যা
Question: A sum of 20,000 Taka is invested at 8% per annum. If the interest is compounded quarterly, what is the amount after 9 months?
Solution:
এখানে, আসল (P) = 20,000 টাকা
বার্ষিক সুদের হার = 8%
সময় = 9 মাস
যেহেতু সুদ ত্রৈমাসিক (quarterly) ভিত্তিতে গণনা করা হয়,
∴ ত্রৈমাসিক সুদের হার = 8% ÷ 4 = 2%
9 মাসে চক্রবৃদ্ধির সংখ্যা = 9 মাস ÷ 3 মাস = 3 বার
প্রথম ত্রৈমাসিক:
সুদ = (20,000 এর 2%) = 400 টাকা
নতুন মূল = 20,000 + 400
= 20,400 টাকা
দ্বিতীয় ত্রৈমাসিক:
সুদ = (20,400 এর 2%) = 408 টাকা
নতুন মূল = 20,400 + 408
= 20,808 টাকা
তৃতীয় ত্রৈমাসিক:
সুদ = (20,808 এর 2%) = 416.16 টাকা
নতুন মূল = 20,808 + 416.16
= 21,224.16 টাকা
∴ 9 মাস পর চক্রবৃদ্ধি মূল হবে 21,224.16 টাকা।
১০,৩৭১.
A coin is tossed five times. What is the probability of getting head on all tosses?
ব্যাখ্যা
Probability of getting head on all tosses = 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/32
১০,৩৭২.
The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, the mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.
ব্যাখ্যা
Question: The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, the mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.
Solution:
Let the daughter’s present age be ‘n’ years.
∴ Mother’s present age = (3n + 3) years
So, according to the question
(3n + 3 + 3) = 2 (n + 3) + 10
⇒ 3n + 6 = 2n + 16
⇒ n = 10
Hence, mother’s present age = (3n + 3) = ((3 × 10) + 3) years = 33 years
Solution:
Let the daughter’s present age be ‘n’ years.
∴ Mother’s present age = (3n + 3) years
So, according to the question
(3n + 3 + 3) = 2 (n + 3) + 10
⇒ 3n + 6 = 2n + 16
⇒ n = 10
Hence, mother’s present age = (3n + 3) = ((3 × 10) + 3) years = 33 years
১০,৩৭৩.
If secA - tanA = 3/5, find the value of secA + tanA?
ব্যাখ্যা
Question: If secA - tanA = 3/5, find the value of secA + tanA?
Solution:
We know that,
sec2A - tan2A = 1
or, (secA + tanA)(secA - tanA) = 1
or, (secA + tanA) = 1/(secA - tanA)
∴ secA + tanA = 5/3
Solution:
We know that,
sec2A - tan2A = 1
or, (secA + tanA)(secA - tanA) = 1
or, (secA + tanA) = 1/(secA - tanA)
∴ secA + tanA = 5/3
১০,৩৭৪.
Two numbers are in the ratio 2 : 3, and their greatest common divisor is 4. Find the difference between the two numbers.
ব্যাখ্যা
Question: Two numbers are in the ratio 2 : 3, and their greatest common divisor is 4. Find the difference between the two numbers.
Solution:
Given,
The ratio of two numbers = 2 : 3
And the GCD (greatest common divisor) = 4
So,
The two numbers are:
(2 × 4) = 8 and
(3 × 4) = 12
∴ The difference between the numbers = (12 − 8) = 4
Solution:
Given,
The ratio of two numbers = 2 : 3
And the GCD (greatest common divisor) = 4
So,
The two numbers are:
(2 × 4) = 8 and
(3 × 4) = 12
∴ The difference between the numbers = (12 − 8) = 4
১০,৩৭৫.
A train 800 meters long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 minute , then the length of the tunnel (in metres) is :
ব্যাখ্যা
Let the length of the tunnel be 'x' meters
Time = (Length of train + Length of tunnel) / speed
Speed = 78 km/h
= 78 × 1000 meter/(60 × 60 sec)
= 78 × 5 / 18
= 65/3 m/sec
Therefore, 60 = (800 + x) / (65/3)
or, 60 × 65 = 2400 + 3x
∴ x = (3900 - 2400) / 3
= 500 meters
Time = (Length of train + Length of tunnel) / speed
Speed = 78 km/h
= 78 × 1000 meter/(60 × 60 sec)
= 78 × 5 / 18
= 65/3 m/sec
Therefore, 60 = (800 + x) / (65/3)
or, 60 × 65 = 2400 + 3x
∴ x = (3900 - 2400) / 3
= 500 meters
১০,৩৭৬.
Find the volume of a cylinder that is 14 cm tall with a base diameter of 6 cm.
ব্যাখ্যা
Question: Find the volume of a cylinder that is 14 cm tall with a base diameter of 6 cm.
Solution:
দেওয়া আছে,
সিলিন্ডারের উচ্চতা, h = 14 সে.মি.
সিলিন্ডারের ভূমির ব্যাস = 6 সে.মি.
∴ সিলিন্ডারের ভূমির ব্যাসার্ধ, r = 6/2 = 3 সে.মি.
আমরা জানি,
সিলিন্ডারের আয়তন, V = πr2h
= π × 32 × 14
= π × 9 × 14
= 126π ঘন সে.মি.
অতএব, সিলিন্ডারটির আয়তন 126π ঘন সে.মি.
১০,৩৭৭.
If x is 90% of y then what percent of x is y?
ব্যাখ্যা
Question: If x is 90% of y then what percent of x is y?
Solution:
x = 90% of y
⇒ x = 90y/100
⇒ x = 9y/10
⇒ y/x = 10/9
= (10/9) × 100%
= 111.1%
১০,৩৭৮.
A piece of wire 78 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?
ব্যাখ্যা
Question: A piece of wire 78 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?
Solution:
Given,
ratio of one of the equal sides to the base is 5:3
Therefore, the sides are 5x, 3x, 5x.
78 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 78 cm.
Therefore, 5x + 3x + 5x = 13x
∴ 13x=78
⇒ x = 6
Thus the length of the base = 3 × 6 = 18 cm.
Solution:
Given,
ratio of one of the equal sides to the base is 5:3
Therefore, the sides are 5x, 3x, 5x.
78 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 78 cm.
Therefore, 5x + 3x + 5x = 13x
∴ 13x=78
⇒ x = 6
Thus the length of the base = 3 × 6 = 18 cm.
১০,৩৭৯.
A, B and C rent a pasture. If A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing and the rent of the pasture is Tk. 175, then how much should C pay as his share of the rent?
ব্যাখ্যা
A : B : C
= 10 × 7 : 12 × 5 : 15 × 3
= (2 × 7) : (12 × 1) : (3 × 3)
= 14 : 12 : 9
Amount that C should pay
= 175 × (9/35)
= 5 × 9
= 45.
১০,৩৮০.
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
ব্যাখ্যা
Question: Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
Solution:
Let,
Ronit's present age be x years.
Then, father's present age =(x + 3x) years
= 4x years.
ATQ,
(4x + 8) = {5(x + 8)}/2
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
∴ x = 8
Now, (4x + 16)/(x + 16) = 48/24 = 2
∴ Father would be 2 times of Ronit's age.
Solution:
Let,
Ronit's present age be x years.
Then, father's present age =(x + 3x) years
= 4x years.
ATQ,
(4x + 8) = {5(x + 8)}/2
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
∴ x = 8
Now, (4x + 16)/(x + 16) = 48/24 = 2
∴ Father would be 2 times of Ronit's age.
১০,৩৮১.
If numbers N and K are added to set X {2, 8, 10, 12}, its mean will increase by 25%. What is the value of N2 + 2NK + K2?
ব্যাখ্যা
Question: If numbers N and K are added to set X {2, 8, 10, 12}, its mean will increase by 25%. What is the value of N2 + 2NK + K2?
Solution:
old mean = (2 + 8 + 10 + 12)/4 = 8
new mean = 8 + 8 × .25
= 8 + 2
= 10
new sum = 10 × 6 = 60
N + K = 60 - 32 = 28
N2 + 2NK + K2
= (N + K)2
= 282
= 784
Solution:
old mean = (2 + 8 + 10 + 12)/4 = 8
new mean = 8 + 8 × .25
= 8 + 2
= 10
new sum = 10 × 6 = 60
N + K = 60 - 32 = 28
N2 + 2NK + K2
= (N + K)2
= 282
= 784
১০,৩৮২.
A policeman sighted a robber from a distance of 300 m. The robber also noticed the policeman and started running at 8 km/hr. The policeman also started running after him at the speed of 10 km/hr. Find the distance that the robber would run before being caught.
ব্যাখ্যা
Question: A policeman sighted a robber from a distance of 300 m. The robber also noticed the policeman and started running at 8 km/hr. The policeman also started running after him at the speed of 10 km/hr. Find the distance that the robber would run before being caught.
Solution:
Since both are running in the same direction, relative speed = 10 - 8 = 2 km/hr
Now, to catch the robber if he were stagnant, the policeman would have to run 300 m. But since both are moving, the policeman needs to finish off this separation of 300 m.
= 300 m (or 0.3 km)is to be covered at the relative speed of 2 km/hr.
∴ Time taken = 0.3/2 = 0.15 hours
Therefore, distance run by robber before being caught = 8 × 0.15 = 1.2 km
Solution:
Since both are running in the same direction, relative speed = 10 - 8 = 2 km/hr
Now, to catch the robber if he were stagnant, the policeman would have to run 300 m. But since both are moving, the policeman needs to finish off this separation of 300 m.
= 300 m (or 0.3 km)is to be covered at the relative speed of 2 km/hr.
∴ Time taken = 0.3/2 = 0.15 hours
Therefore, distance run by robber before being caught = 8 × 0.15 = 1.2 km
১০,৩৮৩.
If the cost of x metres of wire is d Taka, then what is the cost of y metres of wire at the same rate?
ব্যাখ্যা
প্রশ্ন: If the cost of x metres of wire is d Taka, then what is the cost of y metres of wire at the same rate?
সমাধান:
x মিটারের দাম d টাকা
∴ 1 মিটারের দাম d/x টাকা
∴ y মিটারের দাম (yd)/x টাকা
সমাধান:
x মিটারের দাম d টাকা
∴ 1 মিটারের দাম d/x টাকা
∴ y মিটারের দাম (yd)/x টাকা
১০,৩৮৪.
A piece of string 70 cm in length was cut into pieces, the ratio of whose lengths was 3: 7. Find the length of longest piece.
ব্যাখ্যা
Question: A piece of string 70 cm in length was cut into pieces, the ratio of whose lengths was 3: 7. Find the length of longest piece.
Solution:
Total length = 70 cm.
Ratio is 3 : 7
∴ Length of longest piece is (70 × 7)/10 = 49 cm
Solution:
Total length = 70 cm.
Ratio is 3 : 7
∴ Length of longest piece is (70 × 7)/10 = 49 cm
১০,৩৮৫.
A water tank is two-fifth full.Pipe A can fill a tank in 6 minutes and pipe B can empty it in 10 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?
ব্যাখ্যা
Question: A water tank is two-fifth full.Pipe A can fill a tank in 6 minutes and pipe B can empty it in 10 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?
Solution:
ট্যাংকটি ২/৫ পূর্ণ
পাইপ A,
৬ মিনিটে ভর্তি করতে পারে ১ অংশ
∴ ১ মিনিটে ভর্তি করে ১/৬ অংশ
আবার,
পাইপ B,
১০ মিনিটে খালি করতে পারে ১ অংশ
∴ ১ মিনিটে খালি করতে পারে ১/১০ অংশ
একসাথে কাজ করার হার = (১/৬) - (১/১০) = ২/৩০ = ১/১৫ অংশ
অর্থাৎ, একসাথে কাজ করলে প্রতি মিনিটে ট্যাংকের ১/১৫ অংশ ভর্তি হয়।
∴ ট্যাংকটি বাকি থাকে = ১ - (২/৫) = ৩/৫ অংশ
∴ ৩/৫ অংশ ভর্তি হতে সময় লাগে = (১/১৫) × (৫/৩) = ১/৯ মিনিট
∴ ১ বা সম্পূর্ণ অংশ ভর্তি হতে সময় লাগে = ৯ মিনিট
Solution:
ট্যাংকটি ২/৫ পূর্ণ
পাইপ A,
৬ মিনিটে ভর্তি করতে পারে ১ অংশ
∴ ১ মিনিটে ভর্তি করে ১/৬ অংশ
আবার,
পাইপ B,
১০ মিনিটে খালি করতে পারে ১ অংশ
∴ ১ মিনিটে খালি করতে পারে ১/১০ অংশ
একসাথে কাজ করার হার = (১/৬) - (১/১০) = ২/৩০ = ১/১৫ অংশ
অর্থাৎ, একসাথে কাজ করলে প্রতি মিনিটে ট্যাংকের ১/১৫ অংশ ভর্তি হয়।
∴ ট্যাংকটি বাকি থাকে = ১ - (২/৫) = ৩/৫ অংশ
∴ ৩/৫ অংশ ভর্তি হতে সময় লাগে = (১/১৫) × (৫/৩) = ১/৯ মিনিট
∴ ১ বা সম্পূর্ণ অংশ ভর্তি হতে সময় লাগে = ৯ মিনিট
১০,৩৮৬.
A light was seen at intervals of 13 seconds. It was seen for the first time at 1 hr. 54 min 50 secs. a.m. and the last time at 3 hrs. 17 min. 49 secs. a.m. How many times was the light seen?
ব্যাখ্যা
Question: A light was seen at intervals of 13 seconds. It was seen for the first time at 1 hr. 54 min 50 secs. a.m. and the last time at 3 hrs. 17 min. 49 secs. a.m. How many times was the light seen?
Solution:
Solution:
Let us convert Initial time and Final time in seconds.
Intial time = 1hr 54min 50s = 1 × 60 × 60 + 54 × 60 + 50 = 3600 + 3240 + 50 = 6890 sec.
Final Time = 3hr 17min 49s = 3 × 60 × 60 + 17 × 60 + 49 = 10800 + 1020 + 49 = 11869sec
Total Time interval for which light was seen
= 11869 - 6890 = 4979
Time interval at which light seen = 13 sec.
Therefore no. Of time light was seen = 4979/13 = 383
With the light seen first time no. Of time light was seen = 383 + 1 = 384
With the light seen first time no. Of time light was seen = 383 + 1 = 384
১০,৩৮৭.
Two trains of equal length are running on parallel lines in the same direction at 46 km and 36 km per hour. The faster train passes the slower train in 36 seconds. The length of each train is-
ব্যাখ্যা
Question: Two trains of equal length are running on parallel lines in the same direction at 46 km and 36 km per hour. The faster train passes the slower train in 36 seconds. The length of each train is-
Solution:
To cross each other, two trains have to cover a distance equal to the sum of the lengths of the train.
Let the length of the trains be = a m each.
So the distance to be covered = 2a
Now the trains are running int he same direction.
∴ Their relative speed = (46 - 36) km/hr.
= 10km/hr. = 10 × (5/18) km/hr. = (25/9) m/sec.
So, the time taken by the trains to cove 2a m distance
= 2a ÷ (25/9) sec
∴ By the given conditions,
2a ÷ (25/9) = 36
⇒ 2a × (9/25) = 36
⇒ 2a = (36 × 25)/9
⇒ 2a = 100
∴ a = 50
So, the length of each train = 50 m.
Solution:
To cross each other, two trains have to cover a distance equal to the sum of the lengths of the train.
Let the length of the trains be = a m each.
So the distance to be covered = 2a
Now the trains are running int he same direction.
∴ Their relative speed = (46 - 36) km/hr.
= 10km/hr. = 10 × (5/18) km/hr. = (25/9) m/sec.
So, the time taken by the trains to cove 2a m distance
= 2a ÷ (25/9) sec
∴ By the given conditions,
2a ÷ (25/9) = 36
⇒ 2a × (9/25) = 36
⇒ 2a = (36 × 25)/9
⇒ 2a = 100
∴ a = 50
So, the length of each train = 50 m.
১০,৩৮৮.
If 2 tables and 3 chairs cost Tk. 4800 and 3 tables and 2 chairs cost Tk. 5700, then how much does a table cost?
ব্যাখ্যা
Question: If 2 tables and 3 chairs cost Tk. 4800 and 3 tables and 2 chairs cost Tk. 5700, then how much does a table cost?
Solution:
Let
The cost of a table and that of a chair be Tk. x and Tk. y respectively.
Then,
2x + 3y = 4800................(i)
and
3x + 2y = 5700................(ii)
(ii)× 3 - (i) × 2 ⇒
9x + 6y - 4x - 6y = 17100 - 9600
5x = 7500
x = 1500
The cost of a table Tk. 1500
Solution:
Let
The cost of a table and that of a chair be Tk. x and Tk. y respectively.
Then,
2x + 3y = 4800................(i)
and
3x + 2y = 5700................(ii)
(ii)× 3 - (i) × 2 ⇒
9x + 6y - 4x - 6y = 17100 - 9600
5x = 7500
x = 1500
The cost of a table Tk. 1500
১০,৩৮৯.
A trader, while selling a shirt, was asking for such a price that would enable him to offer a 20% discount and still make a profit of 25% on cost. If the cost of the shirt was Tk. 400, what was his asking price?
ব্যাখ্যা
Question: A trader, while selling a shirt, was asking for such a price that would enable him to offer a 20% discount and still make a profit of 25% on cost. If the cost of the shirt was Tk. 400, what was his asking price?
Solution:
প্রথম ধাপে, 25% লাভে শার্টটির বিক্রয় মূল্য নির্ণয় করতে হবে।
ক্রয়মূল্য = 400 টাকা
বিক্রয়মূল্য = ক্রয়মূল্য + ক্রয়মূল্যের 25%
= 400 + (400 × 25/100) = 400 + 100 = 500 টাকা
এখন, দ্বিতীয় ধাপে, 20% ছাড়ের উপর ভিত্তি করে ধার্য মূল্য নির্ণয় করতে হবে।
যদি ধার্য মূল্য 100 টাকা হয়,
তাহলে 20% ছাড়ে বিক্রয়মূল্য হবে 100 - 20 = 80 টাকা।
বিক্রয়মূল্য 80 টাকা হলে, ধার্য মূল্য = 100 টাকা
বিক্রয়মূল্য 1 টাকা হলে, ধার্য মূল্য = 100/80 টাকা
বিক্রয়মূল্য 500 টাকা হলে, ধার্য মূল্য = (100/80) × 500 = 625 টাকা
সুতরাং, শার্টটির ধার্য মূল্য ছিল 625 টাকা।
১০,৩৯০.
If 12 men or 18 boys can make 360 baskets in 15 days, then how many baskets will be made by 10 men and 15 boys in 15 days?
ব্যাখ্যা
Question: If 12 men or 18 boys can make 360 baskets in 15 days, then how many baskets will be made by 10 men and 15 boys in 15 days?
Solution:
Here,
12 men = 18 boys
∴ 1 man = 18/12 boys
= 3/2 boys
∴ 10 men = (3/2) × 10 = 15 boys
∴ 10 men and 15 boys = 15 boys + 15 boys
= 30 boys
18 boys can make 360 baskets in 15 days
∴ 1 boy can make in 15 days = 360/18 = 20 baskets
∴ 30 boys can make in 15 days = 20 × 30 = 600 baskets
∴ 10 men and 15 boys can make 600 baskets.
১০,৩৯১.
For which value of the angle of elevation, length of a stick and length of its' shadow is equal?
ব্যাখ্যা
Question: For which value of the angle of elevation, length of a stick and length of its' shadow is equal?
Solution:
ধরি, খুঁটিটির দৈর্ঘ্য AB, ছায়ার দৈর্ঘ্য BC, উন্নতি কোণ θ
AB = BC
চিত্র হতে,
tanθ = AB/BC = 1
⇒ tanθ = tan45°
⇒ θ = 45°
অর্থাৎ, উন্নতি কোণ 45° হলে, খুঁটির দৈর্ঘ্য ও ছায়ার দৈর্ঘ্য সমান হবে।
Solution:
ধরি, খুঁটিটির দৈর্ঘ্য AB, ছায়ার দৈর্ঘ্য BC, উন্নতি কোণ θ
AB = BC
চিত্র হতে,
tanθ = AB/BC = 1
⇒ tanθ = tan45°
⇒ θ = 45°
অর্থাৎ, উন্নতি কোণ 45° হলে, খুঁটির দৈর্ঘ্য ও ছায়ার দৈর্ঘ্য সমান হবে।
১০,৩৯২.
The sum of the first three of six consecutive integers is 30. Find the sum of the remaining three consecutive integers.
ব্যাখ্যা
Question: The sum of the first three of six consecutive integers is 30. Find the sum of the remaining three consecutive integers.
Solution:
Let the six consecutive integers are,
x, x + 1, x + 2, x + 3, x + 4, x + 5
The first three integers are,
x + (x + 1) + (x + 2) = 3x + 3
Given that,
3x + 3 = 30
⇒ 3x = 27
⇒ x = 9
So, the six consecutive integers are,
9, 10, 11, 12, 13, 14
The remaining three (i.e., the last three) integers are, 12, 13, 14
So their sum is = 12 + 13 + 14 = 39
Solution:
Let the six consecutive integers are,
x, x + 1, x + 2, x + 3, x + 4, x + 5
The first three integers are,
x + (x + 1) + (x + 2) = 3x + 3
Given that,
3x + 3 = 30
⇒ 3x = 27
⇒ x = 9
So, the six consecutive integers are,
9, 10, 11, 12, 13, 14
The remaining three (i.e., the last three) integers are, 12, 13, 14
So their sum is = 12 + 13 + 14 = 39
১০,৩৯৩.
Committee X has 4 members, Committee Y has 5 members, and these committee's have no members in common. If a task force is to be formed consisting of one member of X and one member of Y, how many different task forces are possible
ব্যাখ্যা
Committee X has 4 members, Committee Y has 5 members, and these committee's have no members in common.
A task force is to be formed consisting of one member of X and one member of Y,
So, 4 × 5 or 20 different task forces are possible.
----------------------------------------------------------------------------
Alternative way:
Let, Committee X has 4 members A, B, C, D
and Committee Y has 5 members M, N, P, Q, R
These committee's have no members in common and a task force is to be formed consisting of one member of X and one member of Y,
So, 4 × 5 or 20 different task forces are possible.
20 different task forces are given below:
Task force - (1) with X(A) and Y(M)
Task force - (2) with X(A) and Y(N)
Task force - (3) with X(A) and Y(P)
Task force - (4) with X(A) and Y(Q)
Task force - (5) with X(A) and Y(R)
Task force - (6) with X(B) and Y(M)
Task force - (7) with X(B) and Y(N)
Task force - (8) with X(B) and Y(P)
Task force - (9) with X(B) and Y(Q)
Task force - (10) with X(B) and Y(R)
Task force - (11) with X(C) and Y(M)
Task force - (12) with X(C) and Y(N)
Task force - (13) with X(C) and Y(P)
Task force - (14) with X(C) and Y(Q)
Task force - (15) with X(C) and Y(R)
Task force - (16) with X(D) and Y(M)
Task force - (17) with X(D) and Y(N)
Task force - (18) with X(D) and Y(P)
Task force - (19) with X(D) and Y(Q)
Task force - (20) with X(D) and Y(R)
A task force is to be formed consisting of one member of X and one member of Y,
So, 4 × 5 or 20 different task forces are possible.
----------------------------------------------------------------------------
Alternative way:
Let, Committee X has 4 members A, B, C, D
and Committee Y has 5 members M, N, P, Q, R
These committee's have no members in common and a task force is to be formed consisting of one member of X and one member of Y,
So, 4 × 5 or 20 different task forces are possible.
20 different task forces are given below:
Task force - (1) with X(A) and Y(M)
Task force - (2) with X(A) and Y(N)
Task force - (3) with X(A) and Y(P)
Task force - (4) with X(A) and Y(Q)
Task force - (5) with X(A) and Y(R)
Task force - (6) with X(B) and Y(M)
Task force - (7) with X(B) and Y(N)
Task force - (8) with X(B) and Y(P)
Task force - (9) with X(B) and Y(Q)
Task force - (10) with X(B) and Y(R)
Task force - (11) with X(C) and Y(M)
Task force - (12) with X(C) and Y(N)
Task force - (13) with X(C) and Y(P)
Task force - (14) with X(C) and Y(Q)
Task force - (15) with X(C) and Y(R)
Task force - (16) with X(D) and Y(M)
Task force - (17) with X(D) and Y(N)
Task force - (18) with X(D) and Y(P)
Task force - (19) with X(D) and Y(Q)
Task force - (20) with X(D) and Y(R)
১০,৩৯৪.
I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both way 55 minutes. How long would it take me to ride both ways?
ব্যাখ্যা
Question: I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both way 55 minutes. How long would it take me to ride both ways?
Solution:
Walk + ride back = 37 minutes. ................. (1)
Walk + Walk = 55 minutes.
Or, 2 walk = 55 minutes.
Walk = 55/2 = 27.5 minutes. .......................(2)
Putting equation (2) in equation (1).
27.5 + ride back = 37
Or, Ride back = 37 - 27.5 = 9.5.
∴ 2 × ride back = 9.5 × 2 = 19 minutes
Solution:
Walk + ride back = 37 minutes. ................. (1)
Walk + Walk = 55 minutes.
Or, 2 walk = 55 minutes.
Walk = 55/2 = 27.5 minutes. .......................(2)
Putting equation (2) in equation (1).
27.5 + ride back = 37
Or, Ride back = 37 - 27.5 = 9.5.
∴ 2 × ride back = 9.5 × 2 = 19 minutes
১০,৩৯৫.
A certain machine produces 1,000 units of product P per hour. Working continuously at this constant rate, this machine will produce how many units of product P in 7 days?
ব্যাখ্যা
7 days = 7 × 24 hours = 168 hours
In 1 hour, the machine produces 1,000 units
In 168 hours, the machine produces 1,000 × 168 units = 168000 units
In 1 hour, the machine produces 1,000 units
In 168 hours, the machine produces 1,000 × 168 units = 168000 units
১০,৩৯৬.
The sum of seven consecutive odd numbers exceeds four times the largest by 15. Find the average of average of these numbers.
ব্যাখ্যা
Question: The sum of seven consecutive odd numbers exceeds four times the largest by 15. Find the average of average of these numbers.
Solution:
Let the seven consecutive odd numbers be centered at n,
n - 6, n - 4, n - 2, n, n + 2, n + 4, n + 6
Sum of these consecutive odd numbers = n - 6 + n - 4 + n - 2 + n + n + 2 + n + 4 + n + 6 = 7n
And largest number = n + 6
ATQ,
7n = 4(n + 6) + 15
⇒ 7n = 4n + 24 + 15
⇒ 7n - 4n = 39
⇒ 3n = 39
∴ n = 13
∴ Average = 7n/7 = n = 13
১০,৩৯৭.
Two trains are running in opposite directions at the same speed. The length of each train is 120 metre. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
ব্যাখ্যা
Distance covered = (120 + 120) = 240 metre
Time = 12 seconds
Relative speed = 240/ 12
= 20 m/s
= 20 × 18 /5 km/hr
= 72 km/hr
Relative speed in this case is the sum of the speeds of the trains and each train has same speed,
speed of each train = 72 /2
= 36 km/hr
১০,৩৯৮.
The average (arithmetic mean) of three numbers is 3x + 2. If one of the numbers is x, what is the average of the other two numbers?
ব্যাখ্যা
Question: The average (arithmetic mean) of three numbers is 3x + 2. If one of the numbers is x, what is the average of the other two numbers?
Solution:
The average (arithmetic mean) of three numbers is 3x + 2
∴ The sum of three numbers is 3(3x + 2) = 9x + 6
If one of the numbers is x
∴ Sum of the other two numbers = 9x + 6 - x = 8x + 6
∴ The average of the other two numbers is = (8x + 6)/2 = 4x + 3
Solution:
The average (arithmetic mean) of three numbers is 3x + 2
∴ The sum of three numbers is 3(3x + 2) = 9x + 6
If one of the numbers is x
∴ Sum of the other two numbers = 9x + 6 - x = 8x + 6
∴ The average of the other two numbers is = (8x + 6)/2 = 4x + 3
১০,৩৯৯.
Of three consecutive even numbers, the sum of the 1st and 2nd is 166. The sum of the 2nd and 3rd is 170 and the sum of the 3rd and twice of 1st is 250. The second number is -
ব্যাখ্যা
Question: Of three consecutive even numbers, the sum of the 1st and 2nd is 166. The sum of the 2nd and 3rd is 170 and the sum of the 3rd and twice of 1st is 250. The second number is -
Solution:
১ম জোড় সংখ্যাটি = x - 2
২য় জোড় সংখ্যাটি = x
৩য় জোড় সংখ্যাটি = x + 2
প্রশ্নমতে,
x - 2 + x = 166
2x = 166 + 2
x = 168/2
x = 84
Solution:
১ম জোড় সংখ্যাটি = x - 2
২য় জোড় সংখ্যাটি = x
৩য় জোড় সংখ্যাটি = x + 2
প্রশ্নমতে,
x - 2 + x = 166
2x = 166 + 2
x = 168/2
x = 84
১০,৪০০.
A 80 m long ladder is leaning on a wall. If the ladder makes an angle of 45° with the ground, find the distance of the ladder from the wall.
ব্যাখ্যা
Question: A 80 m long ladder is leaning on a wall. If the ladder makes an angle of 45° with the ground, find the distance of the ladder from the wall.
Solution:
Here,
cosθ = Base/Hypotenuse
⇒ cos45° = Base/80
⇒ Base = 80cos 45° = 80 × (1/√2) = 40√2
∴ Distance of the ladder from the wall = 40√2 m
Solution:
Here,
cosθ = Base/Hypotenuse
⇒ cos45° = Base/80
⇒ Base = 80cos 45° = 80 × (1/√2) = 40√2
∴ Distance of the ladder from the wall = 40√2 m