ব্যাখ্যা
Question: Find the value of x, if (x/7) - (x/9) = 2
Solution:
Given that,
(x/7) - (x/9) = 2
⇒ (9x - 7x)/63 = 2
⇒ 2x = 2 × 63
∴ x = 63
PrepBank · বিষয়ভিত্তিক প্রশ্ন
PrepBank · পাতা ৮৮ / ১৬১ · ৮,৭০১–৮,৮০০ / ১৬,১২৪
Question: Find the value of x, if (x/7) - (x/9) = 2
Solution:
Given that,
(x/7) - (x/9) = 2
⇒ (9x - 7x)/63 = 2
⇒ 2x = 2 × 63
∴ x = 63
(C1 × Q1 + C2 × Q2)/(Q1 + Q2) = P
or, (8 × 124 + 4 × Q2)/(24 + Q2) = 5
or, 192 + 4 × Q2 = 120 + 5 × Q2
or, Q2 = 72
Question: The number of multiples of 4 between 10 and 250 is:
Solution:
10 এবং 250 এর মধ্যে 4-এর প্রথম গুণিতক হলো 12 (যেহেতু 12 > 10)
10 এবং 250 এর মধ্যে 4-এর শেষ গুণিতক হলো 248 (যেহেতু 248 < 250)
এখন, এটি একটি সমান্তর ধারা, যার প্রথম পদ (a) = 12, শেষ পদ (p) = 248 এবং সাধারণ অন্তর (d) = 4।
আমরা জানি,
পদসংখ্যা (n) = {(শেষ পদ - প্রথম পদ)/সাধারণ অন্তর} + 1
= {(248 - 12)/4} + 1
= {236 / 4} + 1
= 59 + 1
∴ n = 60
Question: - 5x - [4y - {9x - (3y - 7x)}] simplifies to
Solution:
- 5x - [4y - {9x - (3y - 7x)}]
= - 5x - [4y - {9x - 3y + 7x}]
= - 5x - [4y - 9x + 3y - 7x]
= - 5x - [7y - 16x]
= - 5x - 7y + 16x
= 11x - 7y
Man walks 20 km in → 5 hours
That means it will take more time to walk 32 km.
This is the case of direct proportion.
20/5 = 32/x
x = 32/4
= 8
Question: If (x + 3)2 = 64, which of the following can be the value of (x + 2)?
Solution:
Given, (x + 3)2 = 64
⇒ (x + 3)2 = 82
∴ x + 3 = ± 8
Case 1: x + 3 = 8
x = 8 - 3 = 5
x + 2 = 5 + 2 = 7
Case 2: x + 3 = - 8
x = - 8 - 3 = - 11
x + 2 = - 11 + 2 = - 9
Possible values of (x + 2) are 7 or - 9.
Question: A number is tripled, then 7 is subtracted from it. If the result is then doubled, it becomes 58. What is the number?
Solution:
Let,
the number be x
ATQ,
2(3x - 7) = 58
⇒ 6x - 14 = 58
⇒ 6x = 58 + 14
⇒ 6x = 72
⇒ x = 72/6
∴ x = 12
So the number is 12.
The question requires you to find a number of the outcomes in which at most 3 coins turn up as heads.
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is,
6C0 = 1 outcome.
The number of outcomes in which 1 coin turns head is,
6C1 = 6 outcomes.
The number of outcomes in which 2 coins turn heads is,
6C2 = 15 outcomes.
The number of outcomes in which 3 coins turn heads is,
6C3
Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.
Question: A buys a product for Tk. 500 and sells it to B at a profit of 25%. B then sells it to C at a profit of 20%. How much does C pay to B?
সমাধান:
A এর 25% লাভে বিক্রয়মূল্য = 500 + 500 এর 25%
= 500 + (500 × 25 / 100)
= 500 + 125
= 625
A এর বিক্রয়মূল্য = B এর ক্রয়মূল্য
B এর 20% লাভে বিক্রয়মূল্য = 625 + 625 এর 20%
= 625 + (625 × 20 / 100)
= 625 + 125
= 750
সুতরাং, B এর বিক্রয়মূল্য = C এর ক্রয়মূল্য = Tk. 750
Question: What is the average of 0.36, 4.6, 0.64, and 2.42?
Solution:
Average = (0.36 + 4.6 + 0.64 + 2.4)/4
= 8.00/4
= 2.00
Price after 1stdiscount of 20% = (100-20)%
= 80% of Marked Price
∴ Price = (80/100) × 12800
= Tk. 10240
SP = (100 - Discount) % of Price
∴ 9216 = (100 - Discount) % x 10240
⇒ (100 - Discount)/100 = 9216/10240
⇒ 100 - Discount = (9216 × 100)/10240
⇒ 100 - Discount = 90
⇒ Discount = 100 - 90 = 10%
10% is the percent of 2nddiscount offered.
Question: A cube with side length 6 cm fits perfectly inside a hollow spherical ball. What is the total surface area of the sphere?
Solution:
If a cube fits perfectly inside a sphere, then the diameter of the sphere = space diagonal of the cube
Diagonal of the cube = √3a = 6√3
So, diameter of sphere = 6√3
Radius = 6√3/2 = 3√3
Surface area of the sphere = 4πr2
= 4π(3√3)2
= 108π cm2
n(S) = 20C2 = 190
n(E) = 15C2 = 105
Therefore,
P(E) = 105/190
= 21/38
Question: A boat sails m miles upstream at r miles/hr. If the speed of the stream is s miles/hr, how long will it take the boat to return to its starting point?
Solution:
মনেকরি
নৌকার গতিবেগ = x কিমি/ঘণ্টা
স্রোতের গতিবেগ = s কিমি/ঘণ্টা
এখানে
x - s = r
x = r + s
আবার
স্রোতের অনুকূলে বেগ = x + s কিমি/ঘণ্টা
= r + s + s কিমি/ঘণ্টা
= r + 2s
স্রোতের অনুকূলে ফিরে আসতে সময় লাগবে = দূরত্ব/বেগ
= m/(r + 2s)
Question: If two coins are tossed, what is the probability of getting at least one head?
Solution:
দুটি মুদ্রা নিক্ষেপ করলে নমুনা ক্ষেত্রটি হলো,
S = {HH, HT, TH, TT}
এখানে মোট ফলাফল সংখ্যা, n(S) = 4
"at least one head" বলতে বোঝায় কমপক্ষে 1টি head অর্থাৎ 1 টি অথবা 2 টিও হতে পারে।
1টি head আছে এমন ফলাফল: {HT, TH}
2টি head আছে এমন ফলাফল: {HH}
∴ অনুকূল ফলাফল, n(E) = {HT, TH, HH}
∴ n(E) = 3
∴ সম্ভাবনা P(E) = n(E)/n(S)
∴ P(E) = 3/4
Question: A, B, and C invest a total of TK. 80000 in a business. A invests TK. 7000 more than B, and B invests TK. 5000 more than C. Out of a total profit of TK. 48000. How much profit does A receive?
Solution:
Let C = x.
Then, B = x + 5000
and A = x + 5000 + 7000
= x + 12000
So,
x + x + 5000 + x + 12000 = 80000
⇒ 3x + 17000 = 80000
⇒ 3x = 80000 - 17000
⇒ 3x = 63000
⇒ x = 21000
∴ x = 21000
so, C = 21000, B = 21000 + 5000 = 26000, A = 21000 + 12000 = 33000
A : B : C = 33000 : 26000 : 21000
= 33 : 26 : 21
∴ Total ratio = 33 + 26 + 21 = 80
So A's Share
= TK. 48000 × (33/80)
= TK. 19800
A = 150% of B
⇒ A = 150/100 B
⇒ A/B = 3/2
⇒ A/B + 1 = 3/2 + 1
⇒ (A + B)/B = 5/2
⇒ B/(A + B) = 2/5
∴ Required percentage:
= {B / (A + B)} × 100 %
= (2/5 × 100) %
= 40%
Let sum be x
T = 20 years
Since the money doubles, simple interest = x
R = (100 × SI)/PT
= (100 × x)/(x × 20)
= 5%
Question: The average temperature for Wednesday, Thursday and Friday was 40°C. The average for Thursday, Friday and Saturday was 41° C. If temperature on Saturday was 44° C, what was the temperature on Wednesday?
Solution:
Average temperature for Wednesday, Thursday and Friday = 40° C
∴ Total temperature = 3 × 40 = 120° C
Average temperature for Thursday, Friday and Saturday = 41° C
∴ Total temperature = 41 × 3 = 123° C
And,
Temperature on Saturday = 44° C
Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120
⇒ Saturday - Wednesday = 3
∴ Wednesday = 44 - 3 = 41° C
প্রতি লাইনে 11 words, প্রতি page এ 36 লাইন বিশিষ্ট 125 pages এ
word আছে 11× 36 × 125
5 দিনে 11 × 36 × 125 words হয়
1 দিনে (11 × 36 × 125)/5 words হয়
∴ 6 দিনে (11 × 36 × 125 × 6)/5 words হয়
= (11 × 36 × 125 × 6)/(5 × 12) lines
= (11 × 36 × 125 × 6)/(5 × 12 ×30) pages
= 165 pages.
There are 15 dots in total,and to make a triangle we need to select any three of those dots.
So, 15C3 = 455
এখানে,
21, 18, 15, 12,
১ম পদ, a = 21
সাধারন অন্তর, d = 18 - 21 = -3
∴ সমষ্টি = n/2{2a + (n - 1)d}
⇒ 0 = n/2{(2 × 21) + (n - 1)(-3)}
⇒ 0 = n/2(42 - 3n + 3)
⇒ 45n - 3n2 = 0
⇒ 3n(n - 15) = 0
⇒ n - 15 = 0
∴ n = 15.
Question: In the xy-plane, a triangle has vertices (0, 0), (k, 0) and (k, - 5k), where k > 0. If the area of the region enclosed by the triangle is 40, what is the value of k?
Solution:
প্রদত্ত ত্রিভুজটির শীর্ষবিন্দুগুলো হলো (0, 0), (k, 0) এবং (k, - 5k)।
যেহেতু B এবং C বিন্দুর x-স্থানাঙ্ক একই (k), তাই BC রেখাটি y-অক্ষের সমান্তরাল।
যেহেতু A এবং B বিন্দুর y-স্থানাঙ্ক একই (0), তাই AB রেখাটি x-অক্ষের সমান্তরাল।
সুতরাং, ত্রিভুজটি B বিন্দুতে একটি সমকোণী ত্রিভুজ।
দুটি বিন্দুর স্থানাঙ্ক (x1, y1) এবং (x2, y2) হলে তাদের মধ্যবর্তী দূরত্ব = √{(x2 - x1)2 + (y2 - y1)2}
∴ ভূমি = (0, 0) এবং (k, 0) বিন্দুর মধ্যবর্তী দূরত্ব = k (যেহেতু k > 0)।
∴ উচ্চতা = (k, 0) এবং (k, - 5k) বিন্দুর মধ্যবর্তী দূরত্ব = 5k (যেহেতু k > 0)।
ত্রিভুজের ক্ষেত্রফল = (1/2) × ভূমি × উচ্চতা
প্রশ্নমতে,
(1/2) × k × 5k = 40
⇒ (5/2)k2 = 40
⇒ 5k2 = 80
⇒ k2 = 80/5
⇒ k2 = 16
⇒ k = √16
⇒ k = ±4
যেহেতু প্রশ্নে দেওয়া আছে k > 0, তাই k এর মান হবে 4।
∴ k এর মান 4।
Question: If x2 - 10x + 25 = 0, then the value of x is:
Solution:
দেওয়া আছে,
x2 - 10x + 25 = 0
⇒ x2 - 2. x. 5 + 25 = 0
⇒ (x - 5)2 = 0
⇒ (x - 5)(x - 5) = 0
∴ x = 5 এবং x = 5 [যেহেতু সমীকরণটি একটি দ্বিঘাত সমীকরণ তাই এর মূল হবে দুইটি]
Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
Solution:
In a simultaneous throw of two dice,
we have n(S) = (6 × 6) = 36
Now, we find the odd product,
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5 ,5)}
∴ n(E) = 9
∴ P(odd product)= 9/36 = 1/4
Now, we find the even product,
probability(product even) = 1 - (1/4) = 3/4
Assume first child (the youngest) get = Tk. x
According to the question ;
each son having Tk. 30 more than the younger one
Second child will get = Tk. x + 30
Third child will get = Tk. x + 30 + 30 = x + 60
Fourth child will get = Tk. x + 30 + 30 + 30 = x + 90
Fifth child will get = Tk. x + 30 + 30 + 30 + 30 = x + 120
Total amount they got = Tk. 2000
x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Tk. 340
So the youngest child will get Tk. 340.
Question: If logxy = 100 and log3x = 20; then the value of y is-
Solution:
Given,
log3x = 20
∴ x = 320
And, logxy = 100
⇒ y = x100
⇒ y = (320)100
∴ y = 32000
Question: The speeds of three cars are the ratio 2 : 3 : 4. The ratio of the taken by these cars to travel the same distance is-
Solution:
Given that,
Speed ratio of three cars,
v1 : v2 : v3 = 2 : 3 : 4
Let,
v1 = 2k, v2 = 3k, v3 = 4k (for some constant k)
We know,
Time = distance/Speed
∴ t1 = d/2k, t2 = d/3k, t3 = d/4k
∴ Ratio of time = t1 : t2 : t3 = d/2k : d/3k : d/4k
= 1/2 : 1/3 : 1/4 ; [Cancel d and k (since d,k ≠ 0)]
= 12/2 : 12/3 : 12/4. ; [LCM of 2, 3, 4 = 12]
= 6 : 4 : 3
So the ratio of the time taken is 6 : 4 : 3
ATQ,
dx + 11 = 2dx + 48
⇒ dx = 37
So, the divisor is 37
Greatest number of four - digits is 9999.
L.C.M. of 4, 7 and 13 = 364
On dividing 9999 by 364, the remainder obtained is 171.
∴ Greatest number of 4 - digits divisible by 4, 7 and 13 = (9999 - 171)
= 9828.
Hence, required number = (9828 + 3)
= 9831.
From M1D1 = M2D2
⇒ M1D1 / M2D2 = 5/6
⇒ (x−1)(x+1)/(x+1)(x+2) = 5/6
⇒ (x−1)/(x+2) = 5/6
⇒ 6x−6 = 5x+10
⇒ x = 16
Question: What is the greatest number that will divide 96, 132, and 150 leaving remainders 6, 9, and 12 respectively?
Solution:
We have to subtract the reminder first:
96 - 6 = 90
132 - 9 = 123
150 - 12 = 138
Using prime factorization:
For 123:
123 = 3 × 41
For 90:
90 = 2 × 32 × 5
For 138:
138 = 2 × 3 × 23
Digit (123, 90, 138) have at least 3 in common.
∴ The greatest number is 3