৯ম - ১৩তম গ্রেড পরীক্ষার প্রস্তুতি

১৯৬৯ সালে ওআইসি প্রতিষ্ঠিত হয়। ওআইসির মূল অঙ্গসংস্থা চারটি। এগুলো হলো-
১. রাষ্ট্র বা সরকার প্রধানের শীর্ষ সম্মেলন।
২. পররাষ্ট্র মন্ত্রীদের সম্মেলন‌।
৩. ইসলামী সচিবালয়।
৪. আন্তর্জাতিক ইসলামী সচিবালয়।
উৎস: পৌরনীতি ও সুশাসন, একাদশ-দ্বাদশ শ্রেণি, প্রফেসর মোঃ মোজাম্মেল হক।

Clearly, (4 - 1) = 3, (5 - 2) = 3, (6 - 3) = 3, (7 - 4) = 3 and (8 - 5) = 3
LCM of 4, 5, 6, 7, 8 = 840
Greatest number of 4 digits = 9999
On dividing 9999 by 840, the remainder is 759
So, required number = (9999 - 759) - 3 = 9237

Let the total number of students in the class be x.
Then, number of boys = 2x/3 & number of girls = x/3.
Number of boys who are over 160 cm tall = (3/4)×(2x/3) = x/2.
So, x/2 = 18 or x = 36
Therefore, Number of girls = 36/3 = 12.

If a month beings with Sunday then there are 5 Sundays in that month.
Total number of visitors come on Sunday = 510×5 = 2550
Total number of visitors come on other days = 240×25 = 6000
Average number of visitors per day = (2550+6000)/30 = 8550/30 = 285

Average of 12 numbers = 15
Total of 12 numbers = 15×12 = 180
Average of first two number = 14
Total of first two number = 14×2 = 28
Total of remaining ten numbers = 180-28 = 152
Required average of remaining ten number
= 152/10 = 76/5 = 15(1/5)

Let the numbers be x, x+1 and x+2
Then,
x²+(x+1)²+(x+2)² = 2030
⇒ 3x²+6x-2025 = 0
⇒ x²+2x-675 = 0
⇒ (x+27) (x-25) = 0
⇒ x = 25
∴ Middle number = (x+1) = 26

Let the present age of Y be a years
Three years ago X's age = 3a years
Then, present age of X is (3a + 3)
Z's present age = 2a
According to question
Now, (3a + 3) - 2a = 12
a = 9 year
∴ Present age of Z = 2a = 2 × 9 = 18 years

Let total population = x. Then, number of males = (5/9)x
Married males = 30% of (5/9)x = (30/100)×(5/9)x = x/6.
Married females = x/6
Number of females = (x - (5/9)x) = 4x/9
Unmarried females = (4x/9)-(x/6) = 5x/18
∴ Required percentage = ((5x/18)×(1/x)×100)%
= 27(7/9)%

Let error = x gms.
Then,
x/(1000−x)×100 = 25 ⇔ 100x/(1000−x) = 25
⇔ 100x = 25000−25x ⇔ 125x = 25000
⇔ x = 200
∴ Weight used = (1000−200) gms
= 800 gms.

When marbles are shared in the ratio 1 : 2 : 3
A's share = 60×(1/6) = 10; B's share = 60×(2/6) = 20;
C's share = 60×(3/6) = 30.
When marbles are shared in the ratio 3 : 4 : 5
A's share = 60×(3/12) = 15; B's share = 60×(4/12) = 20;
C's share = 60×(5/12) = 25.
Clearly, 5 marbles have been transferred from C to A.

Suppose they meet x hrs after 9 a.m.
Then,
Distance traveled by car X in (x-1) hrs + Distance traveled by car Y in x hrs = 700 km
Now, let they will meet at a distance of x km from X's position.
∴ 60(x-1)+60x = 700 ⇒ 120x = 760
⇒ x = 760/120 = 19/3 hrs = 6 hr 20 min.
So, they will cross each other after 6 hr 20 min after 9 a.m. i.e., at 3 : 20 pm.

Let the speed of the boat in still water be x kmph
Then, Speed downstream = (x+4) kmph.
Speed upstream = (x-4) kmph.
∴ (x+4)×6 = (x-4)×8
⇒ 6x+24 = 8x-32 ⇒ 2x = 56 ⇒ x = 28 kmph.

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x−1) hours = 25(x−1) km
∴ 20x+25(x−1) = 110
⇒ 45x = 135
⇒ x = 3
So, they meet at 10 a.m.

Let the quantity of milk and water in initial mixture be 4x and 3x litres.
Quantity of water = 3x litres
On adding 6 litres of water, we get
4x/(3x+6) = 8/7
⇒ 28x = 24x + 48
⇒ 28x - 24x = 48
⇒ 4x = 48
⇒ x = 48/4 = 12.
∴ Required quantity of milk
= 4x = 4×12 = 48 litres.

Let the sum be Tk x and original rate be R%
Then, ((x×(R+1)×3)/100)-((x×R×3)/100) = 5100
⇒ 3Rx+3x-3Rx = 5,10,000
⇒ 3x = 5,10,000 ⇒ x = 1,70,000
Hence, sum = Tk 1,70,000

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

The given sequence has four terms. So the rule must be correct for at least three of them. The first term is 1. The second term is 2 which can either mean 1 is added or 2 has been multiplied. the third term is 6 which can be got from 2 by multiplying it by 3. So thus far we have 1×1, 1×2, 2×3, 6×4, which works perfectly for the series. Thus the rule has been found and the last term will be 24×5 = 120. The answer is thus 120.

In this question, we are required to find out the possible length of the farm. We need to find the inequality formula to find the length. The formula for the perimeter of a rectangle is 2l + 2b, where l is for length and b is for breadth. As given in the question the total perimeter required should be less than or equal to 60, because that is the amount of fencing Harun and Salam has for their farm.

So, the inequality formula here will be, 2l + 2b ≤ 60. But the breadth (width) is given to us in this question. So, we will replace b by 20 feet. Thus, the above formula will be written as,

2l + 40 ≤ 60. Now, we need to solve this equation to find the possible length of the farm. So, the given will be,
2l + 40 ≤ 60
2l + 40 – 40 ≤ 60 – 40
2l ≤ 20
2l/2 ≤ 20/2
l ≤ 10

Thus, the required possible length for the farm is 10 feet.