৯ম - ১৩তম গ্রেড পরীক্ষার প্রস্তুতি

Let the numbers be 13x and 15x
Then, their LCM = 195x
So, 195x = 39780
x = 204
∴ The numbers are 2652 and 3060

We, know that
(a × b) = (HCF and LCM)
396 × 576 = HCF × 6336
HCF = 36

Let the numbers be 2x, 3x, 5x and 7x respectively
Then their LCM = (2×3×5×7)x = 210x. [∴ 2, 3, 5, 7 are prime numbers]
So, 210x = 630 or x = 3
∴ The numbers are 6, 9, 15 and 21
Required difference = 21 - 6 = 15

Other number
= (12×336)/84
= 48

(4848/24)×11−222
=202×11−222
=2222−222
=2000

4 gal = (4×4)qt = 16 qt = (16×12)pt = 32 pt
∴ Number of botteles = 32/(1/2) = 64

Sourov's monthly income = (2,34,000)/12 = Tk 19,500.

Jashim's monthly income = (3/2)×19,500 = Tk 29,250.

Gourob's monthly income = (2×29,250) = Tk 58,500.

Sum of marks in 4 subjects = 75×4 = 300
Sum of marks in 5 subjects = 300+80 = 380
∴ New average = 380/5 = 76

Age of the boy sitting in the middle
= 26×7-(19×3+32×3)
= 180-153
= 29 years

Let N be the no. of persons in the group.
Required number of person is given by;
Member in group × aged increased = difference of replacement
N × 5 = 38 - 18
Or, 5N = 20
Or, N = 4

Let the five numbers be x, (x + 2), (x + 4), (x + 6) and (x + 8)

Then, x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 575
⇔ 5x + 20 = 575 ⇔ 5x = 555 ⇔ x = 111

∴ Required sum = (x + 10) + (x + 12) + (x + 14) + (x + 16) + (x + 18)
= 5x + 70
= 5 × 111 + 70
= 555 + 70
= 625

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y
∴ 10x+y = 7(x+y) ⇔ 3x = 6y ⇔ x = 2y
Number formed by reversing the digits = 10y + x
∴ (10x+y)−(10y+x) = 18 ⇔ 9x−9y = 18 ⇔ x−y = 2 ⇔ 2y−y = 2 ⇔ y = 2
So, x = 2y = 4.

Let the numbers be a and b, where a > b
According to the question,
a+b = 37 and a²−b² = 185
⇒ (a+b)(a−b) = 185
⇒ 37(a−b) = 185
⇒ a−b = 185/37
⇒ a−b = 5

Let's assume the present age of mother be 7x years and daughter be x years.
(7x-4)/(x-4) = 19/1
⇒ 7x-4 = 19(x-4)
⇒ 19x-7x = 76-4
⇒ 12x = 72
⇒ x = 6
∴ Mother's age after 4 years = 7x+4 = 7×6+4 = 46 years.

Let Nishi's age be 6x years. Then, Tanvir's age = 5x years.
∴ (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) = x years = 3 years.

15% of 578 + 22.5% of 644 = ((15/100)×578)+((225/10)×(1/100)×644
= 86.7+144.9
= 231.6

Let the total number of students be x.
Then, (100−76)% of x = 204
⇒ 24% of x = 204
⇒ (24/100)x = 204
⇒ x = 204×(100/24)
⇒ x = 850

B's wage is less than A's by = [20/(100+20)×100]% i.e, 50/3% = 16(2/3)%

Let C.P. = 10x and S.P. = 11x.
Profit = 11x−10x = x

∴ Profit % = ((x/10x)×100)% = 10%.

Let the cost of the article be x.
Then, cost paid by retail merchant = 95% of 120% of x
= ((95/100)×(120/100)×x) = ((114/100)x)
= 114% of x.
∴ Profit = 14%

Rate of discount
= ((12/80)×100)%
= 15%

Let the numbers be x and y respectively.
Then, y−30% of x = (4/5)y
⇒ y-(4/5)y = (30/100)x
⇒ y/5 = 3x/10
⇒ x/y = (1/5)×(10/3) = 2/3
⇒ x:y= 2:3

Let the numbers be 7x and 11x respectively.
Then, (7x+7)/(11x+7) = 23
⇒ 21x+21 = 22x+14
⇒ x = 7
∴ Smaller number = 7×7 = 49

Total number of days = (30 + 31 + 30) = 91
Let the number of bananas be x
More days, More bananas (Direct proportion)
∴ 7:91::651:x
⇒ 7x = 91×651
⇒ x = (91×651)/7
⇒ x = 8463

Let the required number of days be x
Less men, More days (Indirect proportion)
∴ 42:56::24:x
⇒ 42×x = 56×24
⇒ x = (56×24)/42
⇒ x = 32

Let the required number of mats be x.
10 men ≡ 20 boys → 1 man ≡ 2 boys → 8 men ≡ 16 boys
⇒ (8 men + 4 boys) ≡ (16+4) boys = 20 boys
∴ 8 men and 4 boys can make as many as mats as 20 boys i.e., 260 mats

A's 1 day's work = 1/(T+3)
B's 1 day's work = 1/(T+12)
(A + B)'s 1 day's work = 1/T

∴ 1/(T+3) + 1/(T+12) = 1/T ⇒ (2T+15)/((T+3)(T+12)) = 1/T
⇒ 2T²+15T = T²+15T+36
⇒ T² = 36
⇒ T = 6 days

Ratio of times taken by A and B = 100 : 80 = 5 : 4
Suppose B takes x hours to do the work.
Here, 5:4 :: 15/2 : x
⇒ x = (2×15)/3
⇒ x = 6 hours

B's 8 day's work = (1/10)×8 = 4/5
Remaining work = 1−(4/5) = 1/5
Now, 1/15 work is done by A in 1 day.
∴ 1/5 work is done by A in = 15×(1/5) = 3 days.

Ratio of the speed of A, B and C = 6 : 3 : 1
Then, ratio of time taken= 1/6 : 1/3 : 1 = 1 : 2 : 6
Hence, time taken by A = 72/6 = 12 minutes.

Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or, 20×(10/60) = 8/(60(20+x))
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph.

Speed downstream = (5 + 1) kmph = 6 kmph
Speed upstream = (5 - 1) kmph = 4 kmph
Let the required distance be x km
Then,
(x/6)+(x/4) = 1
⇒ 2x + 3x = 12
⇒ 5x = 12
⇒ x = 2.4 km

Distance travelled in 14 sec = 50+l
Distance travelled in 10 sec = l
So, speed of train
= 50/(14−10) m/sec
= (50/4)×(18/5) km/hr
= 45 km/hr

Speed = 60×(5/18) m/sec = 50/3 m/sec
Length of the train = (Speed×Time)
∴ Length of the train = (50/3)×9 m = 150 m

In 20 litres of mixture
Quantity of milk ⇒ (3/4)×20 = 15 litres

Quantity of water ⇒ (1/4)×20 = 5 litres

Let the quantity of milk added be x litres.
According to the question,
⇒ (15+x)/5 = 41
⇒ 15+x = 4×5
⇒ x = 20−15 = 5 litres

Let Milk was 4x and water was 3x.
6L water was added, so
Water = 3x +6
New ratio = 8/7
4x/(3x +6) = 8/7
28x = 24x + 48
4x = 48
x = 12.
Milk = 4x = 4×12 = 48 litres

Let the required time = T years
Then, (500×4×6.25)/100 = (400×5×T)/100
⇒ T = (500×4×6.25)/(400×5) years = 6(1/4) years

πR² = 7×(2πR)
⇒ R = 14
∴ Circumference = (2×(22/7)×14) units
= 88 units

Radius of circle = 7 cm
Given area of rectangle = Area of circle = (22/7)×7×7
= 154 cm²

3×1+1×2 = 3+2 = 5
5×2+2×3 = 10+6 = 16
18×3+3×4 = 48+12 = 60
60×4+4×5 = 240+20 = 260
260×5+5×6 = 1300+30 = 1330

২৮ + ১১ = ৩৯
৩৯ + ১৯ = ৫৮
৫৮ + ২৭ = ৮৫

এখানে,
১১ + ৮ = ১৯
১৯ + ৮ = ২৭

4 → 2²×1
18 → 3²×2
48 → 4²×3
100 → 5²×4
180 → 6²×5
294 → 7²×6

Option 1: The range of values that x takes is infinite.
Option 2: An infinite range of values.
Option 3: An infinite range of values.
Option 4: Finite range of values for 'x'.

Factorizing x² - 4x + 3 < 0 we get, (x - 3)(x - 1) < 0.
This inequality will hold good when one of the terms (x - 3) or (x - 1) is positive and the other is negative.

Possibility 1: (x -3) is positive and (x - 1) is negative.
i.e., x - 3 > 0 AND x -1 < 0
i.e., x > 3 AND x < 1
Such a number DOES NOT exist. It is an infeasible solution.

Possibility 2: (x - 3) is negative and (x - 1) is positive.
i.e., x - 3 < 0 AND x - 1 > 0
i.e., x < 3 AND x > 1
Essentially translates to 1 < x < 3