৯ম - ১৩তম গ্রেড পরীক্ষার প্রস্তুতি

Average of 20 numbers = 0
∴ Sum of 20 numbers (0 × 20) = 0
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20^th number is (-a)

Sum of the present ages of husband, wife and child
= (27 x 3 + 3 x 3) years
= 90 years
Sum of the present ages of wife and child
= (20 x 2 + 5 x 2) years
= 50 years
∴ Husband's present age
= (90 - 50) years
= 40 years

HCF = 17
Let numbers are = 17x, 17y
LCM = 17xy = 714 (given)
xy = 42
Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7)
Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119)
but given that both numbers are of three digits
∴ numbers are = (102, 119)
∴ sum of numbers = 102 + 119 = 221

Let number are = 2x, 3x, 4x
given,
LCM of (2×3×2)x = 12x
12x = 240
x = 20
∴ numbers are 2×20 = 40
3×20 = 60
4×20 = 80
∴ Smaller is 40

Let the loan amount be Tk. x
Then,
⇒ (6x/100)+(7.5x/100)+(9x/100) = 8190
⇒ 22.5x = 819000
⇒ x = 36400

According to question,
√0.00005746
⇒ √5746100000000
⇒ 75.8/10000
⇒ 0.00758

According to question,
√(0.25/0.0009) × √(0.09/0.36)
⇒ √((25/9)×100) × √(9/36)
⇒ ((5×10)/3) × (3/6)
⇒ 25/3
⇒ 8(1/3)

Let the number be x and y
Then,
x²−y² = 63 & x−y = 3
On dividing, we get: x + y = 21
Solving x + y = 21 and x - y = 3,
We get: x = 12 and y = 9
∴ Larger number = 12

Let X be the number which is added to 80
80% of X = 0.8X
Now,
80 + 0.8X = X
0.2X = 80
X = 80/0.2 = 400

Quantity of salt in 6L of sea water,
= (6×4)/100 = 0.24
Percentage of salt in 5L of sea water,
= (0.24×100)/5
= 4(4/5)%

To gain 20% on whole he must sell all good for,
Tk. 400 + 20% of 400 = 480
As he get 5% gain on half of the goods i.e. 200 + 5% of 200 = 210
So required balance = 480 - 210 = 270
He must gain Tk. 70 on rest Tk. 200
% gain on remainder goods = (70×100)/200
= 35%

Let original CP = Rs. 100
Then, the Marked Price = 40% of 100 + 100 = 140
SP = 140 - 25% of 140 = 105
%Profit = (5×100)/100 = 5%

Net Graphic Change Method:
100 == 40% UP ⇒ 140 == 25% discount ⇒ 105 So, % Profit = 5%

Let Nila's age be 5x years and
Shila's age be 6x years

((1/3)×5x):((1/2)×6x) = 5:9
⇒ 5x/(3×3x) = 5/9

Thus, Shila's age cannot be determined

Let the son's age 18 years ago be x years,
Then man's age 18 years ago = 3x years
(3x+18) = 2(x+18)
⇒ 3x+18 = 2x+36
⇒ x = 18
Sum of their present ages
⇒ (3x+18+x+18) years
⇒ (4x+36) years
⇒ (4×18+36) years
⇒ 108 years

Cost of 21 pencils and 9 clippers = Tk. 819
Cost of 7 pencils and 3 clippers = 819/3 = Tk. 273

Cloth is required for 1 shirt
= 2 m, 60 cm or 260 cm
Cloth is required for 7 shirt
= 260 × 7
= 1820 cm or 18 m 20 cm

Let the required number of rounds be x
More radius, Less rounds (Indirect proportion)
∴ 20:14::70:x
⇔ (20×x) = (14×70)
⇔ x = (14×70)/20
⇔ x = 49

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving the two equations, we get
x = 11/400, y = 1/400
∴ 1 woman's 1 day's work = 1/400
⇒ 10 women's 1 day's work = ((1/400)×10) = 1/40
Hence, 10 women will complete the work in 40 days

Let the speed of the stream be x km/hr
Then speed downstream = (10 + x) km/hr
Speed upstream
= (10−x)km/hr
∴ 26/(10+x) = 14/(10−x)
⇒ 260−26x = 140+14x
⇒ 40x = 120
⇒ x = 3km/hr

x:7.5 = 7:17.5
⇒ 17.5x = 7.5×7
⇒ x = (7.5×7)/17.5
= 3

Product of 1^st and 4^th terms (extremes) = product of 2^nd and 3^rd terms (means)
⇒ 2.5x = 40
⇒ x = 40/2.5 = 16

25% of A = 35% of B
⇒ (25/100)A = (35/100)B
⇒ A/4 = 7B/20
⇒ A/B = (7/20)×4 = 7/5
⇒ A:B = 7:5

Let X be the distance, then
(x/5)−(x/8) = 3/2
x = 20km

Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute. Hence, if it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min per hour.
Time taken to travel 300 km with stoppage,
= 300/40 = 7.5 hours
Time taken for stoppage as it stops for 20 min per hour
= (7 × 20 + 10)
= 140 + 10
= 150 minutes.
= 150/60 hours
= 2.5 hours.

By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)
Hence, his time originally must have been 30 minutes.
Thus required distance = 20 kmph × 0.5 hours = 10 km.

Speed in m/sec = 60×(5/18) = (50/3) m/sec
Time taken to cross the man = 6 secs
Therefore, Length of the train = (Speed×Time)
= (50/3)×6 = 100 metres

Distance = 600 metres
Total Speed = 64 + 80 = 144 kmph (added because they are travelling in opposite directions)
In m/sec, speed = 144×(5/18) = 40 m/sec
Distance = Speed×Time
600 = 40×Time
Therefore, Time = 15 seconds

Total Distance = 300 + 150 = 450 m
Speed = 60 kmph = 60×(5/18)=(50/3) m/sec
Distance = Speed×Time
450 =(50/3)×Time
Time = 27 seconds

According to the question,
Acid : Water -
Vessel A - 4 : 3
Vessel B - 2 : 3
Now using alligation,

According to the question,
Chromium : Steel -
Type 1 - 2 : 11
Type 2 - 5 : 21
Now using alligation,

Ratio of quantity → 1 : 2

According to the question,
Alcohol : Water -
Vessel A - 4 : 3
Vessel B - 2 : 3
Now using alligation,

let the width of rectangle be x so the length & breath is increased by 2x.

so, new total area along with walkway is (60+2x)×(20+2x)

so, (60+2x)×(20+2x)-60×20 = 516
⇒ (60+2x)×(20+2x) = 1716
⇒ (30+x)×(10+x) = 429 = 33×13
⇒ x = 3

Let, ABCD be a || gm in which AB = 30 m, BC = 14 m & AC = 40 m.
Clearly, area of || gm ABCD = 2 (area of ∆ABC).
Let, a = 30, b = 14 & c = 40.
Then, s = (1/2)(a+b+c) = 42
Therefore, area of ∆ABC = √s(s-a)(s-b)(s-c)
= √42×12×28×2 = 168 m²
Therefore area of || gm = (2×168) m² = 336 m²

Sum of present dimension 48+30+52 = 130.
New dimension = 156.
Increase in dimension = 26.
Ratio of dimensions = 48:30:52 ⇒ 24:15:26.
Therefore, increase in the shortest side = 15×(26)/(24+15+26) = 6.

Total area to be painted = 25×12 +2(10×12 + 10×25) = 1040 sqr.ft

A paints = 200/5 = 40 sqr.ft per day
B paints = 250/2 = 125 sqr.ft per day

A + B = 40 + 125 = 165 sqr.ft
Number of days = 1040/165 = 6(10/33)