উত্তর
ব্যাখ্যা
Solution:
Milk in the 700 ml of mixture = 700 × (8/10) = 560 ml
So, water in the mixture would be = 700 - 560 = 140 ml
Let water to be added = x ml
Now,
(140 + x)/560 = 3/8
1120 + 8x = 1680
8x = 1680 - 1120
8x= 560
x = 560/8 = 70 ml
PrepBank · বিষয়ভিত্তিক প্রশ্ন
PrepBank · পাতা ১০ / ১১ · ৯০১–১,০০০ / ১,০৮৬
Question: In a mixture of milk and water, the ratio is 5 : 3. If 4 liters of water is added, the new ratio becomes 5 : 4. What was the original amount of milk in the mixture?
Solution:
ধরি, শুরুতে দুধ ছিল = 5x লিটার,
পানি ছিল = 3x লিটার।
এখন ৪ লিটার পানি যোগ করলে,
নতুন পানি = 3x + 4 লিটার
ATQ,
5x/(3x + 4) = 5/4
⇒ 4 × 5x = 5 × (3x + 4)
⇒ 20x = 15x + 20
⇒ 5x = 20
⇒ x = 4
∴ দুধের পরিমাণ = 5x = 5 × 4 = 20 লিটার
Question: In what ratio must sugar at Tk. 12 per kg be mixed with sugar at Tk. 18 per kg so that the mixture be worth Tk. 15 per kg?
Solution:
Let x kg of sugar at Tk. 12 and y kg of sugar at Tk. 18 be mixed.
ATQ,
⇒ 12x + 18y = 15(x + y)
⇒ 12x + 18y = 15x + 15y
⇒ 15x - 12x = 18y - 15y
⇒ 3x = 3y
∴ x : y = 1 : 1
∴ required ratio 1 : 1.
Question: A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?
Solution:
lets, the total cups sold 15
small cups = (3/5) × 15 = 9
large cups = 15 - 9 = 6
let, small cups were sold 6 taka each, then large cups were sold 7 taka each.
large cup's revenue = 7 × 6 = 42 taka
small cup's revenue = 6 × 9 = 54 taka
fraction of Tuesday's total revenue was from the sale of large cups = 42/(42 + 54)
= 42/96
= 7/16
Question: In a map, 2 cm represents 85 km. The distance between two cities is 9.4 cm on the map. The actual distance between the cities is -
Solution:
Since 2 cm = 85 km,
Actual Distance = (85/2) × 9.4
= 799/2
= 399.5 km
CP of first tea = Tk. 192 per kg.
CP of Second tea = Tk. 150 per kg.
The mixture is to be sold in Tk. 194.40 per kg, which has included 20% profit. So,
SP of Mixture = Tk. 194.40 per kg.
Let the CP of Mixture be Tk. X per kg. Therefore,
X + 20% of X = SP
6x/5 = 194.40
6X = 194.40 × 5
X = Tk. 162 per kg.
Let N kg of first tea and M kg of second tea be added.
Now, Using Alligation,
We get,
N/M = 12/30
N/M = 2 : 5
x:7.5 = 7:17.5
⇒ 17.5x = 7.5×7
⇒ x = (7.5×7)/17.5
= 3
W1 : A1 W2 : A2 …… WN : AN
2 : 3 4 : 5 …… 5 : 7
W1/(W1+A1) = 2/5 ; W2/(W2+A2) = 4/9; WN/(WN+AN) = 5/12
= 72/180
= 80/180
= 75/180
=> 5 : 3
Therefore, the ratio is 5 : 3
ATQ,
if the ratio of coins = 4: 6: 9
That means if Tk. 5 coins are 4, Tk. 2 coins are 6, and then Tk. 1 coins are 9.
According to the given ratio, the ratio of amounts = 5 × 4 : 6 × 2: 9 × 1 = 20 : 12 : 9
The sum of the ratios of the amounts = 20 + 12 + 9
= Tk. 41
But ATQ,
it is Tk. 410, which means multiply each ratio by 10
i.e., new ratio = 40 : 60 : 90
Now, 40 × 5 : 60 × 2 : 90 × 1 = 200 : 120 : 90
The total amount in the form of two rupees coins = 120
So, the two rupees coins = 120/2
= 60.
Let the number of boys and girls be 8x and 5x.
Total number of students = 13x
= 13 × 32
= 416.
A:B = 3:2 = 6:4
A:C = 2:1 = 6:3
A:B:C = 6:4:3
∴ B's share = 4/13 × 157300 = 48400
Question: A mixture contains two liquids A and B are in the ratio 2 : 1. If 6 litres of mixture is withdrawn and replaced with 6 litres of B, then the ratio becomes 3 : 2. What was the initial quantity of A?
Solution:
মনে করি,
মিশ্রণের প্রাথমিক পরিমাণ = 3X লিটার
A এর পরিমাণ = 2X লিটার
B এর পরিমাণ = X লিটার
∴ 6 লিটার মিশ্রণ তুলে নেওয়ার পর,
A এর পরিমাণ = 2X - (2/3) × 6 = (2X - 4) লিটার
B এর পরিমাণ = X - (1/3) × 6 = (X - 2) লিটার
আবার,
B তে 6 লিটার যোগ করার পর,
B এর পরিমাণ = X - 2 + 6 = (X + 4) লিটার
প্রদত্ত অনুপাত,
(2X - 4) /(X + 4) = 3/2
বা, 4X - 8 = 3X + 12
বা, X = 12 + 8
∴ X = 20
তাহলে, A এর পরিমাণ = 2 × 20 = 40 লিটার
Question: Two numbers are in the ratio 2 : 3. If 4 is subtracted from the first number, the ratio becomes 1 : 2. What are the numbers?
Solution:
Let the two numbers be: 2x and 3x
According to the question,
(2x - 4)/3x = 1/2
⇒ 2(2x - 4) = 3x
⇒ 4x - 8 = 3x
⇒ x = 8
∴ First number = 2 × 8 = 16
∴ Second number = 3 × 8 = 24
Question: The ratio of milk to water in a mixture is 5 : 3. If 6 liters of milk are added to the mixture, the new ratio of milk to water becomes 8 : 3. Find the final amount of milk in the new mixture.
Solution:
Let the initial amount of milk be 5x liters
and the amount of water 3x liters.
ATQ,
Ratio of milk and water after adding 6 liters of milk
(5x + 6)/3x = 8/3
⇒ 3(5x + 6) = 8 × 3x
⇒ 15x + 18 = 24x
⇒ 18 = 9x
⇒ x = 2
∴ Final amount of milk in mixture = 5x + 6
= (5 × 2) + 6
= 10 + 6 = 16 liters.
Let, John's present age = 6x,
and, Mary’s present age = 4x
Therefore,
(6x - 5)/(4x - 5) = 5/3
Or, 20x - 25 = 18x - 15
Or, x = 5
∴ John’s age = 6 × 5 = 30 years
Given that,
A dog takes 3 leaps for every 5 leaps of a hare
Therefore,
Dog : Hare = 3 : 5
One leap of a dog is equal to 3 leap of the hare
Therefore,
1 leap of dog = 3 leap of hare
Now the ratio becomes,
Dog : Hare = 3 (3) : 5
Dog : Hare = 9 : 5
Thus the ratio of the speed of the dog to that of the hare is 9 : 5
In the original 125 gallons of mixture, 20% is water.
Hence, no. of gallons of other materials in the mixture: 125 x 80% = 100 gallons
In the new mixture, water makes up 25%, thus 75% is other materials.
As no. of gallons of others is unchanged, 100 gallons = 75% in the new mixture volume.
The total volume of the new mixture is : 100 / 75% = 100/ 0.75 = 133.33 gallons.
∴ Required additional amount of water = 133.3 – 125 = 8.33 = 8(1/3) gallons
Question: A sample of 50 liters of glycerine is found to be adulterated to the extent of 20%. How much pure glycerine should be added to it so as to bring down the percentage of impurity to 5%?
Solution:
Initially, the 50 liters of glycerine is 20% adulterated. So, the pure glycerine in it is 80% of 50 liters.
Amount of pure glycerine initially = 80% of 50 liters
= 0.8 × 50
= 40 liters
Let 'x' represent the amount of pure glycerine that needs to be added to reduce the percentage of impurity to 5%.
The total volume of glycerine after adding pure glycerine will be 50 + x liters
After adding 'x' liters of pure glycerine, the total amount of pure glycerine in the mixture will be 40 + x liters (added).
Now, this total amount of pure glycerine should be 95% of the new total volume (50 liters original + x liters added), as the impurity percentage is to be reduced to 5%.
So, we can set up an equation:
Total amount of pure glycerine = 95% of total volume after adding pure glycerine
40 + x = 0.95 × 50 + 0.95 × x
⇒ 40 + x = 47.5 + 0.95x
⇒ x - 0.95x = 47.5 - 40
⇒ 0.05x = 7.5
⇒ x = 7.5 / 0.05
⇒ x = 150 liters
A:: - G:S = 40:60
B:: - G:S:C = 35:40:25
New,
G:S= (1×40 + 4×35) : (40×4 + 1×60)
= 180:220
= 18:22
= 9:11
Let their income be x
and expenditure be y
A's income = 5x
B's income = 4x
A's expenditure = 3y
B's expenditure = 2y
Both saves 1600
Income - expenditure = saves
5x - 3y = 1600 .... (i)
4x - 2y = 1600 .... (ii)
Substract (ii) × 3 from (i) × 2
10x - 6y - ( 12x - 6y ) = 3200 - 4800
10x - 6y - 12x + 6y = - 1600
- 2x = - 1600
x = 800
Putting the value of x in (i)
5x - 3y = 1600
4000 - 3y = 1600
- 3y = - 2400
y = 800
∴ Income of a = 5x = 4000
Question: A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
Solution:
Total quantity of sugar = 1000 kg
Let x kg be sold at 18% profit
Then (1000 - x) kg be sold at 8% profit
Profit from x kg at 18% = 18x/100
Profit from (1000 - x) kg at 8% = 8(1000 - x)/100
Total profit = 14% of 1000 = 14000/100 = 140
According to the question,
18x/100 + 8(1000 - x)/100 = 140
⇒ 18x + 8(1000 - x) = 14000
⇒ 18x + 8000 - 8x = 14000
⇒ 10x = 14000 - 8000
⇒ 10x = 6000
⇒ x = 600
∴ 600 kg of sugar was sold at 18% profit.
বিকল্প সমাধান:
By the rule of alligation,
Ration of 1st and 2nd parts = 4 : 6 = 2 : 3
∴ Quantity of 2nd kind = (3/5) × 1000 = 600 kg
We know,
The ratio of Investment x Time = Ratio of Profit
∴ (A's investment x Time) : (B's investment x Time) = Profit of A : Profit of B
The total value of investment of Arman, Ghalib, and Jishan after 12 months is =
Tk. 40000 x 12 months : Tk. 80000 x 7 months : Tk. 144000 x (12 - 7)months = 480000 : 560000 : 720000
∴ Profit ratio = 240000 : 280000 : 360000 = 6 : 7 : 9
∴ Share of Ghalib = 7/(6 + 7 + 9) x 50600 = Tk. 16100
Parts of B = 72 × (3/24) kg
= 9 kg
Question: A mixture of milk and water has a total volume of 42 liters, where milk and water are mixed in the ratio 3 : 4. How many liters of milk must be added so that the quantities of milk and water become equal?
Solution:
The ratio of milk to water is 3 : 4
Total portion = 3 + 4 = 7
Quantity of milk = 42 × (3/7) = 18 liters.
Quantity of water = 42 × (4/7) = 24 liters.
Let,
Quantity of milk to be added = x liters
According to the question,
(18 + x) : 24 = 1 : 1
⇒ (18 + x)/24 = 1/1
⇒ 18 + x = 24
⇒ x = 24 - 18
⇒ x = 6
∴ Quantity of milk to be added = 6 liters
According to the question,
(x + y)/(x - y) = 4
⇒ x + y = 4x - 4y
⇒ 4x - x = 4y + y
⇒ 3x = 5y
⇒ x/y = 5/3
⇒ x2/y2 = 25/9
Now,
⇒ x2 + y2/x2 - y2
= {(x2 /y2) + 1}/{(x2/y2) - 1}
= {(25/9) +1}/{(25/9 - 1}
(34/9) × (9/16)
= 17/8
Question: A 42-liter mixture contains milk and water in a 3 : 4 ratio. How much milk should be added to the mixture to make the ratio equal?
Solution:
The ratio of milk to water is 3 : 4
Total portion = 3 + 4 = 7
Quantity of milk = 42 × (3/7) = 18 liters.
Quantity of water = 42 × (4/7) = 24 liters.
Let,
Quantity of milk to be added = x liters
According to the question,
(18 + x) : 24 = 1 : 1
⇒ (18 + x)/24 = 1/1
⇒ 18 + x = 24
⇒ x = 24 - 18
⇒ x = 6
∴ Quantity of milk to be added = 6 liters
Question: A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is-
Solution:
A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and gains 25%
We know,
Profit = SP - CP
Let milk bought (cost price) be 1 litres at Tk. 100 and profit = 25
∴ Selling price = 100 + 25 = Tk. 125
As the selling price of 1 litre was Tk. 100(same as cost)
Quantity sold = 125/100 = 5/4 = 1.25 litre
Hence, water added = 1.25 - 1 = 0.25 litres
∴ Percentage of water = (0.25/1.20) × 100 = 20%
Question: The ratio of the present ages of two friends is 3 : 5. After 7 years, the ratio becomes 2 : 3. What will be the ratio of their ages after 10 years?
Solution:
Let
The present age of the friend1 = 3x
The present age of the friend2 = 5x
After 7 years, their ages will be,
⇒ (3x + 7)/(5x + 7) = 2/3
⇒ 10x + 14 = 9x + 21
⇒ 10x - 9x = 21 - 14
∴ x = 7
Present age of friend1 = 3x = 21 years
Present age of friend2 = 5x = 35 years
Now, after 10 years,
friend1 age = 21 + 10 = 31 years
friend2 age = 35 + 10 = 45 years
∴ The ratio of their ages after 10 years,
= 31 : 45
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 us assume that the lotion has 50% alcohol and 50% water.
ratio = 1:1
As the total solution is 9ml
alcohol = water = 4.5ml
Now if we want the quantity of alcohol = 30%
The quantity of water = 70%
The new ratio = 3:7
Let x ml of water be added
We get,
4.5/(4.5 + x) = 3/7
⇒ 9/(9 + 2x) = 3/7
⇒ 63 = 27 + 6x
⇒ 6x = 63 - 27
⇒ 6x = 36
⇒ x = 6
Hence 6ml of water is added.
Question: The ratio of the ages of A and B at present is 3 ∶ 1. Four years earlier the ratio was 4 ∶ 1. The present age of B is-
Solution:
Given that,
Ratio of present ages of A and B = 3 ∶ 1.
Ratio of their ages 4 years ago = 4 ∶ 1.
Let the present ages of A and B be 3x and x, respectively.
Four years ago,
Age of A = 3x - 4
Age of B = x - 4
AQT,
(3x - 4)/(x - 4) = 4/1
⇒ 3x - 4 = 4(x - 4)
⇒ 3x - 4 = 4x - 16
⇒ 3x - 4x = - 16 + 4
⇒ - x = - 12
∴ x = 12
∴ Present age of B = 12 years.
Milk : Water
7 : 5
7 : 8
---------------
3 unit
∴ Remember water is added and not milk, so make milk equal but here milk is already equal
3 units = 15 litres
1 units = 5 litres
8 units = 40 litres
Total quantity of water in the new mixture = 40 litres
Question: In a class, 10% of the girls have blue eyes, and 20% of the boys have blue eyes. If the ratio of girls to boys in the class is 3 : 4, then what is the fraction of the students in the class having blue eyes?
Solution:
Let the number of girls be x
Since the ratio of the girls to boys is 3 : 4, the number of boys = 4x/3
Hence, the number of students in the class = x + (4x/3) = 7x/3
We are given that 10% of girls are blue-eyed,
∴ 10% of x = (10/100)x = x/10
Also, 20% of the boys are blue-eyed,
∴ 20% of 4x/3 = (20/100) × (4x/3) = 4x/15
Hence, the total number of blue-eyed students = (x/10) + (4x/15)
= 11x/30
Hence, the required fraction = (11x/30)/(7x/3)
= (11 × 3)/(30 × 7)
= 11/70
Question: Two vessels A and B contain milk and water mixed in the ratio 4 : 3 and 2 : 3 respectively. What will be the new ratio of milk to water if these two mixtures are mixed together in equal quantities? Solution:
সমাধান:
ধরি, পাত্র A এবং B এর মিশ্রণের পরিমাণ সমান, অর্থাৎ ১ একক।
পাত্র A তে দুধের পরিমাণ = 4/(4 + 3) = 4/7
পাত্র A তে পানির পরিমাণ = 3/(4 + 3) = 3/7
পাত্র B তে দুধের পরিমাণ = 2/(2 + 3) = 2/5
পাত্র B তে পানির পরিমাণ = 3/(2 + 3) = 3/5
নতুন মিশ্রণে মোট দুধের পরিমাণ = (4/7) + (2/5)
= (20 + 14)/35
= 34/35
নতুন মিশ্রণে মোট পানির পরিমাণ = (3/7) + (3/5)
= (15 + 21)/35
= 36/35
∴ নতুন অনুপাত = (34/35) : (36/35)
= 34 : 36
= 17 : 18
Question: In a mixture of milk and water, the ratio is 4 : 3. If 5 liters of water is added, the new ratio becomes 4 : 4. What was the original amount of milk in the mixture?
Solution:
ধরি, শুরুতে দুধ ছিল = 4x লিটার,
পানি ছিল = 3x লিটার।
এখন 5 লিটার পানি যোগ করলে, নতুন পানি = 3x + 5 লিটার
ATQ,
4x/(3x + 5) = 4/4
⇒ 4x/(3x + 5) = 1
⇒ 4x = 1 × (3x + 5)
⇒ 4x = 3x + 5
⇒ 4x - 3x = 5
⇒ x = 5
∴ দুধের পরিমাণ = 4x = 4 × 5 = 20 লিটার