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Let y be the amount of flavour.
(0.2 × 20) + 1 × y = 0.25 (20 + y)
Or, 4 + y = 5 + 0.25y
Or, 0.75y = 1
So, y = 1.33
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Let y be the amount of flavour.
(0.2 × 20) + 1 × y = 0.25 (20 + y)
Or, 4 + y = 5 + 0.25y
Or, 0.75y = 1
So, y = 1.33
Question: If a : b = 3 : 1, find ratio (3a + 5b) : (3a - 5b).
Solution:
(3a + 5b) : (3a - 5b)
= b(3a/b + 5) : b (3a/b - 5)
= (3 × 3/1 + 5) : (3 × 3/1 - 5)
= (9 + 5) : (9 - 5)
= 14 : 4
= 7 : 2
Question: A shopkeeper mixes 15 kg of tea costing Tk. 280 per kg with 10 kg of tea costing Tk. 400 per kg. He then adds some inferior tea costing Tk. 200 per kg so that the average price of the mixture becomes Tk. 300 per kg. How many kg of inferior tea is added?
Solution:
Let the quantity of inferior tea added be X kg.
Total cost = (15 × 280) + (10 × 400) + (X × 200)
= 4200 + 4000 + 200X
= 8200 + 200X Tk.
Total weight = 15 + 10 + X
= 25 + X kg.
∴ Average price = 300 Tk. per kg.
(8200 + 200X)/(25 + X) = 300
⇒ 8200 + 200X = 300 × (25 + X)
⇒ 8200 + 200X = 7500 + 300X
⇒ 8200 - 7500 = 300x - 200X
⇒ 700 = 100X
⇒ X = 7
Thus, 7 kg of inferior tea is added.
Question: A cricket team has a ratio of win to loss 4 : 3. After losing 5 games in a row, the team's ratio of win to loss became 9 : 8. How many games had the team loss before it played the last five games?
Solution:
মনে করি,
দলটির জেতা খেলার সংখ্যা ছিল 4p
দলটির হারা খেলার সংখ্যা ছিল 3p
পরপর 5 টি খেলা হারার পর,
জেতা খেলার সংখ্যা 4p
হারা খেলার নতুন সংখ্যা (3p + 5)
প্রশ্নমতে,
4p : (3p + 5) = 9 : 8
⇒ 4p/(3p + 5) = 9/8
⇒ 32p = 27p + 45
⇒ 32p - 27p = 45
⇒ 5p = 45
⇒ p = 45/5
⇒ p = 9
∴ p = 9
∴ ১ম অবস্থায় হারা খেলার সংখ্যা ছিল = 3 × 9 = 27
অতএব, শেষ 5 টি খেলা খেলার আগে দলটি 27 খেলায় হেরেছিল।
Original ratio of the mixture 3:1
Taking out the mixture actually means just taking out the wine because the water anyways is going to be added back.
If we remove 1 part of wine, it makes the ratio 2:2 (1 part water is added to keep the volume constant )
so, what we have actually done is remove 1 part wine from 3 part Wine. i.e. 1/3.
So, 1/3 mixture drawn.
Question: If P : 6 = Q : 8 = R : 10, then (P + Q + R)/P =?
Solution:
P : 6 = Q : 8
or, P/6 = Q/8
or, 6Q = 8P
or, Q = 8P/6
or, Q = 4P/3
P : 6 = R : 10
or, P/6 = R/10
or, 6R = 10P
or, R = 10P/6
or, R = 5P/3
∴ (P + Q + R)
= P + (4P/3) + (5P/3)
= (3P + 4P + 5P)/3
= 12P/3
= 4P
∴ (P + Q + R)/P
= 4P/P
= 4
Question: Silver is 12 times as heavy as water, and lead is 8 times as heavy as water. In what ratio should these be mixed to get an alloy 9 times as heavy as water?
Solution:
Let silver be 12x times as heavy and lead 8y times as heavy as water.
12x + 8y = 9(x + y)
⇒ 12x + 8y = 9x + 9y
⇒ 12x - 9x = 9y - 8y
⇒ 3x = y
∴ x/y = 1/3 = 1 : 3
ধরি,
ক্ষুদ্রতর সংখ্যাটি 3x এবং বৃহত্তর সংখ্যাটি 4x
প্রশ্নমতে,
3x + 4x = 630
⇒ 7x = 630
⇒ x = 630/7
⇒ x = 90
অতএব, ক্ষুদ্রতর সংখ্যাটি = 3 × 90
= 270.
Answer: 270.
Question: How much water be mixed in 45 litre of milk worth Tk. 6.40 per litre, so that value of mixture is Tk. 4.80 per litre?
Solution:
মনে করি, x লিটার পানি মেশাতে হবে।
মিশ্রণের মোট পরিমাণ = (45 + x) লিটার
45 লিটার দুধের দাম = 45 × 6.40 = 288 টাকা
পানির দাম = 0 টাকা
প্রশ্নমতে, মিশ্রণের মোট দাম = (45 + x) × 4.80
যেহেতু পানি বিনামূল্যে পাওয়া যায়, তাই মোট দাম অপরিবর্তিত থাকবে:
288 = (45 + x) × 4.80
⇒ 288/4.80 = 45 + x [উভয় পক্ষকে 4.80 দ্বারা ভাগ করে পাই]
⇒ 60 = 45 + x
⇒ x = 60 - 45
⇒ x = 15 litres
∴ 15 লিটার পানি মেশাতে হবে।
Question: In a 729 liters mixture of milk and water, the ratio of milk to water 7 : 2. To get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is-
Solution:
Quantity of milk in 729 litre of mixture
= 7 × 729/9 = 567 litres
Quantity of water
= 729 - 567
= 162 litres
Let x litre of water be added to become ratio 7 : 3
According to the question,
7/3 = 567 / (162 + x)
Or, 162 × 7 + 7x = 567 × 3
Or, 7x = 1701 - 1134 = 567
Or, x = 567 / 7 = 81
Therefore, 81 litres of water is to be added.
Let,
Habib’s age is = 6x
And Shikha’s age = 4x
ATQ,
(6x - 5)/(4x - 5) = 5/3
⇒ 18x - 15 = 20x - 25
⇒ 2x = 10
⇒ x = 5
So, Habib’s age is = 6×5 = 30 year
Question: A chemist has two solutions, one containing 40% acid and the other containing 80% acid. How many liters of each solution should be mixed to get 12 liters of a solution containing 60% acid?
Solution:
Let x liters of 40% solution be used. Then (12 - x) liters of 80% solution will be used.
According to the problem:
40% × x + 80% × (12 - x) = 60% × 12
⇒ 40x + 80(12 - x) = 720
⇒ 40x + 960 - 80x = 720
⇒ -40x + 960 = 720
⇒ -40x = -240
⇒ x = 6
∴ 6 liters of 40% solution and 6 liters of 80% solution are needed.
Numbers are in the ratio 4:5
Let the numbers be 4x and 5x
Hence, LCM = 20x
Hence, 20x = 180
Hence, x = 180/20 = 9
Hence the numbers are 36 and 45
Question: 600 grams of sugar solution has 30% sugar in it. How much sugar should be added to make 75% in the solution?
Solution:
Amount of sugar =600 × 30/100
=180 grams
Let,
x gram sugar to be added
ATQ,
(180 + X)/(600 + X) = 75%
⇒ (180 + X)/(600 + X) = 75/100
⇒ (180 + X)/(600 + X) = 3/4
⇒ 4(180 + X) = 3(600 + X)
⇒ 4X + 720 = 3X + 1800
⇒ 4X - 3X= 1800 - 720
∴ X = 1080 grams
Question: If a : b : c = 2 : 3 : 4 and 2a - 3b + 4c = 33, then the value of c is-
Solution:
Let,
a = 2x, b = 3x and c = 4x ; (since a : b : c = 2 : 3 : 4)
Now substitute into the equation,
2a - 3b + 4c = 33
⇒ 2(2x) - 3(3x) + 4(4x) = 33
⇒ 4x - 9x + 16x = 33
⇒ 11x = 33
∴ x = 3
Since c = 4x = 4 × 3 = 12
Question: A vessel contains milk and water in the ratio 7 : 4. If 15 liters of milk are added to it, the ratio of milk to water becomes 10 : 4. Find the final amount of milk in the new mixture.
Solution:
Let the initial amount of milk be 7x liters
and the amount of water 4x liters.
According to the question,
(7x + 15)/4x = 10/4
⇒ 4(7x + 15) = 10 × 4x
⇒ 28x + 60 = 40x
⇒ 60 = 12x
⇒ x = 60/12
⇒ x = 5
∴ Final amount of milk in mixture = 7x + 15
= 7 × 5 + 15
= 35 + 15
= 50 liters.
Question: Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 10 times as heavy as water?
Solution:
let, gold is 19x tme heavy and copper 9y times heavy as water
19x + 9y = 10 (x + y)
⇒ 19x + 9y = 10x + 10y
⇒ 19x - 10x = 10y - 9y
⇒ 9x = y
∴ x/y = 1/9
= 1 : 9
A : B = 1000 : (1000 - 100)
= 1000 : 900
= 10 : 9
B : C = 1000 : (1000 - 60)
= 1000 : 940
100 : 94
= 9 : 846/100
Therefor, A : C = 10 : 846/100
= 10000 : 846
hence A can beat C by = (10000 - 846)
= 154 metres.
Question: In your wallet, there are Tk 1000, Tk 500, and Tk 100 notes in the ratio 3 : 5 : 2, amounting to Tk 22,800. Find the number of each note respectively.
Solution:
Let,
The number of Tk 1000 notes is 3x
The number of Tk 500 notes is 5x
The number of Tk 100 notes is 2x
ATQ,
1000 × 3x + 500 × 5x + 100 × 2x = 22800
⇒ 3000x + 2500x + 200x = 22800
⇒ 5700x = 22800
⇒ x = 22800 / 5700
⇒ x = 4
Number of Tk 1000 note = 3x = 3 × 4 = 12
Number of Tk 500 note = 5x = 5 × 4 = 20
Number of Tk 100 note = 2x = 2 × 4 = 8
Therefore, the number of Tk 1000, Tk 500, and Tk 100 notes are respectively 12, 20, and 8.
Let third proportional be x
⇒ 25 : 30 : : 30 : x
⇒ 25 × x = 30 x 30
⇒ x = (30 x 30)/25
= 36.
ধরি, Nabila's income = T
এবং Nuru's income = N
প্রশ্নমতে, T এর 35% = S এর 25%
⇒ T(35/100) = N (25/100)
⇒ T/N = 25/100 × 100/35 = 5/7
T : N = 5 : 7
Question: In a business, the ratio of the capitals of Arun and Babul is 2 : 1, that of Babul and Chandan is 4 : 3, and that of Dipu and Chandan is 6 : 5. What is the ratio of the capitals of Arun and Dipu?
Solution:
Given,
Arun : Babul = 2 : 1
⇒ Arun/Babul = 2/1
Babul : Chandan = 4 : 3
⇒ Babul/Chandan = 4/3
Dipu : Chandan = 6 : 5
⇒ Chandan/Dipu = 5/6
Now,
Arun/Dipu = (Arun/Babul) × (Babul/Chandan) × (Chandan/D)
= (2/1) × (4/3) × (5/6)
= 20/9
∴ Arun/Dipu = 20 : 9
Question: In what ratio must a grocer mix two varieties of pulses costing Tk. 15 and Tk. 20 per kg respectively so as to get a mixture worth Tk. 16.50 per kg?
Solution:
Let the ratio be x : y
Cost of first variety = Tk. 15 per kg
Cost of second variety = Tk. 20 per kg
Cost of mixture = Tk. 16.50 per kg
For x kg of first variety and y kg of second variety:
Total cost of mixture = (x × 15) + (y × 20) = 15x + 20y
Total quantity of mixture = (x + y) kg
According to the question,
(15x + 20y)/(x + y) = 16.50
⇒ 15x + 20y = 16.50(x + y)
⇒ 15x + 20y = 16.50x + 16.50y
⇒ 20y - 16.50y = 16.50x - 15x
⇒ 3.50y = 1.50x
⇒ y/x = 1.50/3.50
⇒ y/x = 150/350
⇒ y/x = 3/7
Therefore, x : y = 7 : 3
বিকল্প সমাধান:
By the rule of alligation,
∴ Required rate = 3.50 : 1.50 = 7 : 3
Question: In a box, the ratio of red marbles to blue marbles is 7 : 4. Which of the following could be the total number of marbles in the box?
Solution:
18, 19, 21 কোনটিই 7 + 4 বা 11 দ্বারা নি:শেষে বিভাজ্য নয়।
অতএব, সঠিক উত্তর 22
Let the constant be x,
so putting the first scenario in equation
A = x × (B/C)
⇒ 6 = x × 2/3
⇒ 2x = 18
⇒ x = 9
We can find out A in the second scenario by putting the value of x as 9
A = 9 × 8/6
⇒ A = 12.
Question: A seller bought 120 pens for 600 taka. How many pens does he need to sell for 600 taka to make a profit of 20%?
Solution:
Cost price of 120 pens = Tk. 600 Cost price per pen = 600 ÷ 120
= Tk. 5
∴ Selling price per pen for 20% profit = 5 × (120/100)
= 5 × 1.2
= Tk. 6
∴ Number of pens to be sold for Tk. 600
= 600 ÷ 6
= 100 pens
Question: The ratio of X : Y is 3 : 5, and the ratio of Y : Z is 6 : 7. If X = 18, what is the value of Z?
Solution:
Given that,
X : Y = 3 : 5
Y : Z = 6 : 7
And X = 18
Now,
X : Y = 3 : 5
⇒ X/Y = 3/5
⇒ 5X = 3Y
⇒ Y = (5 × 18)/3 ; [X = 18]
∴ Y = 30
And,
Y : Z = 6 : 7
Y/Z = 6/7
⇒ 30/Z = 6/7
⇒ Z = (30 × 7)/6
∴ Z = 35
So the value of Z is 35.
Question: The ratio of A : B is 2 : 3, and the ratio of B : C is 4 : 5. If A 16, what is the value of C?
Solution:
Given that,
A : B = 2 : 3
B : C = 4 : 5
And A = 16
Now,
A : B = 2 : 3
⇒ A/B = 2/3
⇒ 2B = 3A
⇒ B = (16 × 3)/2 ; [A = 16]
∴ B = 24
And,
B : C = 4 : 5
B/C = 4/5
⇒ 24/C = 4/5
⇒ C = (24 × 5) / 4
∴ C = 30
So the value of C is 30.