পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

Solution:
Let,
The quantity of the wine in the cask originally be x litres
Then, quantity of wine left in cask after 4 operations =[x(1 - 8/x)⁴] litres
∴ {x(1 - 8/x)⁴}/x = 16/81
⇒ (1 - 8/x)⁴ = (2/3)⁴
⇒ 1 - 8/x = 2/3
⇒ 8/x = 1  - 2/3
⇒ 8/x = 1/3
⇒ x/8 = 3
∴ x = 24

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:

⇒ Quantity of cheaper : Quantity of Dearer  = (CP of of Dearer - Mean Price) : (Mean Price - CP of Cheaper)
⇒ Quantity of cheaper : Quantity of Dearer = (18 - 14) : ( 14 - 8) = 4 : 6 = 2 : 3

∴ Quantity of suger sold at 18% profit = (3/5) × 1000
= 600 Kg

Question: Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is-

Solution:
Let the third number be x.
Then,
first number = 120% of x = 120x/100 = 6x/5
Second number = 150% of x = 150x/100 = 3x/2

∴ Ratio of first two numbers = (6x/5 : 3x/2) = 12x : 15x = 4 : 5

Question: A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A's share is Tk. 855, the total profit is-

Solution:
Let the total profit be Tk. 100.
After paying to charity,A's share = Tk. 95 × (3/5) = Tk. 57.

If A's share is Tk. 57, total profit = Tk. 100.
If A's share Tk. 855, total profit = (100 × 855)/57 = 1500.

Question: Ricky purchased a shirt for Tk. 2000. The discount offered for the shirt is 10%. Find how many takas has he given to the cashier?

Solution:
Here,
Principal amount, p = Tk. 2000
Discount rate, r = 10%

Total Discount = p × r
= 2000 × 10% = 200

Dress price after discount = Principal amount - Total Discount
= 2000 - 200 = 1800

Ricky has given Tk. 1800 to the cashier.

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 kg?

Solution:
Let
15 Tk kg pulses = x kg and 20 Tk kg pulses = y kg
ATQ,
15x + 20y = 16.5(x + y)
⇒ 15x + 20y = 16.5x + 16.5y
⇒ 1.5x = 3.5y
⇒ x/y = 3.5/1.5
⇒ x/y = 7/3
⇒ x : y = 7 : 3

Question: A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Tk. 1000 more than D, what is B's share?

Solution:
Let the shares of A, B, C and D be Tk. 5x, Tk. 2x, Tk. 4x and Tk. 3x respectively.
Then,
4x - 3x = 1000
∴ x = 1000.

B's share = Tk. 2x = Tk. (2 × 1000) = Tk. 2000.

Question: A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Tk. 6500 for 6 months, B, Tk. 8400 for 5 months and C, Tk. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Tk. 7400. Calculate the share of B in the profit.

Solution:
For managing, A received = 5% of Tk. 7400 = Tk. 370.
Balance = Tk. (7400 - 370) = Tk. 7030.

Ratio of their investments = (6500 × 6) : (8400 × 5) : (10000 × 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10

∴ B's share = Tk. 7030 × (14/37) = Tk. 2660

Question: Kavita bought land for Tk. 50000. She sold the land by offering a 30% discount on the land. Find how much did she lose.

Solution:
Principal amount p = 50000
Discount rate, r = 30%

Total Discount = 50000 × (30/100) 
= 15000

Thus, Selling price = 50000 - 15000 = 35000
Hence, Kavitha lost Tk. 15000

Question: Tea worth Tk. 126 per kg and Tk. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Tk. 153 per kg, the price of the third variety per kg will be-

Solution:
Since first and second varieties are mixed in equal proportions.
So, their average price = Tk. (126 + 135)/2 = Tk. 130.50
So, the mixture is formed by mixing two varieties, one at Tk. 130.50 per kg and the other at say, Tk. x per kg in the ratio 2 : 2 = 1 : 1. We have to find x.
We know,

⇒ Quantity of cheaper : Quantity of Dearer  = (CP of of Dearer - Mean Price) : (Mean Price - CP of Cheaper)
⇒ (CP of of Dearer - Mean Price) : (Mean Price - CP of Cheaper) = Quantity of cheaper : Quantity of Dearer
⇒ (x - 153) : (153 - 130.50) = 1 : 1
⇒ x - 153 = 153 - 130.50
⇒ x = 22.5 + 153
∴ x = 175.5

Question: A and B together have Tk. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?

Solution:
(4/15) A = (2/5) B
⇒ A = (2/5 × 15/4) B
⇒ A = (3/2) B
⇒ A/B = 3/2
∴ A : B = 3 : 2

B's share = 1210 × (2/5) = Tk. 484

Question: A, B and C enter into a partnership in the ratio 7/2 : 4/3 : 6/5 . After 4 months, A increases his share 50%. If the total profit at the end of one year be Tk. 21,600, then B's share in the profit is-

Solution:
Ratio of initial investments = 7/2 : 4/3 : 6/5 = 105 : 40 : 36.
Let the initial investments be 105x, 40x and 36x.

A : B : C = {105x × 4 + (150/100) × 105x × 8} : (40x × 12) : (36x × 12)
= 1680x : 480x : 432x
= 35 : 10 : 9.

Hence, B's share = Tk. 21600 × (10/54) = Tk. 4000.

Question: In a departmental store, an Tk. 400 dress is marked with Save 25%. What will be the sale price of this dress?

Solution:
The phrase, Save 25% (i.e. 0.25) , means discount rate.

The original price of the dress is given as Tk. 400.
The discount amount will be 0.25 × 400 = Tk.100

Sale price = 400 - 100 = Tk. 300

Question: A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

Solution:
Let the cost of 1 litre milk be Tk. 1

Milk in 1 litre mix. in 1st can = 3/4 litre, C.P. of 1 litre mix. in 1st can Tk. 3/4
Milk in 1 litre mix. in 2nd can = 1/2 litre, C.P. of 1 litre mix. in 2nd can Tk. 1/2
Milk in 1 litre of final mix. = 5/8 litre, Mean price = Tk. 5/8

By the rule of alligation, we have:

⇒ Quantity of 2nd can : Quantity of 1st can = (3/4 - 5/8) : (5/8 - 1/2) = (6 - 5)/8 : (5 - 4)/8 = 1/8 : 1/8 = 1 : 1

∴ Quantity of mixure taken from each can = 12 × (1/2) = 6 litres

Question: Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

Solution:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
⇒ (140/100) × 5x , (150/100) × 7x and (175/100) × 8x
⇒ 7x, 21x/2 and 14x.

The required ratio = 7x : 21x/2 : 14x
= 14x : 21x : 28x
= 2 : 3 : 4.

Question: A, B, C subscribe Tk. 50,000 for a business. A subscribes Tk. 4000 more than B and B Tk. 5000 more than C. Out of a total profit of Tk. 35,000, A receives-

Solution:
Let C = x.
Then, B = x + 5000 and
A = x + 5000 + 4000 = x + 9000.

So,
x + x + 5000 + x + 9000 = 50000
⇒ 3x = 36000
∴ x = 12000

A : B : C = 21000 : 17000 : 12000
= 21 : 17 : 12.

A's share = 35000 × (21/50) = Tk. 14,700.

Question: A refrigerator is on sale for 20% off the original price. A store-wide sale results in an additional reduction of 25%. What is the total discount based on the original price?

Solution:
Let
x equal the original price.
First discount: x(1 - 0.20) = 0.80x (sale price).
Second discount: 0.80x(1 - 0.25) = 0.60x (final sale price)

Therefore, the final price is 60% of the original price. To calculate the discount:
1 - 0.60 = 0.40 or 40%

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:
Let C.P. of 1 litre milk be Tk. 1
Then,
S.P. of 1 litre of mixture = Tk. 1,
Gain = 25%.

C.P. of 1 litre mixture = Tk. (100/125) × 1 = Tk. 4/5

By the rule of alligation, we have:

⇒ Quantity of water : Quantity of milk = (1 - 4/5) : (4/5 - 0) = 1/5 : 4/5 = 1 : 4

Hence, percentage of water in the mixture = (1/5) × 100% = 20%

Question: In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is-

Solution:
Quantity of milk = 60 × (2/3) litres = 40 litres.
Quantity of water in it = (60 -  40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then,
milk : water = 40/(20 + x)
Now,
40/(20 + x) = 1/2
⇒ 20 + x = 80
∴ x = 60.
Quantity of water to be added = 60 litres.

Question: Vanessa purchased an mp3 player, originally priced at $290, but discounted by $27. Approximately what percent discount did Vanessa receive on the mp3 player?

Solution:
In 290 discount 27
∴ In 1 discount 27/290
∴ In 100 discount (27 × 100)/290
= 9.31

Question: The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

Solution:
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
⇒ (120/100) × 7x and (110/100) × 8x
⇒ 42x/5 and 44x/5

∴ The required ratio = 42x/5 : 44x/5 = 21 : 22.

Question: A starts business with Tk. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B's contribution in the capital?

Solution:
Let B's capital be Tk. x.
Then,
(3500 × 12)/7x = 2/3
⇒ 14x = 126000
∴ x = 9000.

Question: A clothing store offers a dress initially at $100. Then one week later, the store offers the dress at 15% off. The dress still doesn't sell, so after another week, the store reduces its new price by 20%. How much does the dress cost now?

Solution:
For the first discount, we have 100(1 - 0.15) = 100(0.85) = 85.
So the dress is worth $85
After the second discount, we have 85(1 - 0.20) = 85(0.8) = 85 × (4/5) = 17 × 4 = 68
So the final dress price is $68

Question: A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Solution:
Amount of milk left after 3 operations = 40(1 - 4/40)³ litres
= 40 × 36/40 × 36/40 × 36/40
= 40 × 9/10 × 9/10 × 9/10
= 29.16 litres

Question: Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Tk. 4000, the new ratio becomes 40 : 57. What is Sumit's salary?

Solution:
Let the original salaries of Ravi and Sumit be Tk. 2x and Tk. 3x respectively.
Then,
(2x + 4000)/(3x + 4000) = 40/57
⇒ 57(2x + 4000) = 40(3x + 4000)
⇒ 6x = 68,000
⇒ 3x = 34,000

∴ Sumit's present salary = (3x + 4000) = Tk.(34000 + 4000) = Tk. 38,000.

Question: A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?

Solution:
Let the initial investments of A and B be 3x and 5x.
A : B : C = (3x × 12) : (5x × 12) : (5x × 6) = 36 : 60 : 30 = 6 : 10 : 5.

Question: If a pen cost Tk.50 after 10% discount, then what is the actual price or marked price (MP) of the pen?

Solution:
Since, we know;
MP - D = SP ..................(1)
where MP is marked price, D is discount, SP is selling price.

Percentage discount, D% = D/MP × 100
⇒ D = (D% × MP)/100 ........(2)

Substitute value of D in (1).
MP - (D% × MP)/100 = SP
⇒ MP × (100 - D%)/100 = SP

Putting the given values in formula
MP × (100 - 10)/100 = 50
⇒ MP × (90/100) = 50
⇒ MP = (50 × 100)/90
∴ MP = 55.55

Alternative:
SP Tk. 90 when MP Tk. 100
∴ SP Tk. 1 when MP Tk. (100/90)
∴ SP Tk. 50 when MP Tk. (100 × 50)/90
= Tk. 55.55

Question: Find the ratio in which rice at Tk. 7.20 a kg be mixed with rice at Tk. 5.70 a kg to produce a mixture worth Tk. 6.30 a kg.

Solution:
Quantity of rice at Tk. 7.20 per kg = x
Quantity of rice at Tk. 5.70 per kg = y
By the rule of alligation:

⇒ x : y = (7.20 - 6.30) : (6.30 - 5.70) = 0.9 : 0.6 = 9 : 6 = 3 : 2