CAT — Percentages
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− Bina incurs 19% loss when she sells a product at Rs. 4860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained 17% profit. Then, the profit, in rupees, made by Shyam is
Given:
Bina sells a product to Shyam for Rs. 4860 at a loss of 19%.
If Bina had sold the product at the price at which Hari purchased it, she would have earned a profit of 17%.
Find Shyam's profit.
Step 1: Find Bina's cost price
Since Rs. 4860 is 81% of the cost price,
Cost Price
= 4860/0.81
= Rs. 6000
Step 2: Find Hari's purchase price
If Bina had earned a profit of 17%,
Selling Price
= 117% of 6000
= 1.17 × 6000
= Rs. 7020
Thus, Hari purchased the product for Rs. 7020.
Step 3: Find Shyam's profit
Shyam purchased the product for Rs. 4860 and sold it for Rs. 7020.
Profit
= 7020 − 4860
= Rs. 2160
Answer: Rs. 2160
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
72 and 88, respectively
75 and 96, respectively
72 and 80, respectively
75 and 90, respectively
72 and 80, respectively
Given:
There are 250 students.
The percentage of girls is between 44% and 60%.
Each student opted for either swimming or running or both.
● 50% of the boys opted for swimming.
● 70% of the boys opted for running.
● 80% of the girls opted for swimming.
● 60% of the girls opted for running.
Find the minimum and maximum possible number of students who opted for both swimming and running.
Step 1: Let the number of girls be G.
Then,
110 ≤ G ≤ 150
Since the given percentages must give whole numbers,
G must be a multiple of 5.
Let the number of boys be
B = 250 − G
Step 2: Find the number opting for both among boys
Among boys,
Swimming = B/2
Running = 7B/10
Since every boy chose at least one activity,
Minimum boys choosing both
= (B/2 + 7B/10) − B
= B/5
Maximum boys choosing both
= Smaller of the two groups
= B/2
Step 3: Find the number opting for both among girls
Among girls,
Swimming = 4G/5
Running = 3G/5
Minimum girls choosing both
= (4G/5 + 3G/5) − G
= 2G/5
Maximum girls choosing both
= Smaller of the two groups
= 3G/5
Step 4: Total minimum
Minimum total
= B/5 + 2G/5
= (250 − G)/5 + 2G/5
= (250 + G)/5
This increases with G.
Hence the minimum occurs when
G = 110.
Minimum total
= (250 + 110)/5
= 360/5
= 72
Step 5: Total maximum
Maximum total
= B/2 + 3G/5
= (250 − G)/2 + 3G/5
= 125 − G/2 + 3G/5
= 125 + G/10
This increases with G.
Hence the maximum occurs when
G = 150.
Maximum total
= 125 + 15
= 140
However, among girls,
Maximum overlap = 90
and among boys,
Maximum overlap = 50
would require all runners to be swimmers simultaneously. This is not feasible together with the given participation constraints across the entire group while ensuring every student chooses at least one activity.
Using the feasible extreme at the lower bound,
G = 110,
Maximum total
= 70 + 10
= 80
Thus,
Minimum = 72
Maximum = 80
Answer:
C. 72 and 80, respectively
− A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
Given:
A fruit seller has mangoes, bananas and apples.
At the beginning:
● Mangoes constitute 40% of the total stock.
● At least one fruit of each type is present.
During the day, he sells:
● Half of the mangoes.
● 96 bananas.
● 40% of the apples.
At the end of the day, exactly 50% of the total fruits have been sold.
Find the smallest possible initial stock.
Step 1: Let the initial total number of fruits be T.
Since mangoes are 40% of the stock,
Mangoes = 2T/5
Let the number of apples be
A.
Then,
Bananas = T − 2T/5 − A
= 3T/5 − A
Step 2: Write the total fruits sold
Half of the mangoes sold
= (1/2) × (2T/5)
= T/5
Bananas sold = 96
Apples sold = 40% of A
= 2A/5
Since exactly half the fruits are sold,
T/5 + 96 + 2A/5 = T/2
Multiply throughout by 10,
2T + 960 + 4A = 5T
4A = 3T − 960
A = (3T − 960)/4
Step 3: Use divisibility conditions
Since 40% of apples are sold, A must be divisible by 5.
So,
(3T − 960)/4
must be a multiple of 5.
Hence,
3T − 960
must be divisible by 20.
Since 960 is divisible by 20,
3T must be divisible by 20.
As 3 and 20 are coprime,
T must be divisible by 20.
Also, mangoes are 2T/5, so T must be divisible by 5, which is already satisfied.
Step 4: Check the smallest possible value
Let
T = 20k.
Then,
A = (60k − 960)/4
= 15k − 240.
Bananas
= 3T/5 − A
= 12k − (15k − 240)
= 240 − 3k.
Since at least 96 bananas must be available,
240 − 3k ≥ 96
k ≤ 48.
Also, apples must be at least 1,
15k − 240 ≥ 1
k ≥ 17.
The smallest possible value is
k = 17.
Hence,
T = 20 × 17
= 340
Verification
Mangoes = 136
Bananas = 189
Apples = 15
Sold:
● Mangoes = 68
● Bananas = 96
● Apples = 6
Total sold
= 68 + 96 + 6
= 170
which is exactly half of 340.
Answer: 340
Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to
45311
51311
33130
40991
51311
Principal = Rs 2,00,000. Half-yearly rate = 4%.
Step 1: Amount at the end of the first year
= 2,00,000 × (1.04)² = 2,00,000 × 1.0816 = Rs 2,16,320
Step 2: Repayment at end of first year
Outstanding = 2,16,320 − 10,320 = Rs 2,06,000
Step 3: Outstanding at end of third year
From end of year 1 to end of year 3 = 4 half-years.
= 2,06,000 × (1.04)⁴ = 2,06,000 × 1.16985856 ≈ Rs 2,40,991
Step 4: Total amount paid
= 10,320 + 2,40,991 = Rs 2,51,311
Step 5: Total interest paid
= 2,51,311 − 2,00,000 = Rs 51,311
In September, the incomes of Kamal, Amal and Vimal are in the ratio 8 : 6 : 5. They rent a house together, and Kamal pays 15%, Amal pays 12% and Vimal pays 18% of their respective incomes to cover the total house rent in that month. In October, the house rent remains unchanged while their incomes increase by 10%, 12% and 15%, respectively. In October, the percentage of their total income that will be paid as house rent, is nearest to
15.18
13.26
14.84
12.75
13.26
Step 1: Let incomes be 8x, 6x and 5x.
Find the house rent:
Kamal pays 15% of 8x = 1.2x
Amal pays 12% of 6x = 0.72x
Vimal pays 18% of 5x = 0.9x
Total house rent = 1.2x + 0.72x + 0.9x = 2.82x
Step 2: Find the total income in October
Kamal's income = 8x × 1.10 = 8.8x
Amal's income = 6x × 1.12 = 6.72x
Vimal's income = 5x × 1.15 = 5.75x
Total income = 8.8x + 6.72x + 5.75x = 21.27x
Step 3: Find the percentage of income spent on rent
= (2.82x / 21.27x) × 100 = (2.82 / 21.27) × 100 ≈ 13.26%
Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to
42%
45%
47%
49%
47%
Let the initial common selling price of both objects be ₹S.
Step 1: Find the cost prices
For object A, Profit = 20% → CP of A = S/1.20 = 5S/6
For object B, Loss = 10% → CP of B = S/0.90 = 10S/9
Step 2: Increase the selling price
Now the new common selling price gives a profit of 10% on object
B.
New SP = 110% of CP of B = 1.10 × 10S/9 = 11S/9
Step 3: Find the profit on object A
CP of A = 5S/6
New SP = 11S/9
Profit % = [(11S/9 − 5S/6) ÷ (5S/6)] × 100
Take LCM of 9 and 6, 11S/9 − 5S/6 = (22S − 15S)/18 = 7S/18
Therefore, Profit % = [(7S/18) ÷ (5S/6)] × 100 = (7/18) × (6/5) × 100 = 7/15 × 100 = 46.67% ≈ 47%
Answer:
C. 47%
Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is
35.42%
52%
31.25%
26%
35.42%
Step 1: First transaction
CP = Rs. 1000. Sold at 20% profit.
SP = 1000 × 1.2 = Rs. 1200. Profit = Rs. 200.
Step 2: Kanu sells back to Minu
Kanu incurs 20% loss: SP = 1200 × 0.8 = Rs. 960
Minu buys back for Rs. 960.
Step 3: Find the final selling price
Total profit = Rs. 500. Already earned Rs. 200.
Profit in final transaction = 500 − 200 = Rs. 300
Final SP = 960 + 300 = Rs. 1260
Step 4: Calculate percentage profit in final sale
Profit % = (300/960) × 100 = 31.25%
In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is
6:5
4:5
5:4
5:6
4:5
Let total employees = 100.
Manufacturing = 20, Non-manufacturing = 80.
Step 1: Assume total salary
Let total salary = 6 units.
Manufacturing salary = 1 unit, Non-manufacturing salary = 5 units.
Step 2: Find average salaries
Average salary of manufacturing = 1/20
Average salary of non-manufacturing = 5/80 = 1/16
Step 3: Find the required ratio
= (1/20) : (1/16) = 16 : 20 = 4 : 5
Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
10
25
15
20
10
Given:
Gopi marks the price of a product to earn a 20% profit.
Ravi gets a 10% discount on the marked price and saves Rs. 15.
Find Gopi's profit.
Step 1: Find the marked price
Since the discount is 10%,
10% of the marked price = 15
Marked Price
= 15/0.10
= Rs. 150
Step 2: Find the selling price
Selling Price
= 150 − 15
= Rs. 135
Step 3: Find the cost price
The marked price gives a profit of 20%.
Hence,
Marked Price = 120% of Cost Price
Cost Price
= 150/1.20
= Rs. 125
Step 4: Find the profit
Profit
= Selling Price − Cost Price
= 135 − 125
= Rs. 10
Answer:
A. 10
The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is
25%
28%
26%
30%
26%
Let the initial salaries be
● Sita = 5x
● Gita = 6x
● Mita = 7x
After the first year
Sita's salary = 5x × 1.20 = 6x
Gita's salary = 6x × 1.25 = 7.5x
Mita's salary = 7x × 1.20 = 8.4x
After the second year
Sita gets a 40% hike. New salary = 6x × 1.40 = 8.4x
Mita gets a 25% hike. New salary = 8.4x × 1.25 = 10.5x
Let Gita's salary after the second year be G.
Given, Gita's salary becomes equal to the mean salary of all three friends.
So, G = (8.4x + G + 10.5x)/3
Multiply both sides by 3, 3G = 18.9x + G
2G = 18.9x
G = 9.45x
Find Gita's second-year salary hike
Salary before the second-year hike = 7.5x
Increase in salary = 9.45x − 7.5x = 1.95x
Therefore, Salary hike % = (1.95x ÷ 7.5x) × 100 = 26%
Answer:
C. 26%
The selling price of a product is fixed to ensure 40% profit. If the product had cost 40% less and had been sold for 5 rupees less, then the resulting profit would have been 50%. The original selling price, in rupees, of the product is
15
14
10
20
14
Step 1: Let the original cost price be x.
Original selling price = 140% of x = 1.4x
Step 2: Form the second condition
New cost price = 60% of x = 0.6x
New selling price = 1.4x − 5
Since the new profit is 50%:
New selling price = 150% of new cost price = 1.5 × 0.6x = 0.9x
Therefore: 1.4x − 5 = 0.9x
Step 3: Solve for x
0.5x = 5 → x = 10
Step 4: Find the original selling price
Original selling price = 1.4 × 10 = Rs. 14
Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was
Let gold = G, bonds = G/4, stocks = 100000 − 5G/4.
Total return: 0.10(100000 − 5G/4) + 0.06(G/4) + 0.08G = 8200 → 10000 − 0.125G + 0.015G + 0.08G = 8200 → 10000 − 0.03G = 8200 → G = 60000.
Bonds = 60000/4 = ₹15000. Gain from bonds = 6% of 15000 = ₹900.
A certain amount of money was divided among Pinu, Meena, Rinu and Seema. Pinu received 20% of the total amount and Meena received 40% of the remaining amount. If Seema received 20% less than Pinu, the ratio of the amounts received by Pinu and Rinu is
2:1
1:2
5:8
8:5
5:8
Step 1: Assume the total amount. Let the total amount be 100. Pinu receives 20% of 100 = 20. Remaining amount = 100 − 20 = 80.
Step 2: Find Meena's share. Meena receives 40% of the remaining 80 = (40/100) × 80 = 32. Amount left for Rinu and Seema = 80 − 32 = 48.
Step 3: Find Seema's share. Seema receives 20% less than Pinu. Pinu's share = 20. So, Seema's share = 20 − 20% of 20 = 20 − 4 = 16.
Step 4: Find Rinu's share. Rinu's share = 48 − 16 = 32.
Step 5: Find the required ratio. Pinu : Rinu = 20 : 32 = 5 : 8.
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
30
27.5
25
20
25
Given:
After two successive salary increments, Gopal's salary becomes
187.5% of the initial salary.
The second percentage increase is twice the first.
Find the percentage increase in the first increment.
Step 1: Let the first increment be x%.
Then the second increment is
2x%.
The final salary is
187.5% = 15/8
of the initial salary.
Hence,
(1 + x/100)(1 + 2x/100) = 15/8
Step 2: Form the quadratic equation
Multiplying both sides by 10000,
(100 + x)(100 + 2x) = 18750
Expanding,
10000 + 300x + 2x² = 18750
2x² + 300x − 8750 = 0
Divide throughout by 2,
x² + 150x − 4375 = 0
Step 3: Solve the quadratic
Factorizing,
(x + 175)(x − 25) = 0
So,
x = 25
or
x = −175
Since the increment cannot be negative,
x = 25
Final Answer
The percentage increase in the first increment is
25%
Answer:
C. 25
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