ARITHMETIC REVIEW TEST - 1
Q1–10 of 25A man earns x% on the first ₹ 5,000 of his investment and y% on the rest of his investment. If he earns ₹ 1250 from ₹ 7,000 and ₹ 1750 from ₹ 9,000 invested, find the value of x.
Given: $x\%$ on first ₹5000, $y\%$ on remainder.
Investment ₹7000 $\implies 5000(x\%) + 2000(y\%) = 1250$
Investment ₹9000 $\implies 5000(x\%) + 4000(y\%) = 1750$
Calculation: Subtract equation (1) from (2):
Substitute $y\% = 25\%$ back into equation (1):
Correct Answer: (b) 15%
The price of a television set drops by 30% while the sales of the set goes up by 50% What is the percentage change in the total revenue from the sales of the set?
Calculation: Price drops by $30\%$ (Multiplier = $0.70$), Sales increase by $50\%$ (Multiplier = $1.50$).
Correct Answer: (c) +5%
A person who has a certain amount with him goes to the market. He can buy 100 oranges or 80 mangoes. He retains 20% of the amount for petrol expenses and buys 40 mangoes and of the balance, he purchases oranges. The number of oranges he can purchase is:
Given: Let total amount = ₹100.
Price of 1 orange $= \frac{100}{100} = ₹1$
Price of 1 mango $= \frac{100}{80} = ₹1.25$
Expenditures:
Petrol $= 20\%$ of ₹100 $= ₹20$
40 Mangoes $= 40 \times 1.25 = ₹50$
Remaining amount $= 100 - (20 + 50) = ₹30$
Oranges purchased: $\frac{₹30}{₹1} = 30$ oranges.
Correct Answer: (a) 30
A cloth merchant cheats his supplier and his customer to the tune of 20% while buying and selling cloth respectively. He professes to sell at the cost price but also offers a discount of 20% on cash payment, what is his overall profit percentage?
Cheating Mechanism:
Buys $120$ units of cloth for the price of $100$ units $\implies \text{Cost Price per unit} = \frac{100}{120} = \frac{5}{6}$.
Sells $80$ units of cloth while charging customer for $100$ units $\implies \text{Marked Price per unit} = \frac{100}{80} = \frac{5}{4}$.
Cash discount of $20\%$ $\implies \text{Selling Price per unit} = \frac{5}{4} \times 0.80 = 1$.
Profit Percentage:
Correct Answer: (a) 20%
I sold two horses for ₹ 50000 each, one at the loss of 20% and the other at the profit of 20%. What is the percentage of loss (−) or profit (+) that resulted from the transaction?
Rule: When two items are sold at the same selling price, one at $P\%$ profit and the other at $P\%$ loss, there is an overall loss given by $\frac{P^2}{100}\%$.
Calculation:
Correct Answer: (b) (-) 4
The cost of a diamond varies directly as the square of its weight. A diamond fell and broke into four pieces whose weights were in the ratio 1:2:3:4. As a result the merchant had a loss of ₹ 700000. Find the original price of the diamond.
Given: Weight ratio $= 1:2:3:4$. Total weight $= 1+2+3+4 = 10x$.
Original Price $= k \times (10x)^2 = 100kx^2$
New Price $= k \times (1^2 + 2^2 + 3^2 + 4^2)x^2 = k(1 + 4 + 9 + 16)x^2 = 30kx^2$
Loss $= 100kx^2 - 30kx^2 = 70kx^2 = ₹7,00,000$
$kx^2 = 10,000$
Original Price: $100kx^2 = 100 \times 10,000 = ₹10,00,000 = 10 \text{ lacs}$.
Correct Answer: (c) 10 lacs
Two oranges, three bananas and four apples cost ₹ 25. Three oranges, two bananas and one apple cost ₹ 20. I brought 3 oranges, 3 bananas and 3 apples. How much did I pay?
Given:
$2O + 3B + 4A = 25$
$3O + 2B + 1A = 20$
Calculation: Add both equations:
Cost of $3O + 3B + 3A$:
Correct Answer: (b) ₹27
From each of two given numbers, half the smaller number is subtracted. Of the resulting numbers the larger one is five times as large as the smaller one. What is the ratio of the two numbers?
Given: Let larger number $= L$, smaller number $= S$.
Half the smaller number is subtracted from both:
$$\text{New Larger} = L - \frac{S}{2}$$$$\text{New Smaller} = S - \frac{S}{2} = \frac{S}{2}$$Relation:
$$L - \frac{S}{2} = 5 \left(\frac{S}{2}\right) \implies L = \frac{5S}{2} + \frac{S}{2} = \frac{6S}{2} = 3S$$
Ratio: $\frac{L}{S} = \frac{3}{1} \implies 3:1$.
Directions for Questions 9 and 10: Answer these questions based on the following information.
A watch dealer incurs an expense of ₹ 150 for producing every watch. He also incurs an additional expenditure of ₹ 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for ₹ 250. If he fails to do so, he has to sell each watch for ₹ 100.
Q9. If he is able to sell only 1,000 out of 1,500 watches he has made in the season, then he has made a profit of:
Given: $1500$ watches made. $1000$ sold at ₹250, remaining $500$ sold at ₹100.
Total Production Cost $= 1500 \times 150 + 30,000 = 2,25,000 + 30,000 = ₹2,55,000$
Total Revenue $= (1000 \times 250) + (500 \times 100) = 2,50,000 + 50,000 = ₹3,00,000$
Profit: $3,00,000 - 2,55,000 = ₹45,000$.
Correct Answer: (c) ₹45,000
Directions for Questions 9 and 10: Answer these questions based on the following information.A watch dealer incurs an expense of ₹ 150 for producing every watch. He also incurs an additional expenditure of ₹ 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for ₹ 250. If he fails to do so, he has to sell each watch for ₹ 100.
Q. If he produces 2000 watches, what is the number of watches that he must sell during the season (to the nearest 100) in order to break-even, given that he is able to sell all the watches produced?
Given: Total production $= 2000$. Let $x$ be the number sold during the season at ₹250.
Remaining sold off-season at ₹100 $= 2000 - x$
Total Cost $= (2000 \times 150) + 30,000 = 3,00,000 + 30,000 = ₹3,30,000$
Break-even equation:
Nearest 100: $900$ watches