Explanation for the Arithmetic Quiz 05

This is the detailed explanation for the Arithmetic Quiz 05. Each question’s solution is broken down step-by-step to help you understand the core concepts and calculations.

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1. If the price of the commodity is increased by 50% by what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?

Answer: 1/3

• Step 1: Assume an initial price and consumption. Let the original price be $P=100$ and the original consumption be $C=100$.
• Step 2: Calculate the original expenditure. Original Expenditure = $P\times C=100\times 100=10000$.
• Step 3: Calculate the new price. The price is increased by 50%, so the new price is $100+(50/100)\times 100=150$.
• Step 4: Find the new consumption (x) to maintain the same expenditure.$150\times x=10000$$x=10000/150=66.67$
• Step 5: Calculate the reduction in consumption. Reduction = Original Consumption – New Consumption = $100-66.67=33.33$.
• Step 6: Convert the reduction to a fraction of the new consumption. The reduction is $33.33$. The new consumption is $66.67$.$33.33/100=1/3$. The fraction of the original consumption to be reduced is $33.33/100=1/3$.

• Alternative Method (for percentages): If the price increases by R%, the consumption should be reduced by $[R/(100+R)]\times 100\%$. Here, R = 50%. Reduction fraction = $50/(100+50)=50/150=1/3$.

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2. In an examination 80% candidates passed in English and 85% candidates passed in Mathematics. If 73% candidates passed in both these subjects, then what per cent of candidates failed in both the subjects?

Answer: 8

• Step 1: Find the percentage of students who passed in at least one subject. Use the formula:$P(E\cup M)=P(E)+P(M)-P(E\cap M)$ where $P(E\cup M)$ is the percentage passed in at least one subject, $P(E)$ is the percentage passed in English, and $P(M)$ is the percentage passed in Mathematics, and $P(E\cap M)$ is the percentage passed in both.
• Step 2: Plug in the values.$P(\text{English or Math})=80\%+85\%-73\%=165\%-73\%=92\%$. This means 92% of candidates passed in at least one subject.
• Step 3: Calculate the percentage of candidates who failed in both subjects. Total candidates = 100%. Percentage failed in both = $100\%-92\%=8\%$.

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3. A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3cm. if the length of AC is increased by 6%, the length of CB is decreased by

Answer: 9%

• Step 1: Determine the original length of CB. Total length AB = 5 cm. AC = 3 cm. CB = AB – AC = $5-3=2$ cm.
• Step 2: Calculate the new length of AC after a 6% increase. Increase = $6\%$ of $3$ cm = $(6/100)\times 3=0.18$ cm. New AC = $3+0.18=3.18$ cm.
• Step 3: Calculate the new length of CB. New CB = Total length AB – New AC = $5-3.18=1.82$ cm.
• Step 4: Calculate the decrease in the length of CB. Decrease = Original CB – New CB = $2-1.82=0.18$ cm.
• Step 5: Calculate the percentage decrease in CB’s length. Percentage Decrease = (Decrease / Original CB) $\times 100\%$ Percentage Decrease = $(0.18/2)\times 100\%=0.09\times 100\%=9\%$.

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4. The cost of an article was Rs.75. The cost was first increased by 20% and later on it was reduced by 20%. The present cost of the article is:

Answer: Rs. 72

• Step 1: Calculate the cost after the 20% increase. Increase = $20\%$ of $75=(20/100)\times 75=15$. New cost = $75+15=90$.
• Step 2: Calculate the cost after the 20% reduction. The reduction is on the new cost (Rs. 90). Reduction = $20\%$ of $90=(20/100)\times 90=18$.
• Step 3: Find the final cost. Final cost = New cost – Reduction = $90-18=\text{Rs. }72$.
• Shortcut: A percentage increase followed by the same percentage decrease is a net decrease. The final cost will be P(1−r2) where $r$ is the percentage as a decimal.75(1−0.202)=75(1−0.04)=75×0.96=72.

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5. If the price of a commodity is decreased by 20% and its consumption is increased by 20%, what will be the increase or decrease in expenditure on the commodity?

Answer: 4% decrease

• Step 1: Assume an initial price and consumption. Let the original price be $P=100$ and the original consumption be $C=100$. Original expenditure = $100\times 100=10000$.
• Step 2: Calculate the new price and consumption. New Price = $100-(20/100)\times 100=80$. New Consumption = $100+(20/100)\times 100=120$.
• Step 3: Calculate the new expenditure. New Expenditure = $80\times 120=9600$.
• Step 4: Find the percentage change in expenditure. Change = Original Expenditure – New Expenditure = $10000-9600=400$. Percentage change = (Change / Original Expenditure) $\times 100\%$ Percentage change = $(400/10000)\times 100\%=4\%$. Since the expenditure decreased, it’s a 4% decrease.
• Shortcut: For a percentage increase ($+x\%$) and an equal percentage decrease ($-x\%$), the net change is a decrease of x2/100%. Net change = 202/100=400/100=4%. Since it’s a decrease, the answer is a 4% decrease.

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6. The price of the sugar rise by 25%. If a family wants to keep their expenses on sugar the same as earlier, the family will have to decrease its consumption of sugar by

Answer: 20%

• Step 1: Use the formula for consumption reduction to keep expenditure constant. Reduction percentage = $[R/(100+R)]\times 100\%$ where R is the percentage increase in price.
• Step 2: Plug in the value. R = 25%. Reduction percentage = $[25/(100+25)]\times 100\%=(25/125)\times 100\%=(1/5)\times 100\%=20\%$.

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7. Each side of a rectangular field diminished by 40%. By how much per cent is the area of the field diminished?

 

Answer: 64%

• Step 1: Assume original dimensions. Let the original length be $L=10$ and the original width be $W=10$. Original Area = $L\times W=10\times 10=100$.
• Step 2: Calculate the new dimensions after a 40% decrease. New Length = $10-(40/100)\times 10=10-4=6$. New Width = $10-(40/100)\times 10=10-4=6$.
• Step 3: Calculate the new area. New Area = $6\times 6=36$.
• Step 4: Calculate the percentage diminished. Diminished Area = Original Area – New Area = $100-36=64$. Percentage Diminished = (Diminished Area / Original Area) $\times 100\%=(64/100)\times 100\%=64\%$.
• Shortcut: For a decrease of $x\%$ on both sides, the area decrease is [2x−(x2/100)]%.$x=40$. Decrease = 2(40)−(402/100)=80−(1600/100)=80−16=64%.

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8. 1.14 expressed as a per cent of 1.9 is:

 

Answer: 60%

• Step 1: Set up the fraction. To express ‘A’ as a percentage of ‘B’, the formula is $(A/B)\times 100\%$.
• Step 2: Plug in the values. Percentage = $(1.14/1.9)\times 100\%$ Percentage = $(114/190)\times 100\%=(114/19)\times 10\%=6\times 10\%=60\%$.

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9. Half percent, written as a decimal, is

 

Answer: 0.005

• Step 1: Convert “half percent” into a fraction. Half percent is $0.5\%$.
• Step 2: To convert a percentage to a decimal, divide by 100.$0.5/100=0.005$.

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10. The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

 

Answer: 54,080

• Step 1: Calculate the population after the first year. Increase = $4\%$ of $50,000=(4/100)\times 50000=2000$. Population after 1st year = $50000+2000=52000$.
• Step 2: Calculate the population after the second year. The increase is based on the new population. Increase = $4\%$ of $52,000=(4/100)\times 52000=2080$. Population after 2nd year = $52000+2080=54080$.
• Formula Method: For a constant growth rate, the final population is given by P(1+r/100)t.$P=50000$, $r=4$, $t=2$. Population = 50000(1+4/100)2=50000(1.04)2=50000×1.0816=54080.