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After decreasing by 25%, the cost of a jacket becomes ₹4,500. Find its original cost.
25% की कमी के बाद, एक जैकेट का मूल्य ₹4,500 हो जाता है। इसका मूल मूल्य ज्ञात कीजिए।
(A) (a) ₹5,500
(B) (b) ₹6,000
(C) (c) ₹6,500
(D) (d) ₹5,800
✅ Answer & Explanation
Sahi jawab: B) (b) ₹6,000Explanation: Method 1 (Ratio Method): $25\% = \frac{1}{4}$. Original Cost : Decreased Cost $= 4 : (4 - 1) = 4 : 3$. Given: 3 units $= ₹4500 \implies 1 \text{ unit} = \frac{4500}{3} = ₹1500$. Original Cost = 4 units $= 4 \times 1500 = ₹6000$. Atah sahi vikalp (b) ₹6,000 hai. Method 2 (100-Base Method): Original cost ko $100\%$ maan lo. $25\%$ decrease ke baad cost $= 100\% - 25\% = 75\%$. Given: $75\% = ₹4500$. $1\% = \frac{4500}{75} = ₹60$. Original Cost ($100\%$) $= 60 \times 100 = ₹6000$. Atah sahi uttar (b) hai.
The price of a laptop was ₹80,000 last year. This year, its price decreased by 20%. What is the price (in ₹) of the laptop this year?
पिछले वर्ष एक लैपटॉप का मूल्य ₹80,000 था। इस वर्ष इसके मूल्य में 20% की कमी आई। इस वर्ष लैपटॉप का मूल्य (₹ में) क्या है?
(A) (a) ₹60,000
(B) (b) ₹62,000
(C) (c) ₹64,000
(D) (d) ₹66,000
✅ Answer & Explanation
Sahi jawab: C) (c) ₹64,000Explanation: Method 1 (Ratio Method): $20\% = \frac{1}{5}$. Original Price : New Price $= 5 : 4$. Given: 5 units $= ₹80000 \implies 1 \text{ unit} = \frac{80000}{5} = ₹16000$. New Price = 4 units $= 4 \times 16000 = ₹64000$. Atah sahi vikalp (c) ₹64,000 hai. Method 2 (100-Base Method): Original Price $= 100\% = ₹80000$. Price $20\%$ decrease hone ke baad bacha $= 80\%$. New Price $= 80\% \text{ of } 80000 = \frac{80}{100} \times 80000 = ₹64000$. Atah sahi uttar (c) hai.
Rahul's salary in 2021 was ₹30,000. He gets an increment of 10% every year. What was his salary in 2023?
2021 में राहुल का वेतन ₹30,000 था। उसे प्रत्येक वर्ष 10% की वेतन वृद्धि मिलती है। 2023 में उसका वेतन क्या था?
(A) (a) ₹36,000
(B) (b) ₹36,300
(C) (c) ₹35,800
(D) (d) ₹37,200
✅ Answer & Explanation
Sahi jawab: B) (b) ₹36,300Explanation: Method 1 (Ratio Method): $10\% = \frac{1}{10}$. Har saal salary ka ratio $= 10 : 11$. 2 saal (2021 se 2023) me ratio $= 10^2 : 11^2 = 100 : 121$. Given: 100 units $= ₹30000 \implies 1 \text{ unit} = ₹300$. Salary in 2023 = 121 units $= 121 \times 300 = ₹36300$. Atah sahi vikalp (b) ₹36,300 hai. Method 2 (100-Base / Successive Formula): Two successive hikes of $10\%$: Net hike $= 10 + 10 + \frac{10 \times 10}{100} = 21\%$. Salary in 2023 $= 30000 + (21\% \text{ of } 30000) = 30000 + 6300 = ₹36300$. Atah sahi uttar (b) hai.
A motorcycle is sold for ₹72,000, which is 20% less than its original price. What was the original price of the motorcycle?
एक मोटरसाइकिल ₹72,000 में बेची जाती है, जो इसके मूल मूल्य से 20% कम है। मोटरसाइकिल का मूल मूल्य क्या था?
(A) (a) ₹85,000
(B) (b) ₹88,000
(C) (c) ₹90,000
(D) (d) ₹95,000
✅ Answer & Explanation
Sahi jawab: C) (c) ₹90,000Explanation: Method 1 (Ratio Method): $20\% = \frac{1}{5}$. Original Price : Selling Price $= 5 : 4$. Given: 4 units $= ₹72000 \implies 1 \text{ unit} = \frac{72000}{4} = ₹18000$. Original Price = 5 units $= 5 \times 18000 = ₹90000$. Atah sahi vikalp (c) ₹90,000 hai. Method 2 (100-Base Method): Original price $= 100\%$. Selling price after $20\%$ reduction $= 80\%$. Given: $80\% = ₹72000$. $1\% = \frac{72000}{80} = ₹900$. Original Price ($100\%$) $= 900 \times 100 = ₹90000$. Atah sahi uttar (c) hai.
If 25% of 40% of a number is 180, then 15% of that number is:
यदि किसी संख्या के 40% का 25% मान 180 है, तो उस संख्या का 15% क्या होगा?
(A) (a) 250
(B) (b) 270
(C) (c) 285
(D) (d) 300
✅ Answer & Explanation
Sahi jawab: B) (b) 270Explanation: Method 1 (Ratio Method): $25\% = \frac{1}{4}$ aur $40\% = \frac{2}{5}$. Fractional part $= \frac{1}{4} \times \frac{2}{5} = \frac{1}{10}$. Given: $\frac{1}{10}$ of number $= 180 \implies \text{Number} = 1800$. Ab number ka $15\%$ nikaalna hai: $15\% = \frac{3}{20} \implies \frac{3}{20} \times 1800 = 3 \times 90 = 270$. Atah sahi vikalp (b) 270 hai. Method 2 (100-Base Method): Let the number be $100\%$. $40\% \text{ of } 100\% = 40\%$. $25\% \text{ of } 40\% = 10\%$. Given: $10\% = 180$. $1\% = 18$. Required $15\% = 15 \times 18 = 270$. Atah sahi uttar (b) hai.
During the first year, the population of a town increased by 10% and during the second year it decreased by 10%. At the end of the second year, its population was 59,400. What was the population at the beginning of the first year?
पहले वर्ष के दौरान एक कस्बे की जनसंख्या में 10% की वृद्धि हुई और दूसरे वर्ष के दौरान इसमें 10% की कमी आई। दूसरे वर्ष के अंत में इसकी जनसंख्या 59,400 थी। पहले वर्ष की शुरुआत में जनसंख्या कितनी थी?
(A) (a) 60,000
(B) (b) 62,000
(C) (c) 58,000
(D) (d) 65,000
✅ Answer & Explanation
Sahi jawab: A) (a) 60,000Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10 : 11$. Year 2 (-10%): $10 : 9$. Initial Population : Final Population $= (10 \times 10) : (11 \times 9) = 100 : 99$. Given: 99 units $= 59400 \implies 1 \text{ unit} = \frac{59400}{99} = 600$. Initial Population = 100 units $= 100 \times 600 = 60000$. Atah sahi vikalp (a) 60,000 hai. Method 2 (Successive Formula): Net percentage change $= +10 - 10 - \frac{10 \times 10}{100} = -1\%$. Final population $= 100\% - 1\% = 99\%$. $99\% = 59400 \implies 1\% = 600$. Initial population ($100\%$) $= 600 \times 100 = 60000$. Atah sahi uttar (a) hai.
Pankaj gets a 10% increase in his sales in the first year and a 20% increase in the second year. If his present sales amount is ₹1,58,400, what was his sales amount two years ago?
पंकज की बिक्री में पहले वर्ष 10% और दूसरे वर्ष 20% की वृद्धि होती है। यदि उसकी वर्तमान बिक्री राशि ₹1,58,400 है, तो दो वर्ष पहले उसकी बिक्री राशि क्या थी?
(A) (a) ₹1,10,000
(B) (b) ₹1,15,000
(C) (c) ₹1,20,000
(D) (d) ₹1,25,000
✅ Answer & Explanation
Sahi jawab: C) (c) ₹1,20,000Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10\% = \frac{1}{10} \implies 10 : 11$. Year 2 (+20%): $20\% = \frac{1}{5} \implies 5 : 6$. Initial Sales : Present Sales $= (10 \times 5) : (11 \times 6) = 50 : 66 = 25 : 33$. Given: 33 units $= ₹158400 \implies 1 \text{ unit} = \frac{158400}{33} = ₹4800$. Initial Sales = 25 units $= 25 \times 4800 = ₹120000$. Atah sahi vikalp (c) ₹1,20,000 hai. Method 2 (100-Base Method): Net increase $= 10 + 20 + \frac{10 \times 20}{100} = 32\%$. Present Sales $= 100\% + 32\% = 132\%$. Given: $132\% = ₹158400$. $1\% = \frac{158400}{132} = ₹1200$. Sales 2 years ago ($100\%$) $= 1200 \times 100 = ₹120000$. Atah sahi uttar (c) hai.
After a discount of 15%, an electronic gadget is available for ₹1,19,000. Find its marked price.
15% की छूट के बाद, एक इलेक्ट्रॉनिक उपकरण ₹1,19,000 में उपलब्ध है। इसका अंकित मूल्य ज्ञात कीजिए।
(A) (a) ₹1,35,000
(B) (b) ₹1,40,000
(C) (c) ₹1,42,000
(D) (d) ₹1,45,000
✅ Answer & Explanation
Sahi jawab: B) (b) ₹1,40,000Explanation: Method 1 (Ratio Method): $15\% = \frac{3}{20}$. Marked Price : Discounted Price $= 20 : (20 - 3) = 20 : 17$. Given: 17 units $= ₹119000 \implies 1 \text{ unit} = \frac{119000}{17} = ₹7000$. Marked Price = 20 units $= 20 \times 7000 = ₹140000$. Atah sahi vikalp (b) ₹1,40,000 hai. Method 2 (100-Base Method): Marked Price $= 100\%$. Price after $15\%$ discount $= 85\%$. Given: $85\% = ₹119000$. $1\% = \frac{119000}{85} = ₹1400$. Marked Price ($100\%$) $= 1400 \times 100 = ₹140000$. Atah sahi uttar (b) hai.
Two years ago, the population of a town was 80,000. Due to industrial development, it increases every year at the rate of 5%. The present population of the town is:
दो वर्ष पहले, एक शहर की जनसंख्या 80,000 थी। औद्योगिक विकास के कारण, इसमें प्रत्येक वर्ष 5% की दर से वृद्धि होती है। शहर की वर्तमान जनसंख्या क्या है?
(A) (a) 88,000
(B) (b) 88,200
(C) (c) 89,000
(D) (d) 89,200
✅ Answer & Explanation
Sahi jawab: B) (b) 88,200Explanation: Method 1 (Ratio Method): $5\% = \frac{1}{20}$. Har saal population ratio $= 20 : 21$. 2 saal me compound change $= 20^2 : 21^2 = 400 : 441$. Given: 400 units $= 80000 \implies 1 \text{ unit} = \frac{80000}{400} = 200$. Present population = 441 units $= 441 \times 200 = 88200$. Atah sahi vikalp (b) 88,200 hai. Method 2 (100-Base / Successive Formula): 2 saal ka net percentage increase $= 5 + 5 + \frac{5 \times 5}{100} = 10.25\%$. Present population $= 80000 + (10.25\% \text{ of } 80000) = 80000 + 8200 = 88200$. Atah sahi uttar (b) hai.
Sumit gets a 5% increase in his sale amount in the first year and 10% in the second year. If his present sale is ₹1,38,600, what was his sale (in ₹) two years ago?
सुमित की बिक्री राशि में पहले वर्ष 5% और दूसरे वर्ष 10% की वृद्धि होती है। यदि उसकी वर्तमान बिक्री ₹1,38,600 है, तो दो वर्ष पहले उसकी बिक्री (₹ में) क्या थी?
(A) (a) ₹1,15,000
(B) (b) ₹1,20,000
(C) (c) ₹1,25,000
(D) (d) ₹1,30,000
✅ Answer & Explanation
Sahi jawab: B) (b) ₹1,20,000Explanation: Method 1 (Ratio Method): Year 1 (+5%): $5\% = \frac{1}{20} \implies 20 : 21$. Year 2 (+10%): $10\% = \frac{1}{10} \implies 10 : 11$. Initial Sale : Present Sale $= (20 \times 10) : (21 \times 11) = 200 : 231$. Given: 231 units $= ₹138600 \implies 1 \text{ unit} = \frac{138600}{231} = ₹600$. Sale 2 years ago = 200 units $= 200 \times 600 = ₹120000$. Atah sahi vikalp (b) ₹1,20,000 hai. Method 2 (100-Base Method): Net increase over 2 years $= 5 + 10 + \frac{5 \times 10}{100} = 15.5\%$. Present sale $= 100\% + 15.5\% = 115.5\%$. Given: $115.5\% = ₹138600$. $1\% = \frac{138600}{115.5} = ₹1200$. Original sale ($100\%$) $= 1200 \times 100 = ₹120000$. Atah sahi uttar (b) hai.
The price of an article is increased by k%. The new price was decreased by k% later. Now the latest price is ₹P. What was the original price of the article?
एक वस्तु के मूल्य में k% की वृद्धि की जाती है। बाद में नए मूल्य में k% की कमी की जाती है। अब नवीनतम मूल्य ₹P है। वस्तु का मूल मूल्य क्या था?
(A) (a) \frac{10000 P}{10000 - k^2}
(B) (b) \frac{(10000 - k^2) P}{10000}
(C) (c) \frac{100 P}{100 - k^2}
(D) (d) \frac{100 P}{1 - k^2}
✅ Answer & Explanation
Sahi jawab: A) (a) \frac{10000 P}{10000 - k^2}Explanation: Method 1 (Successive Multiplier Method): Pehle $k\%$ badhane par factor $= \left(1 + \frac{k}{100}\right)$. Fir $k\%$ ghatane par factor $= \left(1 - \frac{k}{100}\right)$. Net Multiplier $= \left(1 + \frac{k}{100}\right)\left(1 - \frac{k}{100}\right) = 1 - \frac{k^2}{10000} = \frac{10000 - k^2}{10000}$. Latest Price $= \text{Original Price} \times \frac{10000 - k^2}{10000} = P$. $\implies \text{Original Price} = \frac{10000 P}{10000 - k^2}$. Atah sahi vikalp (a) \frac{10000 P}{10000 - k^2} hai. Method 2 (100-Base / Direct Loss Formula): Jab kisi value ko pehle $k\%$ badhayein aur fir $k\%$ ghatayein, toh net loss $= \frac{k^2}{100}\%$ hota hai. Final percentage $= 100\% - \frac{k^2}{100}\% = \frac{10000 - k^2}{100}\%$. Ye value $P$ ke barabar hai: $\frac{10000 - k^2}{10000} \times \text{Original} = P \implies \text{Original} = \frac{10000 P}{10000 - k^2}$. Atah sahi uttar (a) hai.
The population of a village was 2,50,000. It increased by 10% in the first year and increased by 20% in the second year. Its population after two years is:
एक गाँव की जनसंख्या 2,50,000 थी। पहले वर्ष में इसमें 10% की वृद्धि हुई और दूसरे वर्ष में 20% की वृद्धि हुई। दो वर्ष बाद इसकी जनसंख्या क्या होगी?
(A) (a) 3,25,000
(B) (b) 3,30,000
(C) (c) 3,35,000
(D) (d) 3,40,000
✅ Answer & Explanation
Sahi jawab: B) (b) 3,30,000Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10 : 11$. Year 2 (+20%): $5 : 6$. Initial Population : Final Population $= (10 \times 5) : (11 \times 6) = 50 : 66 = 25 : 33$. Given: 25 units $= 250000 \implies 1 \text{ unit} = 10000$. Population after 2 years = 33 units $= 33 \times 10000 = 330000$. Atah sahi vikalp (b) 3,30,000 hai. Method 2 (100-Base / Successive Formula): Two successive increases of $10\%$ and $20\%$: Net increase $= 10 + 20 + \frac{10 \times 20}{100} = 32\%$. Population after 2 years $= 250000 + (32\% \text{ of } 250000) = 250000 + 80000 = 330000$. Atah sahi uttar (b) hai.
The population of a town increases by 10% in the first year, decreases by 20% in the second year, and again increases by 30% in the third year. If the population at the beginning was 50,000, what will it be after 3 years?
एक शहर की जनसंख्या पहले वर्ष में 10% बढ़ती है, दूसरे वर्ष में 20% घटती है, और तीसरे वर्ष में पुनः 30% बढ़ती है। यदि शुरुआत में जनसंख्या 50,000 थी, तो 3 वर्ष बाद यह कितनी होगी?
(A) (a) 56,800
(B) (b) 57,200
(C) (c) 57,600
(D) (d) 58,400
✅ Answer & Explanation
Sahi jawab: B) (b) 57,200Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10 : 11$. Year 2 (-20%): $5 : 4$. Year 3 (+30%): $10 : 13$. Initial : Final $= (10 \times 5 \times 10) : (11 \times 4 \times 13) = 500 : 572$. Given: 500 units $= 50000 \implies 1 \text{ unit} = 100$. Population after 3 years = 572 units $= 572 \times 100 = 57200$. Atah sahi vikalp (b) 57,200 hai. Method 2 (100-Base Method): Initial population $= 100$. After Year 1 (+10%): $100 + 10 = 110$. After Year 2 (-20%): $110 - (20\% \text{ of } 110) = 110 - 22 = 88$. After Year 3 (+30%): $88 + (30\% \text{ of } 88) = 88 + 26.4 = 114.4$. Population after 3 years $= 50000 \times \frac{114.4}{100} = 500 \times 114.4 = 57200$. Atah sahi uttar (b) hai.
The population of a city is 60,000. It increases by 10% in the first year and decreases by 5% in the second year. What will be the population of the city after 2 years?
एक शहर की जनसंख्या 60,000 है। पहले वर्ष में इसमें 10% की वृद्धि होती है और दूसरे वर्ष में 5% की कमी होती है। 2 वर्ष बाद शहर की जनसंख्या क्या होगी?
(A) (a) 62,500
(B) (b) 62,700
(C) (c) 63,000
(D) (d) 63,200
✅ Answer & Explanation
Sahi jawab: B) (b) 62,700Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10\% = \frac{1}{10} \implies 10 : 11$. Year 2 (-5%): $5\% = \frac{1}{20} \implies 20 : 19$. Initial : Final $= (10 \times 20) : (11 \times 19) = 200 : 209$. Given: 200 units $= 60000 \implies 1 \text{ unit} = \frac{60000}{200} = 300$. Population after 2 years = 209 units $= 209 \times 300 = 62700$. Atah sahi vikalp (b) 62,700 hai. Method 2 (Successive Formula): Net percentage change $= +10 - 5 + \frac{10 \times (-5)}{100} = 5 - 0.5 = +4.5\%$. Population after 2 years $= 60000 + (4.5\% \text{ of } 60000) = 60000 + 2700 = 62700$. Atah sahi uttar (b) hai.
The price of a digital camera is first reduced by 20% and then increased by 40%. If the resulting price of the digital camera is ₹8,960, then what was its original price (in ₹)?
एक डिजिटल कैमरे के मूल्य में पहले 20% की कमी की जाती है और फिर 40% की वृद्धि की जाती है। यदि डिजिटल कैमरे का परिणामी मूल्य ₹8,960 है, तो इसका मूल मूल्य (₹ में) क्या था?
(A) (a) ₹7,500
(B) (b) ₹8,000
(C) (c) ₹8,200
(D) (d) ₹8,500
✅ Answer & Explanation
Sahi jawab: B) (b) ₹8,000Explanation: Method 1 (Ratio Method): Reduction: $20\% = \frac{1}{5} \implies 5 : 4$. Increase: $40\% = \frac{2}{5} \implies 5 : 7$. Initial Price : Resulting Price $= (5 \times 5) : (4 \times 7) = 25 : 28$. Given: 28 units $= ₹8960 \implies 1 \text{ unit} = \frac{8960}{28} = ₹320$. Original Price = 25 units $= 25 \times 320 = ₹8000$. Atah sahi vikalp (b) ₹8,000 hai. Method 2 (100-Base Method): Initial Price $= 100\%$. Net Change $= -20 + 40 + \frac{(-20) \times 40}{100} = 20 - 8 = +12\%$. Resulting Price $= 100\% + 12\% = 112\%$. Given: $112\% = ₹8960$. $1\% = \frac{8960}{112} = ₹80$. Original Price ($100\%$) $= 80 \times 100 = ₹8000$. Atah sahi uttar (b) hai.
The price of an electronic printer decreased by 10%, 10%, and 20% during the first three years. What will be the price of the printer after three years if the value at the beginning was ₹15,000?
एक इलेक्ट्रॉनिक प्रिंटर के मूल्य में पहले तीन वर्षों के दौरान क्रमशः 10%, 10% और 20% की कमी आई। यदि शुरुआत में इसका मूल्य ₹15,000 था, तो तीन वर्ष बाद प्रिंटर का मूल्य क्या होगा?
(A) (a) ₹9,720
(B) (b) ₹9,850
(C) (c) ₹10,200
(D) (d) ₹10,500
✅ Answer & Explanation
Sahi jawab: A) (a) ₹9,720Explanation: Method 1 (Ratio Method): Year 1 (-10%): $10 : 9$ Year 2 (-10%): $10 : 9$ Year 3 (-20%): $5 : 4$ Initial Value : Value after 3 years $= (10 \times 10 \times 5) : (9 \times 9 \times 4) = 500 : 324$. Given: 500 units $= ₹15000 \implies 1 \text{ unit} = \frac{15000}{500} = ₹30$. Price after 3 years = 324 units $= 324 \times 30 = ₹9720$. Atah sahi vikalp (a) ₹9,720 hai. Method 2 (100-Base Method): Initial Value $= ₹15000$. After Year 1: $15000 \times 0.90 = 13500$. After Year 2: $13500 \times 0.90 = 12150$. After Year 3: $12150 \times 0.80 = ₹9720$. Atah sahi uttar (a) hai.
The population of a town is 20,000. If the population increases by 10% in the first year, by 20% in the second year, and due to migration it decreases by 5% in the third year, what will be its population after 3 years?
एक शहर की जनसंख्या 20,000 है। यदि पहले वर्ष में जनसंख्या में 10% की वृद्धि होती है, दूसरे वर्ष में 20% की वृद्धि होती है, और प्रवास के कारण तीसरे वर्ष में 5% की कमी होती है, तो 3 वर्ष बाद इसकी जनसंख्या क्या होगी?
(A) (a) 24,850
(B) (b) 25,080
(C) (c) 25,240
(D) (d) 26,120
✅ Answer & Explanation
Sahi jawab: B) (b) 25,080Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10 : 11$ Year 2 (+20%): $5 : 6$ Year 3 (-5%): $20 : 19$ Initial : Final $= (10 \times 5 \times 20) : (11 \times 6 \times 19) = 1000 : 1254$. Given: 1000 units $= 20000 \implies 1 \text{ unit} = 20$. Population after 3 years = 1254 units $= 1254 \times 20 = 25080$. Atah sahi vikalp (b) 25,080 hai. Method 2 (100-Base Method): Initial population $= 20000$. After Year 1 (+10%): $20000 + 2000 = 22000$. After Year 2 (+20%): $22000 + 4400 = 26400$. After Year 3 (-5%): $26400 - (5\% \text{ of } 26400) = 26400 - 1320 = 25080$. Atah sahi uttar (b) hai.
Deepak studies the pattern of daily sale of a grocery shop. He finds that on Monday, the sale was decreased by 20% in comparison to the sale on Sunday. On Tuesday, the sale was increased by 20% in comparison to the sale of Monday. If the sale on Tuesday was of ₹2,880, how much was the sale on Sunday?
दीपक एक किराना दुकान की दैनिक बिक्री के पैटर्न का अध्ययन करता है। वह पाता है कि रविवार की बिक्री की तुलना में सोमवार को बिक्री में 20% की कमी आई। सोमवार की बिक्री की तुलना में मंगलवार को बिक्री में 20% की वृद्धि हुई। यदि मंगलवार को ₹2,880 की बिक्री हुई, तो रविवार को बिक्री कितनी थी?
(A) (a) ₹2,900
(B) (b) ₹3,000
(C) (c) ₹3,150
(D) (d) ₹3,200
✅ Answer & Explanation
Sahi jawab: B) (b) ₹3,000Explanation: Method 1 (Equal Change Direct Shortcut): Jab kisi base value ko pehle $x\%$ ghatayein aur fir $x\%$ badhayein, toh net loss $= \frac{x^2}{100}\%$ hota hai. Net loss $= \frac{20^2}{100}\% = 4\%$. Sunday ki sale ke mukable Tuesday ki sale $= 100\% - 4\% = 96\%$. Given: $96\% = ₹2880$. $1\% = \frac{2880}{96} = ₹30$. Sunday ki sale ($100\%$) $= 30 \times 100 = ₹3000$. Atah sahi vikalp (b) ₹3,000 hai. Method 2 (Ratio Method): Monday (-20%): $5 : 4$. Tuesday (+20%): $5 : 6$. Sunday : Tuesday $= (5 \times 5) : (4 \times 6) = 25 : 24$. Given: 24 units $= ₹2880 \implies 1 \text{ unit} = \frac{2880}{24} = ₹120$. Sunday's sale = 25 units $= 25 \times 120 = ₹3000$. Atah sahi uttar (b) hai.
The price of a car increases by 25% in the first year and decreases by 20% in the second year. If its final price becomes ₹4,80,000, find the original price of the car.
एक कार के मूल्य में पहले वर्ष 25% की वृद्धि होती है और दूसरे वर्ष 20% की कमी होती है। यदि इसका अंतिम मूल्य ₹4,80,000 हो जाता है, तो कार का मूल मूल्य ज्ञात कीजिए।
(A) (a) ₹4,50,000
(B) (b) ₹4,60,000
(C) (c) ₹4,80,000
(D) (d) ₹5,00,000
✅ Answer & Explanation
Sahi jawab: C) (c) ₹4,80,000Explanation: Method 1 (Ratio Method): Year 1 (+25%): $25\% = \frac{1}{4} \implies 4 : 5$. Year 2 (-20%): $20\% = \frac{1}{5} \implies 5 : 4$. Original Price : Final Price $= (4 \times 5) : (5 \times 4) = 20 : 20 = 1 : 1$. Since ratio is $1 : 1$, Original Price $=$ Final Price $= ₹480000$. Atah sahi vikalp (c) ₹4,80,000 hai. Method 2 (Successive Formula): Net Change $= +25 - 20 + \frac{25 \times (-20)}{100} = 5 - 5 = 0\%$. Kyunki net change $0\%$ hai, isliye price unaltered rahega. Original Price $= ₹480000$. Atah sahi uttar (c) hai.
10% of the inhabitants of a village having died of an epidemic, a panic set in, during which 20% of the remaining inhabitants left the village. The population was then reduced to 14,400. What was the number of inhabitants initially?
एक महामारी से एक गाँव के 10% निवासियों की मृत्यु हो जाने के बाद, दहशत फैल गई, जिसके दौरान शेष निवासियों में से 20% ने गाँव छोड़ दिया। इसके बाद जनसंख्या घटकर 14,400 हो गई। शुरुआत में निवासियों की संख्या क्या थी?
(A) (a) 18,000
(B) (b) 20,000
(C) (c) 22,500
(D) (d) 24,000
✅ Answer & Explanation
Sahi jawab: B) (b) 20,000Explanation: Method 1 (Ratio Method): Step 1 (-10%): $10 : 9$. Step 2 (-20%): $5 : 4$. Initial Population : Final Population $= (10 \times 5) : (9 \times 4) = 50 : 36 = 25 : 18$. Given: 18 units $= 14400 \implies 1 \text{ unit} = \frac{14400}{18} = 800$. Initial Population = 25 units $= 25 \times 800 = 20000$. Atah sahi vikalp (b) 20,000 hai. Method 2 (100-Base Method): Initial population $= 100\%$. After epidemic deaths: $100 - 10 = 90\%$. After migration of remaining: $90\% - (20\% \text{ of } 90\%) = 90 - 18 = 72\%$. Given: $72\% = 14400$. $1\% = \frac{14400}{72} = 200$. Initial Population ($100\%$) $= 200 \times 100 = 20000$. Atah sahi uttar (b) hai.
On account of a viral outbreak, 10% of the population of a village died. Out of the remaining population, 25% fled due to panic. If the present population is 5,400, then what was the population of the village before the outbreak?
वायरस के प्रकोप के कारण एक गाँव की 10% जनसंख्या की मृत्यु हो गई। शेष जनसंख्या में से 25% दहशत के कारण भाग गए। यदि वर्तमान जनसंख्या 5,400 है, तो प्रकोप से पहले गाँव की जनसंख्या क्या थी?
(A) (a) 7,500
(B) (b) 8,000
(C) (c) 8,400
(D) (d) 8,800
✅ Answer & Explanation
Sahi jawab: B) (b) 8,000Explanation: Method 1 (Ratio Method): Step 1 (-10%): $10 : 9$. Step 2 (-25%): $4 : 3$. Initial Population : Present Population $= (10 \times 4) : (9 \times 3) = 40 : 27$. Given: 27 units $= 5400 \implies 1 \text{ unit} = \frac{5400}{27} = 200$. Initial Population = 40 units $= 40 \times 200 = 8000$. Atah sahi vikalp (b) 8,000 hai. Method 2 (100-Base Method): Initial population $= 100\%$. After deaths: $100 - 10 = 90\%$. After panic exit: $90\% - (25\% \text{ of } 90\%) = 90 - 22.5 = 67.5\%$. Given: $67.5\% = 5400$. $1\% = \frac{5400}{67.5} = 80$. Initial Population ($100\%$) $= 80 \times 100 = 8000$. Atah sahi uttar (b) hai.
A town has a present population of 1,60,000. Every year the population increases by a birth rate of 15% and decreases by a death rate of 5%. Find the population of the town at the end of 3 years.
एक शहर की वर्तमान जनसंख्या 1,60,000 है। प्रत्येक वर्ष जनसंख्या में 15% की जन्म दर से वृद्धि होती है और 5% की मृत्यु दर से कमी होती है। 3 वर्ष के अंत में शहर की जनसंख्या ज्ञात कीजिए।
(A) (a) 2,10,500
(B) (b) 2,12,960
(C) (c) 2,15,400
(D) (d) 2,18,000
✅ Answer & Explanation
Sahi jawab: B) (b) 2,12,960Explanation: Method 1 (Ratio Method): Net annual growth rate $= \text{Birth Rate} - \text{Death Rate} = 15\% - 5\% = 10\%$. $10\% = \frac{1}{10} \implies$ Har saal population ratio $= 10 : 11$. 3 saal baad ratio $= 10^3 : 11^3 = 1000 : 1331$. Initial base 1000 units $= 160000 \implies 1 \text{ unit} = 160$. Population at the end of 3 years $= 1331 \times 160 = 212960$. Atah sahi vikalp (b) 2,12,960 hai. Method 2 (100-Base / Multiplier Method): Har saal net growth $10\%$ hai, isliye multiplier factor $= 1.10$ hoga. Population after 3 years $= 160000 \times (1.10)^3 = 160000 \times 1.331 = 212960$. Atah sahi uttar (b) hai.
The number of items produced by a manufacturing unit in 2020 was 40,000, which increased by 25% in 2021. In 2022, the production was affected by a raw material shortage and it fell by 20%. How many total items were manufactured in these three years combined?
2020 में एक विनिर्माण इकाई द्वारा उत्पादित वस्तुओं की संख्या 40,000 थी, जो 2021 में 25% बढ़ गई। 2022 में, कच्चे माल की कमी से उत्पादन प्रभावित हुआ और यह 20% गिर गया। इन तीन वर्षों में कुल मिलाकर कितनी वस्तुओं का निर्माण किया गया?
(A) (a) 1,20,000
(B) (b) 1,25,000
(C) (c) 1,30,000
(D) (d) 1,35,000
✅ Answer & Explanation
Sahi jawab: C) (c) 1,30,000Explanation: Method 1 (Step-by-Step Production Method): • Production in 2020 $= 40000$. • Production in 2021 ($+25\%$): $40000 + (25\% \text{ of } 40000) = 40000 + 10000 = 50000$. • Production in 2022 ($-20\%$ on 2021): $50000 - (20\% \text{ of } 50000) = 50000 - 10000 = 40000$. Total production in three years $= 40000 + 50000 + 40000 = 130000$. Atah sahi vikalp (c) 1,30,000 hai. Method 2 (Ratio Multiplier Method): Production in 2020 $= 40000$. Production in 2021 $= 40000 \times \frac{5}{4} = 50000$. Production in 2022 $= 50000 \times \frac{4}{5} = 40000$. Total $= 40000 + 50000 + 40000 = 130000$. Atah sahi uttar (c) hai.
During 2022, the population of a town increased by 10% and in 2023, it increased by 20%. At the end of 2023, its population was 66,000. What was the population of the town at the end of 2022?
2022 के दौरान, एक शहर की जनसंख्या में 10% की वृद्धि हुई और 2023 में इसमें 20% की वृद्धि हुई। 2023 के अंत में इसकी जनसंख्या 66,000 थी। 2022 के अंत में शहर की जनसंख्या कितनी थी?
(A) (a) 50,000
(B) (b) 52,500
(C) (c) 55,000
(D) (d) 57,500
✅ Answer & Explanation
Sahi jawab: C) (c) 55,000Explanation: Method 1 (Single Step Backtrack Method): Question me specifically end of 2022 ki population puchi gayi hai. 2022 ke end se 2023 ke end tak sirf $20\%$ ki increase hui hai: $20\% = \frac{1}{5} \implies$ Population at end of 2022 : Population at end of 2023 $= 5 : 6$. Given: 6 units $= 66000 \implies 1 \text{ unit} = 11000$. Population at end of 2022 = 5 units $= 5 \times 11000 = 55000$. Atah sahi vikalp (c) 55,000 hai. Method 2 (100-Base Method): End of 2022 ki population ko $100\%$ maan lo. 2023 ke end me population $= 100\% + 20\% = 120\%$. Given: $120\% = 66000$. $1\% = \frac{66000}{120} = 550$. Population at end of 2022 ($100\%$) $= 550 \times 100 = 55000$. Atah sahi uttar (c) hai.
Two years ago, the population of a locality was 80,000. It increased by 10% in the first year, but due to health issues, it decreased by 5% in the second year. What is its present population?
दो वर्ष पहले, एक इलाके की जनसंख्या 80,000 थी। पहले वर्ष में इसमें 10% की वृद्धि हुई, लेकिन स्वास्थ्य समस्याओं के कारण दूसरे वर्ष में 5% की कमी आई। इसकी वर्तमान जनसंख्या क्या है?
(A) (a) 82,400
(B) (b) 83,600
(C) (c) 84,000
(D) (d) 85,200
✅ Answer & Explanation
Sahi jawab: B) (b) 83,600Explanation: Method 1 (Ratio Method): Year 1 (+10%): $10 : 11$. Year 2 (-5%): $20 : 19$. Initial Population : Present Population $= (10 \times 20) : (11 \times 19) = 200 : 209$. Given: 200 units $= 80000 \implies 1 \text{ unit} = \frac{80000}{200} = 400$. Present population = 209 units $= 209 \times 400 = 83600$. Atah sahi vikalp (b) 83,600 hai. Method 2 (Successive Net Change Method): Net percentage change $= +10 - 5 + \frac{10 \times (-5)}{100} = 5 - 0.5 = +4.5\%$. Present population $= 80000 + (4.5\% \text{ of } 80000) = 80000 + 3600 = 83600$. Atah sahi uttar (b) hai.
A bacteria culture increases by 10% every hour for the first 2 hours, then decreases by 20% in the 3rd hour, and finally increases by 25% in the 4th hour. If the initial count was 10,000, what will be the count after 4 hours?
एक बैक्टीरिया संवर्धन (culture) में पहले 2 घंटों में प्रत्येक घंटे 10% की वृद्धि होती है, फिर तीसरे घंटे में 20% की कमी होती है, और अंत में चौथे घंटे में 25% की वृद्धि होती है। यदि प्रारंभिक संख्या 10,000 थी, तो 4 घंटे बाद संख्या क्या होगी?
(A) (a) 11,800
(B) (b) 12,100
(C) (c) 12,500
(D) (d) 12,800
✅ Answer & Explanation
Sahi jawab: B) (b) 12,100Explanation: Method 1 (Ratio Multiplication Method): 1st hour (+10%): $\frac{11}{10}$ 2nd hour (+10%): $\frac{11}{10}$ 3rd hour (-20%): $\frac{4}{5}$ 4th hour (+25%): $\frac{5}{4}$ Final Count $= 10000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{4}{5} \times \frac{5}{4} = 10000 \times \frac{121}{100} \times 1 = 12100$. Atah sahi vikalp (b) 12,100 hai. Method 2 (100-Base Step Method): Initial count $= 10000$. After 1st hour: $10000 + 1000 = 11000$. After 2nd hour: $11000 + 1100 = 12100$. After 3rd hour: $12100 - (20\% \text{ of } 12100) = 12100 - 2420 = 9680$. After 4th hour: $9680 + (25\% \text{ of } 9680) = 9680 + 2420 = 12100$. Atah sahi uttar (b) hai.
In the first year, the population of a town decreased by 10%. In the next year, it decreased again by 10%, and in the third year, it increased by 10%. At the end of the third year, the population was 89,100. What was the population at the beginning of the first year?
पहले वर्ष में एक कस्बे की जनसंख्या में 10% की कमी आई। अगले वर्ष इसमें फिर से 10% की कमी आई और तीसरे वर्ष इसमें 10% की वृद्धि हुई। तीसरे वर्ष के अंत में जनसंख्या 89,100 थी। पहले वर्ष की शुरुआत में जनसंख्या कितनी थी?
(A) (a) 95,000
(B) (b) 1,00,000
(C) (c) 1,05,000
(D) (d) 1,10,000
✅ Answer & Explanation
Sahi jawab: B) (b) 1,00,000Explanation: Method 1 (Ratio Method): Year 1 (-10%): $10 : 9$ Year 2 (-10%): $10 : 9$ Year 3 (+10%): $10 : 11$ Initial Population : Final Population $= (10 \times 10 \times 10) : (9 \times 9 \times 11) = 1000 : 891$. Given: 891 units $= 89100 \implies 1 \text{ unit} = 100$. Initial Population = 1000 units $= 1000 \times 100 = 100000$. Atah sahi vikalp (b) 1,00,000 hai. Method 2 (100-Base Fraction Method): Final population factor $= \frac{9}{10} \times \frac{9}{10} \times \frac{11}{10} = \frac{891}{1000}$. $\text{Initial Population} \times \frac{891}{1000} = 89100 \implies \text{Initial Population} = 100000$. Atah sahi uttar (b) hai.
The number of students in a school is 50,000 and it is increasing annually by 20%. What will be the number of students in the school at the end of 3 years?
एक विद्यालय में विद्यार्थियों की संख्या 50,000 है और इसमें प्रतिवर्ष 20% की वृद्धि हो रही है। 3 वर्ष के अंत में विद्यालय में विद्यार्थियों की संख्या क्या होगी?
(A) (a) 84,400
(B) (b) 86,400
(C) (c) 88,000
(D) (d) 90,200
✅ Answer & Explanation
Sahi jawab: B) (b) 86,400Explanation: Method 1 (Ratio Method): $20\% = \frac{1}{5} \implies$ Growth ratio $= 5 : 6$. After 3 years, ratio $= 5^3 : 6^3 = 125 : 216$. Given: 125 units $= 50000 \implies 1 \text{ unit} = \frac{50000}{125} = 400$. Students after 3 years = 216 units $= 216 \times 400 = 86400$. Atah sahi vikalp (b) 86,400 hai. Method 2 (100-Base Successive Formula): 3 successive increases of $20\%$: Step 1 (2 years): $20 + 20 + \frac{20 \times 20}{100} = 44\%$. Step 2 (3 years): $44 + 20 + \frac{44 \times 20}{100} = 64 + 8.8 = 72.8\%$. Students after 3 years $= 50000 + (72.8\% \text{ of } 50000) = 50000 + 36400 = 86400$. Atah sahi uttar (b) hai.
The strength of a school increases and decreases in alternate years by 10%. It started with an increase in 2016 and the initial strength was 5,000. Find the strength of the school in 2018 and the net percentage increase from 2016, respectively.
एक विद्यालय की छात्र संख्या में प्रत्येक एकांतर वर्ष में 10% की वृद्धि और कमी होती है। इसकी शुरुआत 2016 में वृद्धि के साथ हुई और प्रारंभिक छात्र संख्या 5,000 थी। वर्ष 2018 में विद्यालय की छात्र संख्या तथा छात्र संख्या में कुल प्रतिशत वृद्धि क्रमशः ज्ञात कीजिए।
(A) (a) 4,950, 1% decrease
(B) (b) 4,950, 1% increase
(C) (c) 5,050, 1% increase
(D) (d) 5,100, 2% increase
✅ Answer & Explanation
Sahi jawab: A) (a) 4,950, 1% decreaseExplanation: Method 1 (Ratio Method): 2016 se 2018 tak kul 2 saal ka badlav hua (Year 1: $+10\%$, Year 2: $-10\%$): Year 1: $10 : 11$, Year 2: $10 : 9$. Initial : Final $= (10 \times 10) : (11 \times 9) = 100 : 99$. 100 units $= 5000 \implies 1 \text{ unit} = 50$. Final strength $= 99 \times 50 = 4950$. Net change $= 99 - 100 = -1\%$ (1% decrease). Atah sahi vikalp (a) 4,950, 1% decrease hai. Method 2 (100-Base Method): Initial strength $= 5000$. 2017 ke end me ($+10\%$): $5000 + 500 = 5500$. 2018 ke end me ($-10\%$): $5500 - 550 = 4950$. Net decrease $= 5000 - 4950 = 50$. Percentage change $= \frac{50}{5000} \times 100\% = 1\%$ decrease. Atah sahi uttar (a) hai.
The population of a city increased to 40,000 from 2018 to 2020 at the rate of 10% per annum, and continued the same trend for the next 2 years. If X is the population in 2018 and Y is the population in 2022, find the value of Y - X (rounded to nearest integer).
एक शहर की जनसंख्या 2018 से 2020 तक 10% वार्षिक दर से बढ़कर 40,000 हो गई, और अगले 2 वर्षों तक यही क्रम जारी रहा। यदि X वर्ष 2018 की जनसंख्या है और Y वर्ष 2022 की जनसंख्या है, तो Y - X का मान ज्ञात कीजिए (निकटतम पूर्णांक तक)।
(A) (a) 15,340
(B) (b) 15,342
(C) (c) 15,350
(D) (d) 15,355
✅ Answer & Explanation
Sahi jawab: B) (b) 15,342Explanation: Method 1 (Ratio Method): Annual growth $= 10\% = \frac{1}{10} \implies$ Multiplier per year $= \frac{11}{10}$. 2020 ki population $= 40000$. 2018 ki population ($X$) $= 40000 \times \left(\frac{10}{11}\right)^2 = \frac{4000000}{121} \approx 33057.85$. 2022 ki population ($Y$) $= 40000 \times \left(\frac{11}{10}\right)^2 = 40000 \times 1.21 = 48400$. Difference $Y - X = 48400 - 33057.85 = 15342.15 \approx 15342$. Atah sahi vikalp (b) 15,342 hai. Method 2 (100-Base Method): $X = \frac{40000}{1.21} = 33057.85$. $Y = 40000 \times 1.21 = 48400$. $Y - X = 48400 - 33057.85 \approx 15342$. Atah sahi uttar (b) hai.