Percentage - Study Mode

[#156] A person who spends $$66frac{2}{3}$$% of his income is able to save Rs. 1200 per month. His monthly expenses (in Rs.) is :
Correct Answer

(B) 2400

Explanation

Solution: $$66frac{2}{3}$$% = $$frac{2}{3}$$ Let the income of the person = 3 units Expenditure = 2 units Savings = (3 - 2) = 1 unit According to the question, 1 unit = Rs. 1200 2 units = 2 × 1200 = Rs. 2400

[#157] The cost of an article worth Rs. 100 is increased by 10% first and again increased by 10%. The total increase in rupees is :
Correct Answer

(B) 21

Explanation

Solution: % increase = 10 + 10 + $$frac{10 × 10}{100}$$ xa0= 21% Total increase = $$frac{100 × 21}{100}$$ xa0= Rs. 21

[#158] Income tax is raised from 4 paise to 5 paise in a rupee but the revenue is increased by 10% only. Find the decrease percent in the amount taxed.
Correct Answer

(A) 12%

Explanation

Solution: Let the original taxed amount be Rs. x and new taxed amount be Rs. y Let the original revenue be Rs. 100 Then, 4% of x = 100 or x = $$left( {frac{{100 imes 100}}{{4}}}
ight)$$ x = Rs. 2500 New revenue = 110% of Rs. 100 = Rs. 110 Then, 5% of y = 110 or y = $$left( {frac{{110 imes 100}}{{5}}}
ight)$$ y = Rs. 2200 Decrease in taxed amount : = Rs. (2500 - 2200) = Rs. 300 ∴ Decrease % : $$=$$$${ ext{ }}left( {frac{{300}}{{2500}} imes 100}
ight)$$ $$= 12$$ %

[#159] The population of a town is 189000. It decreases by 8% in the first year and increases by 5% in the second year. What is the population of the town at the end of 2 year ?
Correct Answer

(A) 182574

Explanation

Solution: Population at the end of 2 years = 189000 $$left( {1 - frac{8}{{100}}}
ight)$$ $$left( {1 + frac{5}{{100}}}
ight)$$ = $$left( {189000 imes frac{{23}}{{25}} imes frac{{21}}{{20}}}
ight)$$ = 182574

[#160] Two vessels contain equal quantities of 40% alcohol. Sachin changed the concentration of the first vessel to 50% by adding extra quantity of pure alcohol. Vivek changed the concentration of the second vessel to 50% replacing a certain quantity of the solution with pure alcohol. By what percentage is the quantity of alcohol added by Sachin more/less than that replaced by Vivek ?
Correct Answer

(D) $$20$$% more

Explanation

Solution: Let each vessel contain 100 litres of 40% alcohol Suppose Sachine added x litres of pure alcohol. Then, $$eqalign{
& Leftrightarrow frac{{40 + x}}{{100 + x}} = frac{{50}}{{100}} cr
& Leftrightarrow frac{{40 + x}}{{100 + x}} = frac{1}{2} cr
& Leftrightarrow 80 + 2x = 100 + x cr
& Leftrightarrow x = 20 cr} $$ Suppose Vivek replaced y litres. Then, alcohol in y litres = 40% of y = $$frac{2y}{5}$$ $$eqalign{
& herefore frac{{40 - frac{{2y}}{5} + y}}{{100}} = frac{{50}}{{100}} cr
& Rightarrow frac{{40 - frac{{2y}}{5} + y}}{{100}} = frac{1}{2} cr
& Rightarrow 80 + frac{{6y}}{{25}} = 100 cr
& Rightarrow y = frac{{20 imes 5}}{6} cr
& Rightarrow y = frac{{50}}{3} cr} $$ Required percentage : $$eqalign{
& = left[ {frac{{left( {20 - frac{{50}}{3}}
ight)}}{{left( {frac{{50}}{3}}
ight)}} imes 100}
ight]\% cr
& = left( {frac{{10}}{3} imes frac{3}{{50}} imes 100}
ight)\% cr
& = 20\% cr} $$