Percentage - Study Mode
[#171] The base of a triangle is increased by 40%. By what percentage (correct to two decimal places) should its height be increased so that the area increases by 60%?
Correct Answer
(A) 14.29%
Explanation
Solution: $$eqalign{
& { ext{Base}} - 40\% = frac{2}{5} cr
& { ext{Area}} - 60\% = frac{3}{5} cr
& { ext{Area}} o { ext{5:8}} cr
& { ext{Base}} o { ext{5:7}} cr
& { ext{Height}} o frac{5}{5}:frac{8}{7} cr
& underbrace {7,,:,,8}_1 cr
& frac{1}{7} imes 100 = 14.29\% cr} $$
[#172] A class has five sections that have 25, 30, 40, 45 and 60 students, respectively. The pass percentage of these section are 20%, 30%, 35%, 40% and 100% respectively. The pass percentage of the entire class is:
Correct Answer
(C) 53%
Explanation
Solution: Number of students of all sections = 25, 30, 40, 45, 60 $$eqalign{
& 25 imes frac{1}{5} = 5 cr
& 30 imes frac{3}{{10}} = 9 cr
& 40 imes frac{7}{{20}} = 14 cr
& 45 imes frac{2}{5} = 18 cr
& 60 imes frac{{100}}{{100}} = 60 cr} $$ 106 → Passed student Total student = 200 $${ ext{Pass}}\% = frac{{106}}{{200}} imes 100 = 53\% $$
[#173] A crate of fruits contains one spoiled fruit for every 25 fruits. 60% of the spoiled fruits were sold. If the seller had sold 48 spoiled fruits, then the number of fruits in the crate were:
Correct Answer
(A) 2000
Explanation
Solution: Let the number of Crate = x Spoiled Fruit = x $$eqalign{
& { ext{Now, }}frac{{x imes 60}}{{100}} = 48 cr
& x = 80 cr} $$ The number of Fruits = 80 × 25 = 2000
[#174] Rice is now being sold at Rs. 29 per kg. During the last month, its cost was Rs. 25 per kg. By how much percentage should a family reduce its consumption, so as to keep the expenditure the same as before? (correct to nearest integer)
Correct Answer
(A) 14%
Explanation
Solution: Given: New selling price = Rs. 29 per kg Old selling price = Rs. 25 per kg Expenditure = constant Formula used: (i) Expenditure = Price × Quantity (ii) Percentage difference = $$frac{{{ ext{New price}} - { ext{Old price}}}}{{{ ext{Old price}}}} imes 100$$ Calculations: Let expenditure, initial selling price and initial quantity be C, P 1 and Q 1 respectively. C = P 1 × Q 1 ⇒ C = 25 × Q 1 ⇒ Q 1 = $$frac{{ ext{C}}}{{25}}$$ Let final selling price and final quantity be P 2 and Q 2 respectively. C = P 2 × Q 2 ⇒ C = 29 × Q 2 ⇒ Q 2 = $$frac{{ ext{C}}}{{29}}$$ Percentage difference $$ = frac{{{{ ext{Q}}_2} - {{ ext{Q}}_1}}}{{{{ ext{Q}}_1}}} imes 100$$ $$eqalign{
& Rightarrow left[ {frac{{frac{{ ext{C}}}{{29}} - frac{{ ext{C}}}{{25}}}}{{frac{{ ext{C}}}{{25}}}}}
ight] imes 100 cr
& Rightarrow left[ {frac{{frac{{25{ ext{C}} - 29{ ext{C}}}}{{25 imes 29}}}}{{frac{{ ext{C}}}{{25}}}}}
ight] imes 100 cr
& Rightarrow frac{{ - left( {frac{{4{ ext{C}}}}{{725}}}
ight)}}{{frac{{ ext{C}}}{{25}}}} imes 100 cr
& Rightarrow frac{{ - left( {4{ ext{C}} imes { ext{25}}}
ight)}}{{725{ ext{C}}}} imes 100 cr
& Rightarrow - 13.79\% approx - 14\% cr} $$ ∴ By 14% should a family reduce its consumption, so as to keep the expenditure the same as before. Alternate solution [x08egin{array}{*{20}{c}}
{}&{{ ext{Price}}}&{{ ext{Consumption}}} \
{{ ext{Old}}}&{25}&{{mathbf{29}}} \
{{ ext{New}}}&{29}&{{mathbf{25}}} \
{{ ext{Expenditure}}}&{25 imes 29}&{29 imes 25}
end{array}] Thus, family should reduce their consumption by 4 kg % Reduction $$ = frac{4}{{29}} imes 100 = 13.79 approx 14\% $$ ∴ The family should reduce the consumption by 14%.
[#175] Some students (only boys and girls) from different schools appeared for an Olympiad exam. 20% of the boys and 15% of the girls failed the exam. The number of boys who passed the exam was 70 more than that of the girls who passed the exam. A total of 90 students failed. Find the number of students that appeared for the exam.
Correct Answer
(C) 500
Explanation
Solution: Let number of boys = B and number of girls = G According to the question, B × 80% - G × 85% = 70 ⇒ 80B - 85G = 7000 . . . . . . (i) Again, B × 20% + G × 15% = 90 ⇒ 20B + 15G = 9000 ⇒ 80B + 60G = 36000 . . . . . . (ii) From equation (i) - (ii) 145G = 29000 ⇒ G = 200 20B + 15 × 200 = 9000 ⇒ 20B = 6000 ⇒ B = 300 ∴ Total students = 300 + 200 = 500