31. Two numbers x & y are such that the sum of 6% of x and 5% of y is \(\frac{3}{5}\) of the sum of 6% of x and 7% of y. Then find the ratio of x and y.
Sol:
6% × x + 5% × y = \(\frac{3}{5}\)(6% × x + 7% × y)
⟹ \(\frac{6}{{100}}x + \frac{5}{{100}}y = \frac{3}{5}(\frac{6}{{100}}x + \frac{7}{{100}}y)\)
⟹ 30x + 25y = 18x + 21y
⟹ 12x = – 4y
3x = -y
∴ \( \boldsymbol {\frac{x}{y} = – \frac{1}{3}} \) Answer
32. The ratio of number of boys to that of girls in a school is 5 : 1. If 70% of boys and 60% of girls are scholarship holders, then find the percentage of students who do not get scholarship.
Sol:
\(\frac{{boys}}{{girls}} = \frac{5}{1} = \frac{{500}}{{100}}\)
∴ total students who do not get scholarship
= 500(100 – 70)% + 100(100 – 60)%
= 500 × \(\frac{{30}}{{100}}\) + 100 × \(\frac{{40}}{{100}}\)
= 150 + 40
= 190
∴ required number of percentage of students who do not get scholarship = \(\frac{{190}}{{600}} \times 100\) = \( \boldsymbol{\frac{{19}}{6}} \)% Answer
33. The cost of manufacture of an article is made up of 4 components A, B, C and D which have a ratio of 1 : 2 : 3 : 4 respectively. If there are respective changes in the cost of 10%, -20%, -30%, -40%, then what will be the percentage change in the cost ?
Sol:
total old cost = 100 + 200 + 300 + 400
= 1000
total new cost = 110 + 160 + 390 + 240
= 900
∴ required % change = \(\frac{{900 – 1000}}{{1000}} \times 100\% \)
= – 10% Answer
34. The height of a triangle is increased by 20%. What could be maximum percentage increase in length of base whose area increases by 60%.
Sol:
\(60 = x + 20 + \frac{{20x}}{{100}}\)
\(40 = x + \frac{x}{5}\)
200 = 6x
⟹ x = \(\frac{{200}}{6} = \frac{{100}}{3}\) = \( \boldsymbol{33\frac{1}{3}} \)% Answer
35. In an examination there are three subjects physics, chemistry and math having max marks 200, 120, 150 respectively. A student gets 50%, 70%, 60% in physics, chemistry, math respectively. If he wants to get 70% marks if four subjects then how many marks he must obtain in biology of max marks 200.
Sol:
Total max marks of 4 subject = 200 + 120 + 150 + 200
= 670
∴ 70% of max marks = \(\frac{{70}}{{100}} \times 670\)
= 469
& total marks obtained in three subjects
= 200 × 50% + 120 × 70% + 150 × 60%
= 100 + 84 + 90
= 274
∴ marks to be obtained in biology = 469 – 274 = 195 Answer
36. In an examination 500 girls and 1000 boys appeared. In which 50% girls and 40% boys passed the exam. Then find the % of failed students ?
Sol:
∴ % of failed students = \(\frac{{250 + 600}}{{500 + 1000}} \times 100\)
= \(\frac{{850}}{{1500}} \times 100\)
= \(\frac{{17}}{3}\)
= \( \boldsymbol{5\frac{2}{3}} \)% Answer
37. In a class, the number of girls is 20% more than that of the boys. The strength of the class is 77. If 6 more girls are admitted to the class then find the ratio of the number of boys to that of the girls.
Sol:
Strength of class = 5 + 6 = 11 ⟶ 77 (Given)
∴ Number of girls = 6 × 7 = 42
& Number of boys = 5 × 7 = 35
∴ \(\frac{{boys}}{{girls}} = \frac{{35}}{{42}}\) = \( \boldsymbol{\frac{5}{6}} \) Answer
38. If 70% of the students in a school are girls and the number of boys is 570, how many girls are in school ?
Sol:
39. Ankit invests 1000 Rs. in some share in the ratio 2 : 3 : 5 which pay dividends of 5%, 10% and 20% (on his investment) for that year respectively. Find the dividend income.
Sol:
Ratio of shares = 2x : 3x : 5x
∴ 2x + 3x + 5x = 1000
x = 100
∴ 1ˢᵗ share = 200
2ⁿᵈ share = 300
3ʳᵈ share = 500
& dividend income = \(200 \times \frac{5}{{100}} + 300 \times \frac{{10}}{{100}} + 500 \times \frac{{20}}{{100}}\)
= 10 + 30 + 100
= 140 Answer
40. A batsman scored 130 runs which included 5 four and 5 sixes. What percentage of his total score did he make by running between the wickets ?
Sol:
The runs scored buy running = 130 – 5 × 4 – 5 × 6 = 80
∴ required % = \(\frac{{80}}{{130}} \times 100\)
= \( \boldsymbol{61\frac{7}{{13}}} \)% Answer