21. A manufacture undertakes to supply 2000 pieces of a particular component at ₹25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However, as it turned out, 50% of the components were rejected. What is the loss to the manufacture?
(a) ₹ 12,000
(b) ₹ 13,000
(c) ₹ 14,000
(d) ₹ 15,000
(e) None of these
Solution: (b)
5% of 2000 = 100
2000 – 100 =1900
If he sells 1900 he will get 25% profit cost per piece ₹25
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{25\times 1900\times 100}}{{125}}=38000\)
CP if 50% rejected, only 1000 pieces sold so
1000 × 25 = 25000 = SP
Loss = CP – SP = 38000 – 25000 = 13000
22. In an examination, 40% of the candidates wrote their answers in Hindi and the others in English. The average marks of the candidates written in Hindi is 74 and the average marks of the candidates written in English is 77. What is the average marks of all the candidates?
(a) 75.5
(b) 75.8
(c) 76.0
(d) 76.8
(e) None of these
Solution: (b)
Let total number of candidates = 100
Therefore, total marks of 40 candidates = 40 × 74
Total marks of 60 candidates = 60 × 77
Therefore, required average marks = \(\displaystyle \frac{{40\times 74+60\times 77}}{{100}}\)
= \(\displaystyle \frac{{2960+4620}}{{100}}=\frac{{7580}}{{100}}=75.80\)
23. Ram spends 50% of his monthly income on household items, 20% of his monthly income on buying clothes, 5% of his monthly income on medicines and saves remaining ₹ 11,250. What is Ram’s monthly income?
(a) ₹ 38,200
(b) ₹ 34,000
(c) ₹ 41,600
(d) ₹ 45,000
(e) None of these
Solution: (d)
Let total income of Ram be x. Then
(100 – 50 – 20 – 5)% of x = 11250
x = 45000.
24. If the numerator of a fraction is increased by 350% and the denominator of the fraction is increased by 300% the resultant fraction is \(\displaystyle \frac{9}{22}\). What is the original fraction?
(a) \(\displaystyle \frac{3}{4}\)
(b) \(\displaystyle \frac{5}{12}\)
(c) \(\displaystyle \frac{7}{9}\)
(d) \(\displaystyle \frac{4}{11}\)
(e) None of these
Solution: (d)
Let the original fraction is \(\displaystyle \frac{a}{b}\)
According to question,
\(\displaystyle \frac{{a+\frac{{350}}{{100}}\times a}}{{b+\frac{{300}}{{100}}\times b}}=\frac{9}{{22}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{4.5a}}{{4b}}=\frac{9}{{22}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{a}{b}=\frac{9}{{22}}\times \frac{4}{{4.5}}=\frac{4}{{11}}\)
25. If the numerator of a fraction is increased by 300% and the denominator is increased by 500%, the resultant fraction is \(\displaystyle \frac{5}{12}\). What was the original fraction?
(a) \(\displaystyle \frac{8}{5}\)
(b) \(\displaystyle \frac{5}{{11}}\)
(c) \(\displaystyle \frac{{12}}{5}\)
(d) \(\displaystyle \frac{5}{7}\)
(e) None of these
Solution: (e)
Let the original fraction be \(\displaystyle \frac{x}{y}\)
According to the question,
\(\displaystyle \frac{{x\times 400}}{{y\times 600}}=\frac{5}{{12}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{x}{y}=\frac{5}{{12}}\times \frac{6}{4}=\frac{5}{8}\)
26. Bina’s monthly income is 90% of Anita’s monthly income. The total of both their monthly incomes is Mr. Sen’s monthly income. Mr. Sen’s annual income is 7,75,200. What is Bina’s monthly income?
(a) ₹ 34,000
(b) ₹ 36,000
(c) ₹ 30,600
(d) ₹ 30,000
(e) None of these
Solution: (c)
Sen’s monthly income = \(\displaystyle \frac{{775200}}{{12}}=64600\)
Let the monthly income of Anita be Rs. x.
Therefore, Bina’s monthly income = \(\displaystyle \frac{{90\times x}}{{100}}=0.9x\)
Now, according to the question,
x+0.9x=64600
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{64600}}{{1.9}}=34000\)
Bina’s monthly income = 34000 × 0.9 = ₹ 30600
27. A has double the money of B and B has 50% more money than C. If average money of all the three persons is 12000, how much money A have?
(a) \(\displaystyle \frac{{211000}}{{11}}\)
(b) \(\displaystyle \frac{{315000}}{{11}}\)
(c) \(\displaystyle \frac{{216000}}{{11}}\)
(d) \(\displaystyle \frac{{316000}}{{11}}\)
(e) None of these
Solution: (c)
Let the money of C be x.
According to the question,
Total money of B = x + x + 50%
=\(\displaystyle x+\frac{{50x}}{{100}}=\frac{{3x}}{2}\)
Total money of A = \(\displaystyle 2\times \frac{{3x}}{2}=3x\)
Average money of three persons = 12000
Therefore, total money to thee 12000 × 3
\(\displaystyle 3x+\frac{{3x}}{2}+x=12000\times 3\)
\(\displaystyle \frac{{6x+3x+2x}}{2}=36000\)
3x=\(\displaystyle 3x=\frac{{3\times 7200}}{{11}}\)
Therefore, x \(\displaystyle \frac{{36000\times 2}}{{11}}=\frac{{72000}}{{11}}\)
Now, money of A
=\(\displaystyle 3x=3\times \frac{{72000}}{{11}}=\frac{{216000}}{{11}}\)
28. Fresh grapes contain 80% water, while dry grapes contain 10% water. If the weight of dry grapes is 500 kg, then what is its total weight when it is fresh?
(a) 2350 kg
(b) 2085 kg
(c) 2255 kg
(d) 2250 kg
(e) None of these
Solution: (d)
Let the weight of fresh grapes be x.
Quantity of water in it = \(\displaystyle \frac{{80}}{{100}}\times x=\frac{{4x}}{5}\)
Quantity of pulp in it = \(\displaystyle x-\frac{{4x}}{5}=\frac{x}{5}\)
Quantity of water in 500 kg dry grapes = \(\displaystyle \frac{{10}}{{100}}\times 500=50kg\)
Therefore, quantity of pulp in it = (500 – 50) = 450 kg
\(\displaystyle \frac{x}{5}=450\)
x = 2250 kg
29. Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys?
(a) 62.5%
(b) 54.8%
(c) 52.6%
(d) 55.8%
(e) 53.5%
Solution: (b)
Boys in class 4 = \(\displaystyle \frac{{60}}{{100}}\times 6x=\frac{{360x}}{{100}}\)
Boys in class 5 = \(\displaystyle \frac{{52}}{{100}}\times 11x=\frac{{572x}}{{100}}\)
So total boys = \(\displaystyle \frac{{360x}}{{100}}+\frac{{572x}}{{100}}=\frac{{932x}}{{100}}=9.32x\)
% of boys = \(\displaystyle \frac{{9.32x}}{{17x}}\times 100=54.8\%\)
30. In a school the number of boys and girls are in the ratio of 4:7. If the number of boys are increased by 25% and the number of girls are increased by 15%. What will be the new ratio of number of boys to that of girls?
(a) 100:131
(b) 100:151
(c) 100:161
(d) 100:181
(e) None of these
Solution: (c)
Boys = 4x and girls = 7x
Ratio =\(\displaystyle 4x\times \frac{{125}}{{100}}:7x\times \frac{{115}}{{100}}=100:161\)
Alterative method,
Let x be the constant ratio.
Boys : Girls = 4 : 7
Boys : Girls = 4x : 7x
The number of boys is increased by 25%.
25% of 4x = 0.25 x 4x = x
Number of boys now = 4x + x = 5x
The number of girls is increased by 15%.
15% of 7x = 0.15 x 7x = 1.05x
Number of girls now = 7x + 1.05x = 8.05x
New Ratio
Boys : Girls = 5x : 8.05x
Boys : Girls = 100(5x) : 100(8.05x)
Boys : Girls =500x : 805x
Boys : Girls =100x : 161x
Boys : Girls =100 : 161
