In an office in Singapore there are 60% female employees. 50 % of all the male employees are computer literate. If there are total 62% employees computer literate out of total 1600 employees, then the no. of female employees who are computer literate ?
(A) 690
(B) 674
(C) 672
(D) 960
Solution:
Total employees = 1600 Female employees, 60% of 1600 $$ = \frac{{60 \times 1600}}{{100}} = 960$$ Then male employees = 640 50% of male are computer literate, = 320 male computer literate 62% of total employees are computer literate, $$ = \frac{{62 \times 1600}}{{100}} = 992$$ computer literate Thus, Female computer literate = 992 - 320 = 672 Alternatively : Let 60% employees are female and 40% are male Then, 20% of male are computer literate and 42% are female computer literate Female computer literate $$ = \frac{{1600 \times 42}}{{100}} = 672$$
112.
A person who spends $$66\frac{2}{3}$$% of his income is able to save Rs. 1200 per month. His monthly expenses (in Rs.) is :
(A) 1200
(B) 2400
(C) 3000
(D) 3200
Solution:
$$66\frac{2}{3}$$% = $$\frac{2}{3}$$ 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
113.
How much $$66\frac{2}{3}$$% of Rs. 312 exceeds Rs. 200 ?
(A) Rs. 96
(B) Rs. 4
(C) Rs. 8
(D) Rs. 104
Solution:
$$66\frac{2}{3}$$% of Rs. 312/- exceeds Rs. 200 by Rs. $$x$$ According to the question, required difference $$\eqalign{ & x = {\text{Rs}}{\text{. }}\left( {312 \times \frac{{200}}{3}\% - 200} \right) \cr & x = {\text{Rs}}{\text{. }}\left( {312 \times \frac{{200}}{{3 \times 100}} - 200} \right) \cr & x = {\text{Rs}}{\text{. }}\left( {312 \times \frac{{200}}{{300}} - 200} \right) \cr & x = {\text{Rs}}{\text{. }}\left( {208 - 200} \right) \cr & x = {\text{Rs}}{\text{. 8}} \cr} $$
114.
A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6:7:8:9:10. In all papers together, the candidate obtained 60% of the total marks then, the number of papers in which he got more than 50% marks is
(A) 1
(B) 3
(C) 4
(D) 5
Solution:
Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and 10x. Total marks obtained = 40x Max. Marks of the five subjects = $$\frac{{40{\text{x}}}}{{0.6}}$$ [40x is 60% of total marks] Max. Marks in each subject = $$\frac{{40{\text{x}}}}{{0.6 \times 5}}$$ = 13.33x Hence, % of each subject = $$\frac{{6{\text{x}} \times 100}}{{13.33}}$$ = 45.01% Or, $$\frac{{7{\text{x}} \times 100}}{{13.33}}$$ = 52.51 In same way other percentage are 60.01%, 67.52%, 75.01%. Hence, number of subjects in which he gets more than 50% marks = 4
115.
An alloy contains the copper and aluminum in the ratio of 7 : 4 While making the weapons from this alloy, 12% of the alloy got destroyed. If there is 12 kg of aluminum in the weapon, then weight of the alloy required is :
(A) 14.4 kg
(B) 37.5 kg
(C) 40 kg
(D) 48 kg
Solution:
Copper : Aluminum = 7 : 4 Let Copper and Aluminum in the weapon be 7x and 4x respectively Given, Aluminum in weapon = 12 kg So, → 4x = 12 → x = 3 Copper = 7x = 7 × 3 = 21 Kg. Total alloy in the weapon = 12 + 21 = 33 kg But 12% alloy get destroyed in making the weapon, i.e. 88% alloy is used in the weapon, so, → 88 % alloy = 33 kg → 100 % alloy = 37.5 kg
116.
25% of 120 + 40% of 380 = ? of 637
(A) $$\frac{2}{7}$$
(B) $$\frac{1}{7}$$
(C) $$\frac{4}{7}$$
(D) $$\frac{3}{7}$$
Solution:
120 × $$\frac{25}{100}$$ + 380 × $$\frac{40}{100}$$ = x × 637 ⇒ 30 + 152 = x × 637 ⇒ $$\frac{182}{637}$$ = x ⇒ x = $$\frac{2}{7}$$ ⇒ Required answer = $$\frac{2}{7}$$
117.
What is to be added to 15% of 180 so that the sum is equal to 20% of 360?
(A) 45
(B) 40
(C) 60
(D) 50
Solution:
$$\eqalign{ & {\text{Let be the number is }}x \cr & 180 \times \frac{{15}}{{100}} + x = \frac{{360 \times 20}}{{100}} \cr & 27 + x = 72 \cr & x = 72 - 27 \cr & x = 45 \cr} $$
Anuja owns $$66\frac{2}{3}\% $$ ÃÂÃÂ of a property. If 30% of the property that she owns is worth Rs. 1,25,000, then 45% of the value (in Rs.) of the property is: