If 25% of 400 + 35% of 1260 + 27% of 1800 = 1020 + x, then the value of x lies between:
(A) 6 to 10
(B) 0 to 5
(C) 11 to 15
(D) 16 to 20
Solution:
400 × 25% + 1260 × 35% + 180 × 27% = 1020 + x 100 + 441 + 486 = 1020 + x 1027 = 1020 + x x = 7
52.
In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared at the examination, how many passed in either subject but not in both ?
(A) 325
(B) 1175
(C) 2125
(D) None of these
Solution:
Failed in 1st subject = $$\frac{35}{100}$$ × 2500 = 875 Failed in 2nd subject = $$\frac{42}{100}$$ × 2500 = 1050 Failed in both = $$\frac{15}{100}$$ × 2500 = 375 Failed in 1st subject only = (875 - 375) = 500 Failed in 2nd subject only = (1050 - 375) = 675 ∴ Passed in 2nd only + Passed in 1st only = (675 + 500) = 1175
53.
Ticket for all but 100 seats in a 10000 seat stadium were sold. Of the ticket sold, 20% were sold at half price and the remaining tickets were sold at the full price of Rs. 20. The total revenue from the ticket sales, (in Rs.) was :
(A) 158400
(B) 178200
(C) 180000
(D) 198000
Solution:
Total seats = 10000 Ticket sold = (10000 - 100) = 9900 According to the question, Total revenue = 9900 × $$\frac{20}{100}$$ × 10 + 9900 × $$\frac{80}{100}$$ × 20 = 9900 × 2 + 9900 × 16 = 9900 (2 + 16) = Rs. 178200
54.
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 $$ = \frac{{1600 \times 42}}{{100}} = 672$$
55.
A shepherd had n goats in the year 2000. In 2001 the no. of goats increased by 40%. In 2002 the no. of goats declined to 70%. In 2003 the no. of goats grew up 30%. In 2004, he sold 10% goats and then he had only 34,398 goats. The percentage increase of the no. of goats in this duration was :
(A) 16.66%
(B) 14.66%
(C) 11.33%
(D) 20%
Solution:
There is no need of the number of goats given i.e. 34,398. Initially, let there be 100 goats. Then 100 == 40% ↑==> 140 == 30%↓(declined to 70%) ==> 98 == 30%↑ ==> 127.4 == 10%↓(sold) ==> 114.66 Hence, % increase = 14.66% [As 100 becomes 114.66]
56.
One-fourth of sixty percent of a number is equal to two-fifth of twenty percent of another number. What is the respective ratio of the first number to the second number ?
(A) 4 : 7
(B) 5 : 9
(C) 8 : 13
(D) Cannot be determined
Solution:
Answer & Solution Answer: Option E Solution: Let the first number be x and the second number be y $$\frac{1}{4}{\text{ of 60% of }}x = \frac{2}{5}{\text{ of 20% of }}y$$ $$\eqalign{ & \Rightarrow \frac{1}{4} \times \frac{{60}}{{100}} \times x = \frac{2}{5} \times \frac{{20}}{{100}} \times y \cr & \Rightarrow \frac{{3x}}{{20}} = \frac{{2y}}{{25}} \cr & \Rightarrow \frac{x}{y} = \frac{2}{{25}} \times \frac{{20}}{3} \cr & \Rightarrow \frac{x}{y} = \frac{8}{{15}} \cr} $$
57.
Asha's monthly income is 60% of Deepak's monthly income and 120% of Maya's income. What is Maya's monthly income if Deepak's monthly income is Rs. 78000 ?
(A) Rs. 36000
(B) Rs. 39000
(C) Rs. 42000
(D) Cannot be determined
Solution:
Asha's monthly income : = 60% of 78000 $$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {\frac{{60}}{{100}} \times 78000} \right) \cr & {\text{ = Rs}}{\text{. 46800}} \cr} $$ Let Maya's monthly income be Rs. x Then, 120% of x = 46800 $$\eqalign{ & \Rightarrow {\text{x}} = \left( {\frac{{46800 \times 100}}{{120}}} \right) \cr & \Rightarrow {\text{x}} = 39000 \cr} $$
A and B are two fixed points 5 cm apart and C is a point an AB such that AC is 3 cm. If the length of AC is increased by 6%, the length CB is decreased by :
(A) 6%
(B) 7%
(C) 8%
(D) 9%
Solution:
After increment of 6% new length of AC = 3 + $$\frac{3 × 6}{100}$$ = 3.18cm Required % decrease = $$\frac{0.18}{2}$$ × 100 = 9%