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61. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Answer: Option A
Explanation:Number of runs made by running = 110 - (3 * 4 + 8 * 6)
= 110 - (60) = 50.
Required percentage = ({50/110}*100)% = 45{5/11}%
Workspace
CGI
62. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
Answer: Option C
Explanation:Let their marks be (x + 9) and x.
Then, x + 9 = (56/100)(x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
Workspace
CGI
63. 1,40,00,000 pencils are put up straight. all the pencils are of length range 3 to 6 inches. 80% of the pencils have average of five inches. so the find out the total length spanned by the pencils.
Answer: Option B
Explanation:length covered by 80% of the pencils =80/100 * 14000000 * 5 = 56000000 inches
Minimum length covered by remaining 20% of the pencils =20/100 * 14000000 * 3 = 8400000 inches
Maximum length covered by remaining 20% of the pencils =20/100 * 14000000 * 6 = 16800000 inches
Hence, total length spanned by the pencils is (56000000 +8400000) to (56000000 +16800000 )
ie, 64400000 inches to 72800000 inches
Workspace
CGI
64. A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday?
Answer: Option A
Explanation:If half of the rolls were sold by noon,
Than remaining are 50 % (40) = 20.
Given 80% of the remaining were sold after the noon to closing time
80% (20) = 16
Unsold rolls = (20 - 16) = 4
Workspace
TCS
65. A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2500 gallons of water evaporate from the tank, the remaining solution will be approximately what percentage of sodium chloride?
Answer: Option B
Explanation:Sodium chloride in the original solution = 5% of 10000 = 500
Water in the original solution = 10000 - 500 = 9500
If 2,500 Liters of the water is evaporated then the remaining water = (9500 - 2500) = 7000
Sodium chloride concentration = 500/(500+7000) * 100 = 6.67 %
Workspace
TCS
66. Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?
Answer: Option A
Explanation:Total seeds germinated in Plot I = 57% of 200 = 114
Total seeds germinated in Plot II = 42% of 300 = 126
So, total germinated seeds = (114 + 126) = 240
Percentage of germinated seeds of the total seeds = 240/500 = 48%
Workspace
TCS
67. If 75% of a class answered the first question on a certain test correctly, 55% answered the second question on the test correctly, and 20% answered neither of the questions correctly, what percentage answered both correctly?
Answer: Option A
Explanation:Given 20% student answered neither question => students who answered atleast one question = 100% - 20% = 80%
Student answered first question = 75%
Student answered second question = 55%
Follow the formula n(A?B) = n(A) + n(B) - n(A?B)
=> 80% = 75% + 55% - n(A?B)
? n(A?B) = 50%
Workspace
TCS
68. There are 120 male and 100 female in a society. Out of 25% male and 20% female are rural. 20% of male and 25% of female rural people passed in the exam. What % of rural students have passed the exam?
Answer: Option B
Explanation:Given 120 male and 100 female society.
Rural male = 25*120/100= 30, Rural female = 20*100/100 = 20.
Passed students from rural:
male = 20*30/100 = 6, female = 25*20/100 = 5
Required percentage = 11/50*100=22%
Workspace
TCS
69. In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?
Answer: Option A
Explanation:Find the percentages. Paper A = 18/70 * 100 = 25.7
Paper B = 14/30 * 100 = 46.6
Workspace
TCS
70. If the value of the numerator is increased by 20% and that of the denominator is decreased by 25% the fraction becomes unity the numerator of original fraction is:
Answer: Option B
Explanation:120n/75d=1
n/d=75/120=5/8
Workspace
Syntel Inc.