Pass percentage =[(252/270)*100] % =280/3 % = 9 %
Let the number of students in the class be 100 and let this required average be x.
Then, (10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
=> 70x = 8000 - (950 + 1800) = 5250
=> x = 75.
Let the original number be x.
Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x.
x-(99/100)x=10
=> x =1000
5 : 4 = 5/4 = ( (5/4) x 100 )% = 125%.
A's Salary = 40% of B = 40% of (25% of C) = 40/100[(52/100)*100] % of C = 10% of C.
Let original consumption = 100 kg and new consumption = x kg,
So, 100 * 6 = x * 7.50 => x = 80kg
Reduction in consumption = 20%.
Total number of students = 1100 + 700 = 1800.
Number of students passed = (42% of 1100 + 30% of 700) - (462 + 210) = 672.
Number of failues = 1800-672 = 1128.
Percentage failure = (1128/1800 * 100 )% = 62 * 2/3 %.
Actual price = Rs. (25 + 2.50) = Rs. 27.50.
Saving = 2.50
Percentage Saving = [(2.50/27.50)*100]
= 2500/275
= %
%
x = 80 % of y
=> x =(80/100 )y
=> y/x = 5/4
Required percentage = [(y/2x)*100] % = (5/8*100) % =62.5%
Let the total original sale be Rs. 100. Then, original number of visitors = 100.
New number of visitors = 120/0.75 = 160.
Increase % = 60 %.
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