Solution :-
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Q1
Q2 Correct Answer is: b)90%
Let, there are 100 days in each semester
then, Jaya’s total attendance for four semesters = 4 × 80 = 320 days
To minimize her attendance in 3rd semester, we must assume 100% attendance in 4th semester.
Thus, minimum attendance required in 3rd semester = 320 – (60 + 70 + 100)
= 90 days
= i.e. 90 %
Q3 Correct Answer is: a)10000
Let x be the total number of voters
Voters promised to A = 2x/5
Voters backed out = 15% of 2x/5
Voters promised to B = 3x/5
Voters backed out = 25% of 3x/5
Total No. of votes for A = 2x/5 – 15% of 2x/5 + 25% of 3x/5
= 49x/100
Total No. of votes for B = 3x/5 – 25% of 3x/5 +15% of 2x/5
= 51x/100
51x/100 – 49x/100 = 200 … (Given)
x = 10000
Q4 Correct Answer is:c)
No. of blue balls = 100
No. of red balls = 50
No. of black ball = 50
Reduction in blue ball is 25%
Remaining blue balls = 75
Now, reduction in red balls = 50%
Remaining red balls = 25
Total remaining balls = 75 + 25+ 50 =150
Percentage of black balls = 50/150 *100 = 100/3 %
Q5 Correct Answer is: d)140000
Total Votes = 260000
Let x voters voted against the party in the Assembly Poll.
Then votes in favor = 260000 – x
Therefore, majority of votes by which party won in previous poll = 260000 – x – x = 260000 – 2x
Next year votes against the KBC party increase by 25%
So, votes against the party in general election = 1.25 x
And votes polled in favor of the party = total votes – votes against = 260000 – 1.25x
Therefore, majority of votes by which party lost in general election
= 1.25x – (260000 – 1.25x) = 2.5x – 260000
It is given that, KBC Party lost by a majority twice as large as that by which it had won the Assembly Polls, Therefore
2.5x – 260000 = 2(260000 – 2 x)
2.5x – 260000 = 2 * 260000 – 4x
6.5x = 3*260000
x = 120000
Therefore, voters polled by the voters for the party in Assembly Polls for previous year
= (260000 – x) = (260000 – 120000) = 140000.
Q6 Correct Answer is: d)80%
Marks of Ketan = 360 and total marks =500
Marks of Karan = 100/90 *360 = 400
Marks of Keshav = 100/125 * 400 = Rs.320
Marks of Kiran = 100/80 *320 = 400
Percentage marks of Kiran = 400/500 *100 = 80%
Q7 Correct Answer is: c)7.6%
Earnings of food items per person = 1.1 times the original
Increase in number of customers = 1.2 times the original
Net figure = 1.2 *1.1 * 0.7 = 0.924
Overall financials of the KPQ mall change = (1 – 0.924)*100% = Decreased by 7.6%
Q8 Correct Answer is: b)108.9
Let the numbers are a,b,c
Now according to the question
6a/11 = 22b/100 …(1)
And b = c/4 …(2)
We are also given by the value of c i.e. 2400
So the value of b would be 600
Now put b = 600 in equation (1)
6a/11 = (22 x 600)/100
6a/11= 132
a = 242
Now 45% of 242 = 108.9
Q9 Correct Answer is: a)12.5%less
Say runs scored by Sehwag = a, Gambhir = b, Ricky = c and Sachin = d.
Then (c + d) = 1.2 (a + b)
Also b = 30, d = 40, c = 80
80 + 40 = 1.2 (a + 30) => 120/1.2 = a + 30 => a = 70
So Sehwag = 70.
Sehwag scored (80 – 70)/80 * 100 = 10/80 * 100 = 12.5 % less than Ricky.
Q10 Correct Answer is: d)3/4
Let there are 40 male and 60 female in the company.
Now out of 40 male, 75% i.e. 30 earn more than Rs .25000 and
45% of the total employees i.e. 45 employees earn more than Rs 25000.
Hence, there are
45 – 30 = 15 female who earn more than Rs25000.
So, 60 – 15 = 45 women earn less than Rs 25000.
Hence, the required fraction = 45/60 = ¾
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