**Solution :-**

*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*