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Solution :-

Q1 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 %

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

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 %

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.

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%

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%

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

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.

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