Percentages - 2 - Free MCQ Practice Test with solutions, UPSC

Let the number of matches to be played be ‘x’
We are given that 40 matches are already played and 30% of them are won.
If 60% of the remaining matches are won, then the overall win percentage is 50%.
30% of 40 + 60% of x = 50% of (40 + x)
0.3 (40) + 0.6 (x) = 0.5 (40 + x)
12 + 0.6 (x) = 20 + 0.5 (x)
0.1 (x) = 8
x = 80
The number of matches to be played is 80.
If the team wins 90% of the remaining matches, it would win 90% of 80 = 72 matches.
Total matches won by the team
= 30% of 40 + 72
= 12 + 72 = 84.

Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.

Number of such number =14

Given A = x% of y, this can be written as:

A = (x/100) × y.

Similarly, B = y% of x translates to:

B = (y/100) × x.

Since multiplication is commutative (x × y = y × x), it follows that:

A = B.

Thus:

• Neither A is smaller than B,
• Nor A is greater than B.

The relationship between A and B can indeed be determined as equal.

Therefore, the correct answer is: D.

Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option D is the correct answer.

The total number of votes is calculated as follows:

• Candidate 1: 1136 votes
• Candidate 2: 7636 votes
• Candidate 3: 11628 votes

Calculate the total votes:

• Total votes = 1136 + 7636 + 11628 = 20400

The winning candidate received 11628 votes. To find the percentage:

• Percentage = (11628 / 20400) × 100
• Percentage = 57%

X+ y=Rs. 550
x = 120% of y
120% * y+ y =550
120/100*y+y = 550
6/5y+y =550
6y+5y/5 = 550
11/5y = 550
11y = 550*5
y = 250

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.