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Practice Test: Percentages - 2 - UPSC MCQ


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10 Questions MCQ Test CSAT Preparation - Practice Test: Percentages - 2

Practice Test: Percentages - 2 for UPSC 2024 is part of CSAT Preparation preparation. The Practice Test: Percentages - 2 questions and answers have been prepared according to the UPSC exam syllabus.The Practice Test: Percentages - 2 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Percentages - 2 below.
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Practice Test: Percentages - 2 - Question 1

Q. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

Detailed Solution for Practice Test: Percentages - 2 - Question 1

Practice Test: Percentages - 2 - Question 2

In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be 
 

Detailed Solution for Practice Test: Percentages - 2 - Question 2

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :


 
24 +1.2x=40+x
0.2x=16
x=80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

Practice Test: Percentages - 2 - Question 3

What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?

Detailed Solution for Practice Test: Percentages - 2 - Question 3

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

Practice Test: Percentages - 2 - Question 4

If A = x% of y and B = y% of x, then which of the following is true?

Detailed Solution for Practice Test: Percentages - 2 - Question 4

x%ofy=x/100×y
=y/100×x
=y%ofx
∴A=B

Practice Test: Percentages - 2 - Question 5

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

Detailed Solution for Practice Test: Percentages - 2 - Question 5

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.

Practice Test: Percentages - 2 - Question 6

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was: 

Detailed Solution for Practice Test: Percentages - 2 - Question 6

Total Valid votes are = 80% of 7500= 6000
If one has 55% vote then the other has left with 45% votes
Hence, 45% of 6000= 2700

Practice Test: Percentages - 2 - Question 7

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

Detailed Solution for Practice Test: Percentages - 2 - Question 7

Practice Test: Percentages - 2 - Question 8

Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

Detailed Solution for Practice Test: Percentages - 2 - Question 8

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

Practice Test: Percentages - 2 - Question 9

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. 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. The strength, in percentage, of the resulting solution in vessel A is

Detailed Solution for Practice Test: Percentages - 2 - Question 9

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.

Practice Test: Percentages - 2 - Question 10

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

Detailed Solution for Practice Test: Percentages - 2 - Question 10

Let the alloy contain x Kg silver and y kg copper 
Now when mixed with 3Kg Pure silver 

we get 10x+30 =9x+9y+27
9y-x=3    (1)
Now as per condition 2
silver in 2nd alloy = 2(0.9) =1.8

we get 21y-4x =3   (2)
solving (1) and (2) we get y= 0.6 and x =2.4
so x+y = 3

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