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Re: 50 tricky questions [#permalink]
08 Dec 2011, 22:16

Expert's post

paata01 wrote:

A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than 1. 6 2. 12 3. 3 4. 9 Soln. (3) — Let the number of correct answers be ‘x’, number of wrong answers be ‘y’ and number of questions not attempted be ‘z’. Thus, x + y + z = 50 … (i) And x – y – z 32 3 6 The second equation can be written as, 6x – 2y – z = 192 … (ii) Adding the two equations we get, 7x – y = 242 or x = 242 + y 7 Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.

can u explain why Y can not be 1 or 2....i tried to make 242/7+Y integer but even 3 is not enough

The equation you get is 7x - y = 242 x = (242 + y)/7 (don't forget that the entire 242+y is divided by 7, not just 242) Since x must be an integer, (242+y) must be divisible by 7. After 242, the closest multiple of 7 is 245 (if you are wondering how to get it, divide 242 by 7. You get 4 as remainder. So you need another 3 to go to the next multiple of 7). So y must be at least 3. Think: Can y take other values? If yes, which ones? _________________

Re: 50 tricky questions [#permalink]
08 Dec 2011, 22:36

VeritasPrepKarishma wrote:

paata01 wrote:

A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than 1. 6 2. 12 3. 3 4. 9 Soln. (3) — Let the number of correct answers be ‘x’, number of wrong answers be ‘y’ and number of questions not attempted be ‘z’. Thus, x + y + z = 50 … (i) And x – y – z 32 3 6 The second equation can be written as, 6x – 2y – z = 192 … (ii) Adding the two equations we get, 7x – y = 242 or x = 242 + y 7 Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.

can u explain why Y can not be 1 or 2....i tried to make 242/7+Y integer but even 3 is not enough

The equation you get is 7x - y = 242 x = (242 + y)/7 (don't forget that the entire 242+y is divided by 7, not just 242) Since x must be an integer, (242+y) must be divisible by 7. After 242, the closest multiple of 7 is 245 (if you are wondering how to get it, divide 242 by 7. You get 4 as remainder. So you need another 3 to go to the next multiple of 7). So y must be at least 3. Think: Can y take other values? If yes, which ones?

Re: 50 tricky questions [#permalink]
08 Dec 2011, 23:33

The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at 1. x = 2.3 2. x = 2.5 3. x = 2.7 4. None of the above Soln. (2) — Case 1: If x < 2, then y = 2 – x + 2.5 – x + 3.6 – x = 8.1 – 3x. This will be least if x is highest i.e. just less than 2. In this case y will be just more than 2.1 Case 2: If 2 x 2.5 , then y = x – 2 + 2.5 – x 3.6 – x = 4.1 – x Again, this will be least if x is the highest case y will be just more than 1.6. Case 3: If 2.5 x 3.6 , then y = x – 2 + x – 2.5 + 3.6 – x = x – 0.9 This will be least if x is least i.e. X = 2.5. Case 4: If In this case y = 1.6 X 3.6 , then y = x – 2 + x – 2.5 + x – 3.6 = 3x – 8.1 The minimum value of this will be at x = 3.6 = 27 Hence the minimum value of y is attained at x = 2.5

Re: Tricky questions [#permalink]
11 Oct 2012, 01:01

ykaiim wrote:

Here is another set on another 123 tricky questions covering allmost all topics in PS and DS.

If you have any doubt then please bring it on the forum. Lets analyse together.

Hello, 89th prob.., we are asked to find out the dollar amount received after sale of suits ,, but the solution ended after finding number of suits!! amount would be <price of each suit * number of suits> i.e. x*(200-x/2).., correct me if m wrong!!

Re: 50 tricky questions [#permalink]
05 Jan 2014, 20:29

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