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60% (01:35) correct 40% (01:42) wrong based on 210 sessions
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When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only B. II only C. III only D. I and II only E. I, II and III
When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only B. II only C. III only D. I and II only E. I, II and III
When \(x\) is divided by \(y\), the quotient is 3 and the remainder is \(z\): \(x=3y+z\), where \(0\leq{z} < y\) (the remainder must be less than the divisor).
When \(z\) is divided by \(y\), the remainder is 2: when divisor (\(y\) in our case) is more than dividend (\(z\) in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, \(z=2\) and \(2 < y\).
So, we have that \(x=3y+2\) and \(2 < y\). This implies that the least value of \(x\) is \(x=3*3+2=11\): \(x\) cannot be 5 or 8.
Could \(x\) be 32? Yes. If \(y=10\), then \(x=3*10+2=32\).
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then. Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no. 5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ... 8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...
thus the only option is III coz we got no "no options possible"
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then. Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no. 5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ... 8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...
thus the only option is III coz we got no "no options possible"
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then. Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no. 5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ... 8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...
thus the only option is III coz we got no "no options possible"
When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?
I. 5
II. 8
III. 32
A. I only B. II only C. III only D. I and II only E. I, II and III
Alternative approach.
Solving problems of such a type we have to remember 2 fundamental rules: 1. Divident is always equal or greater than remainder. 2. Divisor is always greater than remainder.
We have 2 equations: 1. x/y = 3 + z. Means: x ≥ z, y > z, taking into account the quotient, x ≥ y * 3 + z. 2. z/y = ? + 2. Meаns: z ≥ 2, y > 2.
Make conclusion from both 1 and 2: I. x ≥ z ≥ 2. II. y > z ≥ 2; y can be at least 3, because y is integer and y > z = 2. III. x ≥ 11. Because 3 * 3 + 2 = 11.
I think this is a high-quality question and I agree with explanation. Good Question...Quant questions in Club Tests always have a surprise element
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How is this question 700-level? Make no mistake, I get most of my 700 level questions incorrect, but this is surely not more than 500 level. Bunuel?
The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. If you can solve it easily good for you.
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60% (01:35) correct 
