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18 Jan 2021, 09:38
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D
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Difficulty:
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Question Stats:
30% (03:20) correct
70%
(02:56)
wrong
based on 53
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(100x +10y + z) ÷ 5 = 10w + z
If x, y, z, w are all positive integers less than 10 that satisfy the above equation, what is the value of w?
(1) x = y - 6
(2) x + y+ z = 13
Are You Up For the Challenge: 700 Level Questions
If x, y, z, w are all positive integers less than 10 that satisfy the above equation, what is the value of w?
(1) x = y - 6
(2) x + y+ z = 13
(100∂ +10∆ + ¥) ÷ 5 = 10Φ + ¥
If ∂, ∆, ¥, Φ are all positive integers less than 10 that satisfy the above equation, what is the value of Φ?
(1) ∂ = ∆ - 6
(2) ∂ + ∆+ ¥ = 13
If ∂, ∆, ¥, Φ are all positive integers less than 10 that satisfy the above equation, what is the value of Φ?
(1) ∂ = ∆ - 6
(2) ∂ + ∆+ ¥ = 13
Are You Up For the Challenge: 700 Level Questions
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18 Jan 2021, 10:38
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As 100x +10y+z id divisible by 5 and a 3 digit number divided by 5 gives a two-digit number. 100x+10y+z should be less than 5*99 = 495
Now if z = 0 than x+y must be divisible by 5 to give a number ending in 0. So all possible values are 100,150,200,250,300,350,400,450.
If z = 5 than there must be one more 5 in the number to give 5 as a unit digit again. So possible values are 125,175,225,275,325,375,425,475.
Now Statement 1 : x = y-6
Only value that satisfies this is 175. 175 /5 = 35 Hence w =3
Sufficient
Statement 2: x+y+z = 13
So the number that satisfies this is 175.
Sufficient
Hence Answer is D
Now if z = 0 than x+y must be divisible by 5 to give a number ending in 0. So all possible values are 100,150,200,250,300,350,400,450.
If z = 5 than there must be one more 5 in the number to give 5 as a unit digit again. So possible values are 125,175,225,275,325,375,425,475.
Now Statement 1 : x = y-6
Only value that satisfies this is 175. 175 /5 = 35 Hence w =3
Sufficient
Statement 2: x+y+z = 13
So the number that satisfies this is 175.
Sufficient
Hence Answer is D
18 Jan 2021, 10:51
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(100x +10y + z) ÷ 5 = 10w + z
If x, y, z, w are all positive integers less than 10 that satisfy the above equation, what is the value of w?
\(20x +2y + \frac{z}{5} = 10w + z\)
\(20x +2y - \frac{4}{5}z = 10w\)
So, z = 0 OR z = 5 (multiple of 5)
For z = 0, y = 1,2,3,.. depending on values of x and w
For z = 5, y = 2 (w = 2, x =1 e.g.)
or
z = 5, y = 7 (w = 3, x = 1 e.g.)
(1) x = y - 6
\(20(y - 6) +2y - \frac{4}{5}z = 10w\)
\(22y -120 - \frac{4}{5}z = 10w\)
\(22y - \frac{4}{5}z - 10w = 120\)
z ≠ 0 as it would lead to an equation that is not possible with the conditions given(22y is a multiple of 10 only when y = 5 which is not possible).
For z = 5,
\(22y - 10w = 124\)
22 * 7 - 10 * 3 = 124 (only one solution)
w = 3
SUFFICIENT.
(2) x + y+ z = 13
\(20(13 - y - z) +2y - \frac{4}{5}z = 10w\)
\(260 - 20y - 20z +2y - \frac{4}{5}z = 10w\)
\(-18y - \frac{104}{5}z - 10w = -260\)
\(18y + \frac{104}{5}z + 10w = 260\)
Again, z ≠ 0 as it would lead to an equation that is not possible with the conditions given('18y +10w' is not a multiple of 10).
For z = 5,
\(18y + 10w = 260 - 104\)
\(18y + 10w = 156\)
18 * 7 + 10 * 3 = 156 (taking a clue from St. 1 OR from unit digit of 156)
w = 3
SUFFICIENT.
Answer D.
If x, y, z, w are all positive integers less than 10 that satisfy the above equation, what is the value of w?
\(20x +2y + \frac{z}{5} = 10w + z\)
\(20x +2y - \frac{4}{5}z = 10w\)
So, z = 0 OR z = 5 (multiple of 5)
For z = 0, y = 1,2,3,.. depending on values of x and w
For z = 5, y = 2 (w = 2, x =1 e.g.)
or
z = 5, y = 7 (w = 3, x = 1 e.g.)
(1) x = y - 6
\(20(y - 6) +2y - \frac{4}{5}z = 10w\)
\(22y -120 - \frac{4}{5}z = 10w\)
\(22y - \frac{4}{5}z - 10w = 120\)
z ≠ 0 as it would lead to an equation that is not possible with the conditions given(22y is a multiple of 10 only when y = 5 which is not possible).
For z = 5,
\(22y - 10w = 124\)
22 * 7 - 10 * 3 = 124 (only one solution)
w = 3
SUFFICIENT.
(2) x + y+ z = 13
\(20(13 - y - z) +2y - \frac{4}{5}z = 10w\)
\(260 - 20y - 20z +2y - \frac{4}{5}z = 10w\)
\(-18y - \frac{104}{5}z - 10w = -260\)
\(18y + \frac{104}{5}z + 10w = 260\)
Again, z ≠ 0 as it would lead to an equation that is not possible with the conditions given('18y +10w' is not a multiple of 10).
For z = 5,
\(18y + 10w = 260 - 104\)
\(18y + 10w = 156\)
18 * 7 + 10 * 3 = 156 (taking a clue from St. 1 OR from unit digit of 156)
w = 3
SUFFICIENT.
Answer D.
