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D
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Difficulty:
45%
(medium)
Question Stats:
71% (02:32) correct
29%
(02:41)
wrong
based on 1936
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K and L are each four-digit positive integers with thousands, hundreds, tens, and units digits defined as a, b, c, and d, respectively, for the number K, and p, q, r, and s, respectively, for the number L. For numbers K and L, the function W is defined as \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\). The function Z is defined as (K – L) ÷ 10. If W = 16, what is the value of Z?
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.
As the OA is not given this is how I am trying to solve this question, but I am stuck now. Can someone help please?
Digits of K are a,b,c and d i.e. K is 1000 a + 100 b + 10 c + d ----------------------------------------(1)
Digits of L are p, q, r and s i.e. L is 1000 p + 100 q + 10 r + s ----------------------------------------(2)
f(w) = \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\)-----------------------------------------(3)
Also, f(16) = \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\)-----------------------------------(4)
Therefore, \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\) = 16 i.e. \(2^4\)----------------(5)
f(z) = (1000 a + 100 b + 10 c + d) - 1000 p + 100 q + 10 r + s / 10 ----------------------------------(6)
Now, I am stuck and no OA doesn't help either. Therefore, your help will be much appreciated guys.
Digits of K are a,b,c and d i.e. K is 1000 a + 100 b + 10 c + d ----------------------------------------(1)
Digits of L are p, q, r and s i.e. L is 1000 p + 100 q + 10 r + s ----------------------------------------(2)
f(w) = \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\)-----------------------------------------(3)
Also, f(16) = \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\)-----------------------------------(4)
Therefore, \(5^a 2^b 7^c 3^d\) / \(5^p 2^q 7^r 3^s\) = 16 i.e. \(2^4\)----------------(5)
f(z) = (1000 a + 100 b + 10 c + d) - 1000 p + 100 q + 10 r + s / 10 ----------------------------------(6)
Now, I am stuck and no OA doesn't help either. Therefore, your help will be much appreciated guys.
28 Jan 2012, 02:53
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enigma123
Given: \(w=\frac{5^a*2^b*7^c*3^d}{5^p*2^q*7^r*3^s}=16\) --> \(w=5^{a-p}*2^{b-q}*7^{c-r}*3^{d-s}=2^4\) --> the powers of 3, 5, and 7 must be zero and the power of 2 must be 4: \(a=p\), \(b-q=4\), \(c=r\) and \(d=s\)
Now, as thousands, tens, and units digits in K and L are equal and the difference between hundreds' digits is 4, then K-L=400 (for example K=1923 and L=1523 --> K-L=1923-1523=400).
Z=(K-L)/10=400/10=40.
Answer: D.
Also discussed here: functions-concepts-testing-91004.html
Similar questions:
the-function-f-is-defined-for-each-positive-three-digit-100847.html
for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html
Hope it helps.
01 Dec 2015, 01:03
Kudos
Bookmarks
[quote="enigma123"]K and L are each four-digit positive integers with thousands, hundreds, tens, and units digits defined as a, b, c, and d, respectively, for the number K, and p, q, r, and s, respectively, for the number L. For numbers K and L, the function W is defined as \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\). The function Z is defined as (K – L) ÷ 10. If W = 16, what is the value of Z?
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.
Given:
K = abcd = 1000a + 100b + 10c + d
L = pqrs = 1000p + 100q + 10r + s
W = \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\) = \(5^{a-p} 2^{b-q} 3^{c-r} 5^{d-s}\) = 16 = \(2^4\)
W can 16 only when W carries the powers of 2 only.
Hence b - q = 4 (i)
And the rest of the powers will be 0.
a= p, c = r, d = s (ii)
Required: Z = (K – L) ÷ 10 =?
Z = (abcd - pqrs)÷10 = (1000a + 100b + 10c + d) - (1000p + 100q + 10r + s) ÷ 10
Z = 1000 (a - p) + 100(b - q) + 10 (c - r) + 10 (d - s) ÷ 10
From equations (i) and (ii)
Z = 100(b-q) ÷ 10 = 100*4 ÷ 10= 40
Option D
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.
Given:
K = abcd = 1000a + 100b + 10c + d
L = pqrs = 1000p + 100q + 10r + s
W = \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\) = \(5^{a-p} 2^{b-q} 3^{c-r} 5^{d-s}\) = 16 = \(2^4\)
W can 16 only when W carries the powers of 2 only.
Hence b - q = 4 (i)
And the rest of the powers will be 0.
a= p, c = r, d = s (ii)
Required: Z = (K – L) ÷ 10 =?
Z = (abcd - pqrs)÷10 = (1000a + 100b + 10c + d) - (1000p + 100q + 10r + s) ÷ 10
Z = 1000 (a - p) + 100(b - q) + 10 (c - r) + 10 (d - s) ÷ 10
From equations (i) and (ii)
Z = 100(b-q) ÷ 10 = 100*4 ÷ 10= 40
Option D
