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# If r, s, t are different 1-digit positive integer and (ab) is defined

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Joined: 16 Aug 2015
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If r, s, t are different 1-digit positive integer and (ab) is defined  [#permalink]

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06 Apr 2016, 05:33
00:00

Difficulty:

75% (hard)

Question Stats:

52% (02:19) correct 48% (02:10) wrong based on 60 sessions

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If r, s, t are different 1-digit positive integer and (ab) is defined as 2-digit integer such that a is the tens digit and b is the units digit, t=?

1) The product of s and (rs) is (st)
2) The sum of s and (rs) is (rt)

* A solution will be posted in two days.

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Re: If r, s, t are different 1-digit positive integer and (ab) is defined  [#permalink]

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25 Sep 2016, 06:07
1) The product of s and (rs) is (st)

$$s * rs = st$$
==> $$s * ( 10*r + s) = s*10 + t$$
==> $$10rs + s^2 = 10s + t$$
==> $$10rs = 10s , s^2=t$$
==> $$r =1 , t = s^2$$ --{2,4} , { 3,9} , {1,1} -- Insufficient

2) The sum of s and (rs) is (rt)

$$s + rs = rt$$
==> $$s + r*10 + s = r*10 + t$$
==> $$r*10 + 2s = r*10 + t$$
==> $$r=2s$$ --{1,2} , { 2,4} , { 3,6} , { 4,8} -- Insufficient

If we combine (1) and (2) only {2,4} satisfies both the equation. t=4.

Ans. C
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If r, s, t are different 1-digit positive integer and (ab) is defined  [#permalink]

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23 Feb 2017, 03:51
Solution:

Given: r, s and t are different single digit positive integer, “ab” is defined as 2 digit integer. It can be written as $$“10 ×a+b+1".$$
To find: The value of “t”.

Analysis of statement 1: The product of s and (rs) is (st)
We can do this problem by substituting the values ;
Case 1: Let s = 3 and r = 1; we get

$$s ×(10 ×r+s)=(10 ×s+ t)$$

$$3 ×(10 ×1+3)=(10 ×3+9)$$

Case 2: Let s = 2 and r = 1; we get

$$s ×(10 ×r+s)=(10 ×s+ t)$$
$$2 ×(10 ×1+2)=(10 ×2+4)$$
Looking at the above cases, it’s clear that, $$t = s^2$$, as we can have many values for “s”, so we cannot determine the value of “t” as well. So statement 1 is not sufficient to answer. We can eliminate options A and D.

Analysis of statement 2: The sum of s and (rs) is (rt)
We can do this problem by substituting the values ;
Case 1: Let s = 3 and r = 1; we get

$$s+(10 ×r+s)=(10 ×r+t)$$

$$3+(10 ×1+3)=(10 ×1+6)$$

Case 2: Let s = 2 and r = 1; we get

$$s+(10 ×r+s)=(10 ×r+t)$$

$$2+(10 ×1+2)=(10 ×1+4)$$

Looking at the above cases, it’s clear that, $$t = 2s$$, as we can have many values for “s”, so we cannot determine the value of “t” as well.
So statement 2 is not sufficient to answer. We can eliminate option B.

Combining both the statements together; we get:
From statement 1: $$t = s^2$$
From statement 2: $$t = 2s$$
This is true only when, s = 2; therefore t = 4.

So, the correct answer option is “C”.
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If r, s, t are different 1-digit positive integer and (ab) is defined &nbs [#permalink] 23 Feb 2017, 03:51
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