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18 Jun 2025, 10:34
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B
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Five three-digit numbers, including N, were to be added. While adding, the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N’s units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit.
(A) 4
(B) 6
(C) 8
(D) 2
(A) 4
(B) 6
(C) 8
(D) 2
23 Sep 2025, 00:22
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Let N=100h+10t+u.
Reverse — increase
= 99(u-h)=11(h+t+u).
So 9(u-h)=h+t+u
8u=10h+t.
Also 8(u-h)=2h+6
8u=10h+6.
Compare:
10h+t=10h+6
t=6.
**Answer: (B) 6**
Reverse — increase
= 99(u-h)=11(h+t+u).
So 9(u-h)=h+t+u
8u=10h+t.
Also 8(u-h)=2h+6
8u=10h+6.
Compare:
10h+t=10h+6
t=6.
**Answer: (B) 6**
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