Question 92

The sum of the digits of a two-digit number is $$\frac{1}{7}$$ of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:

Solution

Let the number be (10a + b).
ATQ,
a + b = $$\frac{10a + b}{7}$$
7a + 7b = 10a + b
6b = 3a
2b = a ---(1)
a - b = 4 ---(2)
From eq (1) and (2),
2b - b = 4
b = 4
a = 4 $$\times 2$$ = 8
Number = 10a + b = 10 $$\times 8 + 4 = 84
reverse of the number = 48
Remainder after divide by 7 = 48/7 = 6

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SSC CGL Tier-2 11th September 2019 Maths

The sum of the digits of a two-digit number is $$\frac{1}{7}$$ of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:

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