model 2 divisibility, multiples, add and subtract based number system Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

Let the quotient be Q and the remainder be R. Then Divisor = 7 Q = 3 R

∴ Q = $3/7$ R = $3/7$ × 28 = 12

∴ Divisor = 7 Q = 7 × 12 = 84

∴ Dividend = Divisor × Quotient + Remainder

= 84 × 12 + 28 = 1008 + 28 = 1036

Answer: (c)

If two numbers are separately divided by a certain divisor (c) leaving remainders r1 and r2 , then remainder after their sum is divided by the same divisor.

= r1 + r2 – d

= 21 + 28 – 33 = 16

Answer: (a)

By the Binomial expansion we have

$(x + 1)^n$ =$x^n$ +$n_c_1 x^(n–1)$+$n_c_2 x^(n– 2)$ + ..... + $nc_{n-1}$ x +1

Here, each term except the last term contains x. Obviously, each term except the last term is exactly divisible by x.

Following the same logic, $7^19$ = $(6 + 1)^19$ has each term except last term divisible by 6. Hence,$ 7^19$ + 2 when divided by 6 leaves remainder =1 + 2 = 3

Answer: (a)

Quotient = 16; Divisor = 25 × 16 = 400 and remainder = 80

Dividend = Divisor × quotient + Remainder

= 400 × 16 + 80

= 6400 + 80 = 6480

Answer: (b)

Let number (dividend) be X. ∴ X = 296 × Q + 75 where Q is the quotient and can have the values 1, 2, 3 etc.

= 37 × 8 × Q + 37 × 2 + 1

= 37 (8Q + 2) + 1

Thus we see that the remainder is 1.

[Important Note : When the second divisor is a factor of the first divisor, the second remainder is obtained by dividing the first remainder by the second divisor. Hence, divide 75 by 37, the remainder is 1].