Question

$23\times (-99)=a\times (-100+b)=23\times c+23\times 1.$

Find the values of a, b and c.

• A. a = 23
• B. a = 21
• C. b = 1
• D. c = -100

Answer 1

Answer: (A), (C), (D)

The correct options are
A

a = 23

C

b = 1

D

c = -100

Consider $23\times (-99)$.

$23\times (-99)=23\times (-100+1)$

Also, $23\times (-99)=a\times (-100+b)$ (given)

On comparing, we get a = 23 and b = 1

$a\times (-100+b)=23\times c+23\times 1$ (given)

Consider $a\times (-100+b)$.

$a\times (-100+b)$

$=23\times (-100+1)$
(since a = 23 and b = 1)

$=23\times (-100)+23\times 1$
(using distributivity property)

$\Rightarrow 23\times (-100)+23\times 1=23\times c+23\times 1$

On comparing, we get c = -100.

Hence, we have a = 23, b = 1 and c = -100.