Question

the product of tow successive multiples of 10 is 1200 . then find these multiples

(a) 20,30
(b) 48,25
(c) 60,20
(d) 30,40​

Answer 1

Answer:

d) 30, 40

Step-by-step explanation:

30×40= 1200

30 and 40 are successive multiple of 10

Answer 2

$\underline{\textsf{\textbf{\purple{ \mapsto Given:}}}}$

• The product of two succesive multiples of 10 is 1200.

$\underline{\textsf{\textbf{\purple{ \mapsto To\:Find:}}}}$

• The two numbers.

$\underline{\textsf{\textbf{\purple{ \mapsto Concept\:Used:}}}}$

The number is a multiple of 10 , so it will be divisible by 10 . So it will be in the form of 10k , where k is any integer . So if the first number will be 10k , then the second number will be 10k + 10 since they are succesive numbers.

$\underline{\textsf{\textbf{\purple{ \mapsto Answer:}}}}$

$\sf Let \:us\:take:-$

• $\sf \red{Firsrt\: number\:be\:10x.}$
• $\sf \red{Second\: number\:be\:10x+10.}$

$\underline{\underline{\pink{\longmapsto \sf So,\: \mathscr{A}ccording \:to\: the\; \mathscr{Q}uestion ,}}}$

$\tt:\implies 10x(10x+10)=1200$

$\tt:\implies 100x^2+100x=1200$

$\tt:\implies 100x^2+100x-1200=0$

$\tt:\implies 100(x^2+x-12)=0$

$\tt:\implies x^2+x-12=\dfrac{0}{100}$

$\tt:\implies x^2+x-12=0$

$\tt:\implies x^2+4x-3x-12=0$

$\tt:\implies x(x+4)-3(x+4)=0$

$\tt:\implies(x+4)(x-3)=0$

$\underline{\underline{\boxed{\red{\tt\longmapsto x=(-4),3}}}}$

$\bf Hence\: values\:of\:x\:is\:(-4)\:\&\:3$

$\rule{200}3$

$\underline{\green{\orange{\mapsto}\tt\:When\:x\:is\:(-4).}}$

• $\sf First\: Number\:=\:10x\:=\pink{(-40)}$
• $\sf Second\: Number\:=\:10x+10\:=\pink{(-30)}$

$\rule{200}3$

$\underline{\green{\orange{\mapsto}\tt\:When\:x\:is\:(3).}}$

• $\sf First\: Number\:=\:10x\:=\pink{(30)}$
• $\sf Second\: Number\:=\:10x+10\:=\pink{(40)}$