Question

please help me fast​

Answer 1

(i) Largest 5-digit number ⇒ $\sf 99999$

Largest 3-digit number ⇒ $\sf 999$

Distributive Law ⇒ $\sf a\times (b-c) = (a\times b)- (a\times c)$

⇒ $\sf 99999\times 999 = 99999\times (1000-1)$

⇒ $\sf ( 99999 \times 1000)- (99999 \times 1)$

⇒ $\sf 99999000 - 99999$

⇒ $\sf 99899001$

$\sf 99999\times 999 = 99899001$

∴The product of the largest 5-digit number and the largest 3-digit number using Distributive Law is 99899001.

(ii)

$\sf 267 \big) 53968 \big(202 \\ \sf \: \: \: \underline{ \: \: \: \: \: \: \: \: 534 \: \: } \\ \: \: \: \: \: \sf \: \: \: \: \: \: 568 \\ \sf \: \: \: \underline{ 534 \: \: } \\ \sf \: \: \: \underline{ \: \: 34 }$

Division Algorithm ⇒ $\sf Divisor \times Quotient + Remainder = Divident$

⇒ $\sf 267\times 202+34$

⇒ $\sf 53934+34$

⇒$\sf 53968$

∴It is proved.

(iii) Largest 6-digit number ⇒ $\sf 999999$

When we divide 999999 by 16 we get the remainder 15.

Hence, we need to subtract 15 from 999999 to obtain the largest 6-digit number that is divisible by 16

$\sf 999999-15 = 999984$

∴ 999984 is the largest 6-digit number that is divisible by 16.

(iv) The cost price of 23 TV sets is ₹570055. Find the cost of each such sets.

Cost Price of 23 TV sets ⇒ $\sf Rs. \ 570055$

Cost of 1 TV set ⇒ $\sf 570055 \div 23 = 24785$

∴The cost of each set is ₹24785

(v) Number given ⇒ 13801

It shall be divisible by ⇒ 87

Least number that must be subtracted by 13801 to make it divisible by 87.

After dividing 13801 and 87 we get the remainder of 55.

∴We must subtract 55 from 13801 to make it divisible by 87.