Question 129143
Question 1

Let the number be {{{x}}}.
Then 2/5 of the number is {{{(2/5)*x}}}

{{{x+(2/5)*x=210}}}
{{{(1+2/5)*x=210}}}
{{{5/5+2/5)*x=210}}}
{{{(7/5)*x=210}}}
{{{x=210/(7/5)}}}
{{{x=210*(5/7)}}}
{{{x=5*(210/7)=5*30=150}}}

The number is {{{150}}}.

Question 2

Since there are 3 consecutive odd integers, there is a smallest one, a largest one, and one in between. Let the integer in between be {{{x}}}. Then the smallest integer will be {{{x-2}}} and the largest will be {{{x+2}}}.

Their sum is equal to 201, so:
{{{(x-2)+(x)+(x+2)=201}}}
{{{3x=201}}}
{{{x=201/3=67}}}

Since the largest integer is {{{x+2}}},
{{{x+2=67+2=69}}}

The largest integer is {{{69}}}.

Question 3

Let the smaller number be S and the larger number be L.

The larger number is twice the smaller number:
{{{L=2*S}}}

The larger number diminished by 10 is 2 more than the smaller number:
{{{L-10=S+2}}}

Substituting {{{L=2*S}}} into this equation, we have:
{{{2*S-10=S+2}}}
{{{2*S-S=2+10}}}
{{{S=12}}}

Since {{{L=2*S}}},
{{{L=2*12=24}}}

The smaller number is {{{12}}} and the larger number is {{{24}}}.

Question 4

Let A be the number of 20-pound notes and B be the number of 50-pound notes.

{{{A+B=31}}}
{{{20*A+50*B=1830}}}

Substituting {{{A=31-B}}} into the second equation,
{{{20*(31-B)+50*B=1830}}}
{{{620-20*B+50*B=1830}}}
{{{(50-20)*B=1830-620}}}
{{{30*B=1210}}}
{{{B=1210/30=121/3=40+1/3}}}

Then {{{A=31-B=31-(40+1/3)=-(9+1/3)}}}.

Since A and B must necessarily be non-negative integers, there are no valid solutions to this problem. Also, by observation, the total amount of money in the cash box can only range from {{{31*20=620}}} to {{{31*50=1550}}}, and {{{1830}}} as given in the question is outside that range.

Question 5

Let the smallest integer be {{{x}}}. Then the middle-valued integer will be {{{x+2}}}, and the largest integer will be {{[x+4}}}.

{{{x+(x+2)+(x+4)=x+50}}}
{{{3x+6=x+50}}}
{{{2x=50-6=44}}}
{{{x=44/2=22}}}
{{{x+2=22+2=24}}}
{{{x+4=22+4=26}}}

The integers are {{{22}}}, {{{24}}} and {{{26}}}.