Question

Abhimanyu has Rs300 in denominations of Rs1, Rs2, and Rs5 notes. The number of Rs2 notes is 3 times the number of Rs5 notes. If the total number of notes is 160, how many notes of each denomination are with him?

Answer 1

Answer:

Number of Rs.5 note=20

Number of Rs.2 note=3x=3×20=60

Number of Re.1 note=160-4x=160-4×20=160-80=80

Step-by-step explanation:

Given:

• Abhimanyu has Rs300 in denominations of Rs1, Rs2, and Rs5 notes. The number of Rs2 notes is 3 times the number of Rs5 notes. If the total number of notes is 160

To Find:

• How many notes of each denomination are with him

Solution:

Let the number of Rs.5 notes be x

Number of Rs 2 notes=3x

number of Re. 1 note=[160-(x+3x)]=160-4x

ACQ,

5(x)+2(3x)+1(160-4x)=300

5x+6x+160-4x=300

5x+6x-4x=300-160

7x=140

x=20

Number of Rs.5 note=20

Number of Rs.2 note=3x=3×20=60

Number of Re.1 note=160-4x=160-4×20=160-80=80

Answer 2

$\sf\small\underline\red{Let:-}$

$\rm{\leadsto The\: number\:_{(Rs.1\:notes)}=x}$

$\rm{\leadsto The\: number\:_{(Rs.2\:notes)}=3y}$

$\rm{\leadsto The\: number\:_{(Rs.5\:notes)}=y}$

$\sf\small\underline\red{Given:-}$

$\rm{\leadsto Amount\:_{(Abhimanyu\:has)}=Rs.300}$

$\rm{\leadsto Number\:_{(Notes)}=160}$

$\sf\small\underline\red{Solution:-}$

To calculate the denomination of Rs.2 and Rs.5 , Simply by setting up equation. As per the given clue we will solve the problem which is given in the question.

$\sf\small\underline\red{Calculation\:Begins:-}$

$\rm{\leadsto The\:Sum\:of\: Amount\:_{(Rs.1+Rs.2+Rs.5)}=300}$

$\rm{\leadsto x(1)+3y(2)+y(5)=300}$

$\rm{\leadsto x+11y=300-----(i)}$

$\rm{\leadsto The\:sum\:of\: denomination=160}$

$\rm{\leadsto x+3y+y=160}$

$\rm{\leadsto x+4y=160------(ii)}$

• By Substracting eq (I) and (ii) we get:-]

$\rm{\leadsto x+11y=300}$

$\rm{\leadsto x+4y=160}$

• By Solving we get here :-]

$\rm{\leadsto 7y=140}$

$\rm{\leadsto y=20}$

• Putting the value of y=20 in eq (ii):-

$\rm{\leadsto x+4y=160}$

$\rm{\leadsto x+4(20)=160}$

$\rm{\leadsto x=160-80}$

$\rm{\leadsto x=80}$

$\sf\large{Hence,}$

$\rm{\leadsto The\: number\:_{(Rs.1\:notes)}=80}$

$\rm{\leadsto The\: number\:_{(Rs.2\:notes)}=60}$

$\rm{\leadsto The\: number\:_{(Rs.5\:notes)}=20}$