\displaystyle \textbf{Question 1: }\text{Half of the difference of two numbers is }2.\text{ The sum of the}
\displaystyle \text{greater number and twice the smaller number is }13.\text{ Find the numbers.}
\displaystyle \hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the greater number be }x\text{ and the smaller number be }y.
\displaystyle \text{Half of their difference is }2.
\displaystyle \therefore \frac{x-y}{2}=2
\displaystyle \Rightarrow x-y=4\qquad\ldots\text{(i)}
\displaystyle \text{The sum of the greater number and twice the smaller number is }13.
\displaystyle \therefore x+2y=13\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 3y=9
\displaystyle \Rightarrow y=3
\displaystyle \text{Substituting }y=3\text{ in equation (i),}
\displaystyle x-3=4
\displaystyle \Rightarrow x=7
\displaystyle \therefore \text{The numbers are }7\text{ and }3.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A number consists of two digits whose sum is five. When the}
\displaystyle \text{digits are reversed, the number becomes greater by nine. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The number}=10x+y.
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{The sum of the digits is }5.
\displaystyle \therefore x+y=5\qquad\ldots\text{(i)}
\displaystyle \text{The reversed number is }9\text{ greater than the original number.}
\displaystyle \therefore 10y+x=10x+y+9
\displaystyle \Rightarrow 9y-9x=9
\displaystyle \Rightarrow y-x=1\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2y=6
\displaystyle \Rightarrow y=3
\displaystyle \text{Substituting }y=3\text{ in equation (i),}
\displaystyle x+3=5
\displaystyle \Rightarrow x=2
\displaystyle \therefore \text{The required number}=10(2)+3=23.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The sum of digits of a two digit number is }15.\text{ The number}
\displaystyle \text{obtained by reversing the order of digits of the given number exceeds the}
\displaystyle \text{given number by }9.\text{ Find the given number.}\hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{The sum of the digits is }15.
\displaystyle \therefore x+y=15\qquad\ldots\text{(i)}
\displaystyle \text{The reversed number exceeds the given number by }9.
\displaystyle \therefore 10y+x=10x+y+9
\displaystyle \Rightarrow 9y-9x=9
\displaystyle \Rightarrow y-x=1\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2y=16
\displaystyle \Rightarrow y=8
\displaystyle \text{Substituting }y=8\text{ in equation (i),}
\displaystyle x+8=15
\displaystyle \Rightarrow x=7
\displaystyle \therefore \text{The given number}=10(7)+8=78.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The sum of two numbers is }1000\text{ and the difference between}
\displaystyle \text{their squares is }256000.\text{ Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the greater number be }x\text{ and the smaller number be }y.
\displaystyle \text{The sum of the two numbers is }1000.
\displaystyle \therefore x+y=1000\qquad\ldots\text{(i)}
\displaystyle \text{The difference between their squares is }256000.
\displaystyle \therefore x^2-y^2=256000
\displaystyle \Rightarrow (x+y)(x-y)=256000
\displaystyle \text{Using }x+y=1000,
\displaystyle 1000(x-y)=256000
\displaystyle \Rightarrow x-y=256\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=1256
\displaystyle \Rightarrow x=628
\displaystyle \text{Substituting }x=628\text{ in equation (i),}
\displaystyle 628+y=1000
\displaystyle \Rightarrow y=372
\displaystyle \therefore \text{The required numbers are }628\text{ and }372.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A two-digit number is 4 times the sum of its digits.}
\displaystyle \text{If }18\text{ is added to the number, the digits are reversed. Find the number.}
\displaystyle \hfill\text{[CBSE 2001C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The number is 4 times the sum of its digits.}
\displaystyle \therefore 10x+y=4(x+y)
\displaystyle \Rightarrow 6x-3y=0
\displaystyle \Rightarrow 2x-y=0\qquad\ldots\text{(i)}
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{On adding }18\text{ to the number, the digits are reversed.}
\displaystyle \therefore 10x+y+18=10y+x
\displaystyle \Rightarrow 9x-9y=-18
\displaystyle \Rightarrow x-y=-2\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle x=2
\displaystyle \text{Substituting }x=2\text{ in equation (i),}
\displaystyle 4-y=0
\displaystyle \Rightarrow y=4
\displaystyle \therefore \text{The required number}=10(2)+4=24.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The sum of a two-digit number and the number}
\displaystyle \text{formed by reversing the order of digits is }66.\text{ If the two digits differ}
\displaystyle \text{by }2\text{, find the number. How many such numbers are there?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{Their sum is }66.
\displaystyle \therefore (10x+y)+(10y+x)=66
\displaystyle \Rightarrow 11x+11y=66
\displaystyle \Rightarrow x+y=6\qquad\ldots\text{(i)}
\displaystyle \text{Since the digits differ by }2,\text{ there are two cases.}

\displaystyle \text{Case I: }x-y=2\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=8
\displaystyle \Rightarrow x=4
\displaystyle \therefore y=2
\displaystyle \therefore \text{The number}=10(4)+2=42.

\displaystyle \text{Case II: }y-x=2\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle 2y=8
\displaystyle \Rightarrow y=4
\displaystyle \therefore x=2
\displaystyle \therefore \text{The number}=10(2)+4=24.
\displaystyle \therefore \text{The required numbers are }42\text{ and }24.
\displaystyle \therefore \text{There are two such numbers.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A two-digit number is 4 times the sum of its digits}
\displaystyle \text{and twice the product of the digits. Find the number.}\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The number is 4 times the sum of its digits.}
\displaystyle \therefore 10x+y=4(x+y)
\displaystyle \Rightarrow 6x-3y=0
\displaystyle \Rightarrow y=2x\qquad\ldots\text{(i)}
\displaystyle \text{The number is twice the product of its digits.}
\displaystyle \therefore 10x+y=2xy\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting }y=2x\text{ in equation (ii),}
\displaystyle 10x+2x=2x(2x)
\displaystyle \Rightarrow 12x=4x^2
\displaystyle \Rightarrow 4x(x-3)=0
\displaystyle \text{Since }x\ne0\text{ for a two-digit number, }x=3.
\displaystyle \therefore y=2(3)=6
\displaystyle \therefore \text{The required number}=10(3)+6=36.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A two-digit number is such that the product of}
\displaystyle \text{its digits is }20.\text{ If }9\text{ is added to the number, the digits interchange}
\displaystyle \text{their places. Find the number.}\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The product of the digits is }20.
\displaystyle \therefore xy=20\qquad\ldots\text{(i)}
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{On adding }9\text{ to the number, the digits interchange their places.}
\displaystyle \therefore 10x+y+9=10y+x
\displaystyle \Rightarrow 9x-9y=-9
\displaystyle \Rightarrow y-x=1
\displaystyle \Rightarrow y=x+1\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting }y=x+1\text{ in equation (i),}
\displaystyle x(x+1)=20
\displaystyle \Rightarrow x^2+x-20=0
\displaystyle \Rightarrow (x-4)(x+5)=0
\displaystyle \Rightarrow x=4\text{ or }x=-5
\displaystyle \text{Since a digit cannot be negative, }x=4.
\displaystyle \therefore y=4+1=5
\displaystyle \therefore \text{The required number}=10(4)+5=45.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The sum of the digits of a two-digit number is }9.\text{ Also,}
\displaystyle \text{nine times this number is twice the number obtained by reversing the}
\displaystyle \text{order of the digits. Find the number.} 
\displaystyle \text{Answer:}
\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.
\displaystyle \text{The given number}=10x+y.
\displaystyle \text{The number obtained by reversing the digits}=10y+x.
\displaystyle \text{The sum of the digits is }9.
\displaystyle \therefore x+y=9\qquad\ldots\text{(i)}
\displaystyle \text{Nine times the number is twice the reversed number.}
\displaystyle \therefore 9(10x+y)=2(10y+x)
\displaystyle \Rightarrow 90x+9y=20y+2x
\displaystyle \Rightarrow 88x=11y
\displaystyle \Rightarrow y=8x\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting }y=8x\text{ in equation (i),}
\displaystyle x+8x=9
\displaystyle \Rightarrow 9x=9
\displaystyle \Rightarrow x=1
\displaystyle \therefore y=8
\displaystyle \therefore \text{The required number}=10(1)+8=18.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two numbers are in the ratio of }5:6.\text{ If }8\text{ is}
\displaystyle \text{subtracted from each of the numbers, the ratio becomes }4:5.\text{ Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }x\text{ and }y.
\displaystyle \text{The numbers are in the ratio }5:6.
\displaystyle \therefore \frac{x}{y}=\frac{5}{6}
\displaystyle \Rightarrow 6x-5y=0\qquad\ldots\text{(i)}
\displaystyle \text{On subtracting }8\text{ from each number, the ratio becomes }4:5.
\displaystyle \therefore \frac{x-8}{y-8}=\frac{4}{5}
\displaystyle \Rightarrow 5(x-8)=4(y-8)
\displaystyle \Rightarrow 5x-4y=8\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 24x-20y=0\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }5,
\displaystyle 25x-20y=40\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iii) from equation (iv),}
\displaystyle x=40
\displaystyle \text{Substituting }x=40\text{ in equation (i),}
\displaystyle 6(40)-5y=0
\displaystyle \Rightarrow 5y=240
\displaystyle \Rightarrow y=48
\displaystyle \therefore \text{The required numbers are }40\text{ and }48.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A two-digit number is obtained by either multiplying}
\displaystyle \text{the sum of the digits by }8\text{ and then subtracting }5\text{ or by multiplying}
\displaystyle \text{the difference of the digits by }16\text{ and then adding }3.\text{ Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.
\displaystyle \text{The given number}=10y+x.
\displaystyle \text{The number is }5\text{ less than }8\text{ times the sum of its digits.}
\displaystyle \therefore 10y+x=8(x+y)-5
\displaystyle \Rightarrow 7x-2y=5\qquad\ldots\text{(i)}
\displaystyle \text{The number is }3\text{ more than }16\text{ times the difference of its digits.}
\displaystyle \therefore 10y+x=16(y-x)+3
\displaystyle \Rightarrow 17x-6y=3\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 21x-6y=15\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle 4x=12
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (i),}
\displaystyle 21-2y=5
\displaystyle \Rightarrow 2y=16
\displaystyle \Rightarrow y=8
\displaystyle \therefore \text{The required number}=10(8)+3=83.
\displaystyle \\

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