Solution:- Let us assume the two numbers be, P and Q, Then, $P = 10P$ … [because it comes in tens digit] As per the condition given in the question, $PQ = 8$ … [equation (i)] $10P + Q – 18 = 10Q +...
Solution:- let us assume the two natural numbers be $y$, $y – 3$ As per the condition given in the question, \[{{y}^{2}}~+\text{ }{{\left( y-3 \right)}^{2}}~=\text{ }117\] Then, \[{{y}^{2}}~+\text{...
Solution:- let us assume the two numbers be y, $2y – 3$ As per the condition given in the question, \[{{y}^{2}}~+\text{ }{{\left( 2y-3 \right)}^{2}}~=\text{ }233\] Then, \[{{y}^{2}}~+\text{...
Solution:- let us assume the three consecutive natural number be $P – 1$, $P$ and $P + 1$ As per the condition given in the question, \[{{\left( P-1 \right)}^{2}}~+\text{ }\left( P \right)\text{...
Solution:- let us assume the two consecutive odd positive number be P and P + 2 As per the condition given in the question, \[{{P}^{2}}~+\text{ }{{\left( P\text{ }+\text{ }2 \right)}^{2}}~=\text{...
Solution:- Let us assume the two numbers be, P and Q. As per the condition given in the question, \[{{P}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }656\] … [equation (i)] P – Q = 4 P = 4 + Q … [equation (ii)]...
Solution:- Let us assume the two numbers be, P and Q, Then, $P = 10P$ … [because it comes in tens digit] As per the condition given in the question, $PQ = 4(P + Q)$ $4(P + Q) = 10P +Q$ … [equation...
Solution:- Let us assume the two numbers be, P and Q, Q being the bigger number. As per the condition given in the question,\[{{Q}^{2}}~-{{P}^{2}}~=\text{ }45\] … [equation (i)] \[{{P}^{2}}~=\text{...
Solution:- Let us assume the two numbers be, P and Q. As per the condition given in the question, \[{{P}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }208\] … [equation (i)] \[{{Q}^{2}}~=\text{ }18P\] …...
Solution:- Let us assume the two numbers be A and B, B being the bigger number. As per the condition given in the question, $A + B = 25$ … [equation (i)] \[2{{B}^{2}}~=\text{ }3{{A}^{2}}~+\text{...
Solution:- Let us assume the two digits be MN As per the condition given in the question, Product of $2$ digits is $18$, $MN = 14$ … [equation (i)] $45$ is added to the number, then the digit are...
Solution:- let us assume two digits be PQ. As per the condition given in the question, Product of $2$ digits is $18$, $PQ = 18$ … [equation (i)] 63 is subtracted from the number, the digits...
Solution:- Let us assume the number be B. As per the condition given in the question, B + 1/B =\[2\frac{9}{40}\] So, B + 1/B = 89/40 \[({{B}^{2}}~+\text{ }1)/B\text{ }=\text{ }89/40\] Cross...
Solution:- Let us assume the numbers be P and Q. As per the condition given in the question, The sum of $2$ numbers is $18$, $P + Q = 18$ … [equation i] the sum of their reciprocals is $¼$, $1/P +...