5 8. X =7 and y = 3 is the solution of equations ax + by = 18 and bx + ay = 12 then abis (a) -7 (b)-3 (d) 7 (c) 3
Answers
Answer:
a+b = 3
Step-by-step explanation:
Given x =7 and y = 3 is the solution of equations ax + by = 18 and bx + ay = 12
So, 7a + 3b = 18 (i)
and 7b + 3a = 12 (ii)
After doing (i)×7 - (ii)×3 we get:
7(7a + 3b) - 3(7b + 3a) = (7×18) - (3×12)
=> 49a + 21b - 21b - 9a = 126 - 36
=> 40a = 90
=> a = 90/40
=> a = 9/4
Putting the value of b in (i) we get:
7(9/4) + 3b = 18
=> 63/4 + 3b = 18
=> 21/4 + b = 6 [Dividing both sides by 3]
=> b = 6 - 21/4
=> b = 3/4
So, a+b = (9/4)+(3/4) = 12/4 = 3 (Option c)
Step-by-step explanation:
Given : Mahesh invests Rs.3000 for 3 years at rate of 10th p. a. Compound interest.Find the amount and the compound interest that Makesh will get after 3 years.
Solution : According to the question, Mahesh invests ₹3000 for 3 years at 10% p.a. Compound Interest. We need to find the amount and the compound Interest after 3 yrs.
★ Formula Used :
\odot \quad \underline{\boxed{ \green{ \tt{A = P(1 +} \sf{\frac{r}{100})}^{n}}}}⊙
A=P(1+
100
r
)
n
Now, getting the amount and compound Interest
\begin{gathered} \leadsto \tt{A = P(1 +} \sf{\frac{r}{100})}^{n} \\ \\ \leadsto\tt{A = 3000(1 +} \sf{\frac{10}{100})}^{3} \\ \\ \leadsto\tt{A = 3000(1 +} \sf{\frac{1}{10})}^{3} \\ \\ \leadsto\tt{A = 3000( \frac{11}{10} )}^{3} \\ \\ \leadsto\tt{A = 3000 \times \frac{1331}{1000}} \\ \\ \star \quad \underline{ \blue{\tt{A} = 3993}}\end{gathered}
⇝A=P(1+
100
r
)
n
⇝A=3000(1+
100
10
)
3
⇝A=3000(1+
10
1
)
3
⇝A=3000(
10
11
)
3
⇝A=3000×
1000
1331
⋆
A=3993
Hence, the amount is ₹3993
Getting the compound Interest
\begin{gathered} : \implies \sf{Compound \: Interest = Amount - Principal} \\ \\ : \implies \sf{Compound \: Interest =3993 - 3000} \\ \\ \star \quad \underline{ \blue{\frak{Compound \: Interest =993}}}\end{gathered}
:⟹CompoundInterest=Amount−Principal
:⟹CompoundInterest=3993−3000
⋆
CompoundInterest=993
The compound Interest is ₹993 after 3 years