Question
A sum of ₹25000 is invested in two parts, one at 8%
and the other at 10%. The total interest earned after a year is ₹2300. How much was invested at each rate?Solution
Let x be the amount invested at 8%, so 25000−x is invested at 10%. Equation for total interest: 0.08x+0.10(25000−x)=2300 Simplifying: 0.08x+2500−0.10x=2300⇒−0.02x+2500=2300 Solve for x =−0.02x=−200 ⇒ x = 10000 So: ₹10,000 was invested at 8%. ₹15,000 was invested at 10%.
I. x= √(20+ √(20+ √(20+ √(20…………….∞)) ) )
II. y= √(5√(5√(5√(5……….∞)) ) )
...I. x2 + 13x + 42 = 0
II. y² + 13y + 40 = 0
I). p2 + 22p + 72 = 0,
II). q2 - 24q + 128 = 0
I. p²= ∛1331
II. 2q² - 21q + 55 = 0
I. 27x6-152x3+125=0
II. 216y6 -91y3+8=0
I. √(17x) + √51 = 0
II. √(4y) + 3 = 0
I. x2 - 5x - 14 = 0
II. y2 - 16y + 64 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 6x² - 24x + 18 = 0
Equation 2: 5y² - 20y + 15 = 0
I. 63x² + 146x + 80 = 0
II. 42y² + 109y + 70 = 0
I. x² - 33x + 270 = 0
II. y² - 41y + 414 = 0
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