Question
A sum of ₹4,360 was to be divided among A, B, C, and D
in the ratio 3:4:5:8, but it was divided in the ratio 1/3:1/4:1/5:1/8. Was divided by mistake: which of the following statements will hold as a resultSolution
ATQ, Sum was to be divided in the ratio 3:4:5:8 Let the sum of A,B,C and D is 3x,4x,5x and 8x respectively According to question, we get 20x = 4360 x=218 so we get A = 654 ; B = 872 ; C = 1090; D =1744 But by mistake it was divided in the ratio; 1/3 : 1/4 : 1/5 : 1/8 By Equating , the ratio will be ; 40 : 34 : 24 : 15 New sum of A, B, C and D will be 40x, 30x, 24x and 15x respectively Solving in the same way, we get A = 1600 , B = 1200, C = 960 and D = 600 So we can say D got 1144 less
Given a triangle with two sides measuring 12 cm and 16 cm, which of the following cannot be a possible length for the third side?
- Find the remainder on dividing (5x 3 + 4x 2 – 10x + 1) by (2x – 4).
The area (in sq, units) of the triangle formed by the graphs of 8x + 3y = 24, 2x+8 = y and the x – axis is:
If x −y =4and xy =45,thenthevalueof x 3 −y3is:
If x + y = 2 and 1/x + 1/y = 18/15, then find the value of (x3 + y3)?
22% of ‘x’ is equal to 66% of ‘y’, then find the value of x: y.
If y + 1/y = 2 then find y117 + 1/(y117)
if a2 + b2 + c2 = 2(4a -5b -9c)-122 , then a-b-c = ?
If a – b – c = 0, then the value of (a2 + b2 + c2)/(b + ac) is:
If 5x - 13xy + 6y = 0, find x : y.
Relevant for Exams:
