(Solved) The average weight of A, B, C is 45 kg.if the average weight of A and B is 40 kg and that of B And C is 43 kg, then the weight of B is:

-

 Question :

The Average weight of A, B, C is 45 kg.if the average weight of A and B is 40 kg and that of B And C is 43 kg, then the weight of B is:


Solution :

Sum of average weight = A+B+C  = 45

                                                   3

                                              A+B+C = 45×3

                                              A+B+C = 135......1)


Sum of average weight of A and B

A+B = 40

   2

A+B = 40×2

A+B = 80......2)


Sum of average weight of B and C

B+C= 43

   2

B+C = 43×2

B+C= 86.......3)


equation 2) + equation 3)

A+B+B+C =80 +86 = 166

A+2B + C = 166


=A+2B+C  - A+B+C

=166 - 135

= 31

B= 31 Kg














Any queries about Solution please comment below. 

Comments

Post a Comment

Popular posts from this blog

(Solved) A step down DC chopper has a resistive load of R = 15 ohm and input Vdc = 200 V. When the chopper remain ON, it's voltage drop is 2.5V. The chopper frequency is 1 kHz. If the duty cycle is 50%, determine: a) Average output voltage b) RMS output voltage c) Chopper efficiency d) The effective input resistance of the chopper

-
Image
Question:   A step down DC chopper has a resistive load of R = 15 ohm and input Vdc = 200 V. When the chopper remain ON, it's voltage drop is 2.5V. The chopper frequency is 1 kHz. If the duty cycle is 50%, determine:  a) Average output voltage  b) RMS output voltage  c) Chopper efficiency  d) The effective input resistance of the chopper Answer: We can get answers directly by applying relevant formulas as shown below.  (a) Average output voltage = 98.75 V (b) RMS output voltage = 139.6536 V (c) Chopper efficiency = 98.75% (d) The effective input resistance = 30.3799 ohm Additional information : Waveforms for step down chopper are shown below : For any queries regarding above solution please comment below. Thank you! 
Read more

(Solved) When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature.

-
  Question : When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature. Solution : Given Io = 0.3 * 10-6A and VF = 0.15v The current flowing through the PN diode under forward bias is I = Io (e^(40VF) -1) for germanium diode = 0.3 * 10^(-6) (e^(40*0.15) -1) = 120.73mA. For any further queries please comment below.  
Read more

(Solved) An LVDT is connected to an amplifier of gain 250 the output terminals of which are connected to a 5 V voltmeter. The meter's scale can be read to 1/5th of a division and has 100 divisions in all. Given that a core displacement of 0.5 mm provides an output of 2 mV from the LVDT, find, 1. The sensitivity of the LVDT 2. The sensitivity of the whole system 3. The resolution of the instrument in centimeters

-
Image
 Question : An LVDT is connected to an amplifier of gain 250 the output terminals of which are connected to a 5 V voltmeter. The meter's scale can be read to 1/5th of a division and has 100 divisions in all. Given that a core displacement of 0.5 mm provides an output of 2 mV from the LVDT, find,  1. The sensitivity of the LVDT  2. The sensitivity of the whole system  3. The resolution of the instrument in centimeters Explanation : (a) Sensitivity of LVDT = Output voltage /displacement  Sensitivity of LVDT = 2mV/0.5 mm = 4 mV/mm (b)Sensitivity of whole setup = amplification factor*sensitivity of LVDT  Sensitivity of whole setup = 250*(4 mV/mm) = 1000 mV/mm (c) The resolution of the instrument in centimeters = minimum voltage that can be read by instrument /sensitivity of whole setup  There are 100 divisions to measure upto 5V.  So 1 division can read 5/100 = 0.05V Scale can read 1/5 of a division.  So minimum voltage that can be read = 0.05/5 = 0.01V = 10mV So resolution = (10 mV) /
Read more