Question

In an examination, Ram obtained 9 marks more
than Shyam and Ram's marks was 56% of the
sum of the marks obtained by them. Find the
marks obtained by them separately.​

Answer 1

Answer:

Let the students be A and B and the marks obtained by them be Am and Bm.

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or!4Bm-11Bm = 3Bm = 99, or

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or!4Bm-11Bm = 3Bm = 99, orBm = 99/3 = 33.

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or!4Bm-11Bm = 3Bm = 99, orBm = 99/3 = 33.So Am = 33+9 = 42.

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or!4Bm-11Bm = 3Bm = 99, orBm = 99/3 = 33.So Am = 33+9 = 42.A scored 42 marks while B scored 33. Answer.

Let the students be A and B and the marks obtained by them be Am and Bm.Am = Bm+9 …(1)Am = (56/100)(Am+Bm), or100Am = 56(Am+Bm) = 56Am + 56Bm, or100Am-56Am = 44Am = 56Bm orAm = (56/44)Bm = (14/11)Bm …(2)Equate (1) and (2)Bm+9 = (14/11)Bm, or11(Bm+9) = 14Bm, or11Bm+99 = 14Bm, or!4Bm-11Bm = 3Bm = 99, orBm = 99/3 = 33.So Am = 33+9 = 42.A scored 42 marks while B scored 33. Answer.Check. From (2), Am = (14/11)Bm = (14*33)/11 = (14*3) = 42. Correct

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Answer 2

Answer:

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Step-by-step explanation: