Two students appeared at an examination. One of them secured $9$ marks more than the other and his marks was $56 %$ of the sum of their marks. The marks obtained by them are:

Open in App

Solution

$\begin{matrix} L e t t h e i r m a r k s b e (x + 9) a n d x T h e n, x + 9 = \frac{56}{100} (x + 9 + x) 25 (x + 9) = 14 (2 x + 9) 3 x = 99 x = 33 S o, t h e i r m a r k s a r e 42 a n d 33. \end{matrix}$ .

Suggest Corrections

Similar questions

Question 54. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

Two students appeared at an examination full stop one of them scored 5 marks more than the other and his marks was 56% of the dum of their marks.the marks obtained by them are-

1)39,30

2)41,32

3)42,33

4)43,34

Jared and Leto appeared for an examination. Jared secured 50 marks more than Leto. 50% of the sum of the marks secured by both Jared and Leto is equal to 100. What are the individual marks secured by them respectively?

Two students appeared at an examination. A secured 50 marks more than B. 50% of the sum of marks secured by both A and B is equal to 100. The individual marks secured by A and B is respectively:

Q. Three students Ankit, Abhash and Ajay appeared for an exam. If Ankit and Abhash secured 70 marks and 120 marks respectively, the average marks secured by all of them were

60 %

of the maximum marks. The passing marks in it were 40 less than their average marks and

40 %

of the maximum marks. The marks secured by Ajay___

Join BYJU'S Learning Program

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

Lettheirmarksbe(x+9)andxThen,x+9=56100(x+9+x)25(x+9)=14(2x+9)3x=99x=33So,theirmarksare42and33..

Two students appeared at an examination. One of them secured $9$ marks more than the other and his marks was $56 %$ of the sum of their marks. The marks obtained by them are:

$\begin{matrix} L e t t h e i r m a r k s b e (x + 9) a n d x T h e n, x + 9 = \frac{56}{100} (x + 9 + x) 25 (x + 9) = 14 (2 x + 9) 3 x = 99 x = 33 S o, t h e i r m a r k s a r e 42 a n d 33. \end{matrix}$ .