
In a test, Rakhesh got 50% marks and scored 10 marks more than the pass marks. In the same test, Suresh secured 55% marks and scored 20 marks more than the pass marks. Find the pass marks.
Answer
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Hint: Assume that the pass marks be x and the total marks be y. Using the statement from the question, 50% marks are 10 marks more than pass marks, and 55% marks are 20 marks more than pass marks form two equations. Solve the equations using any of the known methods like elimination method, substitution method etc. Hence find the value of x and y. Hence find the pass marks (x).
Complete step-by-step answer:
Let the pass marks be x, and the total marks be y.
Hence the marks secured by Rakhesh $=\dfrac{50}{100}y=\dfrac{y}{2}$.
Since the marks secured by Rakhesh are 10 more than pass marks, we have
$\dfrac{y}{2}=x+10$
Multiplying both sides by 2, we get
$y=2x+20\text{ }\left( i \right)$
Also, the marks secured by Suresh $=\dfrac{55}{100}y=\dfrac{11}{20}y$
Since the marks secured by Suresh are 20 more than the total marks, we have
$\dfrac{11}{20}y=x+20$
Multiplying both sides by 20, we get
$11y=20x+400\text{ }\left( ii \right)$
Substituting the value of y from equation (i) in equation (ii), we get
$11\left( 2x+20 \right)=20x+400$
Using distributive law of multiplication over addition, i.e. a(b+c) = ab+ac , we get
$22x+220=20x+400$
Subtracting 20x from both sides, we get
2x+220=400
Subtracting 220 from both sides, we get
2x=180
Dividing both sides by 2, we get
x =90
Substituting the value of x in equation (i), we get
$y=2\left( 90 \right)+20=180+20=200$
Hence the pass marks are 90, and the total marks are 200.
Note: Verification:
Marks secured by Rakhesh $=\dfrac{50}{100}\times 200=100$, which are clearly 10 more than the pass marks.
Marks secured by Suresh $=\dfrac{55}{100}\times 200=110$, which are clearly 20 more than the pass marks.
Hence our answer is verified to be correct.
Complete step-by-step answer:
Let the pass marks be x, and the total marks be y.
Hence the marks secured by Rakhesh $=\dfrac{50}{100}y=\dfrac{y}{2}$.
Since the marks secured by Rakhesh are 10 more than pass marks, we have
$\dfrac{y}{2}=x+10$
Multiplying both sides by 2, we get
$y=2x+20\text{ }\left( i \right)$
Also, the marks secured by Suresh $=\dfrac{55}{100}y=\dfrac{11}{20}y$
Since the marks secured by Suresh are 20 more than the total marks, we have
$\dfrac{11}{20}y=x+20$
Multiplying both sides by 20, we get
$11y=20x+400\text{ }\left( ii \right)$
Substituting the value of y from equation (i) in equation (ii), we get
$11\left( 2x+20 \right)=20x+400$
Using distributive law of multiplication over addition, i.e. a(b+c) = ab+ac , we get
$22x+220=20x+400$
Subtracting 20x from both sides, we get
2x+220=400
Subtracting 220 from both sides, we get
2x=180
Dividing both sides by 2, we get
x =90
Substituting the value of x in equation (i), we get
$y=2\left( 90 \right)+20=180+20=200$
Hence the pass marks are 90, and the total marks are 200.
Note: Verification:
Marks secured by Rakhesh $=\dfrac{50}{100}\times 200=100$, which are clearly 10 more than the pass marks.
Marks secured by Suresh $=\dfrac{55}{100}\times 200=110$, which are clearly 20 more than the pass marks.
Hence our answer is verified to be correct.
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