Question

5.) In a class test, the sum of the marks obtained by P in mathematics and
science is 28. Had he got 3 more marks in mathematics and 4 marks less
in science, the product of marks obtained in the two subjects would
have been 180. Find the marks obtained by him in the two subjects
separately.
[CBSE 2008
frana for 180 If he had bought 3 more nane​

Answer 1

Answer:

(16,12) OR (19,9)

Step-by-step explanation:

HEY MATE, HERE'S THE ANSWER:-

SUPPOSE M ARE THE MARKS OBTAINED IN MATHEMATICS AND S ARE THE MARKS OBTAINED IN SCIENCE.

M+S=28

M=28-S

GIVEN THAT,

(3+M)×(S-4)=180

THEREFORE,(3+28-S)(S-4)=180

(31-S)(S-4)=180

31(S-4)-S(S-4)=180

31S-124-S^2+4S=180

-S^2+35S-124=180

S^2-35S+304=0

S^2-16S-19S+304=0

S(S-16)-19(S-16)=0

(S-16)(S-19)=0

S=16 OR S=19

IF S=16 THEN M=28-16=12

IF S=19 THEN M=28-19=9

Answer 2

Appropriate Question:

In the class test the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subject would have been 180. Find the marks obtained by him in the two subject separately

Answer:

$\begin{gathered}\boxed{\begin{array}{c|c} \sf Marks\:obtained\:in\:maths & \sf Marks\:obtained\:in\:science \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 9 & \sf 19 \\ \\ \sf 12 & \sf 16 \end{array}} \\ \end{gathered}$

Step-by-step explanation:

Given that, marks obtained by P in mathematics and science is 28.

Let assume that

$\boxed{\begin{aligned}&\:\sf \:Marks\:obtained\:in\:maths=x \: \\ \\& \:\sf \: Marks\:obtained\:in\:science=28 -x \end{aligned}} \qquad \:$

Further given that, had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subject would have been 180.

So, we have

$\boxed{\begin{aligned}&\:\sf \:Marks\:obtained\:in\:maths=x + 3 \: \\ \\& \:\sf \: Marks\:obtained\:in\:science=28 -x - 4 = 24 - \end{aligned}} \qquad \:$

According to statement, we get

$\sf \: (x + 3)(24 - x) = 180$

$\sf \: 24x - {x}^{2} + 72 - 3x = 180$

$\sf \: 21x - {x}^{2} + 72 = 180$

$\sf \: {x}^{2} - 21x + 108 = 0$

$\sf \: {x}^{2} - 9x - 12x+ 108 = 0$

$\sf \: x(x - 9) - 12(x - 9) = 0$

$\sf \: (x - 9)(x - 12) = 0$

$\implies\sf \: x = 9 \: \: or \: \: x = 12$

Hence,

$\begin{gathered}\boxed{\begin{array}{c|c} \sf Marks\:obtained\:in\:maths & \sf Marks\:obtained\:in\:science \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 9 & \sf 19 \\ \\ \sf 12 & \sf 16 \end{array}} \\ \end{gathered}$