Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
3.3k views
in Mathematics by (15.4k points)
closed by

5 pencils and 7 pens together cost Rs 195 while 7 pencils and 5 pens together cost Rs 153. Find the cost of each one of the pencil and the pen.

2 Answers

+2 votes
by (15.9k points)
selected by
 
Best answer

Let the cost of one pencil is Rs x and the cost of one pen is Rs y. 

Given that together cost of 5 pencils and 7 pens is Rs 195. 

Therefore, 5x+ 7y = 195. … (1) 

And together cost of 7 pencils and 5 pens is Rs 153. 

Therefore, 7x + 5y = 153. … (2) 

Now, multiplying equation (1) by 7, 

we get 35x + 49 y = 1365. … (3) 

Now, multiplying equation (2) by 5, we get 

35x + 25 y = 765. … (4)

Now, subtracting equation (4) from equation (3), we get 

(35x + 49y) – (35x + 25y) = 1365 – 765 

⇒ 24y = 600 

⇒ y = \(\frac{600}{24}\) = 25. 

Now, putting y = 25, in equation (1), we get 

5x+ 175 = 195 

⇒ 5x = 195 – 175 = 20 

⇒ x = \(\frac{20}{5}\) = 4. 

Hence, the cost of one pencil is RS 4 and the cost of one pen is Rs 25.

+2 votes
by (710 points)

Let pencils be x and pens be y 

CASE 1 

No. Of pencils = 5 

No. Of pens = 7

Cost of pens and pencils = Rs. 195

= 5x + 7y = 195   -------(i) 

CASE 2 

No. Of pencils = 7 

No. Of pens = 5

Cost of pens and pencils = Rs .153

= 7x + 5y = 153   -------(ii)

Now ,multiplying (ii) by 5 and (i) by 7

7 ( 5x + 7y =195 )

= 35x + 49y =1365   ------(iii)

5 ( 7x + 5y = 153 )

= 35x + 25y = 765   -------(iv) 

Subtracting eq. (iv) from (iii)             we get,

  24y = 600

⇒ y = 600 / 24    = 25 

Putting value of y in eq (ii)

7x + 5 × 25 = 153

= 7x + 125 = 153

= 7x = 153-125

= x = 28 / 7  = 4

Therefore,  cost of each pencil (x) =   Rs.4

Cost of each Pen (y) = Rs. 25

Hence cost of each pen is Rs.25 and cost of each pencil is Rs.4 .

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...