arrow
arrow
arrow
What would be the maximum value of Q in the following equation? 5P9 + 3R7+2Q8= 1114
Question

What would be the maximum value of Q in the following equation?
5P9 + 3R7+2Q8= 1114

A.

8

B.

9

C.

7

D.

5

Correct option is B

Given:

5P9+3R7+2Q8=11145P9 + 3R7+2Q8= 1114 

Maximum value of Q = ?

Solution:

5P9+3R7+2Q8=11145P9+3R7+2Q8=1114 

It can be written as: 

(500 + 10P + 9) + (300 + 10R + 7) + (200 + 10Q + 8) = 1114

Now,

500 + 300 + 200 + 10P + 10R + 10Q + 9 + 7 + 8 = 1114

1000 + 10P + 10R + 10Q + 24 = 1114

1024 + 10P + 10R + 10Q = 1114

10P + 10R + 10Q = 90

P + R + Q = 9

The smallest possible values for P and R are both 0 (since they are single digits).

So, if P=0 and R=0, then: 

Q = 9

Thus the maximum value of Q is 9.

Free Tests

Free
Must Attempt

CBT-1 Full Mock Test 1

languageIcon English
  • pdpQsnIcon100 Questions
  • pdpsheetsIcon100 Marks
  • timerIcon90 Mins
languageIcon English
Free
Must Attempt

RRB NTPC Graduate Level PYP (Held on 5 Jun 2025 S1)

languageIcon English
  • pdpQsnIcon100 Questions
  • pdpsheetsIcon100 Marks
  • timerIcon90 Mins
languageIcon English
Free
Must Attempt

CBT-1 General Awareness Section Test 1

languageIcon English
  • pdpQsnIcon40 Questions
  • pdpsheetsIcon30 Marks
  • timerIcon25 Mins
languageIcon English

Similar Questions

test-prime-package

Access ‘RRB NTPC’ Mock Tests with

  • 60000+ Mocks and Previous Year Papers
  • Unlimited Re-Attempts
  • Personalised Report Card
  • 500% Refund on Final Selection
  • Largest Community
students-icon
354k+ students have already unlocked exclusive benefits with Test Prime!
Our Plans
Monthsup-arrow