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06 Mar 2020, 04:43
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
88% (01:08) correct
12%
(01:35)
wrong
based on 90
sessions
History
Date
Time
Result
Not Attempted Yet
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
06 Mar 2020, 04:47
Kudos
Bookmarks
Bunuel
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
Markers = 5x
Pens = 7x
Pen remaining = 7x - 6
Markers, 5x = 35 i.e. x = 7
i.e. Pens = 7x = 49
ie.. Pen remaining = 49-6 = 43
Answer: Option B
06 Mar 2020, 05:11
Kudos
Bookmarks
Bunuel
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
(A) 29
(B) 43
(C) 49
(D) 50
(E) 51
Solution:
- • The ratio of markers to pens = \(5: 7\)
• Let \(5x\), and \(7x\) be the number of markers and pens in the drawer.
- o Number of pens after removal of 6 pens = \(7x – 6\)
- o \(5x =35\)
o \(x =7\).
