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Verbal Question of the Day Chat Group 28 Nov 2024, 01:36
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Problem Solving Butler: November 2024
November 28PS 1PS 2
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Verbal Question of the Day Chat Group 28 Nov 2024, 01:37
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November 28CR 1CR 2
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Verbal Question of the Day Chat Group 28 Nov 2024, 01:38
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November 28DS 1DS 2
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Verbal Question of the Day Chat Group 28 Nov 2024, 09:00
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November 28th Verbal Question of the Day





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Verbal Question of the Day Chat Group 28 Nov 2024, 20:27
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If x2-3x <40 and (x+1)(x-1)>8 what will be the minimum value of x that satisfy both of the inequalities?

(A)-2

(B) -3

(C) -4

(D) 5

(E) none

Why is the answer not -4 can anyone explain
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Verbal Question of the Day Chat Group 28 Nov 2024, 23:05
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Arishaoboni
If x2-3x <40 and (x+1)(x-1)>8 what will be the minimum value of x that satisfy both of the inequalities?

(A)-2

(B) -3

(C) -4

(D) 5

(E) none
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For x = -4, second part becomes -(4^2 - 1^2) = -15 which is less than 8

SathishThandhari
For x = -4, second part becomes -(4^2 - 1^2) = -15 which is less than 8
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Ignore that thats incorrect
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Verbal Question of the Day Chat Group 28 Nov 2024, 23:09
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x>3 or x<-3 and x(x-3)<40

x=-4 should be the minimum here
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Verbal Question of the Day Chat Group 29 Nov 2024, 01:32
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November 29PS 1PS 2
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Verbal Question of the Day Chat Group 29 Nov 2024, 01:33
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Verbal Question of the Day Chat Group 29 Nov 2024, 01:35
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November 29DS 1DS 2
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Verbal Question of the Day Chat Group 29 Nov 2024, 04:20
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Arishaoboni
If x2-3x <40 and (x+1)(x-1)>8 what will be the minimum value of x that satisfy both of the inequalities?

(A)-2

(B) -3

(C) -4

(D) 5

(E) none
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What’s the answer?
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Verbal Question of the Day Chat Group 29 Nov 2024, 09:00
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November 29th Verbal Question of the Day





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Verbal Question of the Day Chat Group 30 Nov 2024, 09:00
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November 30th Verbal Question of the Day





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Verbal Question of the Day Chat Group 30 Nov 2024, 09:08
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which RCs to practice for XAT exam?
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Verbal Question of the Day Chat Group 30 Nov 2024, 09:12
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which RCs to practice for XAT exam?
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Below 655 level rcs from og or GMAT club question bank

Their poems are also quite peculiar.... So read vast variety of poem of British poets across genres
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Verbal Question of the Day Chat Group 01 Dec 2024, 09:00
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December 1st Verbal Question of the Day





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Verbal Question of the Day Chat Group 01 Dec 2024, 12:13
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Hello pls help me to understand this task.
"City A and City B are located 610 kilometers (km) apart from each other on Highway X. At 8:00 a.m., Alonzo began driving along Highway X from City A to City B at an average driving speed of 90 kilometers per hour (km/h). At 9:30 a.m., on the same day, Ryan began driving along Highway X from City 8 to City A at an average driving speed of 100 km/h. How long, in hours and minutes to the nearest minute, did Ryan drive before meeting Alonzo on Highway X?

2 hours 30 minutes

2 hours 50 minutes

3 hours 13 minutes

3 hours 55 minutes

4 hours 0 minutes
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Verbal Question of the Day Chat Group 01 Dec 2024, 12:15
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Hello pls help me to understand this task.
"City A and City B are located 610 kilometers (km) apart from each other on Highway X. At 8:00 a.m., Alonzo began driving along Highway X from City A to City B at an average driving speed of 90 kilometers per hour (km/h). At 9:30 a.m., on the same day, Ryan began driving along Highway X from City 8 to City A at an average driving speed of 100 km/h. How long, in hours and minutes to the nearest minute, did Ryan drive before meeting Alonzo on Highway X?

2 hours 30 minutes

2 hours 50 minutes

3 hours 13 minutes

3 hours 55 minutes

4 hours 0 minutes
Show more

This question is discussed here: https://gmatclub.com/forum/city-a-and-c ... 16656.html Hope it helps.
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Verbal Question of the Day Chat Group 01 Dec 2024, 12:16
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Sorry accidentally confused chats
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Verbal Question of the Day Chat Group 02 Dec 2024, 00:49
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If Alonso began driving at 8 AM at speed of 90 kmph, he will cover 90 X 1.5 = 135 kms.
Distance left = 610 -135 = 475 kms

now, Ryan begin driving from B to A at speed of 100kmph

Distance = 475 km
Speed = 100 + 90 ( since both of them are driving towards each other)

Time = Distance / speed
= 475 / 190 = 2.5 hrs
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