11th Grade > Mathematics
INEQUALITIES MODULUS AND LOGARITHMS MCQs
:
A
Step 1 : Underline the keywords
Henko is renting a car. The rental charge is $17.50 per day plus $0.16 per mile. Henko can spend at most $33 for the cost of the car rental. If Henko rents the car for one day, which of the following is one possible number of miles Henko can drive the rental car?
Step 2 : Recognize the variable
Number of miles ,is what the question is looking for.
Let number of miles be x
Step 3 : Decipher the relation
Total cost = no. of days rented × 17.50 + number of miles × 0.16
As henko rented the car for only one day
Total cost =17.50+x×0.16
Step 4 : Decide the inequality
Look for cues : Henko can spend at most 33$ for the car rent. Therefore, we get
33≥17.50+x×0.16
As, the question is looking for possible value of x, simplify the given inequality
96.87≥x
Only, option A satisfies the given inequality.
:
D
Substitute each of the points in the inequalities to check if the inequality is satisfied.
(-2,-1) does not satisfy either inequality
(-1, 3) does not satisfy either inequality
(1,5) does not satisfy either inequality
(2,-1) satisfies both equalities
Alternatively, plotting the graph of the inequalities,
It can be seen that, only (2,-1) belongs to the solution set. Hence D is the correct choice.
:
B
−5<p<−2 and 7<q<9
Adding two inequalities with same natured inequalities will preserve the sign.
So adding the two inequalities:
−5+7<p+q<−2+9
⇒2<p+q<7
:
D
Let y = log2536
= loge36loge25=loge62loge52=2loge62loge5=loge6loge5
= log56
log2536 can be written as,
log5262=2log526
or 12log562
:
A
Cost of entry (dollars) = 5
Cost of playing n games (dollars) = 2n
Total cost (dollars) = 2n + 5
Total cost can be at most $25
Therefore, 2n+5≤25
:
D
3+12x4≥2(4x+1)5
(3+12x)×5≥2(4x+1)×4
15+60x≥32x+8
60x−32x≥8−15
28x≥−7
x≥−728
x≥−14
:
A
Let examine the given inequality 21bx−28>49
Dividing both side by 7
21bx7−287>497
3bx−4>7
Multiply -1 on both side [remember to flip the sign]
−1(3bx−4)<7(−1)
4−3bx<−7
The correct answer is option A.
:
D
log1000 x2
=log103x2
=2 log103x
=23log10x
=23y
Maria plans to rent a boat. The boat rental costs $60 per hour, and she will also have to pay for a water safety course that costs $10. Maria wants to spend no more than $280 for the rental and the course. If the boat rental is available only for a whole number of hours, the maximum number of hours for which Maria can rent the boat is ___
:
Let n be the number of hours for which Maria rents the boat. Given that rental cost is $60 per hour. Also she has to pay $10 for a water safety course. Hence, the total cost is 60n + 10 . Given that she will spend a maximum of $280. Hence,
60n+10≤280
⇒60n≤270
n≤4.5
Since the boat rental is available only for whole number of hours (n is a whole number), the maximum number of hours for which Maria can rent the boat is 4.