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Algebra Level 2 Test 5
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Question 1 |
If log7log5 {√(x + 5) + √x}= 0
1 | |
0 | |
2 | |
None of these |
Question 1 Explanation:
log7 log5 { √(x + 5) + √x }= 0
log √(x + 5)+ √x = 70 = 1,
or
√(x + 5)+ √x = 51 = 5.
Therefore 2√x = 0. or
x = 0.
log √(x + 5)+ √x = 70 = 1,
or
√(x + 5)+ √x = 51 = 5.
Therefore 2√x = 0. or
x = 0.
Question 2 |
If a + b + c = 0, where a ≠ b ≠ c, then {a2/(2a2 +bc)} + {b2/(2b2 +ac)} +{c2/(2c2 +ab)}
zero | |
1 | |
–1 | |
abc |
Question 2 Explanation:
To solve this question, we use the help of values.
Assuming such values of a, b & c so that a + b + c =0
and a ≠ b ≠ c, we can find the value of the expression.
Assuming a = 1, b = -1 and c = 0 , we find that {a2/(2a2 +bc)} + {b2/(2b2 +ac)} +{c2/(2c2 +ab)}
½ + ½ + 0 = 1.
Hence we get our answer without any sweat.
Assuming such values of a, b & c so that a + b + c =0
and a ≠ b ≠ c, we can find the value of the expression.
Assuming a = 1, b = -1 and c = 0 , we find that {a2/(2a2 +bc)} + {b2/(2b2 +ac)} +{c2/(2c2 +ab)}
½ + ½ + 0 = 1.
Hence we get our answer without any sweat.
Question 3 |
If the harmonic mean between two positive numbers is to their geometric mean as 12 : 13; then the numbers could be in the ratio
12 : 13 | |
1/12 : 1/13 | |
4 : 9 | |
2 : 3 |
Question 3 Explanation:
Since the harmonic mean of two numbers x and y is 2xy/(x + y) and the geometric mean is √xy.
So from the question 2xy /(x + y)/√xy = 12/13,
Further solving this
Squaring both sides we get
Therefore 2 x3 y3 / (x + y) 2 = 144/169.
But we want the values for x : y so
= 2xy/(x + y) = 12/13
= 2√xy / (x + y ) √xy
= {(x + y ) √xy / 2√xy}
For further solving we have to get back in our school days for the concept of component and dividendo
2√x/2√y = 6/4 = x / y = 9/4 or y / x = 4/9
So from the question 2xy /(x + y)/√xy = 12/13,
Further solving this
Squaring both sides we get
Therefore 2 x3 y3 / (x + y) 2 = 144/169.
But we want the values for x : y so
= 2xy/(x + y) = 12/13
= 2√xy / (x + y ) √xy
= {(x + y ) √xy / 2√xy}
For further solving we have to get back in our school days for the concept of component and dividendo
2√x/2√y = 6/4 = x / y = 9/4 or y / x = 4/9
Question 4 |
If one root of x2 + px + 12= is 4, while the equation x2 + px + q = has equal roots, then the value of q is
49/4 | |
4/49 | |
4 | |
¼ |
Question 4 Explanation:
If one root of x2 + px + 12 = 0 is 4, then
42 + 4p + 12 = 0,
From thr above expression
p = -7.
Put this value in the second given equation x2 – 7x + q = 0 has equal roots.
Therefore If the roots are equal then b2 = 4ac
4q = 49
q = 49/4
42 + 4p + 12 = 0,
From thr above expression
p = -7.
Put this value in the second given equation x2 – 7x + q = 0 has equal roots.
Therefore If the roots are equal then b2 = 4ac
4q = 49
q = 49/4
Question 5 |
Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?
7 | |
64 | |
56 | |
Cannot be determined |
Question 5 Explanation:
Let’s take an example to understand a trick that can help us solve this question.
Let us suppose we have a series 1 + 2 + 3 and the middle term is 2.
We are asked for the sum of this series so a simple shortcut that can be used is that:
sum of series = middle term x number of terms
i.e the sum of the supposed series is 2 x 3 = 6
Using the above, we are given by the fourth term which is the middle term of the series. So the sum of the series is 8 x 7 = 56.
Let us suppose we have a series 1 + 2 + 3 and the middle term is 2.
We are asked for the sum of this series so a simple shortcut that can be used is that:
sum of series = middle term x number of terms
i.e the sum of the supposed series is 2 x 3 = 6
Using the above, we are given by the fourth term which is the middle term of the series. So the sum of the series is 8 x 7 = 56.
Question 6 |
What is the value of m which satisfies 3m2 – 21m + 30 < 0?
m < 2 or m > 5 | |
m > 2 c. | |
2 < m < 5; | |
Both a and c |
Question 6 Explanation:
3m2 – 21m + 30 < 0
Dividing by 3
Therefore m2 – 7m + 10 < 0,
Now on solving this quadratic equation
(m – 5)(m – 2) < 0.
Now there are two cases
Either (m – 5) < 0 and (m – 2) > 0 or (m – 2) < 0 and (m –5) > 0.
Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.
Dividing by 3
Therefore m2 – 7m + 10 < 0,
Now on solving this quadratic equation
(m – 5)(m – 2) < 0.
Now there are two cases
Either (m – 5) < 0 and (m – 2) > 0 or (m – 2) < 0 and (m –5) > 0.
Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.
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