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Algebra Level 2 Test 5

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Question 1
If log7log5 {√(x + 5) + √x}= 0
A
1
B
0
C
2
D
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.
Question 2
If a + b + c = 0, where a ≠ b ≠ c, then {a2/(2a2 +bc)} + {b2/(2b2 +ac)} +{c2/(2c2 +ab)}
A
zero
B
1
C
–1
D
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.
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
A
12 : 13
B
1/12 : 1/13
C
4 : 9
D
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
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
A
49/4
B
4/49
C
4
D
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
Question 5
Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?
A
7
B
64
C
56
D
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.
Question 6
What is the value of m which satisfies 3m2 – 21m + 30 < 0?
A
m < 2 or m > 5
B
m > 2 c.
C
2 < m < 5;
D
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.
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