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Algebra Level 3 Test 10

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Question 1
When you reverse the digits of the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?
A
5
B
6
C
7
D
8
Question 1 Explanation: 
Let the number be 10x + y so when number is reversed
the number because 10y + x. So, the number increases by 18
Hence (10y + x) - (10x + y) = 9 (y - x) = 18
y - x = 2
So, the possible pairs of (x, y) is (3, 1) (4, 2) (5, 3) (6, 4), (7, 5) (8, 6) (9, 7)
But we want the number other than 13 so, there are 6 possible numbers are there i.e. 24, 35, 46, 57, 68, 79.
So total possible numbers are 6.
Question 2
What are the values of x and y that satisfy both the equations?
(1). 20.7p x 3-1.25q = 8√6/27
(2). 40.3p x 90.2q = 8 x (81)1/5
A
x = 2,y = 5
B
x = 2.5, y = 6
C
x = 3, y = 5
D
x = 5, y = 2
Question 2 Explanation: 
40.3p x 90.2q
= 8 x (81)1/5 (22)0.3p x (32)0.2q
= 23 x (34)1/5
=  23 x 34/5 = 0.6p = 3
P = 5
And 0.4q = 4/5
q = 2
If we put the values of x and y in first equation these values satisfy the first equation also.
So the answer is x = 5, y = 2
Hence, option (d)
Question 3
For which value of k does the following pair of equations yield a unique solution of x such that the solution is positive?
1. x2 –y2 = 0
2. (x – k)2 + y2  = 1
A
2
B
0
C
√2
D
+√2
Question 3 Explanation: 
y2 = x2
2x2 – 2kx + k2 – 1 = 0
D = 0
4k2 = 8k2 – 8
4k2 = 8
k2 = 2 ⇒ k = ±√2 with k = + √2 gives
The equation = 2x2 – 2√2x +1 = 0;
Sum of roots = -b/a = 1/ √2 with k = -√2
The equation is = 2x2 + 2√2x + 1 = 0 root is = -1/√2
As this is in negative form, so we reject this value and the ans would be +√2
Question 4
For a positive integer n, let pn denote the product of the digits of n and sn denote the sum of the digits of n. The number of integers between 10 and 1000 for which pn + sn = n is
A
81
B
16
C
18
D
9
Question 4 Explanation: 
10 < n < 1000
Let n is two digit number.
n = 10a + b ⇒ pn = ab, sn = a + b
Then ab + a + b = 10a + b
⇒ ab = 9a ⇒ b = 9
There are 9 such numbers 19, 29, 33, … 99
Then Let n is three digit number
⇒ n = 100a + 10b + c ⇒ pn = abc, sn= a + b + c
then abc + a + b + c = 100a + 10b + c
⇒ abc = 99a + 9b
bc = 99 + 9b/a
But the maximum value for bc = 81 (when both b & c are 9 i.e.)
And RHS is more than 99. Hence no such number is possible.
Hence option (d).
Question 5
Consider the set S = {2, 3, 4, ……, 2n + 1}, where ‘n’ is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y?
A
0
B
1
C
(1/2n)
D
(n +1)
Question 5 Explanation: 
Sum of the odd integers in the set
S = n/2[2 x 3 +(n-1) x2]
n/2(2n +4) = n x (n+2)
Sum of the even integers in the set S
n/2[2 x 2 +(n-1) x2]
= n/2(2n +2) = n x (n+1)
Therefore the average of the even integers in the set
S = n + 1 , Therefore X – Y = (n + 2) – (n + 1) = 1
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