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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?

5 | |

6 | |

7 | |

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.

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). 2

(2). 4

(1). 2

^{0.7p }x 3^{-1.25q}= 8âˆš6/27(2). 4

^{0.3p }x 9^{0}^{.2q}= 8 x (81)^{1/5} x = 2,y = 5 | |

x = 2.5, y = 6 | |

x = 3, y = 5 | |

x = 5, y = 2 |

Question 2 Explanation:

4

= 8 x (81)

= 2

=Â 2

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)

^{0.3p }x 9^{0.2q}= 8 x (81)

^{1/5}(2^{2})0^{.3p }x (3^{2})^{0.2q}= 2

^{3}x (3^{4})^{1/5}=Â 2

^{3}x 3^{4/5}= 0.6p = 3P = 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. x

2. (x â€“ k)

1. x

^{2}â€“y^{2}= 02. (x â€“ k)

^{2}+ y^{2 }Â = 1 2 | |

0 | |

âˆš2 | |

+âˆš2 |

Question 3 Explanation:

y

2x

D = 0

4k

4k

k

The equation = 2x

Sum of roots = -b/a = 1/ âˆš2 with k = -âˆš2

The equation is = 2x

As this is in negative form, so we reject this value and the ans would be +âˆš2

^{2}= x^{2}2x

^{2}â€“ 2kx + k^{2}â€“ 1 = 0D = 0

4k

^{2}= 8k^{2}â€“ 84k

^{2}= 8k

^{2}= 2 â‡’ k = Â±âˆš2 with k = + âˆš2 givesThe equation = 2x

^{2}â€“ 2âˆš2x +1 = 0;Sum of roots = -b/a = 1/ âˆš2 with k = -âˆš2

The equation is = 2x

^{2}+ 2âˆš2x + 1 = 0 root is = -1/âˆš2As this is in negative form, so we reject this value and the ans would be +âˆš2

Question 4 |

For a positive integer n, let p

_{n}denote the product of the digits of n and s_{n}denote the sum of the digits of n. The number of integers between 10 and 1000 for which p_{n}+ s_{n}= n is81 | |

16 | |

18 | |

9 |

Question 4 Explanation:

10 < n < 1000

Let n is two digit number.

n = 10a + b â‡’ p

Then ab + a + b = 10a + b

â‡’ ab = 9a â‡’ b = 9

There are 9 such numbers 19, 29, 33, â€¦ 99

Then Let n is three digit number

â‡’ n = 100a + 10b + c â‡’ p

then abc + a + b + c = 100a + 10b + c

â‡’ abc = 99a + 9b

bc = 99 + 9

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).

Let n is two digit number.

n = 10a + b â‡’ p

_{n}= ab, s_{n}= a + bThen ab + a + b = 10a + b

â‡’ ab = 9a â‡’ b = 9

There are 9 such numbers 19, 29, 33, â€¦ 99

Then Let n is three digit number

â‡’ n = 100a + 10b + c â‡’ p

_{n}= abc, s_{n}= a + b + cthen abc + a + b + c = 100a + 10b + c

â‡’ abc = 99a + 9b

bc = 99 + 9

^{b}/_{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, â€¦â€¦, 2n + 1}, where â€˜nâ€™ 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 â€“ Y?

0 | |

1 | |

(1/2n) | |

(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 â€“ Y = (n + 2) â€“ (n + 1) = 1

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 â€“ Y = (n + 2) â€“ (n + 1) = 1

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