Grade 7 MTAP 2015 Questions with Solutions Part 4

This is the fourth part of the Metrobank-MTAP Math Challenge Questions for Grade 7. In this post, we discuss the solutions to number 21-30. You can also read the solutions to 1-10, 11-20, and 21 – 30.

31.) If n is a prime number not equal to 2, which of the following can be a prime:

n + 3, 2^n - 1, n^2 + 1.

Solution

All prime numbers not equal to 2 are odd. If n is odd, then n + 3 and n^2 + 1 are both even, while 2^n - 1 is odd. Therefore, only 2^n - 1 can be a prime.

Answer: 2^n - 1

32.) What must be added to the product of 2x - 1 and 4x +5 to obtain 4x^2 + 8x + 1

Solution

The product of 2x - 1 and 4x + 5 is (2x - 1)(4x + 5) = 8x^2 + 6x - 5.
Now, we subtract the result from 4x^2 + 8x + 1.

4x^2 + 8x + 1 - (8x^2 + 6x - 5) = 4x^2 + 8x + 1 - 8x^2 - 6x + 5 = -4x^2 + 2x + 6

Answer: -4x^2 + 2x + 6

33.) If the greatest common divisor of two numbers is 12 and their least common multiple is 180, what are the possible pairs of numbers?

Solution:

The product of the gcd and lcm of two numbers is equal to the product of the numbers. So, we need to examine the factors of 12(180) = 2160 where the gcd is 12 and lcm is 180. We only have two pairs of such numbers, 12 and 180 and 36 and 60.

Answer: {12, 180} and {36, 60}

34.) Given a positive integer n, arrange the following from least to greatest:

\sqrt{n} + 1, \sqrt{n + 1}, n + 1

Solution

Squaring all the expressions, it is easy to see that

n + 1 > \sqrt{n} + 1 > \sqrt{n + 1}.

The details are left as an exercise.

Answer: n + 1 > \sqrt{n} + 1 > \sqrt{n + 1}.

35.) What is the largest integer which is always a factor on the sum of three consecutive even integers?

Solution

Let 2n, 2n + 21, and 2n + 2 be three consecutive even integers. Their sum is 6n + 6 = 6(n + 1). Therefore, 6 is always a factor of their sum.

Answer: 6

36.) If the sum of the first positive 100 integers is 5050, what is the sum of the next 100 integers?

Solution: The next 100 integers are 101, 102, 103, …, 200. That is, we add 100 to each one of the first 100 integers. In effect, we added (100) (100) + 5050 = 15050.

Answer: 15050

37.) With what polynomial must3x^4 - 3x^3 + 5x^2 + 2x be divided to get a quotient 3x^2 +2 and a remainder of 4x - 2?

Solution

Consider dividing 12 by 7. We get a quotient of 1 and a remainder of 5. From this, we can form the equation, 12/7 = 1 + 5/7. In general, if we divide a/b and get a quotient q and a remainder of r, we can have the equation

a/b = q + r/q (1).

In the equation above, we are looking for the value for b, the divisor. Therefore, multiplying the equation in (1) by q, we get a = bq + r, giving us  b = (a - r)/q (*). If we leta = 3x^4 - 3x^3 + 5x^2 + 2x, q = 3x^2 + 2, and r = 4x - 2, substituting the values in (*), we get x^2 - x + 1.

Answer: x^2 - x + 1

38.) If 7n leaves a remainder of 8 when divided by 9, what is the remainder when 5n is divided by 9?

Solution

The modular inverse of 7 \mod 9 is equal to 1 since (7)(4) \equiv 1 (\mod 9). This means that \frac{1}{7} \equiv 4 \mod 9. Multiplying by 8, we have \frac{8}{7} \equiv 5 \mod 9. Therefore, 5n = 25 \equiv 7 \mod 9.

Answer: 7

39.) Find the solution set: |3 - x| = 11.

Solution

3 - x = 11, -x = 8, x = -8
-(3 - x) = 11, x - 3 = 11, x = 14.

Answer: -8 or 14

40.) Compute: \left ( 1 - \frac{2}{3} \right ) \left ( 1 - \frac{2}{5} \right ) \left ( 1 - \frac{2}{7} \right ) \left ( 1 - \frac{2}{9} \right ) \left ( 1 - \frac{2}{75} \right )

Solution

\left ( 1 - \dfrac{2}{3} \right ) \left ( 1 - \dfrac{2}{5} \right ) \left ( 1 - \dfrac{2}{7} \right ) \left ( 1 - \dfrac{2}{9} \right ) \left ( 1 - \dfrac{2}{75} \right )

= \left (\dfrac{1}{3} \right ) \left (\dfrac{3}{5} \right ) \left (\dfrac{5}{7} \right )\left (\dfrac{7}{9} \right ) \left (\dfrac{73}{75} \right )

= \dfrac{73}{(9)(75)}

= \dfrac{73}{675}

Answer: \dfrac{73}{675}

17 thoughts on “Grade 7 MTAP 2015 Questions with Solutions Part 4”

    1. Actually, this can be done easily using modulo division. I just want to have a solution accessible to high school students. As of now, I have no time to go back to it yet. Still typing questions from 2015. I will complete the solutions once I have uploaded everything. I’ll try to look at your solution later. Thank you.

  1. This has a lot of grammatical mistakes (or maybe the test paper has. I don’t know, it’s not the same) so I was confused. All in all, this was very very helpful and thank you for providing this info. 🙂

  2. Hello. I’m a student from A science school. I’m sorry to say but the answer in no.35 is 6 coz it says that “three consecutive even integers”, not just integers. Try to solve it again.

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