Let n be a positive integer such that both 2n+12n+1 and 3n+13n+1 are perfect squ
Let n be a positive integer such that both 2n+12n+1 and 3n+13n+1 are perfect squares. Find all possible values of n.
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. Which is greater than 4?
(a) 5,
(b) -5,
(c) -1/2,
(d) -25.
Solution:
5 greater
. Which is greater than 4?
(a) 5,
(b) -5,
(c) -1/2,
(d) -25.
Solution:
5 greater than 4.
Answer: (a)
2. Which is the smallest?
(a) -1,
(b) -1/2,
(c) 0,
(d) 3.
Solution:
The smallest number is -1.
Answer: (a)
3. Combine terms: 12a + 26b -4b – 16a.
(a) 4a + 22b,
(b) -28a + 30b,
(c) -4a + 22b,
(d) 28a + 30b.
Solution:
12a + 26b -4b – 16a.
= 12a – 16a + 26b – 4b.
= -4a + 22b.
Answer: (c)
4. Simplify: (4 – 5) – (13 – 18 + 2).
(a) -1,
(b) –2,
(c) 1,
(d) 2.
Solution:
(4 – 5) – (13 – 18 + 2).
= -1-(13+2-18).
= -1-(15-18).
= -1-(-3).
= -1+3.
= 2.
Answer: (d)
5. What is |-26|?
(a) -26,
(b) 26,
(c) 0,
(d) 1
Solution:
|-26|
= 26.
Answer: (b)
6. Multiply: (x – 4)(x + 5)
(a) x2 + 5x – 20,
(b) x2 – 4x – 20,
(c) x2 – x – 20,
(d) x2 + x – 20.
Solution:
(x – 4)(x + 5).
= x(x + 5) -4(x + 5).
= x2 + 5x – 4x – 20.
= x2 + x – 20.
Answer: (d)
7. Factor: 5×2 – 15x – 20.
(a) 5(x-4)(x+1),
(b) -2(x-4)(x+5),
(c) -5(x+4)(x-1),
(d) 5(x+4)(x+1).
Solution:
5×2 – 15x – 20.
= 5(x2 – 3x – 4).
= 5(x2 – 4x + x – 4).
= 5{x(x – 4) +1(x – 4)}.
= 5(x-4)(x+1).
Answer: (a).
8. Factor: 3y(x – 3) -2(x – 3).
(a) (x – 3)(x – 3),
(b) (x – 3)2,
(c) (x – 3)(3y – 2),
(d) 3y(x – 3).
Solution:
3y(x – 3) -2(x – 3).
= (x – 3)(3y – 2).
Answer: (c).
9. Solve for x: 2x – y = (3/4)x + 6.
(a) (y + 6)/5,
(b) 4(y + 6)/5,
(c) (y + 6),
(d) 4(y – 6)/5.
Solution:
2x – y = (3/4)x + 6.
or, 2x – (3/4)x = y + 6.
or, (8x -3x)/4 = y + 6.
or, 5x/4 = y + 6.
or, 5x = 4(y + 6).
or, 5x = 4y + 24.
or, x = (4y + 24)/5.
Therefore, x = 4(y + 6)/5.
Answer: (b).
10. Simplify:(4×2 – 2x) – (-5×2 – 8x).
Solution:
(4×2 – 2x) – (-5×2 – 8x)
= 4×2 – 2x + 5×2 + 8x.
= 4×2 + 5×2 – 2x + 8x.
= 9×2 + 6x.
= 3x(3x + 2).
Answer: 3x(3x + 2)
11. Find the value of 3 + 2 • (8 – 3)
(a) 25,
(b) 13,
(c) 17,
(d) 24,
(e) 15.
Solution:
3 + 2 • (8 – 3)
= 3 + 2 (5)
= 3 + 2 × 5
= 3 + 10
= 13
Answer: (d)
12. Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?
Solution:
33/4 ÷ 4 pounds.
= (4 × 3 + 3)/4 ÷ 4 pounds.
= 15/4 ÷ 4 pounds.
= 15/4 × 1/4 pounds.
= 15/16 pounds.
Now we know that, 1 pound = 16 ounces.
Therefore, 15/16 pounds = 15/16 × 16 ounces.
= 15 ounces.
Answer: 15 ounces.
13. Factor: 16w3 – u4w3
Solution:
16w3 – u4w3.
= w3(16 – u4).
= w3(42 – ((u2)2).
= w3(4 + u2)(4 – u2).
= w3(4 + u2)(22 – u2).
= w3(4 + u2)(2 + u)(2 – u).
Answer: w3(4 + u2)(2 + u)(2 – u).
14. Factor: 3x4y3 – 48y3.
Solution:
3x4y3– 48y3.
= 3y3(x4 – 16).
= 3y3[(x2)2 – 42].
= 3y3(x2 + 4)(x2 – 4).
= 3y3(x2 + 4)(x2 – 22).
= 3y3(x2 + 4)(x + 2)(x -2).
Answer: 3y3(x2 + 4)(x + 2)(x -2)
15. What is the radius of a circle that has a circumference of 3.14 meters?
Solution:
Circumference of a circle = 2πr.
Given, circumference = 3.14 meters.
Therefore,
2πr = Circumference of a circle
or, 2πr = 3.14.
or, 2 × 3.14r = 3.14,[Putting the value of pi (π) = 3.14].
or, 6.28r = 3.14.
or, r = 3.14/6.28.
or, r = 0.5.
Answer: 0.5 meter.