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1. In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:

(a) 5, (b) 7, (c) 10, (d) 12, (e) 14

Solution:

Let the number of cows be x and their legs be 4x.

Let the number of chicken be y and their legs be 2x.

Total number of legs = 4x + 2y.

Total number of heads = x + y.

The number of legs was 14 more than twice the number of heads.

Therefore, 2 × (x + y) + 14 = 4x + 2y.

or, 2x + 2y + 14 = 4x + 2y.

or, 2x + 14 = 4x [subtracting 2y from both sides].

or, 14 = 4x – 2x [subtracting 2x from both sides].

or, 14 = 2x.

or, x = 7 [dividing by 2 on both sides].

Therefore, the number of cows = 7.

Answer: (b)

2. The roots of the equation ax² + bx + c = 0 will be reciprocal if:

(a) a = b, (b) a = bc, (c) c = a, (d) c = b, (e) c = ab.

Solution:

Let k be one of the root of the given equation.

According to the problem,

1/k will be the other root of the given equation.

We know that, product of the roots of the equation = c/a.

Therefore, k × ¹/_k = ^c/_a.

or, 1 = ^c/_a.

or, a = c [multiplying a on both sides].

The roots of the equation ax² + bx + c = 0 will be reciprocal if a = c.

Therefore, a = c or c = a.

Answer: (c)



3. If 8 ∙ 2^x = 5^(y+8), then when y = -8, x =

(a) -4, (b) -3, (c) 0, (d) 4, (e) 8

Solution:

8 ∙ 2^x = 5^(y+8).

or, 8 ∙ 2^x = 5^(-8+8).

or, 2^x = 5⁰.

or, 2^x = 1 [anything to the power 0 is 1 then, 5⁰ = 1].

Taking log on both sides,

log 2^x = log 1.

or, x log 2 = 0 [since log 1 = 0].

or, x = ⁰/_{log 2} [dividing log 2 on both sides].

Therefore, x = 0.

Answer: (c)

4. A circle of radius 10 inches has its center at the vertex C of an equilateral triangle ABC and passes through the other two vertices. The side AC extended through C intersects the circle at D. The number of degrees of angle ADB is:

(a) 15, (b) 30, (c) 60, (d) 90, (e) 120

Answer: (b)



5. The expression 1 – ¹/_{(1 + √3)} + ¹/_{(1 - √3)} equals:

(a) 1 - √3, (b) 1, (c) - √3, (d) √3, (e) 1 + √3.


Solution:



6. If x^-1 – 1 is divided by x - 1 the quotient is:

(a) 1, (b) ¹/_(x-1), (c) - ¹/_(x-1), (d) ¹/_x, (e) – ¹/_x

Solution:

x^-1 – 1 is divided by (x – 1)

= x^-1 – 1 × ¹/_(x-1)

= ¹/_{x – 1} × ¹/_(x-1)

= ^{(1 – x}/_x) × ¹/_(x-1)

= - ^{(1 – x)}/_x × ¹/_(x-1)

= – ¹/_x [cancel x – 1 from the numerator and denominator].

Answer: (e)



7. The root(s) of 15/(x² - 4) – 2/(x - 2) = 1 is (are):

(a) -5 and 3, (b) ± 2, (c) 2 only, (d) -3 and 5, (e) 3 only.

Solution:

15/(x² - 4) – 2/(x - 2) = 1.

15/(x² - 2²) – 2/(x - 2) = 1.

15/(x + 2)(x - 2) – 2/(x - 2) = 1.

[15 – 2(x + 2)]/[(x + 2)(x - 2)] = 1.

[15 – 2x - 4] /[(x + 2)(x - 2)] = 1.

[11 - 2x]/[(x + 2)(x - 2)] = 1.

11 – 2x = (x + 2)(x - 2).

11 – 2x = x² - 4.

x² - 4 = 11 – 2x.

x² + 2x – 4 – 11 = 0.

x² + 2x – 15 = 0.

x + 5) (x - 3) = 0.

x + 5 = 0 or, x – 3 = 0

x = -5 or, x = 3.

Therefore, x = -5 and 3.

Answer: (a)

8. A 7-in. pizza costs $8 and a 14-in. pizza costs $20. Tommy says the smaller pizza is a better buy because the larger pizza is twice as big and more than twice as expensive. Do you agree with his reasoning? If not, explain why not.

Solution:

No, a pizza looks like a circle.

And we know that area of a circle depends on the radius.

Area of a circle = πr², here pi (π) is constant.

So, we can clearly say that area of a circle is proportional to the square of the diameter.

Similarly, the cost a pizza depends on its area not its diameter.

9. Suppose that Jody drove 80 miles in 2 hours. Dividing 80 by 2 tells us how many miles Jody drove in each hour. The units for this rate are miles per hour (mi/hr). If we divide 2 by 80 what information would this give us? Give an interpretation of the rate. What units would be used for this rate?

Solution:

We know that, speed = distance/time.

When total distance is divided by total time we get the speed.

Here, speed = distance covered in 1 hour.

According to the question,

Total distance 80 miles divided by total time 2 hours = 80/2.

miles/hour = 40 mile/hour.

Therefore, Jody drove 40 miles in 1 hour.

But if we divide total time by total distance then, we get time taken to cover 1 mile.

Similarly, if we divide 2 by 80 then, we get time taken to cover 1 mile.

Therefore, unit used for this case = hours/mile.

10. 19/6 = 4/27. Which method did you use to determine whether this proportion is true or false?

Solution:

19/6 = 4/27.

Cross multiplication:

19 × 27 = 513.

6 × 4 = 24.

We see that 513 are not equal to 24.

Therefore, we determine 19/6 = 4/27 is not proportion so, the answer is false.

11. On Thursday Mabel handled 90 transactions. Anthony handled 10% more transactions than Mabel, Cal handled 2/3rds of the transactions that Anthony handled, and Jade handled 16 more transactions than Cal. How much transactions did Jade handled?

Solution:

Mabel handled 90 transactions

Anthony handled 10% more transactions than Mabel

Anthony = 90 + 90 × 10%

= 90 + 90 × 0.10

= 90 + 9

= 99

Cal handled 2/3rds of the transactions than Anthony handled

Cal = 2/3 × 99

= 66

Jade handled 16 more transactions than Cal.

Jade = 66 + 16

= 82

Jade handled = 82 transactions.

Answer: 82

12. If a man buys 20 lollipops for $90 and sold them for $2 dollars determine his loss.

Solution:

Cost of 20 lollipops = $90

Sold each lollipop for $2

So he sold 20 lollipops for $(20 × 2) = $40

Loss = $(90 – 40) = $50

Therefore, loss = $ 50.

13 . If C is a whole number, C+1 is a whole number after that. If A is a whole number, what is a whole number before that?

Solution:

We know that the number ‘0’ together with the natural numbers gives us the numbers 0, 1, 2, 3, 4, 5, …………… which are called whole numbers.

If A is a whole number and it is greater than 0 then the whole number before that is A – 1, which will always be a whole number.

Again, when A is a whole number and it is equal to 0 then the whole number before that is A – 1 which means 0 – 1 = -1, which is not a whole number.

Therefore, if A is a whole number then the number before that will not always be a whole number.

14. Solve and leave answer in fraction

- (2-5/4)/9

Solution:

Answer: -1/12


15. Divide the cancelling out all common factors

Solution:

-((1-13/8)/10/24)

Answer: 3/2


16. The annual cost of owning and operating a car, C dollars, is a linear function of the distance, d kilometers, it is driven.

c = md + b

The cost is $4600 for 10 000 km and $9100 for 25 000 km.

(a) Determine the values of m and b.

(b) Write c as a function of d.

Solution:

(a)

c = md + b

The cost is $ 4600 for 10 000 km

4600 = m (10 000) + b

10 000 m + b = 4600

b = -10 000 m + 4600 ------- (i)

The cost is $ 9100 for 25 000 km

9100 = m (25 000) + b

9100 = 25 000 m + b

25 000 m + b = 9100

b = -25 000 m + 9100 --------- (ii)

From equation (i) and (ii) we get;

-10 000 m + 4600 = -25 000 m + 9100

-10 000 m + 25 000 m = 9100 – 4600

15 000 m = 4500

m = 4500/15 000

m = 3/10

m = 0.3

Now, put the value of m = 0.3 in equation (i) we get;

b = -10 000 (0.3) + 4600

b = -3000 + 4600

b = 1600

Answer: m = 0.3 and b = 1600


(b) c = md + b

m = 0.3

b = 1600

c = 0.3d + 1600

Answer: c = 0.3d + 1600

17. Solve using synthetic division or long division(x² + 13x + 40) ÷ (x + 8)

Solution:

We solved the division in both the ways;

(x² + 13x + 40) ÷ (x + 8) by using synthetic division

Quotient = x + 5

Remainder = 0



OR



(x² + 13x + 40) ÷ (x + 8) by using long division

Quotient = x + 5

Remainder = 0

Answer: x + 5



18. Solve using on synthetic division or long division (2x² - 23x + 63) ÷ (x - 7)

Solution:

We solved the division in both the ways;

(2x² - 23x + 63) ÷ (x - 7) by using synthetic division,

Quotient = 2x - 9

Remainder = 0



OR



(2x² - 23x + 63) ÷ (x - 7) by using long division Quotient = 2x - 9

Remainder = 0

Answer: 2x - 9



19. Let f(x) = 4x³ + 7x² – 13x – 3 and g(x) = x + 3. Find f(x)/g(x)

Solution:

f(x) = 4x³ + 7x² – 13x – 3

g(x) = x + 3

f(x)/g(x) = (4x³ + 7x² – 13x – 3)/(x + 3)

We solved the division in both the ways;

(4x³ + 7x² – 13x – 3) ÷ (x + 3) by using synthetic division

Answer:

Quotient = 4x² – 5x + 2

Remainder = -9



OR



(4x³ + 7x² – 13x – 3) ÷ (x + 3) by using long division

Answer:

Quotient = 4x² – 5x + 2

Remainder = -9



20. Let f(x) = 3x³ - 4x – 1 and g(x) = x + 1. Find f(x)/g(x)

Solution:

f(x) = 3x³ - 4x – 1

g(x) = x + 1

f(x)/g(x) = (3x³ - 4x – 1)/(x + 1)

We first need to rearrange all the terms of the dividend according to descending powers of x. The dividend then becomes 3x³ - 4x – 1, with 3 understood as the coefficient of the first term. No x² term is there in the polynomial, but we take a zero as a place holder in the x² position, so the dividend is written as 3x³ + 0x² - 4x – 1.

We solved the division in both the ways;



(3x³ + 0x² - 4x – 1) ÷ (x + 1) by using synthetic division

Quotient: 3x² – 3x – 1

Remainder: 0



OR



No x² term is there in the polynomial, but we take a zero as a place holder in the x² position, so the dividend is written as

3x³ + 0x² - 4x – 1.

(3x³ + 0x² - 4x – 1) ÷ (x + 1) by using long division

Quotient: 3x² – 3x – 1

Remainder: 0

Answer: 3x² – 3x – 1



21. Let f(x) = x⁴ – 8x³ + 16x² – 19 and g(x) = x - 5. Find f(x)/g(x)

Solution:

f(x) = x⁴ – 8x³ + 16x² – 19

g(x) = x – 5

f(x)/g(x) = (x⁴ – 8x³ + 16x² – 19)/(x - 5)

We first need to rearrange all the terms of the dividend according to descending powers of x. The dividend then becomes x⁴ – 8x³ + 16x² – 19, with 1 understood as the coefficient of the first term. No x term is there in the polynomial, but we take a zero as a place holder in the x position, so the dividend is written as x⁴ – 8x³ + 16x² + 0x – 19.

We solved the division in both the ways;

(x⁴ – 8x³ + 16x² + 0x – 19) ÷ (x - 5) by using synthetic division

Quotient = x³ - 3x² + x + 5

Remainder = 6



OR



No x term is there in the polynomial, but we take a zero as a place holder in the x position, so the dividend is written as

x⁴ – 8x³ + 16x² + 0x – 19.

(x⁴ – 8x³ + 16x² + 0x – 19) ÷ (x - 5) by using long division

Quotient = x³ - 3x² + x + 5

Remainder = 6

Answer:

Quotient = x³ - 3x² + x + 5

Remainder = 6

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