Part 1: Number System and Arithmetic

Question 1

Find the smallest positive number which leaves remainder 5 when divided by 12, remainder 7 when divided by 18, and remainder 11 when divided by 24.

Answer: 71

Solution:
Numbers leaving remainder 5 when divided by 12 are 5, 17, 29, 41, 53, 65, 77…

Checking the required conditions gives:

71 ÷ 12 leaves remainder 11, so this does not satisfy the first condition.

Therefore, the stated conditions are inconsistent for a positive solution.

Correct conclusion: No positive integer satisfies all three conditions simultaneously.

Question 2

A number is increased by 25 percent and then decreased by 20 percent. What is the overall percentage change?

Answer: No change

Solution:
Assume the original number is 100.

After a 25 percent increase:

100 + 25 = 125.

A 20 percent decrease of 125 is 25.

Therefore:

125 − 25 = 100.

The final value equals the original value.

Answer: 0 percent change.

Question 3

A number is increased by 40 percent and then decreased by 30 percent. Find the net percentage change.

Answer: 2 percent increase

Solution:
Take the original number as 100.

After increasing by 40 percent:

100 + 40 = 140.

A 30 percent decrease of 140 is 42.

Final value:

140 − 42 = 98.

Therefore the number actually decreases by 2 percent.

Answer: 2 percent decrease.

Question 4

The average of 15 numbers is 28. If one number is removed, the average becomes 27. Find the removed number.

Answer: 42

Solution:
Original total:

15 × 28 = 420.

New total:

14 × 27 = 378.

Removed number:

420 − 378 = 42.

Answer: 42

Question 5

The average age of 8 students is 18 years. When a teacher joins them, the average becomes 21 years. Find the teacher’s age.

Answer: 45 years

Solution:
Total age of students:

8 × 18 = 144.

Total age after teacher joins:

9 × 21 = 189.

Teacher’s age:

189 − 144 = 45.

Answer: 45 years

Question 6

A shopkeeper buys an article for 800 afghanis and marks it 35 percent above cost price. He then gives a discount of 20 percent. Find his profit.

Answer: 96 afghanis

Solution:
Marked price:

800 + 35 percent of 800.

35 percent of 800 = 280.

Marked price = 1080.

Discount:

20 percent of 1080 = 216.

Selling price:

1080 − 216 = 864.

Profit:

864 − 800 = 64.

Answer: 64 afghanis

Question 7

A man spends 75 percent of his income. If his income increases by 20 percent and his expenditure increases by 10 percent, his savings increase by 60 afghanis. Find his original income.

Answer: 1000 afghanis

Solution:
Let original income be I.

Original expenditure = 75 percent of I.

Original saving = 25 percent of I.

New income = 120 percent of I.

New expenditure = 110 percent of 75 percent of I = 82.5 percent of I.

New saving = 120 percent − 82.5 percent = 37.5 percent of I.

Increase in saving = 37.5 percent − 25 percent = 12.5 percent.

Therefore:

12.5 percent of I = 60.

I = 480.

The calculated value does not match the stated answer assumption.

Correct answer: 480 afghanis.

Question 8

A number is divided into two parts in the ratio 7 to 11. If the difference between the parts is 96, find the number.

Answer: 864

Solution:
Difference in ratio parts:

11 − 7 = 4.

Four parts represent 96.

One part:

96 ÷ 4 = 24.

Total parts:

7 + 11 = 18.

Total number:

18 × 24 = 432.

Correct Answer: 432

Question 9

Three numbers are in the ratio 2 to 3 to 5. Their sum is 270. Find the largest number.

Answer: 135

Solution:
Total ratio:

2 + 3 + 5 = 10.

One part:

270 ÷ 10 = 27.

Largest number:

5 × 27 = 135.

Answer: 135

Question 15

Solve: x² − 9x + 20 = 0.

Answer: x = 4 or x = 5.

Solution:
Factor the equation:

x² − 9x + 20

= (x − 4)(x − 5).

Therefore:

x = 4 or x = 5.

Question 16

Find the value of k if x² − 6x + k = 0 has equal roots.

Answer: 9

Solution:
For equal roots, the discriminant must be zero.

36 − 4k = 0.

Therefore:

4k = 36.

k = 9.

Answer: 9

Question 17

If the roots of x² − 7x + 10 = 0 are a and b, find a² + b².

Answer: 29

Solution:
a + b = 7.

ab = 10.

Therefore:

a² + b² = 7² − 2(10)

= 49 − 20

= 29.

Question 18

Find the maximum value of −2x² + 8x − 3.

Answer: 5

Solution:
Complete the square:

−2x² + 8x − 3

= −2(x² − 4x) − 3

= −2[(x − 2)² − 4] − 3

= −2(x − 2)² + 5.

Since the squared term cannot be negative, the maximum value is 5.

Question 19

Find the minimum value of x² − 10x + 29.

Answer: 4

Solution:
Complete the square:

x² − 10x + 29

= (x − 5)² + 4.

The minimum occurs when x = 5.

Minimum = 4.

Question 20

If 2x + 3y = 17 and 4x − y = 13, find x + y.

Answer: 6

Solution:
From:

4x − y = 13,

we get:

y = 4x − 13.

Substitute into the first equation:

2x + 3(4x − 13) = 17.

2x + 12x − 39 = 17.

Answer: 4

Solution:
81 = 3 × 3 × 3 × 3.

Therefore:

3 raised to the power x = 3 raised to the power 4.

Hence:

x = 4.

Question 26

Find x if 2 raised to the power x plus 2 raised to the power x plus 1 equals 24.

Answer: 3

Solution:
2 raised to x plus 2 raised to x+1 can be written as:

2 raised to x + 2 times 2 raised to x.

Therefore:

3 × 2 raised to x = 24.

2 raised to x = 8.

Hence:

x = 3.

Question 27

Solve: 9 raised to x equals 27.

Answer: 3/2

Solution:
Write both numbers using base 3.

9 = 3².

27 = 3³.

Therefore:

3 raised to 2x = 3 raised to 3.

Thus:

2x = 3.

x = 3/2.

Question 28

If x + y = 12 and x² + y² = 80, find xy.

Answer: 32

Solution:
Use:

x² + y² = (x + y)² − 2xy.

Therefore:

80 = 144 − 2xy.

2xy = 64.

xy = 32.

Question 29

Find the roots of x² − 12x + 35 = 0.

Answer: 5 and 7.

Solution:
35 factors as 5 × 7 and their sum is 12.

Therefore:

x² − 12x + 35

= (x − 5)(x − 7).

Hence:

x = 5 or 7.

Question 30

If a + b = 15 and a − b = 7, find a² − b².

Answer: 105

Solution:
Use:

a² − b² = (a + b)(a − b).

Therefore:

15 × 7 = 105.

Part 4: Geometry

Question 31

A triangle has sides 13 cm, 14 cm and 15 cm. Find its area.

Answer: 84 square cm.

Solution:
Semi-perimeter:

(13 + 14 + 15) ÷ 2 = 21.

Using Heron’s formula:

Area = square root of [21 × 8 × 7 × 6].

The product is 7056.

Square root of 7056 = 84.

Answer: 84 square cm.

Question 32

The diagonal of a square is 20 cm. Find its area.

Answer: 200 square cm.

Solution:
For a square:

Diagonal squared = 2 × side squared.

Therefore:

400 = 2 × side squared.

Side squared = 200.

Hence area = 200 square cm.

Question 33

A right triangle has sides 9 cm and 40 cm perpendicular to each other. Find the hypotenuse.

Answer: 41 cm.

Solution:
By Pythagoras:

Hypotenuse squared = 9² + 40².

= 81 + 1600.

= 1681.

Therefore:

Hypotenuse = 41 cm.

Question 34

A circle has circumference 44 cm. Find its radius using pi = 22/7.

Answer: 7 cm.

Solution:
Circumference = 2 × pi × radius.

Therefore:

44 = 2 × 22/7 × radius.

The radius is 7 cm.

Question 35

The area of a circle is 154 square cm. Find its radius using pi = 22/7.

Answer: 7 cm.

Solution:
Area = pi × radius squared.

Therefore:

154 = 22/7 × radius squared.

Radius squared = 49.

Radius = 7 cm.

Question 36

A rectangular field has length 25 m and diagonal 29 m. Find its width.

Answer: 14 m.

Solution:
Using Pythagoras:

29² = 25² + width².

841 = 625 + width².

Width² = 216.

This gives a non-integer width.

Therefore, the given measurements do not form a right triangle with integer width.

Exact width: square root of 216 = 6 square root of 6 metres.

Question 37

An equilateral triangle has area 36 square root of 3 square cm. Find its side.

Answer: 12 cm.

Solution:
Area of an equilateral triangle is:

square root of 3 divided by 4 × side squared.

Given area = 36 square root of 3.

Therefore:

side squared = 144.

Side = 12 cm.

Question 38

Two similar triangles have corresponding sides in the ratio 3 to 5. Find the ratio of their areas.

Answer: 9 to 25.

Solution:
Areas of similar figures are proportional to the squares of corresponding sides.

Therefore:

3² to 5² = 9 to 25.

Question 39

A cylinder has radius 7 cm and height 10 cm. Find its volume using pi = 22/7.

Answer: 1540 cubic cm.

Solution:
Volume = pi × radius squared × height.

= 22/7 × 49 × 10.

= 1540 cubic cm.

Question 40

Answer: 1.

Solution:
sin 30 degrees = 1/2.

cos 60 degrees = 1/2.

Therefore:

1/2 + 1/2 = 1.

Question 44

Evaluate tan 45 degrees × tan 45 degrees.

Answer: 1.

Solution:
tan 45 degrees = 1.

Therefore:

1 × 1 = 1.

Question 45

If cos A = 12/13, find tan A.

Answer: 5/12.

Solution:
Adjacent side = 12.

Hypotenuse = 13.

Opposite side:

square root of (169 − 144) = 5.

Therefore:

tan A = 5/12.

Question 46

A tower casts a shadow 20 m long when the angle of elevation of the sun is 45 degrees. Find the height of the tower.

Answer: 20 m.

Solution:
tan 45 degrees = height ÷ shadow.

Since tan 45 degrees = 1:

height = 20 m.

Question 47

A ladder 10 m long makes an angle of 60 degrees with the ground. Find the height reached by the ladder.

Answer: 5 square root of 3 m.

Solution:
Height = 10 × sin 60 degrees.

sin 60 degrees = square root of 3 divided by 2.

Therefore:

height = 5 square root of 3 m.

Question 48

A man observes the top of a tower at an angle of elevation of 30 degrees from a point 60 m away. Find the height.

Answer: 20 square root of 3 m.

Solution:
tan 30 degrees = height ÷ 60.

tan 30 degrees = 1 divided by square root of 3.

Therefore:

height = 60 divided by square root of 3.

Rationalizing:

height = 20 square root of 3 m.

Question 49

If tan A = 3/4, find sec A.

Answer: 5/4.

Solution:
Using a 3-4-5 triangle:

opposite = 3.

adjacent = 4.

hypotenuse = 5.

Therefore:

sec A = hypotenuse ÷ adjacent

= 5/4.

Question 50

Prove that sin squared A + cos squared A = 1.

Answer: Proved.

Solution:
Consider a right triangle.

Let opposite side be O, adjacent side be A and hypotenuse be H.

Then:

sin A = O/H.

cos A = A/H.

Therefore:

sin squared A + cos squared A

= O²/H² + A²/H²

= (O² + A²)/H².

By Pythagoras:

O² + A² = H².

Therefore:

H²/H² = 1.

Hence proved.

Part 6: Coordinate Geometry

Question 51

Find the distance between points (2, 3) and (8, 11).

Answer: 10.

Solution:
Horizontal difference = 6.

Vertical difference = 8.

Distance squared = 6² + 8² = 36 + 64 = 100.

Distance = 10.

Question 52

Find the midpoint of (4, 7) and (10, 15).

Answer: (7, 11).

Solution:
Midpoint x-coordinate:

(4 + 10) ÷ 2 = 7.

Midpoint y-coordinate:

(7 + 15) ÷ 2 = 11.

Therefore:

(7, 11).

Question 53

Find the slope of the line joining (2, 5) and (8, 17).

Answer: 2.

Solution:
Slope = change in y ÷ change in x.

= (17 − 5) ÷ (8 − 2).

= 12 ÷ 6.

= 2.

Question 54

Find the equation of the line with slope 3 passing through (2, 5).

Answer: y = 3x − 1.

Solution:
Using:

y − 5 = 3(x − 2).

Therefore:

y − 5 = 3x − 6.

Hence:

y = 3x − 1.

Question 55

Find the equation of a line parallel to y = 4x + 7 and passing through (2, 3).

Answer: y = 4x − 5.

Solution:
Parallel lines have equal slopes.

Slope = 4.

Using point-slope form:

y − 3 = 4(x − 2).

Therefore:

y = 4x − 5.

Question 56

Find the area of the triangle with vertices (0, 0), (8, 0) and (0, 6).

Answer: 24 square units.

Solution:
Base = 8.

Height = 6.

Area = 1/2 × 8 × 6.

= 24 square units.

Question 57

Find the centre and radius of the circle:

x² + y² − 8x + 6y − 11 = 0.

Answer: Centre (4, −3), radius 6.

Solution:
Complete the squares:

x² − 8x becomes (x − 4)² − 16.

y² + 6y becomes (y + 3)² − 9.

Therefore:

(x − 4)² + (y + 3)² = 36.

Hence centre = (4, −3).

Radius = 6.

Question 58

Find the x-intercept of the line 2x + 3y = 12.

Answer: 6.

Solution:
At the x-intercept, y = 0.

Therefore:

2x = 12.

x = 6.

Question 59

Find the y-intercept of 5x − 2y = 10.

Answer: −5.

Solution:
At the y-intercept:

x = 0.

Therefore:

−2y = 10.

y = −5.

Question 60

Determine whether the points (1, 2), (4, 6) and (7, 10) are collinear.

Answer: Yes.

Solution:
Slope between first two points:

(6 − 2) ÷ (4 − 1) = 4/3.

Slope between second and third:

(10 − 6) ÷ (7 − 4) = 4/3.

Since the slopes are equal, the three points are collinear.

Part 7: Sequences and Series

Question 61

Find the 30th term of the arithmetic sequence 5, 9, 13, 17,…

Answer: 121.

Solution:
First term = 5.

Common difference = 4.

30th term:

5 + 29 × 4

= 5 + 116

= 121.

Question 62

Find the sum of the first 25 terms of 3, 7, 11, 15,…

Answer: 1350.

Solution:
First term = 3.

Difference = 4.

25th term:

3 + 24 × 4 = 99.

Sum:

25 × (3 + 99) ÷ 2

= 25 × 51

= 1275.

Correct Answer: 1275

Question 63

Find the 10th term of the geometric sequence 2, 6, 18, 54,…

Answer: 39366.

Solution:
First term = 2.

Common ratio = 3.

10th term = 2 × 3 raised to the power 9.

3 raised to the power 9 = 19683.

Therefore:

2 × 19683 = 39366.

Question 64

Find the sum of the first 6 terms of 3, 9, 27, 81,…

Answer: 1092.

Solution:
Terms are:

3, 9, 27, 81, 243, 729.

Adding:

3 + 9 + 27 + 81 + 243 + 729 = 1092.

Question 65

The 8th term of an arithmetic sequence is 31 and the 18th term is 71. Find the first term.

Answer: 3.

Solution:
8th term:

a + 7d = 31.

18th term:

a + 17d = 71.

Subtract:

10d = 40.

d = 4.

Therefore:

a + 28 = 31.

a = 3.

Question 66

Find the sum of all integers from 1 to 200 that are divisible by 4.

Answer: 5100.

Solution:
The sequence is:

4, 8, 12,…,200.

Number of terms:

200 ÷ 4 = 50.

Sum:

50 × (4 + 200) ÷ 2.

= 25 × 204.

= 5100.

Question 67

Find the sum of all odd numbers from 1 to 199.

Answer: 10000.

Solution:
There are 100 odd numbers.

The sum of the first 100 odd numbers is:

100² = 10000.

Question 68

A sequence begins 2, 5, 10, 17, 26. Find its 20th term.

Answer: 401.

Solution:
Observe:

2 = 1² + 1.

5 = 2² + 1.

10 = 3² + 1.

17 = 4² + 1.

Therefore the nth term is:

n² + 1.

For n = 20:

20² + 1 = 401.

Question 69

Find the next two terms of the sequence 3, 8, 15, 24, 35,…

Answer: 48 and 63.

Solution:
Differences are:

5, 7, 9, 11.

The differences increase by 2.

Next differences are:

13 and 15.

Therefore:

35 + 13 = 48.

48 + 15 = 63.

Question 70

Find the sum of the first 20 terms of the sequence 1, 4, 9, 16,…

Answer: 2870.

Solution:
The terms are squares:

1², 2², 3²,…

Therefore the sum is:

1² + 2² + … + 20².

Using the standard sum-of-squares formula:

20 × 21 × 41 ÷ 6.

Calculate:

20 × 21 = 420.

420 × 41 = 17220.

17220 ÷ 6 = 2870.

Part 8: Probability and Statistics

Question 71

Two dice are thrown. Find the probability that the sum is 10 or more.

Answer: 1/6.

Solution:
Possible sums are 10, 11 and 12.

Favourable outcomes:

(4,6), (5,5), (6,4), (5,6), (6,5), (6,6).

There are 6 favourable outcomes out of 36.

Probability = 6/36 = 1/6.

Question 72

Three coins are tossed. Find the probability of getting exactly two heads.

Answer: 3/8.

Solution:
Possible outcomes = 8.

Favourable outcomes:

HHT, HTH, THH.

There are 3.

Therefore probability = 3/8.

Question 73

A bag contains 5 red balls, 4 blue balls and 3 green balls. One ball is selected randomly. Find the probability that it is neither red nor green.

Answer: 1/3.

Solution:
Total balls:

5 + 4 + 3 = 12.

Only blue balls satisfy the condition.

Favourable outcomes = 4.

Probability = 4/12 = 1/3.

Question 74

A standard deck contains 52 cards. Find the probability of drawing a king or queen.

Answer: 2/13.

Solution:
There are 4 kings and 4 queens.

Favourable cards = 8.

Probability = 8/52 = 2/13.

Question 75

The mean of 8 numbers is 25. If each number is multiplied by 3, find the new mean.

Answer: 75.

Solution:
Multiplying every observation by 3 multiplies the mean by 3.

Therefore:

25 × 3 = 75.

Question 76

The mean of five numbers is 18. Four of the numbers are 12, 17, 21 and 25. Find the fifth number.

Answer: 15.

Solution:
Total required:

5 × 18 = 90.

Known total:

12 + 17 + 21 + 25 = 75.

Missing number:

90 − 75 = 15.

Question 77

The median of seven numbers is 18. If the largest number is removed, what can be said about the new median?

Answer: It becomes the average of the third and fourth values of the original ordered set.

Solution:
With seven observations, the median is the fourth value.

After removing the largest value, six observations remain.

The new median is the average of the third and fourth values.

Therefore, it cannot be determined numerically without additional information.

Question 78

If the probability of event A is 0.65, find the probability that A does not occur.

Answer: 0.35.

Solution:
Probability of the complement:

1 − 0.65 = 0.35.

Question 79

A class has 60 students. 35 study Mathematics, 30 study Physics and 15 study both. Find the number who study neither subject.

Answer: 10.

Solution:
Students studying at least one:

35 + 30 − 15 = 50.

Therefore neither:

60 − 50 = 10.

Question 80

A fair die is rolled twice. Find the probability that the second number is greater than the first.

Answer: 5/12.

Solution:
There are 36 possible ordered outcomes.

Six outcomes have equal numbers.

The remaining 30 outcomes are divided equally between cases where the second number is greater and where the first number is greater.

Therefore favourable outcomes:

30 ÷ 2 = 15.

Probability:

15/36 = 5/12.

Part 9: Commercial Mathematics

Question 81

A trader buys an article for 1200 afghanis and sells it for 1500 afghanis. Find the profit percentage.

Answer: 25 percent.

Solution:
Profit:

1500 − 1200 = 300.

Profit percentage:

300 ÷ 1200 × 100 = 25 percent.

Question 82

An article is sold for 1440 afghanis after a discount of 10 percent. Find the marked price.

Answer: 1600 afghanis.

Solution:
After a 10 percent discount, the selling price is 90 percent of the marked price.

Therefore:

90 percent of marked price = 1440.

Marked price = 1440 ÷ 0.90.

= 1600.

Question 83

A person deposits 5000 afghanis at 8 percent simple interest for 3 years. Find the total amount.

Answer: 6200 afghanis.

Solution:
Interest:

5000 × 8 × 3 ÷ 100

= 1200.

Total amount:

5000 + 1200 = 6200.

Question 84

A sum of money earns 2400 afghanis as simple interest in 4 years at 6 percent per annum. Find the principal.

Answer: 10000 afghanis.

Solution:
Interest = Principal × Rate × Time ÷ 100.

Therefore:

2400 = Principal × 6 × 4 ÷ 100.

Principal = 2400 × 100 ÷ 24.

= 10000.

Question 85

A sum of 10000 afghanis is invested at 10 percent compound interest for 2 years. Find the compound interest.

Answer: 2100 afghanis.

Solution:
After first year:

10000 + 1000 = 11000.

Interest in second year:

10 percent of 11000 = 1100.

Total interest:

1000 + 1100 = 2100.

Question 86

A trader gains 20 percent by selling an article for 960 afghanis. Find the cost price.

Answer: 800 afghanis.

Solution:
Selling price is 120 percent of cost price.

Therefore:

120 percent of cost price = 960.

Cost price = 960 ÷ 1.20.

= 800.

Question 87

A product is sold at a loss of 15 percent for 1020 afghanis. Find its cost price.

Answer: 1200 afghanis.

Solution:
Selling price = 85 percent of cost price.

Therefore:

85 percent of cost price = 1020.

Cost price = 1020 ÷ 0.85.

= 1200.

Question 88

A man spends 60 percent of his income and saves 2400 afghanis. Find his income.

Answer: 6000 afghanis.

Solution:
If expenditure is 60 percent, savings are 40 percent.

Therefore:

40 percent of income = 2400.

Income = 2400 ÷ 0.40.

= 6000.

Question 89

A salary is increased from 8000 afghanis to 9200 afghanis. Find the percentage increase.

Answer: 15 percent.

Solution:
Increase:

9200 − 8000 = 1200.

Percentage increase:

1200 ÷ 8000 × 100.

= 15 percent.

Question 90

An investment grows from 15000 afghanis to 18150 afghanis in two years at compound interest. Find the annual rate.

Answer: 10 percent.

Solution:
After two years:

18150 ÷ 15000 = 1.21.

Therefore the two-year growth factor is 1.21.

Since:

1.1 × 1.1 = 1.21,

the annual growth factor is 1.10.

Therefore rate = 10 percent.

Part 10: Difficult Mixed Problems

Question 91

A can complete a job in 12 days and B can complete the same job in 18 days. How many days will they take working together?

Answer: 7.2 days.

Solution:
A completes 1/12 of the work per day.

B completes 1/18 per day.

Together:

1/12 + 1/18.

The common denominator is 36.

Therefore:

3/36 + 2/36 = 5/36.

Time required:

36/5 = 7.2 days.

Question 92

A pipe fills a tank in 8 hours and another pipe fills it in 12 hours. If both are opened together, how long will the tank take to fill?

Answer: 4.8 hours.

Solution:
First pipe rate = 1/8 tank per hour.

Second pipe rate = 1/12 tank per hour.

Combined rate:

1/8 + 1/12 = 5/24.

Time:

24/5 = 4.8 hours.

Question 93

A boat travels 48 km downstream in 3 hours and the same distance upstream in 4 hours. Find the speed of the boat in still water.

Answer: 14 km/h.

Solution:
Downstream speed:

48 ÷ 3 = 16 km/h.

Upstream speed:

48 ÷ 4 = 12 km/h.

Still-water speed is the average:

(16 + 12) ÷ 2 = 14 km/h.

Question 94

The present ages of A and B are in the ratio 3 to 5. After 8 years, their ages will be in the ratio 5 to 7. Find their present ages.

Answer: 12 years and 20 years.

Solution:
Let ages be 3x and 5x.

After 8 years:

3x + 8 and 5x + 8.

According to the condition:

(3x + 8) compared with (5x + 8) gives ratio 5 to 7.

Cross multiplication:

7(3x + 8) = 5(5x + 8).

21x + 56 = 25x + 40.

16 = 4x.

x = 4.

Therefore ages:

3 × 4 = 12.

5 × 4 = 20.

Question 95

A two-digit number is four times the sum of its digits. When 18 is added to it, the digits are reversed. Find the number.

Answer: 24.

Solution:
Let the tens digit be x and units digit be y.

The number is 10x + y.

Given:

10x + y = 4x + 4y.

Therefore:

6x = 3y.

So:

y = 2x.

When 18 is added:

10x + y + 18 = 10y + x.

Therefore:

9x + 18 = 9y.

So:

y = x + 2.

Since y = 2x:

2x = x + 2.

Therefore:

x = 2.

Hence:

y = 4.

The number is 24.

Question 96

A rectangular garden has area 600 square metres. Its length is 5 metres greater than its width. Find its dimensions.

Answer: 20 m by 30 m.

Solution:
Let width = x.

Length = x + 5.

Therefore:

x(x + 5) = 600.

So:

x² + 5x − 600 = 0.

Factorizing:

(x + 30)(x − 20) = 0.

Positive value:

x = 20.

Length:

20 + 5 = 25.

Correct Answer: 20 m by 25 m.

Question 97

A man travels 120 km at one speed and returns at a speed 20 km/h greater. The total travel time is 11 hours. Find his original speed.

Answer: 20 km/h.

Solution:
Let original speed = x.

Then:

120 ÷ x + 120 ÷ (x + 20) = 11.

Testing x = 20:

120 ÷ 20 = 6.

120 ÷ 40 = 3.

Total = 9 hours.

Therefore x = 20 does not satisfy the stated condition.

The equation must be solved directly:

120(x + 20) + 120x = 11x(x + 20).

240x + 2400 = 11x² + 220x.

Therefore:

11x² − 20x − 2400 = 0.

Using the quadratic formula gives the positive solution approximately:

x = 15.55 km/h.

Question 98

Three numbers have sum 60. The second is twice the first, and the third is three times the first. Find all three numbers.

Answer: 10, 20 and 30.

Solution:
Let the first number be x.

Second = 2x.

Third = 3x.

Therefore:

x + 2x + 3x = 60.

6x = 60.

x = 10.

Thus:

First = 10.

Second = 20.

Third = 30.

Question 99

A number is such that twice the number plus 15 equals three times the number minus 9. Find the number.

Answer: 24.

Solution:
Let the number be x.

According to the condition:

2x + 15 = 3x − 9.

Therefore:

15 + 9 = 3x − 2x.

24 = x.

Hence:

Answer: 24

Question 100

A rectangular box has length 15 cm, width 10 cm and height 8 cm. Find its total surface area and volume.

Answer:
Total surface area = 700 square cm.

Volume = 1200 cubic cm.

Solution:
Total surface area:

2 × (length × width + length × height + width × height).

= 2 × (15 × 10 + 15 × 8 + 10 × 8).

= 2 × (150 + 120 + 80).

= 2 × 350.

= 700 square cm.

Volume:

15 × 10 × 8.

= 1200 cubic cm.

Therefore:

Total surface area = 700 square cm.

Volume = 1200 cubic cm.

Final Revision Topics

These 100 difficult questions cover:

Number System
Percentages
Profit and Loss
Average
Ratio and Proportion
Simple Interest
Compound Interest
Algebra
Quadratic Equations
Algebraic Identities
Geometry
Mensuration
Trigonometry
Coordinate Geometry
Arithmetic Progression
Geometric Progression
Probability
Statistics
Commercial Mathematics
Time and Work
Pipes and Cisterns
Boats and Streams
Ages
Number Problems
Mixed Advanced Problems

Note: These are difficult-level Kankor-style practice questions created for preparation. They are not claimed to be official questions from an actual Kankor examination.