Mathematics Classes VI-X for School Assistants and TET Paper 2A · Chapter 14

Mixed Problem Bank: Classes VI to X

What to remember

Read the question for what is asked. Write the given data, the unknown and the formula before calculating. Check units (cm, m, km/h, m/s) first.
Use short checks. Substitute the answer back, check that a probability is between 0 and 1, and check that a length is positive.
Link topics. A good problem often joins two ideas, for example Pythagoras with area, or an AP with a word problem. Know the formula sheets and apply them without delay.

A. Arithmetic problems

Ratio. Divide 64 in the ratio 3 : 5. Total parts = 8. One part = 8. Shares are 24 and 40.

Percentage and discount. A shirt is marked at 800. A discount of 15% gives 800 × 85/100 = 680. Two successive discounts of 10% and 20% give 800 × 0.9 × 0.8 = 576. This equals one discount of 28%, not 30%.

Profit and loss. An article bought for 400 is sold at a profit of 25%. SP = 400 × 125/100 = 500. If sold for 340, loss = 60 and loss% = 60/400 × 100 = 15%.

Simple and compound interest. SI on 5000 at 8% per year for 3 years = 5000 × 8 × 3 / 100 = 1200. CI on 10000 at 10% for 2 years: A = 10000 × 1.1 × 1.1 = 12100, so CI = 2100. SI for the same case would be 2000. The difference between CI and SI for 2 years is P (R/100)² = 10000 × 0.01 = 100.

Time, speed and distance. A train 120 m long runs at 54 km/h. Speed = 54 × 5/18 = 15 m/s. It crosses a pole in 120/15 = 8 s. To cross a platform 180 m long, the distance is 300 m and time = 20 s.

Time and work. A does a job in 12 days and B in 6 days. In one day together they do 1/12 + 1/6 = 3/12 = 1/4. They finish in 4 days.

HCF and LCM. For 36 and 48: HCF = 12, LCM = 144. Check: 12 × 144 = 1728 = 36 × 48.

Average. The mean of 12, 15, 18, 20, 25 is 90/5 = 18. If the average of 5 numbers is 18 and one number 25 is replaced by 15, the new average is (90 − 10)/5 = 16.

B. Algebra problems

Quadratic. x² − 5x + 6 = 0 factorises to (x − 2)(x − 3) = 0, so x = 2 or 3. Two numbers have sum 15 and product 56: they are roots of x² − 15x + 56 = 0, which gives 7 and 8.

Discriminant. For 2x² − 4x + 3 = 0, D = 16 − 24 = −8 < 0, so there are no real roots. For x² − 6x + 9 = 0, D = 0, so equal roots x = 3.

Linear pair. x + y = 10 and x − y = 2. Adding: 2x = 12, x = 6; y = 4.

Age problem. A father is 3 times as old as his son. After 5 years the sum of their ages is 70. Let the son be x years. Then (3x + 5) + (x + 5) = 70, so 4x = 60 and x = 15. Father = 45.

AP. For 3, 7, 11, …: d = 4. 10th term = 3 + 9 × 4 = 39. Sum of 10 terms = 10/2 × (3 + 39) = 210.

Identities. If a + b = 7 and ab = 10, then a² + b² = (a + b)² − 2ab = 49 − 20 = 29. Also a − b = √(49 − 40) = 3 (taking a > b), so a = 5, b = 2.

Remainder theorem. The remainder when p(x) = x³ − 2x² + x − 5 is divided by (x − 2) is p(2) = 8 − 8 + 2 − 5 = −3.

C. Geometry problems

Pythagoras. A right triangle has sides 6 cm and 8 cm. Hypotenuse = √(36 + 64) = 10 cm. Triples to remember: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25).

Angle sum. Angles in the ratio 2 : 3 : 4 share 180°. One part = 20°, so the angles are 40°, 60°, 80°.

Polygon. Each interior angle of a regular hexagon = (6 − 2) × 180/6 = 120°. Each exterior angle of a regular polygon with 10 sides = 36°.

BPT. In triangle ABC, DE is parallel to BC, D on AB and E on AC. If AD = 3, DB = 6 and AE = 2, then EC = 4, because AD/DB = AE/EC.

Similar triangles. Two similar triangles have sides in the ratio 2 : 3. Their areas are in the ratio 4 : 9. If the smaller area is 20 cm², the larger is 45 cm².

Circle. A tangent from a point 13 cm from the centre of a circle of radius 5 cm has length √(169 − 25) = 12 cm. A chord of length 8 cm in a circle of radius 5 cm is at distance √(25 − 16) = 3 cm from the centre.

D. Mensuration problems

Problem	Working	Answer
Circle radius 7 cm: area and circumference	πr² = 22/7 × 49; 2πr = 2 × 22/7 × 7	154 cm²; 44 cm
Cuboid 5 × 4 × 3 cm: volume and TSA	5 × 4 × 3; 2(20 + 12 + 15)	60 cm³; 94 cm²
Cylinder r = 7, h = 10 cm: volume	22/7 × 49 × 10	1540 cm³
Cone r = 3, h = 4 cm: slant height, CSA, volume	l = √(9 + 16) = 5; π × 3 × 5; ⅓ π × 9 × 4	5 cm; 15π cm²; 12π cm³
Sphere r = 3 cm: volume and surface area	4/3 π × 27; 4π × 9	36π cm³; 36π cm²
Cube of side 5 cm: diagonal	5√3	5√3 cm

Melting and recasting. A solid metal sphere of radius 3 cm is melted and made into small spheres of radius 1 cm. Number = (3³)/(1³) = 27, because the volume of a sphere is proportional to r³.

Rate problem. A rectangular room 5 m by 4 m needs a carpet. Cost at 120 per m² = 20 × 120 = 2400.

E. Trigonometry and coordinate problems

Height and distance. A tower stands on level ground. From a point 30 m from its foot the angle of elevation of the top is 30°. Height = 30 × tan 30° = 30/√3 = 10√3 m. If the angle were 45° the height would be 30 m.

Value problems. sin²30° + cos²60° = 1/4 + 1/4 = 1/2. 2 tan 45° − sin 90° = 2 − 1 = 1. sin 60° cos 30° + cos 60° sin 30° = 3/4 + 1/4 = 1.

Identity. If sin θ = 3/5 (θ acute) then cos θ = 4/5 and tan θ = 3/4.

Coordinates. Distance from (1, 2) to (4, 6) = √(9 + 16) = 5. Mid-point of (2, 3) and (8, 7) = (5, 5). Point dividing (0, 0) and (9, 6) in the ratio 1 : 2 = (3, 2). Area of triangle with vertices (0, 0), (4, 0), (0, 3) = ½ × 4 × 3 = 6.

F. Statistics and probability

Mean from frequency. Values 2, 4, 6 with frequencies 3, 4, 3: Σfx = 6 + 16 + 18 = 40, Σf = 10, mean = 4.

Median. 3, 9, 5, 7, 11 arranged: 3, 5, 7, 9, 11 gives median 7.

Probability. Two dice: P(sum 9) = 4/36 = 1/9. A card is drawn from a pack: P(king or queen) = 8/52 = 2/13. A letter is chosen from "PROBABILITY": there are 11 letters and B appears twice, so P(B) = 2/11.

G. A method for a mixed test

Step	What to do
1	Read once, underline the unknown, write given values.
2	Pick the formula; draw a figure for geometry and mensuration.
3	Keep exact values (π, √3) until the last step.
4	Convert units before calculating.
5	Check the answer by substitution or estimation.

Exam traps

Successive discounts of 10% and 20% are not a 30% discount.
Speed in km/h must be changed to m/s (multiply by 5/18) when length is in metres.
"Together" in time-and-work problems means adding the rates, not the times.
For CI, the amount includes the principal; subtract P to get the interest.
Slant height of a cone comes from l² = r² + h², and CSA uses l, not h.
A quadratic with D < 0 has no real roots, but it still has a definite sign for all x.
Melting a solid keeps the volume equal, not the surface area.
tan 30° = 1/√3 and tan 60° = √3. Swapping them gives the wrong height.

One-liners

1. 54 km/h = 15 m/s.
2. Discounts of 10% then 20% equal one discount of 28%.
3. CI − SI for 2 years = P (R/100)².
4. Two workers together: rate = sum of the individual rates.
5. (3, 4, 5) and (5, 12, 13) are Pythagorean triples.
6. a² + b² = (a + b)² − 2ab.
7. Similar triangles: area ratio = square of the side ratio.
8. A cylinder r = 7, h = 10 has volume 1540 (using π = 22/7).
9. Melting a sphere into n spheres of equal size: n = R³/r³.
10. sin 60° cos 30° + cos 60° sin 30° = 1.
11. Mid-point of (2, 3) and (8, 7) is (5, 5).
12. The probability of getting a sum of 9 with two dice is 1/9.

Practice questions

64 is divided in the ratio 3 : 5. The larger share is
1. 24
2. 32
3. 45
4. 40
Answer
D. 40
One part = 64/8 = 8; larger share = 5 × 8 = 40.
A shirt is marked at 800 and sold at a discount of 15%. The selling price is
1. 720
2. 680
3. 700
4. 660
Answer
B. 680
800 × 85/100 = 680.
Two successive discounts of 10% and 20% are given on a price of 800. The final price is
1. 576
2. 560
3. 640
4. 600
Answer
A. 576
800 × 0.9 × 0.8 = 576.
An article bought for 400 is sold at a profit of 25%. The selling price is
1. 520
2. 425
3. 500
4. 480
Answer
C. 500
400 × 125/100 = 500.
An article bought for 400 is sold for 340. The loss percentage is
1. 15%
2. 17.6%
3. 12%
4. 60%
Answer
A. 15%
Loss = 60; 60/400 × 100 = 15%.
The simple interest on 5000 at 8% per year for 3 years is
1. 400
2. 1200
3. 1500
4. 1000
Answer
B. 1200
5000 × 8 × 3 / 100 = 1200.
The compound interest on 10000 at 10% per year for 2 years, compounded yearly, is
1. 1210
2. 2000
3. 2200
4. 2100
Answer
D. 2100
Amount = 10000 × 1.1 × 1.1 = 12100; CI = 2100.
A train 120 m long runs at 54 km/h. The time taken to cross a pole is
1. 6 s
2. 10 s
3. 8 s
4. 12 s
Answer
C. 8 s
54 km/h = 15 m/s; 120/15 = 8 s.
The same train (120 m long, 54 km/h) crosses a platform 180 m long. The time taken is
1. 20 s
2. 12 s
3. 24 s
4. 16 s
Answer
A. 20 s
Distance 300 m at 15 m/s gives 20 s.
A can finish a job in 12 days and B in 6 days. Working together they finish it in
1. 3 days
2. 6 days
3. 4 days
4. 9 days
Answer
C. 4 days
Combined rate 1/12 + 1/6 = 1/4 per day.
The HCF and LCM of 36 and 48 are respectively
1. 6 and 288
2. 4 and 144
3. 12 and 72
4. 12 and 144
Answer
D. 12 and 144
36 = 2² × 3², 48 = 2⁴ × 3; HCF = 12, LCM = 144.
The average of 5 numbers is 18. If one number 25 is replaced by 15, the new average is
1. 14
2. 16
3. 15
4. 17
Answer
B. 16
New sum = 90 - 10 = 80; 80/5 = 16.
The roots of x² - 15x + 56 = 0 are
1. 4 and 11
2. 5 and 10
3. 7 and 8
4. 6 and 9
Answer
C. 7 and 8
7 + 8 = 15 and 7 × 8 = 56.
The discriminant of 2x² - 4x + 3 = 0 is
1. 16
2. -8
3. 8
4. -24
Answer
B. -8
D = 16 - 4 × 2 × 3 = 16 - 24 = -8.
If x + y = 10 and x - y = 2, then xy equals
1. 24
2. 16
3. 20
4. 12
Answer
A. 24
x = 6, y = 4, so xy = 24.
A father is 3 times as old as his son. After 5 years the sum of their ages is 70. The father's present age is
1. 40 years
2. 48 years
3. 50 years
4. 45 years
Answer
D. 45 years
(3x + 5) + (x + 5) = 70 gives x = 15; father = 45.
The sum of the first 10 terms of the AP 3, 7, 11, … is
1. 210
2. 190
3. 200
4. 220
Answer
A. 210
S = 10/2 × (2 × 3 + 9 × 4) = 5 × 42 = 210.
If a + b = 7 and ab = 10, then a² + b² equals
1. 9
2. 49
3. 29
4. 39
Answer
C. 29
(a + b)² - 2ab = 49 - 20 = 29.
The remainder when p(x) = x³ - 2x² + x - 5 is divided by (x - 2) is
1. 3
2. -5
3. -3
4. 0
Answer
C. -3
p(2) = 8 - 8 + 2 - 5 = -3.
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is
1. 72°
2. 60°
3. 90°
4. 80°
Answer
D. 80°
One part = 180/9 = 20°; largest = 80°.
The interior angle of a regular hexagon is
1. 135°
2. 120°
3. 108°
4. 60°
Answer
B. 120°
(6 - 2) × 180 / 6 = 120°.
In triangle ABC, DE is parallel to BC with D on AB and E on AC. If AD = 3, DB = 6 and AE = 2, then EC is
1. 3
2. 4
3. 6
4. 1
Answer
B. 4
AD/DB = AE/EC gives 3/6 = 2/EC, so EC = 4.
Two similar triangles have sides in the ratio 2 : 3. If the smaller has area 20 cm², the larger has area
1. 45 cm²
2. 30 cm²
3. 40 cm²
4. 60 cm²
Answer
A. 45 cm²
Area ratio 4 : 9; 20 × 9/4 = 45.
A tangent is drawn from a point 13 cm from the centre of a circle of radius 5 cm. The length of the tangent is
1. 8 cm
2. 18 cm
3. 10 cm
4. 12 cm
Answer
D. 12 cm
√(13² - 5²) = √144 = 12.
A chord of length 8 cm is in a circle of radius 5 cm. The distance of the chord from the centre is
1. √41 cm
2. 4 cm
3. 3 cm
4. 2 cm
Answer
C. 3 cm
The perpendicular bisects the chord: √(25 - 16) = 3.
The area of a circle of radius 7 cm (π = 22/7) is
1. 154 cm²
2. 49 cm²
3. 44 cm²
4. 308 cm²
Answer
A. 154 cm²
22/7 × 49 = 154.
The total surface area of a cuboid 5 cm × 4 cm × 3 cm is
1. 60 cm²
2. 120 cm²
3. 47 cm²
4. 94 cm²
Answer
D. 94 cm²
2(20 + 12 + 15) = 94.
The volume of a cone of radius 3 cm and height 4 cm is
1. 36π cm³
2. 12π cm³
3. 15π cm³
4. 24π cm³
Answer
B. 12π cm³
⅓ × π × 9 × 4 = 12π.
A solid sphere of radius 3 cm is melted and recast into small spheres of radius 1 cm. The number of small spheres is
1. 27
2. 9
3. 3
4. 81
Answer
A. 27
Volume ratio = 3³/1³ = 27.
The cost of carpeting a room 5 m by 4 m at 120 per square metre is
1. 4800
2. 1080
3. 2160
4. 2400
Answer
D. 2400
Area 20 m² × 120 = 2400.
A point on level ground is 30 m from the foot of a tower and the angle of elevation of its top is 30°. The height of the tower is
1. 30√3 m
2. 10√3 m
3. 30 m
4. 15 m
Answer
B. 10√3 m
30 × tan 30° = 30/√3 = 10√3.
The value of sin 60° cos 30° + cos 60° sin 30° is
1. 1/2
2. √3/2
3. 3/4
4. 1
Answer
D. 1
3/4 + 1/4 = 1.
If sin θ = 3/5 and θ is acute, then tan θ equals
1. 4/5
2. 5/3
3. 3/4
4. 4/3
Answer
C. 3/4
cos θ = 4/5, so tan θ = 3/4.
The distance between the points (1, 2) and (4, 6) is
1. 5
2. 25
3. 7
4. √7
Answer
A. 5
√(9 + 16) = 5.
The point dividing the segment from (0, 0) to (9, 6) internally in the ratio 1 : 2 is
1. (2, 3)
2. (4.5, 3)
3. (3, 2)
4. (6, 4)
Answer
C. (3, 2)
((1 × 9 + 0)/3, (1 × 6 + 0)/3) = (3, 2).
Two dice are thrown. The probability that the sum is 9 is
1. 5/36
2. 1/9
3. 1/12
4. 1/6
Answer
B. 1/9
Pairs (3,6), (4,5), (5,4), (6,3): 4/36 = 1/9.
A card is drawn from a pack of 52 cards. The probability that it is a king or a queen is
1. 2/13
2. 1/13
3. 4/13
4. 1/26
Answer
A. 2/13
8 favourable cards out of 52 = 2/13.
The volume of a cylinder of radius 7 cm and height 10 cm (π = 22/7) is
1. 440 cm³
2. 154 cm³
3. 3080 cm³
4. 1540 cm³
Answer
D. 1540 cm³
22/7 × 49 × 10 = 1540.
When two persons work together, their combined rate of work is
1. the sum of their times
2. the sum of their individual rates
3. the difference of their rates
4. the product of their rates
Answer
B. the sum of their individual rates
Work done per day adds up; times do not.
To change a speed from km/h to m/s, multiply by
1. 3.6
2. 1000/60
3. 5/18
4. 18/5
Answer
C. 5/18
1 km/h = 1000/3600 = 5/18 m/s.
Consider the statements. 1. The slant height of a cone satisfies l² = r² + h². 2. The curved surface area of a cone is πrh. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
CSA of a cone is πrl, using slant height l.
Consider the statements. 1. When a solid is melted and recast, its surface area stays the same. 2. When a solid is melted and recast, its volume stays the same. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Volume is conserved; surface area usually changes.
Consider the statements. 1. Two successive discounts of 10% and 20% equal a single discount of 30%. 2. They equal a single discount of 28%. Which is/are correct?
1. 1 only
2. Both 1 and 2
3. 2 only
4. Neither 1 nor 2
Answer
C. 2 only
1 - 0.9 × 0.8 = 0.28.
Consider the statements. 1. tan 30° = 1/√3. 2. tan 60° = √3. Which is/are correct?
1. 1 only
2. 2 only
3. Neither 1 nor 2
4. Both 1 and 2
Answer
D. Both 1 and 2
Both are standard values.
Consider the statements. 1. A quadratic equation with a negative discriminant has two real roots. 2. (a + b)² = a² + b². Which is/are correct?
1. Neither 1 nor 2
2. 1 only
3. 2 only
4. Both 1 and 2
Answer
A. Neither 1 nor 2
Negative D gives no real roots and (a + b)² = a² + 2ab + b².

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1.Three classes have 60, 40, and 50 students respectively. Their average marks are 65, 72, and 60. What is the average mark of all 150 students combined?
2.Four classes have a ratio of boys to girls as 4 : 5, 3 : 2, 6 : 5 and 5 : 6, respectively. Which class has a higher proportion of boys?
3.In a school, Class A has 24 students with an average height of 160 cm, and Class B has 20 students with an average height of 168 cm. The average height of students in Class C is 165 cm. If the overall average height of all students from Classes A, B and C is 164 cm, find the number of students in Class C.
4.The charts below show the distribution of students in two classes (A and B) across different subjects. Study the charts and answer the question. If there are 200 students in Class B, what is the total number of students in Chemistry and Biology in Class B?
5.The given bar graph represents the number of students in different classes in a school. How many more students are in class 10 compared to class 8?
6.The number of children in five different classes are given below in a bar graph. Observe the graph and answer the given question: What is the ratio of the number of students in Class 8th to that in Class 10th?
7.The pie charts below show the distribution of students in two classes (A and B) across different subjects. Study the pie charts and answer the question. Class A has 200 students and Class B has 300 students. What is the difference in the number of students studying Maths in the two classes?
8.On mixing two classes of students having average marks 20 and 35 respectively, the overall average obtained is 25 marks. Find the ratio of students in the classes.
9.Observe the graph given below and answer the question. The graph below represents the number of students in various classes (1 to 5) at a school. Assume all numbers here are multiples of 10. What is the difference between the number of students in Class 3 and Class 4 together and those in Class 1 and Class 5 together?
10.The following bar graph represents the annual result of five classes (VI to X) of a school. Study the bar graph carefully and answer the question given below: What is the average number of students in the given five classes?

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