IMO Maths Olympiad Class 10 Mixed Practice Questions MCQs with Answers

Answer: (b) 1, 5, 6; 2, 3, 4; 7, 8, 9

Explanation:
1. Figures 1, 5 and 6 each consist of two identical figures, one placed inside the other (triangle in triangle, circle in circle, rectangle in rectangle).
2. Figures 2, 3 and 4 are each divided into equal parts by straight lines passing through the centre.
3. Figures 7, 8 and 9 each consist of two different figures, one placed inside the other. So the classes are 1, 5, 6; 2, 3, 4; 7, 8, 9.

Teacher's Note:
1. Look for what is inside each figure and how it is divided, not for the outer shape alone.
2. Each figure must be used once, so every class has three figures.

Answer: (c) M

Explanation:
1. S-T-E-T-H-O-S-C-O-P-E: the third, fifth, eighth and ninth letters are E, H, C and O.
2. These letters make ECHO and also OCHE (the line behind which a darts player stands), so more than one word can be formed.
3. Hence the answer is M.

Teacher's Note:
1. Number the letters of the word carefully before picking them.
2. Try the different arrangements of E, H, C, O: ECHO is the obvious word, but a second valid word changes the answer.

Answer: (b) 40

Explanation:
1. The large grid is 4 × 4: squares = 16 + 9 + 4 + 1 = 30.
2. Each small square sits on a grid point and is cut into 4 tiny squares by the grid lines, so it gives 1 + 4 = 5 squares. There are two of them: 10 squares.
3. Total = 30 + 10 = 40.

Teacher's Note:
1. Count the squares size by size so none is missed.
2. A small square crossed by two lines also contains four smaller squares.

Answer: (b)

Explanation:
1. Folding along the diagonal places the lower-left half on the upper-right half. As the sheet is transparent, the pattern of both halves is seen together.
2. The semicircle lies wholly in the lower-left half, so after folding it appears on the right edge, bulging inwards. This rules out (d).
3. The triangle is cut by the fold line: its upper part stays, and its lower part is reflected in the line. Only option (b) shows this zigzag outline correctly, so the answer is (b).

Teacher's Note:
1. Reflect every part that lies on one side of the fold line in that line.
2. Because the sheet is transparent, the parts on the other half are still visible.

Answer: (c)

Explanation:
1. In Fig. (X) one dot lies in the circle and the wide rectangle only, one dot lies in the two rectangles only, and one dot lies in the square and the triangle only.
2. So the figure needs regions: circle ∩ one rectangle only, rectangle ∩ rectangle only, and the second rectangle ∩ triangle only.
3. (a) has no overlap between its rectangles, (b) has the triangle inside the circle and (d) has no rectangle meeting the triangle in this way. Option (c) has all three regions, so the answer is (c).

Teacher's Note:
1. Name the shapes that overlap at each dot, not the positions on the page.
2. Check each option for all three regions before choosing.

Answer: (c)

Explanation:
1. A mirror on the right gives a left-right (lateral) reversal: what is on the right of the figure appears on the left of the image.
2. So the symbol at the top right goes to the top left, the arrow at the bottom left goes to the bottom right and the slanting lines turn the other way, while up and down stay unchanged.
3. Only option (c) has all these changes, so the answer is (c).

Teacher's Note:
1. With a vertical mirror, left and right are interchanged but top and bottom stay.
2. Check the position and the direction of each small symbol.

Answer: (a) 4

Explanation:
1. Convert the letters to their positions in the English alphabet: H = 8, X = 24; L = 12, B = 2; O = 15, C = 3; T = 20, E = 5.
2. The rule: (position of the farther letter) ÷ (position of the nearer letter) = number: 24 ÷ 8 = 3, 12 ÷ 2 = 6, 15 ÷ 3 = 5.
3. So the missing number is 20 ÷ 5 = 4.

Teacher's Note:
1. When letters and numbers are mixed, change the letters to their alphabet positions first.
2. Test the rule on every complete part before using it.

Answer: (a)

Explanation:
1. Problem Figures 1 and 5 have the same layout: one symbol in each of the top two rows and two symbols in the bottom row.
2. From Figure 1 to Figure 2: the bottom-left symbol moves to the top, the bottom-right symbol moves to the middle and is turned round, the middle symbol moves to the bottom-left and becomes shaded, and a new symbol appears at the bottom right.
3. Applying the same changes to Figure 5: the hexagon goes to the top, the symbol at the bottom right goes to the middle as ‘(’, the square goes to the bottom left as a shaded square, and a new symbol (a circle) appears at the bottom right. This is option (a).

Teacher's Note:
1. Follow each symbol from one figure to the next and note its move and change.
2. Figures with the same layout usually undergo the same set of changes.

Answer: (a)

Explanation:
1. A poet can be a niece or a father, so the circle for Poet overlaps both of the other circles.
2. A niece (a girl) cannot be a father, so the circles for Niece and Father do not overlap.
3. The diagram with two separate circles, both cut by a third circle, is option (a).

Teacher's Note:
1. Decide first which pairs can overlap and which cannot.
2. Two classes with no common member are shown by separate circles.

Answer: (d)

Explanation:
1. Fig. (X) is a square with a vertical line from its top-right corner to one square above, a short line at the top going left and a horizontal line from the top edge going to the right.
2. Without turning or flipping it, this shape fits exactly into the middle part of option (d), where the three small squares and the line segments join in the same way.
3. The other options do not contain the shape in this position, so the answer is (d).

Teacher's Note:
1. In embedded figures the shape must keep its size and direction.
2. Try to match the most distinctive part of the shape first.

Answer: (a)

Explanation:
1. From (i) to (ii), the line carrying the cup and arrow turns through 45° so that it changes from vertical to slanting, with its end symbols turning with it.
2. The line carrying the triangle and circle stays in place, its two end symbols exchange places and each is changed (a filled symbol becomes open and the triangle becomes a square).
3. The same changes take (iii) to (iv) when (iii) has the cup–fork line slanting from top left and the circle–triangle line with a filled circle at the lower left and an open triangle at the upper right. This is option (a).

Teacher's Note:
1. Study each line of the figure separately and note how its ends change.
2. Check the answer by applying the same changes to the option and comparing with (iv).

Answer: (c) 62°, 31°

Explanation:
1. The radius OQ is perpendicular to the tangent, so ∠OQR = 90°. In △OQR: x = 180° − 90° − 28° = 62°.
2. ∠SOQ = 180° − 62° = 118° (linear pair, as S, O, R are in a line).
3. OS = OQ (radii), so in △OQS: y = (180° − 118°) ÷ 2 = 31°. Hence x = 62° and y = 31°.

Teacher's Note:
1. The radius to the point of contact is always perpendicular to the tangent.
2. A triangle with two radii as sides is isosceles.

Answer: (d) −a, −b

Explanation:
1. Move terms: \( \dfrac{1}{a+b+x} - \dfrac{1}{x} = \dfrac{1}{a} + \dfrac{1}{b} \), so \( \dfrac{-(a+b)}{x(a+b+x)} = \dfrac{a+b}{ab} \).
2. Since a + b ≠ 0, divide by it: −ab = x(a + b + x), i.e. x² + (a + b)x + ab = 0.
3. Factorising, (x + a)(x + b) = 0, so the roots are −a, −b.

Teacher's Note:
1. Group the fractions with the same denominator pattern to reduce the work.
2. Dividing by a + b is allowed only because a + b ≠ 0.

Answer: (d) (i) \( \dfrac{1}{26} \), (ii) \( \dfrac{9}{13} \)

Explanation:
1. Total outcomes = 52.
2. (i) There are 2 black kings, so the probability is \( \dfrac{2}{52} = \dfrac{1}{26} \).
3. (ii) Hearts: 13 cards. Kings that are not hearts: 3. Cards that are a heart or a king = 16, so the required cards = 52 − 16 = 36 and the probability is \( \dfrac{36}{52} = \dfrac{9}{13} \).

Teacher's Note:
1. The king of hearts belongs to both groups, so count it only once.
2. Subtract the favourable-to-exclude cards from 52 before forming the fraction.

Answer: (a) 196 cm²

Explanation:
1. Area of trapezium = \( \dfrac{1}{2}(18+32) \times 14 \) = 350 cm².
2. The four arcs are sectors of radius 7 cm whose angles are the four angles of the quadrilateral, which add up to 360°. Together they make one full circle: \( \dfrac{22}{7} \times 7^2 \) = 154 cm².
3. Shaded area = 350 − 154 = 196 cm².

Teacher's Note:
1. The angles at the four vertices of a quadrilateral add up to 360°, so the four sectors form one circle.
2. Remember to subtract the sectors from the trapezium.

Answer: (d) 2.5 cm

Explanation:
1. Volume of the cube = 50³ = 125000 cm³.
2. If the side of a die is a, then 8000a³ = 125000, so a³ = 15.625.
3. a = \( \sqrt[3]{\dfrac{125}{8}} = \dfrac{5}{2} \) = 2.5 cm.

Teacher's Note:
1. Melting and recasting keeps the volume the same.
2. Take the cube root of 125 and 8 separately.

Answer: (d) 11

Explanation:
1. In △BGE and △BDC (GE ∥ DC): \( \dfrac{EG}{CD} = \dfrac{BE}{BC} = \dfrac{4}{7} \), so EG = \( \dfrac{4}{7} \)CD = \( \dfrac{8}{7} \)AB.
2. In △DGF and △DBA (GF ∥ BA): \( \dfrac{FG}{AB} = \dfrac{DF}{DA} = \dfrac{CE}{CB} = \dfrac{3}{7} \), so FG = \( \dfrac{3}{7} \)AB.
3. EF = EG + FG = \( \dfrac{11}{7} \)AB, so 7EF = 11AB and k = 11.

Teacher's Note:
1. Lines parallel to the bases of a trapezium cut the legs in the same ratio.
2. Split EF at G and find each part with a pair of similar triangles.

Answer: (a) 8, 7

Explanation:
1. Total frequency: 45 + x + y = 60, so x + y = 15.
2. The median 28.5 lies in the class 20 – 30, with cumulative frequency before it = 5 + x, frequency 20 and class width 10: \( 28.5 = 20 + \dfrac{30 - (5+x)}{20} \times 10 \).
3. So 8.5 = (25 − x) ÷ 2, giving x = 8 and y = 15 − 8 = 7. Hence 8, 7.

Teacher's Note:
1. Find the median class from the value of the median, then use the formula.
2. Use the total frequency to get a second equation.

Answer: (b) 6\( \dfrac{2}{3} \) cm

Explanation:
1. In right △TPO: PT² = 13² − 5² = 144, so PT = 12 cm. Also ET = 13 − 5 = 8 cm.
2. Tangents from a point are equal: AE = AP = 12 − AT. In right △AET (AB ⊥ OT): AT² = (12 − AT)² + 8², giving 24AT = 208 and AT = \( \dfrac{26}{3} \) cm.
3. AE = 12 − \( \dfrac{26}{3} = \dfrac{10}{3} \), and AB = 2AE = \( \dfrac{20}{3} \) = 6\( \dfrac{2}{3} \) cm.

Teacher's Note:
1. Tangents from an external point are equal, and the tangent at E is perpendicular to OE.
2. By symmetry E is the mid-point of AB.

Answer: (c) 3

Explanation:
1. The centroid is \( \left(\dfrac{a+b+c}{3}, \dfrac{b+c+a}{3}\right) = (0,0) \), so a + b + c = 0.
2. Then a³ + b³ + c³ = 3abc.
3. The expression = \( \dfrac{a^3+b^3+c^3}{abc} = \dfrac{3abc}{abc} \) = 3.

Teacher's Note:
1. If a + b + c = 0, then a³ + b³ + c³ = 3abc.
2. Put the three fractions over the common denominator abc.

Answer: (c) 4480 m

Explanation:
1. Half the vertical angle is 30°. At the middle (10 cm from the vertex): \( r_1 = 10\tan30^\circ = \dfrac{10}{\sqrt{3}} \). At the base (20 cm): \( r_2 = \dfrac{20}{\sqrt{3}} \). The frustum's height is 10 cm.
2. Volume of the frustum = \( \dfrac{1}{3}\pi \times 10\left(\dfrac{100}{3}+\dfrac{400}{3}+\dfrac{200}{3}\right) = \dfrac{7000\pi}{9} \) cm³.
3. The wire has radius \( \dfrac{1}{24} \) cm: \( \pi\left(\dfrac{1}{24}\right)^2 h = \dfrac{7000\pi}{9} \), so h = 448000 cm = 4480 m.

Teacher's Note:
1. The volume does not change when the frustum is drawn into a wire, and the wire is a thin cylinder.
2. Convert the final length from cm to m by dividing by 100.

Answer: (d) All of these

Explanation:
1. (a) is the tangent–radius theorem, which is true.
2. (b) is the definition of a secant, which is true.
3. (c) is true: tangents from an external point are equal in length. So all the statements are correct and the answer is (d).

Teacher's Note:
1. Each statement is a standard result or definition for circles.
2. When one statement is not enough, always check whether all of them are true.

Answer: (d) 80 minutes

Explanation:
1. They begin new cards together after a time that is a multiple of 10, 16 and 20 minutes.
2. 10 = 2 × 5, 16 = 2⁴, 20 = 2² × 5, so L.C.M. = 2⁴ × 5 = 80.
3. They start a new card together after 80 minutes.

Teacher's Note:
1. ‘Together again’ points to the LCM.
2. Take the highest power of each prime factor.

Answer: (b) ₹ 1,66,800

Explanation:
1. Cash paid = ₹60000, balance = ₹60000. First instalment = 5000 + 12% of 60000 = ₹12200.
2. Each year the unpaid amount falls by ₹5000, so the interest falls by ₹600: the instalments 12200, 11600, 11000, ... form an A.P. with a = 12200, d = −600, n = 12.
3. Sum = \( \dfrac{12}{2}[2(12200) + 11(-600)] = 106800 \). Total cost = 106800 + 60000 = ₹ 1,66,800.

Teacher's Note:
1. The interest is charged on the amount still unpaid, which reduces after each instalment.
2. Add the cash payment to the total of all the instalments.

Answer: (d) 36.60 m

Explanation:
1. Let the tree RS have height h. From Q the angle is 45°, so SQ = h. From P the angle is 30°, so PS = h cot 30° = h√3.
2. PS + SQ = 100: h(√3 + 1) = 100.
3. h = \( \dfrac{100}{\sqrt{3}+1} = 50(\sqrt{3}-1) \) ≈ 36.60 m.

Teacher's Note:
1. The foot of the tree lies on the line PQ, so the two horizontal distances add up to 100 m.
2. Rationalise the denominator to get 50(√3 − 1).

Answer: (c) 1910.85 cm

Explanation:
1. In 2 days the short (hour) hand makes 4 rounds and the long (minute) hand makes 48 rounds.
2. Short tip: \( 4 \times 2\pi \times 4 = 4 \times 2 \times \dfrac{22}{7} \times 4 = \dfrac{704}{7} \) cm. Long tip: \( 48 \times 2\pi \times 6 = \dfrac{12672}{7} \) cm.
3. Total = \( \dfrac{13376}{7} \approx 1910.86 \) cm, so the closest option is 1910.85 cm.

Teacher's Note:
1. Distance travelled in one round is the circumference of the circle made by the tip.
2. The hour hand turns twice a day and the minute hand 24 times a day.

Answer: (a) 9735 m²

Explanation:
1. Radius = 52.5 m. Canvas = curved surface of the cone + curved surface of the cylinder = πrl + 2πrh = πr(l + 2h).
2. \( \dfrac{22}{7} \times 52.5 \times (53 + 6) = 165 \times 59 \).
3. Canvas = 9735 m².

Teacher's Note:
1. The tent has no base, so only the two curved surfaces are needed.
2. 52.5 × 22 ÷ 7 = 165 saves a lot of work.

Answer: (a) (P) → (iv), (Q) → (iii), (R) → (i), (S) → (ii)

Explanation:
1. Three collinear points give zero area. (P) \( \tfrac{1}{2}[k(-2-4) + 6(4-3) + (-3)(3+2)] = 0 \) gives −6k − 9 = 0, so k = −3/2 → (iv).
2. (Q) \( 7(1-2k) + 5(2k+2) + 3(-2-1) = 0 \) gives −4k + 8 = 0, so k = 2 → (iii). (R) \( -\dfrac{2}{5}(3-8) + m(8-6) + 2(6-3) = 0 \) gives 2 + 2m + 6 = 0, so m = −4 → (i).
3. (S) \( \tfrac{1}{2}[a(6-1) + (-2)(1-2a) + 3(2a-6)] = 10 \) gives 15a − 20 = 20, so a = 8/3 → (ii). Hence P → (iv), Q → (iii), R → (i), S → (ii).

Teacher's Note:
1. Collinear points form a triangle of area zero.
2. For (S) the area can also be −10 inside the bracket, which gives a = 0, but only 8/3 is listed in Column-II.

Answer: (b) Statement I is true but Statement II is false.

Explanation:
1. Statement I: \( \tan x = \dfrac{TF}{200} = \dfrac{2}{5} \) gives TF = 80 m. Then \( \tan y = \dfrac{80}{80} = 1 \), so y = 45°. True.
2. Statement II: let the angles be θ and 90° − θ. Then h = m² tan θ = n² cot θ, so tan²θ = n²/m² and tan θ = n/m.
3. Height h = m² × (n/m) = mn, not m/n, so Statement II is false.

Teacher's Note:
1. For complementary angles of elevation the height is the geometric mean of the two distances: √(m²·n²) = mn.
2. Compute the height in Statement I before judging the angle.

Answer: (c) \( \mathbf{P} = ax,\ \mathbf{Q} = \dfrac{a^2}{4},\ \mathbf{R} = 2p^2 \)

Explanation:
1. (i) \( b^2 = p^2 + \mathbf{P} + \dfrac{a^2}{4} \) (ii) \( c^2 - p^2 + ax = \mathbf{Q} \) (iii) \( b^2 + c^2 = \mathbf{R} + \dfrac{a^2}{2} \).
2. DC = a/2. In right △AEC: b² = h² + (a/2 + x)², and in right △AED: h² = p² − x². So b² = p² + ax + a²/4: P = ax.
3. In right △AEB: c² = h² + (a/2 − x)² = p² − ax + a²/4, so c² − p² + ax = a²/4 = Q.
4. Adding the results for b² and c²: b² + c² = 2p² + a²/2, so R = 2p². This is option (c).

Teacher's Note:
1. This is Apollonius' theorem: the sum of the squares of two sides is twice the square of the median plus half the square of the third side.
2. Use Pythagoras in the three right triangles AED, AEC and AEB.