Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
Try each problem first, then open the solution to check your working.
Frame the converse of “If a number ends in 0, then it is divisible by 5”. Test both.
Converse: if a number is divisible by 5, then it ends in 0.
P is true (it is a multiple of 10). Q is false: 15 is divisible by 5 but ends in 5.
Answer: P true, Q false (15)
“If n is even then n² is even.” Frame the converse and test.
n = 2k gives n² = 4k², even: P true.
Converse: if n² is even then n is even. If n were odd, n² would be odd. So Q is true.
Answer: both true
“If a triangle is equilateral, then it is isosceles.” Test the converse.
P true (three equal sides include two equal sides).
Converse false: a triangle with sides 5, 5, 8 is isosceles but not equilateral.
Answer: P true, Q false
Test “If n is divisible by 12, then it is divisible by 3 and 4” and its converse.
P: 12k = 3 × 4k = 4 × 3k, true.
Q: if 3 and 4 divide n then n = 3a = 4b. As 3 and 4 share no factor, 12 divides n. True.
Answer: both true
“If n is divisible by 6 and by 9, then n is divisible by 54.” True?
False: n = 18 is divisible by 6 and by 9 but not by 54.
The converse (54 ⇒ 6 and 9) is true.
Answer: false (18)
“The diagonals of a rhombus are perpendicular.” Test the converse about quadrilaterals.
P true.
Converse: if the diagonals of a quadrilateral are perpendicular, it is a rhombus. False: a kite has perpendicular diagonals but is not a rhombus.
Answer: P true, Q false (kite)
“If a parallelogram has a right angle, then it is a rectangle.” Test its converse.
P true: opposite angles are equal and adjacent ones supplementary, so all are 90°.
Q: if a parallelogram is a rectangle, it has a right angle: true by definition.
Answer: both true
Show that n² − n + 41 is not prime for every positive integer n.
n = 41: 41² − 41 + 41 = 41² = 1681 = 41 × 41, composite.
Answer: n = 41
Give a counterexample to “All prime numbers are odd”. Write the converse and a counterexample.
2 is prime and even.
Converse: every odd number is prime: false, 9 = 3 × 3.
Answer: 2; 9
“If two angles are supplementary, they are adjacent.” True?
False: 70° and 110° in different places on a page are supplementary but not adjacent.
Answer: false
Why is 2³² + 1 an important counterexample?
Fermat believed all 2^(2ⁿ) + 1 are prime. For n = 5, 2³² + 1 = 641 × 6700417.
So the pattern 3, 5, 17, 257 does not continue.
Answer: composite
“The sum of two primes is a prime.” True?
False: 3 + 5 = 8, not prime.
Answer: false (3 + 5)
Are 8, 15, 17 and 7, 8, 9 the sides of right-angled triangles?
8² + 15² = 64 + 225 = 289 = 17²: right-angled.
7² + 8² = 113 ≠ 81: no.
Answer: yes; no
Triangle with sides 20, 21, 29: which angle is a right angle?
20² + 21² = 400 + 441 = 841 = 29².
The right angle is opposite the longest side 29.
Answer: angle opposite the side 29
Write “A number has an odd number of factors when it is a perfect square” in if-then form and give its converse.
If a number is a perfect square, then it has an odd number of factors.
Converse: if a number has an odd number of factors, then it is a perfect square (also true).
Answer: both true
Give a proposition and its converse that are both false.
P: if a quadrilateral has equal diagonals then it is a rectangle (false: an isosceles trapezium).
Q: if a quadrilateral is a rectangle then its diagonals are perpendicular (false: a 2 × 3 rectangle).
Answer: e.g. above
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True. In ΔABC with ∠B = ∠C draw the altitude AD to BC. In ΔABD and ΔACD: ∠B = ∠C, ∠ADB = ∠ADC = 90° and AD is common. By AAS the triangles are congruent, so AB = AC.
P: n = 6k = 3(2k), a multiple of 3. Q is false: 9 is a multiple of 3 but not of 6 (also 3, 15, 21, …).
Both are true. The argument starts from “n has an odd number of factors” and ends at “n is a perfect square”, so it proves Q only. For P: n = f² gives the repeated pair (f, f), and there is at most one such pair, so every other factor pairs up with a different partner: the count is odd.
P is false: a 6 × 4 rectangle and an 8 × 3 rectangle (or triangles with base 6, height 4 and base 8, height 3) have the same area 24 but are not congruent. Q is true: congruent triangles can be placed exactly on each other, so their areas are equal.
Yes. P: “If n is prime, then n is odd” (false: n = 2). Q: “If n is odd, then n is prime” (false: n = 9).
Converse: if a² + b² = c², the triangle is right-angled. Build ΔXYZ right-angled at Z with YZ = a, XZ = b. By the theorem XY² = a² + b² = c², so XY = c. ΔABC and ΔXYZ have equal sides (a, b, c), so ΔABC ≅ ΔXYZ by SSS. Thus ΔABC is right-angled (at C). The converse is true.
“If I drop a glass on a stone floor, it breaks.” (Converse: if the glass broke, I dropped it on a stone floor: false.) “If a person lives in Mumbai, they live in India.” (Converse false.) “If a number is a multiple of 10 it is a multiple of 5.” (Converse false: 15.)
Geometry: if a quadrilateral is a square then it is a rectangle (converse false: a 2 × 3 rectangle). If a quadrilateral is a rhombus then its diagonals are perpendicular (converse false: kite). Numbers: if n is divisible by 12 then by 6 (converse false: 18); if n = p² for a prime then n has 3 factors (converse true).
P is true (a property of parallel lines). Converse: if corresponding angles formed by a transversal are equal, then the lines are parallel. True: if the lines were not parallel they would meet and form a triangle in which an exterior angle equals the interior opposite angle, which is impossible.
P true: each angle of a square is 90°. Converse: if all angles of a quadrilateral are equal, it is a square. False: a 2 × 3 rectangle has all angles 90° but is not a square.
P true (adding the same number to equal numbers). Converse: if a + x = a + y then x = y. True (subtract a from both sides). Both are used in solving equations.
P true: a = m², b = n² gives ab = (mn)². Converse: if ab is a perfect square then a and b are perfect squares. False: a = 2, b = 8: ab = 16 is a perfect square but neither 2 nor 8 is.
P true. Converse: if x² = y² then x = y. False: x = 3, y = −3.
P true. Converse: if x³ = y³ then x = y. True for real numbers: x³ − y³ = (x − y)(x² + xy + y²) and the second factor is positive unless x = y = 0, so x − y = 0.
P true: 24k = 4 × 6k = 6 × 4k. Converse: if n is divisible by 4 and 6 then by 24. False: n = 12.
P true: 60k = 5 × 12k = 12 × 5k. Converse: if n is divisible by 5 and 12, then by 60. True: n = 12m and 5 divides 12m, so 5 divides m (5 and 12 share no factor): m = 5t, n = 60t.
P true: p² has factors 1, p, p². Converse: if n has exactly 3 factors, then n is the square of a prime. True: an odd number of factors means n = m². If m were composite, say m = ab with 1 < a < m, then 1, a, m, m² (and more) would give more than 3 factors. So m is prime.
P true: n = pq has divisors 1, p, q, pq. Converse: if n has exactly 4 divisors then n is a product of two unequal primes. False: n = 8 has divisors 1, 2, 4, 8 (also 27).
P true: if 3 divides n, then 3 divides n + 3, and they would have a common factor 3. Converse: if n is not a multiple of 3, then n and n + 3 have no common factor. True: a common factor divides (n + 3) − n = 3, so it is 1 or 3; it cannot be 3 since 3 ∤ n.
(i) n = 4: 4 × 16 + 1 = 65 = 5 × 13. (ii) n = 10: 100 + 10 + 11 = 121 = 11². (iii) n = 4: 256 + 3 = 259 = 7 × 37.
(i) n = 11: 2¹¹ − 1 = 2047 = 23 × 89. (ii) n = 6: 2⁶ + 1 = 65 = 5 × 13.
(i) 8k = 2 × 4k = 4 × 2k. (ii) No. The converse is false: 4, 12, 20 are divisible by 2 and 4 but not by 8. For 8 check whether the last three digits form a multiple of 8.
If a number is divisible by 3, then the sum of its digits is a multiple of 3. And if the sum of the digits of a number is a multiple of 3, then the number is divisible by 3. Both are true.
(a) Yes: any quadrilateral of type Q has equal diagonals, so equal sticks are needed. (b) Yes, it can matter: the property only says Q ⇒ equal diagonals; it does not say every quadrilateral with equal diagonals is of type Q. Other conditions (such as the sticks bisecting each other or crossing at right angles) may be needed.
(a) Use equal sticks (this is how we make sure the quadrilateral has equal diagonals). (b) No: however the equal sticks are put together, the quadrilateral has equal diagonals and so is of type Q. (Q may also contain quadrilaterals with unequal diagonals, but this property does not guarantee them.)