CBSE Class 10 Maths Advanced Real Numbers Short Answer Questions Set 02

Short Answer Questions

Question 1. 96 defective pens are accidentally mixed with 105 good pens (a) What is LCM of 96 and 105? (b) the shopkeeper draws a pen and finds it to be defective. The shopkeeper did not sell and kept the pen aside. Which value is shown by the shopkeeper?
Answer:
(a) 96 = 2⁵ × 3 and 105 = 3 × 5 × 7, so L.C.M. = 2⁵ × 3 × 5 × 7 = 3360.
(b) By not selling the defective pen, the shopkeeper shows honesty (and loyalty towards the customers).
In simple words: The LCM is 3360, and the shopkeeper's value is honesty.

Teacher's Note:
1. Prime-factorise both numbers and take the highest powers.
2. For value-based parts, name the value and say why in one phrase.

Question 2. Aakash, Kushal and Harish go for a morning walk. They step off together and their steps measure 40cm, 42cm and 45cm, respectively. (a) What is the minimum distance each should walk so that each can cover the same distance in complete steps? (b) How is morning walk useful?
Answer:
(a) The minimum distance is the L.C.M. of 40, 42 and 45.
40 = 2³ × 5, 42 = 2 × 3 × 7, 45 = 3² × 5, so L.C.M. = 2³ × 3² × 5 × 7 = 2520 cm.
(b) A morning walk is good for health: it strengthens the heart, improves stamina, helps keep weight in check and gives a fresh start to the day.
In simple words: They first land together after 2520 cm, the LCM of the three step lengths.

Teacher's Note:
1. The three of them each take a whole number of steps: 63, 60 and 56.
2. Mention a health benefit in part (b).

Question 3. Show that cube of the number of the form 4q + 3 is of the form 4q + 3, for some integer q.
Answer:
(4q + 3)³ = 64q³ + 3 × 16q² × 3 + 3 × 4q × 9 + 27 = 64q³ + 144q² + 108q + 27.
Take out 4: = 4(16q³ + 36q² + 27q + 6) + 3.
Put m = 16q³ + 36q² + 27q + 6, an integer. Then the cube is 4m + 3, of the same form.
In simple words: Expand the cube and split off 27 = 24 + 3 so that 4 can be taken out.

Teacher's Note:
1. Use (a + b)³ with a = 4q and b = 3.
2. Check with q = 0: 27 = 4 × 6 + 3.

Question 4. Out of the numbers n, n + 1 and n + 2, show that only one number is divisible by 3.
Answer:
Write n = 3q + r where r = 0, 1 or 2.
r = 0: n = 3q is divisible by 3, while n + 1 = 3q + 1 and n + 2 = 3q + 2 are not.
r = 1: n + 2 = 3q + 3 = 3(q + 1) is divisible; n and n + 1 = 3q + 2 are not.
r = 2: n + 1 = 3q + 3 = 3(q + 1) is divisible; n = 3q + 2 and n + 2 = 3(q + 1) + 1 are not.
So exactly one of the three consecutive numbers is a multiple of 3.
In simple words: Three consecutive numbers leave three different remainders on division by 3, so exactly one leaves 0.

Teacher's Note:
1. Consider each possible remainder of n.
2. This also proves that n(n + 1)(n + 2) is divisible by 3.

Question 5. 105 goats, 140 sheeps have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip.
Answer:
The number of animals per trip must divide both 105 and 140 (so that each kind is carried in whole trips), and we want the largest such number: the H.C.F.
140 = 105 × 1 + 35
105 = 35 × 3 + 0
H.C.F. = 35, so 35 animals went in each trip (3 trips for the goats and 4 for the sheep).
In simple words: Largest equal load that splits both herds exactly is the HCF, 35.

Teacher's Note:
1. Turn the story into a greatest common divisor problem.
2. Check: 105 ÷ 35 = 3 trips and 140 ÷ 35 = 4 trips.

Question 6. Express the following numbers as a product of prime factors: (a) 1771   (b) 8232   (c) 10584
Answer:
(a) 1771 = 7 × 253 = 7 × 11 × 23.
(b) 8232 = 2 × 4116 = 2² × 2058 = 2³ × 1029 = 2³ × 3 × 343 = 2³ × 3 × 7³.
(c) 10584 = 2³ × 1323 = 2³ × 3 × 441 = 2³ × 3 × 3² × 7² = 2³ × 3³ × 7².
In simple words: Divide repeatedly by the smallest prime that goes in.

Teacher's Note:
1. Test 2, 3, 5, 7, 11, ... in order.
2. Multiply the factors back to check, e.g. 8 × 3 × 343 = 8232.

Question 7. Find the HCF and LCM of the following pairs of numbers by prime factorization and verify that LCM × HCF = product of the numbers. (a) 1080 and 252   (b) 252 and 294
Answer:
(a) 1080 = 2³ × 3³ × 5 and 252 = 2² × 3² × 7. H.C.F. = 2² × 3² = 36. L.C.M. = 2³ × 3³ × 5 × 7 = 7560. Check: 36 × 7560 = 272160 = 1080 × 252.
(b) 252 = 2² × 3² × 7 and 294 = 2 × 3 × 7². H.C.F. = 2 × 3 × 7 = 42. L.C.M. = 2² × 3² × 7² = 1764. Check: 42 × 1764 = 74088 = 252 × 294.
In simple words: Smallest common powers give the HCF, largest powers give the LCM, and their product equals the product of the two numbers.

Teacher's Note:
1. Write each product out to verify the relation.
2. 1080 × 252 = 272160 and 252 × 294 = 74088.

Question 8. If the LCM of two numbers is 252, HCF is 2 and one of the numbers is 28 find the other number.
Answer:
H.C.F. × L.C.M. = product of the two numbers.
2 × 252 = 28 × other, so other = 504 ÷ 28 = 18.
In simple words: Use HCF × LCM = product of the numbers to get the missing one.

Teacher's Note:
1. Check: HCF(28, 18) = 2 and LCM(28, 18) = 252.
2. Divide 504 by 28 carefully: 28 × 18 = 504.

Question 9. Leena has music class on alternate days, dance class once in 3 days and yoga once in 5 days. On the 1^st of January she had all the three classes. When will she have all the 3 classes again?
Answer:
Her classes repeat every 2 days, 3 days and 5 days. All three coincide after a number of days that is a multiple of 2, 3 and 5; the first is the L.C.M.
L.C.M.(2, 3, 5) = 30 days.
30 days after 1 January is 31st January.
In simple words: All three classes fall on the same day again after LCM(2, 3, 5) = 30 days.

Teacher's Note:
1. Add the 30 days to the 1st of January, not to the 0th.
2. The cycle lengths 2, 3, 5 are co-prime, so the LCM is their product.

Question 10. There is an oval shaped park with a pathway running round it. Babu and Raju start jogging at P at the same time and jog in the same direction. If Babu can complete a full round in 16 minutes while Raju can do it in 20 minutes, after how many minutes will they meet at P again?
Answer:
They are both at P together at times that are multiples of 16 and 20 minutes; the first such time is the L.C.M.
16 = 2⁴ and 20 = 2² × 5, so L.C.M. = 2⁴ × 5 = 80.
They meet at P again after 80 minutes.
In simple words: The first moment both are back at P is the LCM of 16 and 20 minutes.

Teacher's Note:
1. Babu completes 5 rounds and Raju 4 rounds in 80 minutes.
2. Just like the circular-track problem in the chapter.

Question 11. A rope of length 140 cm has to be cut into 2 pieces in the ratio 3 : 4. What is the maximum length of the measuring stick which should be used to measure both the lengths ?
Answer:
The two pieces are 3/7 × 140 = 60 cm and 4/7 × 140 = 80 cm.
The stick must measure both lengths exactly, so its length divides 60 and 80. The maximum is H.C.F.(60, 80) = 20.
The longest stick is 20 cm.
In simple words: Find the two pieces first, then take the HCF of their lengths.

Teacher's Note:
1. Share 140 in the ratio 3 : 4 to get 60 and 80.
2. The stick fits 3 times into one piece and 4 times into the other.

Question 12. Show that \( \sqrt{11} \) is an irrational number.
Answer:
Suppose, to the contrary, that \( \sqrt{11} \) is rational. Then \( \sqrt{11} = \frac{a}{b} \) where a and b are co-prime positive integers.
Squaring: \( a^2 = 11b^2 \), so 11 divides a². Since 11 is prime, it divides a. Write a = 11c.
Then \( 11^2c^2 = 11b^2 \), so \( b^2 = 11c^2 \). Now 11 divides b², hence divides b.
So 11 is a common factor of a and b, contradicting that they are co-prime. Therefore \( \sqrt{11} \) is irrational.
In simple words: Assuming √11 is rational forces a and b to share the factor 11, which cannot be.

Teacher's Note:
1. Replace 3 by 11 in the standard proof.
2. 11 is prime, so 11 | a² implies 11 | a.

Question 13. Show that \( 4-\sqrt3 \) is an irrational number.
Answer:
Suppose \( 4-\sqrt3 \) is rational, equal to r.
Then \( \sqrt3 = 4-r \), which is a rational number.
That contradicts the fact that \( \sqrt{3} \) is irrational. So \( 4-\sqrt3 \) is irrational.
In simple words: A rational number minus an irrational number cannot be rational.

Teacher's Note:
1. Isolate √3 on one side.
2. Use the known irrationality of √3.

Question 14. Show that \( 2\sqrt5 \) is an irrational number.
Answer:
Suppose \( 2\sqrt5 \) is rational, equal to \( \frac{a}{b} \) with integers a, b (b ≠ 0).
Then \( \sqrt5 = \frac{a}{2b} \), a ratio of two integers, so \( \sqrt{5} \) would be rational.
But \( \sqrt{5} \) is irrational. Hence \( 2\sqrt5 \) is irrational.
In simple words: If 2√5 were a fraction, dividing by 2 would make √5 a fraction too.

Teacher's Note:
1. Rational × irrational (non-zero rational) is irrational.
2. Use the known result for √5.

Question 15. Show that \( \sqrt{8} \) is an irrational number.
Answer:
√8 = √(4 × 2) = 2√2.
Suppose 2√2 is rational, equal to a/b. Then \( \sqrt2 = \frac{a}{2b} \), which would be rational.
But \( \sqrt{2} \) is irrational (shown in Example 14). Hence \( \sqrt{8} \) is irrational.
In simple words: √8 simplifies to 2√2, and a non-zero rational multiple of an irrational number is irrational.

Teacher's Note:
1. Simplify the surd first.
2. The proof for √2 appears in the solved examples.

Question 16. Show that \( \sqrt5-\sqrt3 \) is an irrational number.
Answer:
Suppose \( \sqrt5-\sqrt3 \) is a rational number r. Since √5 > √3, r > 0.
Multiply by the conjugate: \( (\sqrt5-\sqrt3)(\sqrt5+\sqrt3) = 5-3 = 2 \), so \( \sqrt5+\sqrt3 = \frac{2}{r} \), which is rational.
Adding and subtracting: \( 2\sqrt5 = r+\frac2r \), so \( \sqrt5 = \frac12\left(r+\frac2r\right) \) is rational, contradicting the irrationality of \( \sqrt{5} \).
Hence \( \sqrt5-\sqrt3 \) is irrational.
In simple words: If the difference of the roots were rational, so would their sum be, and then each root would be rational.

Teacher's Note:
1. The product of the sum and difference is the rational number 2.
2. Alternatively isolate one root and square, as in Example 17.