FOUNDATION & OLYMPIAD CLASS – 10

solutions

MATHEMATICS CLASS - X SOLUTIONS School Edition

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First Edition: 2014

Printed at:

1.

Class-X

Number Systems .....................................

5 - 15

2. Polynomials-III .........................................

16 - 24

3. Quadratic Equations-II ............................

25 - 36

4. Progressions ............................................

37 - 51

5. Mensuration-III ........................................

52 - 59

6. Co-ordinate Geometry-III ........................

60 - 78

7. Plane Geometry ......................................

79 - 90

8. Trigonometry-II ........................................

91 - 106

9. Probability ..............................................

107 - 116

10. Functions .................................................

117 - 127

11. Limits ......................................................

128 - 135

12. Bionomial Theorem .................................

136 - 144

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1. Number System

IIT Foundation & Olympiad Explorer

Chapter



NUMBER SYSTEMS

SOLUTIONS BASIC PRACTICE 1.

(i)

3.

105 245 (2 – 210

If n = 3q + 1 (i.e., n not divisible by 3)

35)105(3 – 105

⇒ n + 2 = 3q + 3 ⇒ n + 2 is divisible by 3.

(remainder)

0

If n = 3q + 2 (i.e., neither n nor n + 2 is divisble by 3)

∴ H. C. F. (105, 245) is 35. (ii)

⇒ n + 4 = 3q + 6 ⇒ n + 4 is divisible by 3.

305 793 (2 – 610 183) 305 (1 – 183 122) 183 (1 – 122 61) 122 (2 – 122 0

Here, one and only one out of n, n+2, n + 4 is divisible by 3. 4. (i)

where m = 3p3 (3p + 1)3 = 27p3 + 9p (3p + 1) + 1 = 9 × [3p3 + 3p2 + p] + 1 = 9m + 1. where m = 3p3 + 3p2 + p. (3p + 2)3 = 27p3 + 18p (3p + 2) + 8 = 9[3p3 + 6p2 + 4p) + 8 = 9m + 8

(iii) 410 3444 (8 – 3280

where m = 3p3 + 6p2 + 4p

164) 410 (2 – 328 82)164 (2 – 164 0

∴n3 is expressed in the form of

9m, 9m + 1, 9m + 8. 5.

Let if possible the number 7n ends with 0. ⇒10 is a factor of 7n

∴ H.C.F. (3444, 410) is 82. H.C.F. (120, 114) = 6

⇒ 7n = 10 × q for some natural number 2.

⇒ 10 is a factor of 7n i.e., {7 × 7 × 7 × .... × 7 (n factors)}

Hence, the maximum number of sections = 6 Mathematics/Solutions

We can express ‘n’ in either of the form 3p, 3p + 1, 3p + 2 for some integer p. (3p)3 = 27 p3 = 9 × 3p3 = 9m

∴ H.C.F of (305, 793) is 61.

2.

We can express n in either of the form 3q, 3q + 1, 3q + 2. (Use Euclid’s algorithm to prove it)

Class - X

5

1. Number System

IIT Foundation & Olympiad Explorer

Hence, it is impossible that 7n to have 10 as factor.

8.

∴ 7n cannot end with the digit 0.

6. (i)

⇒ q = (10)5 = (2 × 5)5 = 25 × 55

2 cannot be factor of ‘p’. ∴ p = 56125 and q = 100000 are co-prime.

k = 1650

Hence, q = 25 × 55.

(ii)

9.

12576 48624 (3 – 37728 10896) 12576 (1 – 10896 1680) 10896 (6 10080 816) 1680 (2 1632 48) 816 (17 816 0

a = 0.515151 .... ⇒ 100 a = 51.5151 .... = 51 + a ⇒ 99a = 51 ⇒ a=

51 p = 99 q

p = 51 and q = 99 are co-prime numbers. Hence, q = 32 × 11. 10. Let n and n + 1 be two consecutive integers.

∴ H. C. F. (12576, 48624) = 48

Let (n, n + 1) = d

⇒ 12k = 48 ⇒ k = 4

∴ d|n and d| n + 1

Perimeter of rectangular field = 2(100 + 80) = 360 m

d|(n + 1) – n or d|1

First cylist will complete one round in

∴ (n, n + 1) = 1

360 = 15 min 24

∴ d=1 i.e., n and (n + 1) are relatively prime. 11. a2 – b2 = (a – b) (a + b)

Second cyclist will complete one round

Since, a and b are odd primes.

360 in = 12 min 30

So let

Thirs cyclist will complete one round in

∴ a – b = 2k + 1 – 2k' – 1

a = 2k + 1 b = 2k' + 1

360 = 10 min 36

= 2(k – k') is even

∴ 15 = 3 × 5

a + b = 2k + 1 + 2k' + 1

= 2k – 2k'

12 = 2 × 2 × 3 = 22 × 3

= 2k + 2k' + 2

10 = 2 × 5

= 2(k + k' + 1) is even

∴ L.C.M. (15, 12, 10) = 22 × 3 × 5 = 60 minutes

6

56125 100000

p = 56125 and q = 100000

H.C.F. × L.C.M.= Product of two numbers ⇒ 40 × 252 × k = 2520 × 6600

7.

0.56125 =

Class - X

∴ Neither (a–b) nor (a+b) is equal to 1. ∴ Neither of the two divisors (a –b)

Mathematics/Solutions

1. Number System

IIT Foundation & Olympiad Explorer

and (a + b) of (a2 – b2) is equal to 1. ∴

n=

(a2 – b2) is composite.

[since, out of the two divisors of a prime number p, one must be equal to 1.]

Where p ∈ N

= n(n + 1) (n + 2) + 3n(n + 1)

∴ (2m – p) (2m + p) = 7 ∴ 2m + p being positive therefore

Now, n(n + 1) (n + 2) being the product of three consecutive integers is divisible by 3! = 6.

(2m + p) = 7 and 2m – p = 1 Hence, 4m = 8 ⇒ m = 2

n(n + 1) being the product of two consecutive integers is divisible by 2! =2

Thus, we have n2 + 19n + 92 = 4

⇒ n = – 8 or –11

∴ 3n(n + 1) is divisible by 6 ∴ n(n + 1) (n+ 5)

16. n4 – 20n2 + 4 = (n4 – 4n2 + 4) – 16n2

= n(n + 1) (n + 2) + 3n(n + 1) is divisible by 6.

= (n2 – 4n – 2) (n2 + 4n – 2)

... (1)

Note: It can be easily seen that none of the factors n2 – 4n – 2, n2 + 4n – 2 can have the value ± 1, whatever integral value n may have. Here four cases arises.

∴ n(n + 1) (n + 5) = M(6) 13. Let ‘n’ be any odd prime. If we divide any ‘n’ by 4, we get n = 4k + r where 0 ≤ r < 4 i.e., r = 0, 1, 2, 3 n = 4k + 3

4 ± 28 2

(i)

n2 – 4n – 2 = 1 ⇒ n =

(ii)

n2 – 4n – 2 = –1 ⇒ n =

⇒ either n = 4k or n = 4k + 1 or

)

∴ 4m2 – 7 is a square i.e., 4m2 – 7 = p2

12. n(n + 1) (n + 5) = n(n + 1) [(n + 2) + 3]

or n = 4k + 2

(

1 −19 ± 4m2 − 7 2

Clearly, 4n is never prime and 4n + 2 = 2(2n + 1)

(iii) n2 + 4n – 2 = 1 ⇒ n =

cannot be prime unless n = 0

(iv) n2 + 4n – 2 = –1 ⇒ n =

4k + 1 or 4k + 3 But 4k + 3 = 4(k + 1) – 4 + 3 = 4k' – 1 (where k' = k + 1)

∴ An odd prime n is either of the form 4k + 1 or (4k + 3) i.e., 4k' – 1 3(2 +1)

dk(12) =

2

17. This is same as asking what is remainder when 31997 ÷ 100

3(1 +1)

−1 3 −1 × 3 = 2044 23 − 1 3 −1

34 ≡ 81 mod 100

15. Let n2 + 19n + 92 = m2, m is a non negative integer. Then n2 + 19n + 92 – m2 = 0.

38 ≡ 61 mod 100 312 ≡ 41 mod 100 316 ≡ 21 mod 100

Solving for n, we get Mathematics/Solutions

−4 ± 28 2

−4 ± 20 2 From the above four cases, we find that whatever integral value n may have, n4 – 20n2 + 4 is the product of the integers n2 – 4n – 2 and n2 + 4n – 2 neither of which equals ± 1.

∴ An odd prime n is either of the form

14. 12 = 22 × 3

4 ± 20 2

Class - X

7

1. Number System

IIT Foundation & Olympiad Explorer

320 ≡ 1 mod 100

21. 39312 = 24 × 33 × 7 × 13

Now, 340, 360, 380, 3100, ..., 31980 all are ≡ 1 mod 100

There are four distinct primes in factorization of the given number. The required number of ways 23 = 8.

We know 316 ≡ 21 mod 100 317 ≡ 21 × 3 mod 100 3

17

FURTHER PRACTICE

≡ 63 mod 100

∴ 31997 ≡ 31980 × 317 since, 31980 ≡ 1 mod 100 and 317 ≡ 63 mod 100

∴ 31997 ≡ 63 mod 100 ∴ Last two digit is 63.

Let

2. (C)

225 – 1

H.C.F. of 225 –1 and 2115 –1 is of the form 2k – 1 where

∴ n = 8k + 3 for some positive integer k. ∴ n2 = 64k2 + 48k + 9 = 1 + a multiple of 8 19. 1000000. . . . . . .0000

k = H.C.F. of 25 and 115 = 5

∴ H.C.F. is 25 – 1 = 31 3. (C)

– 99

999999. . . . . . .9901 (1 followed by 99 zeros) The result is a 99 digit number having 97 nines followed by a zero and a 1. = 874 20. By Fermat’s theorem 10 ≡ 1 mod 7 Hence, 106m ≡ 1 mod 7 for all m. Now 10 ≡ 4 mod 6, 102 ≡ 40 ≡ 4 mod 6

≡ 4 mod (7)

consequently

...(3)

10 = 5 × 2 + 0

...(4)

From (1), 45 = 210 – 55 × 3

...(5)

From (2), 10 = 55 – 45 × 1

...(6)

From (3), 5 = 45 – 10 × 4

...(7)

≡ 4 × 10 mod 7

∴ y2 = (–19) × (–19) = 361

4. (D) 12 – 22 + 32 –42 + .....+ n2 = n is odd.

∴ The remainder is 5. 8

45 = 10 × 4 + 5

∴ H.C.F of (210, 55) expressed in 210 × x + 55 y, where x = 5 and y = –19

and 10(10n) = 106m . 104 ≡ 104 mod 7

≡ 5 mod 7

...(2)

5 = 210 × 5 + 55 (–19)

Thus, 10n = 6m + 4

+ ...+ 10

55 = 45 × 1 + 10

∴ 5 = 45 – (55 – 45 × 1) × 4

By induction 10n ≡ 4 mod 6 for all n.

10 + 10

...(1)

Express H.C.F in the form 210 x + 55 y.

6

(1010 )

210 = 55 × 3 + 45

∴ H.C.F. (210, 55) = 5

∴ The sum of the digits = (97 × 9)+0+1

(102 )

= (25)5 – 1

Also 2115 – 1 = (25)23 – 1

18. Since 640640640000 is a multiple of 8 643 ≡ 3 (mod 8)

10

p is a non terminating repeating q decimal expansion of rational number, for 2m × 5m.

1. (A)

n (n + 1) 2

if

Here n = 21, which is an odd number. Class - X

Mathematics/Solutions

1. Number System

IIT Foundation & Olympiad Explorer

∴ The required sum = 5. (C)

= L.C.M. × H.C.F. = 1080 × 180 = 194400

21 × 22 = 231 2

H.C.F × L.C.M. = Product of numbers

12. (C) m + n = 240 Given, m is a prime, n is a composite.

⇒ 12 × L.C.M. = 1800

6. (B)

1800 = 150 12 H.C.F. of co-prime numbers is 1.

7. (C)

(x – 4x + 4) (x + 3) = (x–2) (x + 3)

∴ L.C.M. (m, n) = m × n

L.C.M. = 2

⇒ 4199 = m × n ⇒ m=

2

4199 + n = 240 n ⇒ (n – 19) (n – 221) = 0

= (x – 2) (x – 2) (x +3)

∴ m + n = 240 ⇒

(x2 + 2x – 3)(x – 2) = (x + 3)(x – 1)(x – 2)

8. (B)

∴ Common factors are (x – 2) (x + 3).

∴ n = 19, n = 221

∴ H.C.F. of given polynomals are (x – 2) (x + 3).

for n = 19 ⇒ m =

4199 = 221 19 4199 = 19 for n = 221 ⇒ m = 221 ∴ The values of m and n are 19 and 221.

The given number is 337317. The units place digit is 7, by applying the method we divide the power by 4. 337 ÷ 4 = 84R1 Since, the remainder is 1, the new power is 1.

13. (C) xn + an is divisible by (x + a) when n is odd.

∴ 13 + 11 = 24

∴ (337)337 = (337)1

Hence, the units place digit is 7. 9. (B)

The number 99918 is exactly divisible by both 9 and 13.

14. (A) %#!' 

− !#

10. (D) N = (555 + 445) [2 × (555–445)] + 30

"

[By N = dq + r]

%" $

Given, L.C.M. + H.C.F. = 1260 ...(1)

−"

L.C.M. = 900 + H.C.F. ⇒ L.C.M – H.C.F. = 900

From, (1) and (2), we get 2(L.C.M.) = 1260 + 900 = 2160

 ...(2) 15. The unit digit of each term successively 1, 9, 1, 9, 1, 9, ..... The unit digit of sum of first two terms is 0.

⇒ L.C.M. = 1080

and H.C.F. = 1080 – 900 = 180

The unit digit of sum of first three terms is 1.

∴ Product of two numbers

Mathematics/Solutions

%# " −$&

= 1000 × 220 + 30 = 220030 11. (B)

4199 n

Class - X

9

1. Number System

IIT Foundation & Olympiad Explorer

The unit digit of sum of first four terms is 0. Hence, the digit in units place is 0 or 1 depending on number of terms i.e., even or odd respectively. So, the unit digit of the sum of 2008 terms is 0.

A

6. D

C

BRAIN WORKS 1. (a)

The triangles are, ABC, ADC, ADB and BCD.

200 ÷ (3 + 4 + 5) = 200 ÷ 12 = 16 R 8 Hence, 8th bulb will be the 200th i.e., green.

(b)

555 ÷ (3 + 4 + 5) = 555 ÷ 12 = 46 R 3

Hence, the number of triangles is 4. 7.

3.

th

The number 714285 repeats itself after the decimal point.

st 2nd 1 vertex vertex

3000 ÷ 6 = 500 R 0

A

B

E,F,G or H

Number of s 4

Sum = 500 × (7 + 1 + 4 + 2 + 8 + 5) = 500 × 27 = 13500

A

C

E, F, G or H

4

A B

D C

E, F, G or H E, F, G or H

4 4

B C

D D

E, F, F or H E, F, G or H

4 4

Total

4 6 = 24

Assume B dogs ran into B. A

B

2 dogs

18 dogs

10 sheep

10 sheep

10 cows

10 cows

Hence, the number of triangles = 24 × 2 = 48.

Answer : 10 + 10 + 7 = 27 Suppose Ashwin transferred 4 chocolate ice-creams from Box A to Box B.

8.

∴ No. of ice-creams transfered from Box B to Box A = 7 + 8 + 2 = 17 5.

Note : ‘Non-zero’ means that the number cannot be zero; ‘District’ means that the 6 numbers are all different. For D + E + F to be the largest,

Let a = an . . . a3a2a1 be an integer [Note a is not the product of a1, a2, a3, .... an but a1, a2, a3, ...,an are digits in the value of a. For example 368 is not the product of 3, 6 and 8 rather 3, 6, 8 are digits in value of 368 = 8 + 6 × 10 + 3 × (10)2] a = an ...a3a2a1

take A = 1, B = 2 and C = 3.

= a1 + (10)1a2 + (10)2a3 + (10)3a4 +... + (10)n–1 an

A + B + C + 3 (D + E + F)

= a1 + 10a2 + 100a3 + 1000a4 + ....

= 20 + 21 + 22 = 63

= a1 + (a2 + 9a2) + (a3 + 99a3) +

D + E + F = (63 – 6) + 3 = 19

10

3rd vertex

Case 2: Pick 2 points from line EH and 1 point from line AD. Another 24 triangles can be formed.

Worst case scenario, all sheep and cows return before the dogs return. 4.

Listing systematically Case 1: Pick 2 points from line AD and 1 point from line EH.

Hence, 3 bulb will be the 555 bulb, i.e., red. rd

2.

B

(a4 + 999a4) + ... Class - X

Mathematics/Solutions

1. Number System

IIT Foundation & Olympiad Explorer

= (a1 + a2 + a3 + a4 + ... ) + (9a2 + 99a3 + 999a4 + ...)

m is even, say 2p. ...(1)

Now, (k – 2p) (k + 2p) = 3n

or a = S + 9(a2 + 11a3 + 111a4 + ...)

⇒ k – 2p = 1 and k + 2p = 3n

where S = a1 + a2 + a3 + a4 + ...

⇒ 2p+1 + 1 = 3n

is the sum of digits in the value of ‘a’

Since (–1)n ≡ 3n (mod 4) = 2p+1 + 1 ≡ 1, n is even, say 2q.

∴ a – S = 9(a2 + 11a3 + 111a4 + ...) ∴ 9|(a – S)

Part 1:

Now (3q – 1) (3q + 1) = 2p+1 ⇒ 3q – 1 = 2

...(2)

a is divisible by 9

i.e., 9|a

⇒ 3q = 3 ⇒ q = 1 and hence p = 2 ...(3)

∴ 9|[a – (a – S)] [From (2) and (3)]

So, we have only one solution (4, 2). 11. If p = 2, 2p + 3p = 22 + 32 = 13 (not a perfect power).

i.e., 9|S i.e., sum of digits is divisible by 9.

Now, let now p be a prime > 2.

Part 2: S(sum of digits) is divisible by 9 i.e., 9|S

x + a divides xp + ap, whenever p is odd (factor theorem)

... (4)

∴ 2p + 3p is divisible by 2 + 3 = 5.

From (2) and (4), 9|[(a – S) + S]

We shall show that 2p + 3p is not divisible by 52.

i.e, 9|a i.e., the integer a is divisible by 9. 9.

xp + 3p = (x + 3) (xp–1 – 3xp–2 + 32xp–3 + ....

Since, m is an integer

+ (–3)p–1)

∴ Either m is even or m is odd Case-1: m is even

When x = –3 then, xp–1 – 3xp–2 + . . . + (–3)p–1

So, let m = 2k

= (–3)p–1 – 3(–3)p–2 + . . . . + (–3)p–1

∴ m + 2 = (2k) + 2 2

2

= p3p–1

= 4k + 2 2

Showing that x + 3 does not divide

= 2(2k2 + 1)

xp–1 – 3xp–2 + 32xp–3 + . . . + (–3)p–1

= (2 × an odd integer)

Consequently (x + 3)2 does not divide

Which is not divisible by 4.

xp + 3p

Case-2: m is odd

So, (2 + 3)2 does not divide 2p + 3p since 2p + 3p is a multiple of 5 but is not a multiple of 52.

Let m = 2k + 1

∴ m2 + 2 = (2k + 1)2 + 2

∴ It cannot be a perfect power. 12. Consider the product

= 4k2 + 1 + 4k + 2 = (2 + 4k + 4k2) + 1 Which being an odd integer is not divisible by 4. 10. Let 2m + 3n = k2

If possible, that P is a perfect cube = k3

Since (–1) ≡ 2 ≡ k ≡ 1 (mod 3) m

m

P = n(n + 1) (n + 2) (n + 3), where n is a natural number.

2

Two cases arises.

(since, 3|k) Mathematics/Solutions

Class - X

11

1. Number System

IIT Foundation & Olympiad Explorer

Case-1: If ‘n’ is odd.

or N = (n2 + 3n + 1)2

n, (n + 1), (n + 3) are all prime to n + 2

14. 1988 – 1 = (1 + 18)88 – 1 = (1 – 20)88 – 1

Now, we know that every common divisor of n + p and n + q must divide q – p.

In the binomial expansion of (1 + 18)88 first term is 1 and second term is 18 all other terms are divisible by 182.

∴ n + 2 and n(n + 1) (n + 3) are relatively prime.

∴ Maximum power of 3 divides (1 + 18)88 – 1 is 32. In the expansion of

Since, their product is a perfect cube, each of them must be a perfect cube.

(1 – 20)88, first term is 1 and second term is – 20 × 88 = – 25.55 and third term is

Since n < n(n + 1) (n + 3) < (n + 3) 3

3

∴ n(n + 1) (n + 3) = (n + 1)3 or (n + 2)3

88.87 = 26.52.11.87 and therefore 2 (1.20) 88 – 1 divisible by 2 6 . Hence maximum power of 2 which divides (1 – 20)88 – 1 is 25. 202 ×

As n(n +1) (n + 3) and (n + 2) are relatively prime, so second possibility ruled out. 3

Also n(n + 1) (n + 3) = (n + 1)3 ⇒ n = 1. Since P = 24, when n = 1 which is not a perfect cube. So the possibility n = 1 is also ruled out. So n cannot odd. Case-2: If ‘n’ is even. Then n + 1 is prime to n, n + 2 and n + 3. Consequency n + 1 is relatively prime to n(n + 2) (n + 3). Since the product of relatively prime numbers n + 1 and n(n + 2) (n + 3) is a perfect cube, each of them must be perfect cube.

Hence the sum of factors of 1988 – 1 which are of the form a2a. 3b where a, b > 0 is (2 + 22 + 23 + 24 + 25) (3 + 32) = 744 15. Since TTT = T × 111 = T × 37 × 3, therefore one of the numbers on the left-hand side must be a multiple of 37 and the other must be a multiple of 3. Since the only two-digit numbers which are multiples of 37 are 37 and 74, therefore one of the numbers must be either 37 or 74.

n3 < n (n + 2) (n + 3) < (n + 3)3

∴ n(n + 2) (n + 3) = either (n + 1)3 or (n + 2)3 since n(n + 2) (n + 3) and n + 1 are relatively prime ∴ First possiblity ruled out.

Also n(n + 2) (n + 3) = (n + 2)3

⇒ n+4=0 which is out of question. Consequently n cannot be even. 13. Let the consecutive positive integers be n, n + 1, n + 2 and n + 3. Consider the expression, N = n(n + 1) (n + 2) (n + 3) + 1 = (n2 + 3n) (n2 + 3n + 2) + 1 = (n2 + 3n)2 + 2(n2 + 3n) + 1 = [(n2 + 3n) + 1]2 12

Case 1: One of the numbers on L.H.S. is 37. The other number must be a two-digit number ending in 7, and must be a multiple of 3. The only possible such numbers are 27, 57, 87. Out of these 57, and 87 are too big in the sense that when either of these is multiplied by 37, we get a four digit number. Thus we find that the only choice for the other number is 27, and in this case we find that E = 7, T = 9, and Y + M = 5 (because either YE = 37, ME = 27, or ME = 37, YE = 27), so that E + M + Y + T = 21. Case 2: One of the numbers on the L.H.S. is 74. Arguing as in case I we find that the other number can be one of the numbers 24, 54, 84. Since all of them are too big, therefore we find that it is not possible for either number on LHS to be 74. Thus we find that E + M + Y + T = 21

Class - X

Mathematics/Solutions

1. Number System

IIT Foundation & Olympiad Explorer

When any whole number is divided by '2', then the remainder is '0' or '1'.

MULTIPLE ANSWER QUESTIONS 1.

(A, D) a=

12, b =

8.

3

−2 −8 −1 −4 4 4 × × = ; = 5 4 20 5 4 20

a 12 12 = = = 4 =2 b 3 3 2.

−1 5 −5 −3 2 −6 ; ; × = × = 4 5 20 10 2 20 −7 1 −7 × = 20 1 20

(B, C, D) 17 = 0.085 → terminating decimal 23 × 5 2

Obviously,

25 = 0.3472 ..... → Non-terminating 3 × 23 68 = ..... 22 × 5 2 × 7 2 0.0138

Hence,

→ Non- terminating 125 = 0.283 .... → Non-terminating 33 × 7 2

∴ (B),(C),(D) are non-terminating decimals. (C, D) 64 = 26 If the last six digits are divisible by 64, the entire number is divisible by 64. Rule : If the last n digits of a number is divisible by 2n, then the whole number is divisible by 2n. 4.

(B, C)

9.

between

= 1.4142135 ..... (irrational)

= 1.7320508 ..... (irrational) 4 = 2 (rational) 5 = 2.236 ..... (irrational) 10. (B, D) a×c a c a c × = ÷ b d b × d and b d a 1 a d × × = = c b b c d Hence, (B) and (D) are true.

CONCEPT DRILL 1.

(B, D)

Since 4x – y is a multiple of 3.

∴ 4x – y = 3m ∴ y = 4x – 3m On putting value of y in 4x2 + 7xy – 2y2

N ⊂ W ⊂ Z ⊂ Q ⊂ R Hence, R ⊃ Q ⊃ Z ⊃ W ⊃ N

= 4x2 + 7x(4x – 3m) – 2(4x – 3m)2

Supersets of whole numbers are rational numbers and real numbers. 7.

lie

3

(A, B) A six digit number consisting of one digit is always divisible by 7 and 11. Since, it satisfies divisibility rule of 7 and 11.

6.

−1 −7 −3 , and 4 20 10

−2 −1 and . 5 5 (B,C,D) 2

It is an integer and an irrational number. 5.

−5 −6 −7 , , are in between 20 20 20

−8 −4 and . 20 20

2

3.

(A,B,C)

= 4x2 + 28x2 – 21xm – 2(16x2 + 9m2 – 24xm)

(A, B)

Mathematics/Solutions

Class - X

13

1. Number System

IIT Foundation & Olympiad Explorer

S = (1 + 2 + 22 + ... + 2m–1) + p(1 + 2 + 22 + ... + 2m – 2)

= 4x2 + 28x2 – 21xm – 32x2 – 18m2 + 48xm = 27mx – 18m2 = 9m (3x – 2m)

1(2m − 1) p[1(2m −1 − 1)] + 2 −1 2 −1 = 2m – 1 + p(2m – 1 – 1)

∴ 4x2 – 7x – 2y2 is divisible by 9. 2.

=

Since n4 – n2 + 64 > n4 – 2n2 + 1 = (n 2 – 1) 2 for some non-negative integer k, (n4 – n2 + 64) = (n2 + k)2 = n4 + 2n2k + k2

= 2m – 1 (2m – 1) = n 6.

64 − k 2 from which we find 2k + 1 that the possible values 64, 1, 0 for n2 are obtained when k = 0, 7, 8 respectively. i.e., n2 =

Writting A ≡ 1n + 8n – 3n – 6n = (8n – 3n) – (6n – 1n)

Hence, n ∈ (0, ± 1, ± 8). 3.

we find that 8n – 3n and 6n – 1n are both divisible by 5, and consequently A is divisible by 5(= 8 – 3 = 6 – 1). Again, writing A = (8n – 6n) – (3n – 1n), we find that A is divisible by 2 (= 8 – 6 = 3 – 1) Hence, A is divisible by 10.

39 + 312 + 315 + 3n = 39(1 + 34 + 36 + 3n – 9) = (33)3 {1 + 3.32 + 3(32)2 + (32)3 + 3n – 9 – 3(32)2} = (33)3 (1 + 32)3 provided 3n – 9 – 35 = 0 = (270)3 provided 3n – 9 = 35 i.e., provided n = 14 So given number is a perfect cube when n = 14.

4.

If N is a multiple of 17, then N8 is also a multiple of 17, and therefore N8 is of the form 17m for some positive integer m.

PARAGRAPH QUESTIONS Passage – I 1.

∴ Aiswarya’s score = 9 correct + 16 wrong = (9 × 4) + (16 × (–2)) = 36 – 32 = 4 marks

i.e., (N8 – 1) (N8 + 1) ≡ 0 (mod. 17).

5.

As we know by the definition of perfect numbers that if the sum of the divisors of a number n, other than itself, is equal to n, then n is called a perfect number. Let n = 2 number.

m–1

× p where p = 2 – 1 is prime m

2.

(A) Gayatri’s score = 5 correct + 20 wrong = (5 × 4) + (20 × (–2)) = 20 – 40 = (–20) marks

Passage – II 1.

Divisors of 2m –1 × p are 1, 2, 22, 23, ..., 2m–1, p, 2p, 22 p, ..., 2m–2p, 2m – 1p. Now we should sum all these divisors except the last one, i.e., 2m – 1p 14

(C) Correct answer = +4 Wrong answer = (–2)

If N is prime to 17, then by Fermat’s theorem, N16 – 1 ≡ 0 (mod. 17), i.e., N8 = 17m ± 1, for some positive integer m.

Since 10 is the product of two primes 2 and 5, it will suffice to show that the given expression is divisible both by 2 and 5. To do so, we shall use the simple fact that if a and b be any positive integers, then an – bn is always divisible by a – b.

Class - X

(D)

44n − 53n = (256)n − (125)n

= 256 – 125 | 44n − 53n = 131 | 44n − 53n (Since a – b is one of the factors of an – bn) Hence, it is divisible by ‘131’. Mathematics/Solutions

1. Number System

IIT Foundation & Olympiad Explorer

2.

(A)

32n + 7n is divisible by 16.

3.

(C)

24n − 1 = (24 )n − (1)n

Hence, option (A) is the correct answer.

= 16n − 1n

⇒ –5 < n < 1 (n ∈ Z)

= 15|16n − 1n Hence, it is divisible by ‘15’.

⇒ n = {–4, –3, –2, –1, 0}

⇒ Possible values for n = 5.

PREVIOUS CONTEST QUESTIONS Total numbers in the set

Hence, (C) is the correct option. 4.

= (800 – 200) + 1

Since a, c are neither prime nor composite it means a = c = 1 and b = 2,

= 601

3, 5, 7; d = 2, 3, 5, 7 but b ≠ d.

Total numbers which are divisible by

Now ab + 4 = cd

5 = (800 − 200) +1 = 121 5

⇒ ab = 13 and cd = 17 is the only possible combination.

Total numbers which are divisible by

∴ ba = 31 and dc = 71

7 = (798 − 203) +1 = 86 7

⇒ 31 + 40 = 71

Total numbers which are divisible by both 5 and 7

⎛ 770 − 210 ⎞ =⎜ ⎟ +1 = 17 35 ⎝ ⎠

⇒

unit 4 + 92 + 43 + 94 + 45 + 96 + ...... + 499 + 9100 " " !  " " ! " " ! digit  !  ↓ ↓ ↓ ↓ 5 5 5 5

Hence, numbers in the given set which are neither divisible by 5 nor by 7 = 601 – 190 = 411. The value of the expression 2 45 – 15 x − will be maximum only 7 2 when the value of 15 x − is minimum. 7 But the minimum possible value of

2 = 0 (because absolute value 7 will never be negative.)

Thus 5 + 5 + 5 + ..... + 5(50 times) = 5 × 50 = 250. Hence, units digit is 0. 6.

Mathematics/Solutions

3640 – 40 = 3600 ⇒

3600 = 60

Therefore, there are 60 plants in each row. 7.

1 1 1 1 + (n +1) + (n +2) + (n +3) in i i i

1 1 1 1 n + n − n − n = 0. i i i i Such prime numbers are 11, 13, 31, 17, 71, 37, 73, 79 and 97. =

15x −

⇒ 45 − 15x −

ab + ba 13 + 31 44 1 = = = cd + dc 17 + 71 88 2

5. (A) The unit digit of each pair is 5 and there are 50 such pairs in all.

Total numbers which are either divisible by 5 or 7 or both = (121 + 86) – 17 = 190.

2.

2

⇒ (n + 2) 2 < 9

= (16 − 1)|16 n − 1n

1.

9 − (n + 2)2 > 0 ⇒ − ( n + 2) > −9

3.

8.

2 = 45 − 0 = 45 7

They are 9 in number. Class - X

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