IUC AcademyHome561. If , the values of a and b are
- A. 1, 2
- B. 2, 3
- C. 1, 3
- D. None of these
Answer: 1, 2
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562. A 3–digit number 4a3 is added to another 3–digit number 984 to give the four-digit number 13b7, which is divisible by 11. Then, (a + b) is
- A. 10
- B. 11
- C. 12
- D. 15
Answer: 10
563. C takes the value
- A. 0
- B. 2
- C. 2 or 3
- D. 5
Answer: 2
564. B takes the value
- A. 0
- B. 5
- C. 9
- D. 0 or 9
Answer: 9
565. In the following sum, '?' stands for which digit? ? + 1? + 2? +? 3 +? 1 = 21?
- A. 4
- B. 6
- C. 8
- D. 9
Answer: 8
Explanation: Let the required digit be x. Then x + 1x + 2x + x3 + x1 = 21x ⇒ x + 10 + x + 20 + x + 10x + 3 + 10x + 1 = 200 + 10 + x ⇒ 22x = 210 – 34 = 176 ⇒ x = 8. Hence, the required digit is 8.
Explanation: Let the required digit be x. Then x + 1x + 2x + x3 + x1 = 21x ⇒ x + 10 + x + 20 + x + 10x + 3 + 10x + 1 = 200 + 10 + x ⇒ 22x = 210 – 34 = 176 ⇒ x = 8. Hence, the required digit is 8.
566. What should be the maximum value of Q in the following equation? 5P9 – 7Q2 + 9R6 = 823
- A. 5
- B. 6
- C. 7
- D. 9
Answer: 7
Explanation: 5 P 9 – 7 Q 2 + 9 R 6 = 823 ⇒ (500 + 10P + 9) – (700 + 10Q + 2) + (900 + 10R + 6) = 823 ⇒ (500 + 900 – 700) + 10(P + R – Q) + (9 + 6 – 2) = 823 ⇒ 700 + 10(P + R – Q) = 810 = 700 + 110 ⇒ 10 (P + R – Q) = 110 ⇒ P + R – Q = 11 ⇒ Q = (P + R – 11). To get maximum value of Q we take P = 9 and R = 9. This gives Q = (9 + 9 – 11) = 7. Hence, the maximum value of Q is 7.
Explanation: 5 P 9 – 7 Q 2 + 9 R 6 = 823 ⇒ (500 + 10P + 9) – (700 + 10Q + 2) + (900 + 10R + 6) = 823 ⇒ (500 + 900 – 700) + 10(P + R – Q) + (9 + 6 – 2) = 823 ⇒ 700 + 10(P + R – Q) = 810 = 700 + 110 ⇒ 10 (P + R – Q) = 110 ⇒ P + R – Q = 11 ⇒ Q = (P + R – 11). To get maximum value of Q we take P = 9 and R = 9. This gives Q = (9 + 9 – 11) = 7. Hence, the maximum value of Q is 7.
567. If 6*43 – 46@9 = 1904, which of the following should come in place of *?
- A. 4
- B. 6
- C. 9
- D. None of these
Answer: None of these
Explanation: Let 6x43 – 46y9 = 1904. Clearly, y = 3 and x = 5. Hence * must be replaced by 5.
Explanation: Let 6x43 – 46y9 = 1904. Clearly, y = 3 and x = 5. Hence * must be replaced by 5.
568. What should come in place of * mark in the following equation? 1*5$4 ÷ 148 = 78
- A. 1
- B. 4
- C. 6
- D. 8
Answer: 1
Explanation: Let = 1 54 78 148 xy . Then, 10000 + 1000 x + 500 + 10y + 4 = 148 × 78 ⇒ 10000 + 1000x + 500 + 10y + 4 = 11544 = 10000 + 1000 + 500 + 40 + 4 ⇒ 1000x = 1000 ⇒ x = 1.
Explanation: Let = 1 54 78 148 xy . Then, 10000 + 1000 x + 500 + 10y + 4 = 148 × 78 ⇒ 10000 + 1000x + 500 + 10y + 4 = 11544 = 10000 + 1000 + 500 + 40 + 4 ⇒ 1000x = 1000 ⇒ x = 1.
569. What would be the maximum value of Q in the following equation? 5P7 + 8Q9 + R32 = 1928
- A. 6
- B. 8
- C. 9
- D. Data inadequate
Answer: 9
570. If p and q represent digits, what is the maximum possible value of q in the statement (S.S.C., 2010) 5p9 + 327 + 2q8 = 1114?
- A. 6
- B. 7
- C. 8
- D. 9
Answer: 7
Explanation: 5p9 + 327 + 2q8 = 1114 ⇒ (500 + 10p + 9) + (327) + (200 + 10q + 8) = 1114 ⇒ 10(p + q) + 1044 = 1114 ⇒ 10 (p + q) = 70 ⇒ (p + q) = 7 ⇒ Maximum value of q is 7 [As minimum value of p = 0]
Explanation: 5p9 + 327 + 2q8 = 1114 ⇒ (500 + 10p + 9) + (327) + (200 + 10q + 8) = 1114 ⇒ 10(p + q) + 1044 = 1114 ⇒ 10 (p + q) = 70 ⇒ (p + q) = 7 ⇒ Maximum value of q is 7 [As minimum value of p = 0]
571. What number should replace M in this multiplication problem? 3 M 4 × 4 1 2 1 6
- A. 0
- B. 2
- C. 4
- D. 8
Answer: 0
572. Find the missing number in the following addition problem: 8 3 5 4 * 8 + 9 * 4 2 2 * 7
- A. 0
- B. 4
- C. 6
- D. 9
Answer: 6
Explanation: Let the missing digit be x. Then, 1 (carried over) + 3 + x + x = 10 + x ⇒ x = 6.
Explanation: Let the missing digit be x. Then, 1 (carried over) + 3 + x + x = 10 + x ⇒ x = 6.
573. Both addition and multiplication of numbers are operations which are
- A. neither commutative nor associative
- B. associative but not commutative
- C. commutative but not associative
- D. commutative and associative
Answer: commutative and associative
Explanation: Both addition and multiplication of numbers are com-mutative and associative.
Explanation: Both addition and multiplication of numbers are com-mutative and associative.
574. If a = 7, b = 5, c = 3, then the value of a2 + b2 + c2 – ab – bc – ca is
- A. – 12
- B. 0
- C. 8
- D. 12
Answer: 12
Explanation: (a2 + b2 + c2 – ab – bc – ca) = (a + b + c) 2 – 3 (ab + bc + ca) = (7 + 5 + 3)2 – 3 (35 + 15 + 21) = (15)2 – 3 × 71 = (225 – 213) = 12.
Explanation: (a2 + b2 + c2 – ab – bc – ca) = (a + b + c) 2 – 3 (ab + bc + ca) = (7 + 5 + 3)2 – 3 (35 + 15 + 21) = (15)2 – 3 × 71 = (225 – 213) = 12.
575. If a + b + c = 0, (a + b) (b + c) (c + a) equals
- A. ab (a + b)
- B. (a + b + c)2
- C. – abc
- D. a2 + b2 + c2
Answer: – abc
Explanation: a + b + c = 0 ⇒ a + b = – c, (b + c) = – a and (c + a) = – b ⇒ (a + b) (b + c) (c + a) = (– c) × (– a) × (– b) = – (abc).
Explanation: a + b + c = 0 ⇒ a + b = – c, (b + c) = – a and (c + a) = – b ⇒ (a + b) (b + c) (c + a) = (– c) × (– a) × (– b) = – (abc).
576. If a and b are positive integers, a > b and (a + b) 2 – (a – b)2 > 29, then the smallest value of a is
- A. 3
- B. 4
- C. 6
- D. 7
Answer: 4
Explanation: Since (a + b) 2 – (a – b)2 = 4ab, so the given expression should be a multiple of 4. So, the least value of 4ab is 32 and so the least value of ab is 8. Hence, the smallest value of a is 4 and that of b is 2. Hence, a = 4.
Explanation: Since (a + b) 2 – (a – b)2 = 4ab, so the given expression should be a multiple of 4. So, the least value of 4ab is 32 and so the least value of ab is 8. Hence, the smallest value of a is 4 and that of b is 2. Hence, a = 4.
577. If a and b be positive integers such that a2 – b2 = 19, then the value of a is
- A. 9
- B. 10
- C. 19
- D. 20
Answer: 10
Explanation: (a2 – b2) = 19 ⇒ (a + b) (a – b) = 19. Clearly, a = 10 and b = 9.
Explanation: (a2 – b2) = 19 ⇒ (a + b) (a – b) = 19. Clearly, a = 10 and b = 9.
578. ? × (|a| × |b|) = – ab
- A. 0
- B. –1
- C. 1
- D. None of these
Answer: –1
579. 74844 ÷? = 54 × 63
- A. 22
- B. 34
- C. 42
- D. 54
Answer: 22
580. From the sum of 17 and – 12, subtract 48.
- A. – 43
- B. – 48
- C. – 17
- D. – 20
Answer: – 43
Explanation: Given Expression = 17 + (– 12) – 48 = 17 – 60 = – 43.
Explanation: Given Expression = 17 + (– 12) – 48 = 17 – 60 = – 43.
581. (46351 – 36418 – 4505) ÷? = 1357
- A. 2
- B. 3
- C. 4
- D. 6
Answer: 4
582. Product of 82540027 and 43253 is
- A. 3570103787831
- B. 3570103787832
- C. 3570103787833
- D. 3570103787834
Answer: 3570103787831
583. 414 ×? × 7 = 127512
- A. 36
- B. 40
- C. 44
- D. 48
Answer: 44
584. 8888 + 848 + 88 –? = 7337 + 737
- A. 1450
- B. 1550
- C. 1650
- D. 1750
Answer: 1750
585. 10531 + 4813 – 728 =? × 87
- A. 168
- B. 172
- C. 186
- D. 212
Answer: 168
586. 884697 – 773697 – 102479 =?
- A. 8251
- B. 8512
- C. 8521
- D. 8531
Answer: 8521
587. 38649 – 1624 – 4483 =?
- A. 32425
- B. 32452
- C. 34522
- D. None of these
Answer: None of these
Explanation: Given Expression = 38649 – (1624 + 4483) = (38649 – 6107) = 32542.
Explanation: Given Expression = 38649 – (1624 + 4483) = (38649 – 6107) = 32542.
588. 5566 – 7788 + 9988 =? + 4444
- A. 3223
- B. 3232
- C. 3322
- D. 3333
Answer: 3322
589. 1234 + 2345 – 3456 + 4567 =?
- A. 4590
- B. 4670
- C. 4680
- D. 4690
Answer: 4690
590. Let Xk = (p1p2 .......pk) + 1, where p1, p2, ........., pk are the first k primes. Consider the following: 1. Xk is a prime number. 2. Xk is a composite number. 3. Xk + 1 is always an even number. Which of the above is/are correct?
- A. 1 only
- B. 2 only
- C. 3 only
- D. 1 and 3
Answer: 1 and 3
Explanation: X2 = (2 × 3) + 1 = 7, which is prime and X2 + 1 = 8, which is even. X3 = (2 × 3 × 5) + 1 = 31, which is prime and X3 + 1 = 32, which is even. X4 = (2 × 3 × 5 × 7) + 1 = 211, which is prime and X4 + 1 = 212, which is even and so on. Thus Xk is prime and (Xk + 1) is even. Hence, 1 and 3 are true statements.
Explanation: X2 = (2 × 3) + 1 = 7, which is prime and X2 + 1 = 8, which is even. X3 = (2 × 3 × 5) + 1 = 31, which is prime and X3 + 1 = 32, which is even. X4 = (2 × 3 × 5 × 7) + 1 = 211, which is prime and X4 + 1 = 212, which is even and so on. Thus Xk is prime and (Xk + 1) is even. Hence, 1 and 3 are true statements.
591. What is the first value of n for which n2 + n + 41 is not a prime?
- A. 1
- B. 10
- C. 20
- D. 40
Answer: 40
Explanation: n = 1 ⇒ (n2 + n + 41) = (1 + 1 + 41) = 43, which is prime. n = 10 ⇒ (n2 + n + 41) = (100 + 10 + 41) = 151, which is prime. n = 20 ⇒ (n2 + n + 41) = (400 + 20 + 41) = 461, which is prime. n = 40 ⇒ (n2 + n + 41) = (1600 + 40 + 41) = 1681, which is divisible by 41. Thus, 1681 is not a prime number. Hence n = 40 for which (n2 + n + 41) is not prime.
Explanation: n = 1 ⇒ (n2 + n + 41) = (1 + 1 + 41) = 43, which is prime. n = 10 ⇒ (n2 + n + 41) = (100 + 10 + 41) = 151, which is prime. n = 20 ⇒ (n2 + n + 41) = (400 + 20 + 41) = 461, which is prime. n = 40 ⇒ (n2 + n + 41) = (1600 + 40 + 41) = 1681, which is divisible by 41. Thus, 1681 is not a prime number. Hence n = 40 for which (n2 + n + 41) is not prime.
592. The number of prime numbers between 301 and 320 are
- A. 3
- B. 4
- C. 5
- D. 6
Answer: 5
Explanation: Each of the numbers 302, 303, 304, 305, 306, 308, 309, 310, 312, 314, 315, 316 and 318 is clearly a composite number. Out of 307, 311, 313, 317 and 319 clearly every one is prime. Hence, there are 5 prime numbers between 301 and 320.
Explanation: Each of the numbers 302, 303, 304, 305, 306, 308, 309, 310, 312, 314, 315, 316 and 318 is clearly a composite number. Out of 307, 311, 313, 317 and 319 clearly every one is prime. Hence, there are 5 prime numbers between 301 and 320.
593. The smallest prime number, that is the fifth term of an increasing arithmetic sequence in which all the four preceding terms are also prime, is
- A. 17
- B. 29
- C. 37
- D. 53
Answer: 29
Explanation: The required arithmetic sequence of five prime numbers is 5, 11, 17, 23, 29 and therefore, the required 5th term is 29.
Explanation: The required arithmetic sequence of five prime numbers is 5, 11, 17, 23, 29 and therefore, the required 5th term is 29.
594. Which one of the following is a prime number?
- A. 161
- B. 221
- C. 373
- D. 437
Answer: 373
Explanation: 437 > 20 All prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19. 161 is divisible by 7, 221 is divisible by 13 and 437 is divisible by 19. 373 is not divisible by any of the above prime numbers. ∴ 373 is prime.
Explanation: 437 > 20 All prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19. 161 is divisible by 7, 221 is divisible by 13 and 437 is divisible by 19. 373 is not divisible by any of the above prime numbers. ∴ 373 is prime.
595. The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is
- A. 7
- B. 29
- C. 41
- D. 67
Answer: 67
Explanation: Let the required prime numbers be x, y, y + 36. Then, x + y + y + 36 = 100 ⇒ x + 2y = 64 Let x = 2. Then, 2y = 62 ⇒ y = 31. So, these prime numbers are 2, 31 and 67. In given choices 67 is the answer.
Explanation: Let the required prime numbers be x, y, y + 36. Then, x + y + y + 36 = 100 ⇒ x + 2y = 64 Let x = 2. Then, 2y = 62 ⇒ y = 31. So, these prime numbers are 2, 31 and 67. In given choices 67 is the answer.
596. Four prime numbers are arranged in ascending order. The product of first three is 385 and that of last three is 1001. The largest prime number is
- A. 9
- B. 11
- C. 13
- D. 17
Answer: 13
597. How many of the integers between 110 and 120 are prime numbers?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 1
Explanation: Each one of 112, 114, 116, 118 is divisible by 2. So, none is prime. Each one of 111, 114, 117 is divisible by 3. So, none is prime. Clearly, 115 is divisible by 5. So, it is not prime. Each one of 112 and 119 is divisible by 7. So, none is prime. Hence, there is only 1 prime number between 110 and 120, which is 113.
Explanation: Each one of 112, 114, 116, 118 is divisible by 2. So, none is prime. Each one of 111, 114, 117 is divisible by 3. So, none is prime. Clearly, 115 is divisible by 5. So, it is not prime. Each one of 112 and 119 is divisible by 7. So, none is prime. Hence, there is only 1 prime number between 110 and 120, which is 113.
598. The smallest three-digit prime number is
- A. 101
- B. 103
- C. 107
- D. None of these
Answer: 101
Explanation: Clearly, 100 is divisible by 2. So, 100 is not prime. (101) < (11)2 and prime numbers less than 11 are 2, 3, 5, 7. Clearly, 101 is not divisible by any of 2, 3, 5 and 7. Hence, 101 is the smallest 3-digit prime number.
Explanation: Clearly, 100 is divisible by 2. So, 100 is not prime. (101) < (11)2 and prime numbers less than 11 are 2, 3, 5, 7. Clearly, 101 is not divisible by any of 2, 3, 5 and 7. Hence, 101 is the smallest 3-digit prime number.
599. The smallest value of natural number n, for which 2n + 1 is not a prime number, is
- A. 3
- B. 4
- C. 5
- D. None of these
Answer: 4
Explanation: Putting n = 1, 2, 3, 4 respectively in (2n + 1) we get: (2 × 1 + 1) = 3, (2 × 2 + 1) = 5, (2 × 3 + 1) = 7 and (2 × 4 + 1) = 9, where 3, 5, 7 are prime numbers. ∴ The smallest value of n for which (2n + 1) is not prime, is n = 4.
Explanation: Putting n = 1, 2, 3, 4 respectively in (2n + 1) we get: (2 × 1 + 1) = 3, (2 × 2 + 1) = 5, (2 × 3 + 1) = 7 and (2 × 4 + 1) = 9, where 3, 5, 7 are prime numbers. ∴ The smallest value of n for which (2n + 1) is not prime, is n = 4.
600. If x and y are natural numbers such that x + y = 2017, then what is the value of (-1) x + (-1)y?
- A. 2
- B. -2
- C. 0
- D. 1
Answer: 0
601. Which of the following is not a prime number?
- A. 21
- B. 23
- C. 29
- D. 43
Answer: 21
Explanation: Clearly, 21 = 3 × 7, so 21 is not prime.
Explanation: Clearly, 21 = 3 × 7, so 21 is not prime.
602. The remainder obtained when any prime number greater than 6 is divided by 6 must be
- A. either 1 or 2
- B. either 1 or 3
- C. either 1 or 5
- D. either 3 or 5
Answer: either 1 or 5
603. A prime number N, in the range 10 to 50, remains unchanged when its digits are reversed. The square of such a number is
- A. 121
- B. 484
- C. 1089
- D. 1936
Answer: 121
Explanation: Clearly, 11 is a prime number which remains unchanged when its digits are interchanged. And, (11)2 = 121. Hence, the square of such a number is 121.
Explanation: Clearly, 11 is a prime number which remains unchanged when its digits are interchanged. And, (11)2 = 121. Hence, the square of such a number is 121.
604. The sum of all the prime numbers from 1 to 20 is
- A. 75
- B. 76
- C. 77
- D. 78
Answer: 77
Explanation: Sum of all prime numbers from 1 to 20 = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19) = 77.
Explanation: Sum of all prime numbers from 1 to 20 = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19) = 77.
605. The sum of the first four primes is
- A. 10
- B. 11
- C. 16
- D. 17
Answer: 17
Explanation: Sum of first four prime numbers = (2 + 3 + 5 + 7) = 17.
Explanation: Sum of first four prime numbers = (2 + 3 + 5 + 7) = 17.
606. The prime numbers dividing 143 and leaving a remainder of 3 in each case are
- A. 2 and 11
- B. 11 and 13
- C. 3 and 7
- D. 5 and 7
Answer: 5 and 7
Explanation: Each of the required numbers must divide (143 – 3) = 140 exactly. Now, 140 = 5 × 7 × 4. Hence, the required prime numbers are 5 and 7.
Explanation: Each of the required numbers must divide (143 – 3) = 140 exactly. Now, 140 = 5 × 7 × 4. Hence, the required prime numbers are 5 and 7.
607. The number of prime numbers between 0 and 50 is
- A. 14
- B. 15
- C. 16
- D. 17
Answer: 15
Explanation: Prime numbers between 0 and 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. Their number is 15.
Explanation: Prime numbers between 0 and 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. Their number is 15.
608. Consider the following statements: 1. If x and y are composite numbers, then x + y is always composite. 2. There does not exist a natural number which is neither prime nor composite. Which of the above statements is/are correct?
- A. 1 only
- B. 2 only
- C. Both 1 and 2
- D. Neither 1 nor 2
Answer: Neither 1 nor 2
Explanation: Statement 1. Let x = 4 and y = 15. Then, each one of x and y is a composite number. But, x + y = 19, which is not composite. ∴ Statement 1 is not true. Statement 2. We know that 1 is neither prime nor composite. ∴ Statement 2 is not true. Thus, neither 1 nor 2 is correct.
Explanation: Statement 1. Let x = 4 and y = 15. Then, each one of x and y is a composite number. But, x + y = 19, which is not composite. ∴ Statement 1 is not true. Statement 2. We know that 1 is neither prime nor composite. ∴ Statement 2 is not true. Thus, neither 1 nor 2 is correct.
609. The least prime number is
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 2
Explanation: The least prime number is 2.
Explanation: The least prime number is 2.
610. Which of the following is a factor of the given number?
- A. 2
- B. 3
- C. 4
- D. 9
Answer: 3
Explanation: The number is odd. So, it is not divisible by 2 or 4. Sum of digits = 9 + 7 + 5 + 2 + 1 = 24, which is divisible by 3 but not by 9. So, the given number is divisible by 3.
Explanation: The number is odd. So, it is not divisible by 2 or 4. Sum of digits = 9 + 7 + 5 + 2 + 1 = 24, which is divisible by 3 but not by 9. So, the given number is divisible by 3.
611. The largest digit in the number is
- A. 5
- B. 7
- C. 8
- D. 9
Answer: 9
Explanation: The largest digit in the number is 9.
Explanation: The largest digit in the number is 9.
612. The last digit of the number is
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 1
Explanation: The last digit of the number is 1.
Explanation: The last digit of the number is 1.
613. The second digit of the number is
- A. 5
- B. 7
- C. 9
- D. Cannot be determined
Answer: Cannot be determined
Explanation: The second digit of the number is either 5 or 7.
Explanation: The second digit of the number is either 5 or 7.
614. In the relation x > y + z, x + y > p and z < p, which of the following is necessarily true?
- A. y > p
- B. x + y > z
- C. y + p > x
- D. Insufficient data
Answer: x + y > z
615. If X < Z and X < Y, which of the following is necessarily true? I. Y < Z II. X2 < YZ III. ZX < Y + Z
- A. Only I
- B. Only II
- C. Only III
- D. None of these
Answer: Only II
Explanation: (X < Z and X < Y) ⇒ X2 < YZ.
Explanation: (X < Z and X < Y) ⇒ X2 < YZ.
616. If p > q and r < 0, then which is true?
- A. pr < qr
- B. p – r < q – r
- C. p + r < q + r
- D. None of these
Answer: pr < qr
Explanation: (p > q and r < 0) ⇒ pr < qr is true.
Explanation: (p > q and r < 0) ⇒ pr < qr is true.
617. If x is a real number, then x2 + x + 1 is
- A. less than 3 4
- B. zero for at least one value of x
- C. always negative
- D. greater than or equal to 3 4
Answer: greater than or equal to 3 4
618. If p is a positive fraction less than 1, then
- A. 1 is less than 1 p
- B. 1 is a positive integer p
- C. p2 is less than p
- D. 2 – is a positive number p p
Answer: 2 – is a positive number p p
619. If 0 < x < 1, which of the following is greatest?
- A. x
- B. x2
- C. 1 x
- D. 2 1 x
Answer: 2 1 x
620. If B > A, then which expression will have the highest value (given that A and B are positive integers)?
- A. A – B
- B. AB
- C. A + B
- D. Can't say
Answer: Can't say
Explanation: B > A ⇒ A < B ⇒ (A – B) < 0. Since A and B are both positive integers, we have (A + B) > 0 and AB > 0. If (A = 1 and B = 2), we have AB < (A + B). If (A = 2 and B = 3), we have (A + B) < AB. Thus, we cannot say which one of A + B and AB has the highest value.
Explanation: B > A ⇒ A < B ⇒ (A – B) < 0. Since A and B are both positive integers, we have (A + B) > 0 and AB > 0. If (A = 1 and B = 2), we have AB < (A + B). If (A = 2 and B = 3), we have (A + B) < AB. Thus, we cannot say which one of A + B and AB has the highest value.
621. Between two distinct rational numbers a and b, there exists another rational number which is
- A. 2 a
- B. 2b
- C. 2 ab
- D. + 2 ab
Answer: + 2 ab
Explanation: If a and b are two rational numbers, then + 2 ab is a rational number lying between a andb.
Explanation: If a and b are two rational numbers, then + 2 ab is a rational number lying between a andb.
622. The difference between the square of any two con-secutive integers is equal to
- A. sum of two numbers
- B. difference of two numbers
- C. an even number
- D. product of two numbers
Answer: sum of two numbers
Explanation: Let x and (x + 1) be two consecutive integers. Then (x + 1)2 – x2 = (x + 1 + x) (x + 1 – x) = (x + 1 + x) × 1 = (x + 1 + x) = sum of given numbers.
Explanation: Let x and (x + 1) be two consecutive integers. Then (x + 1)2 – x2 = (x + 1 + x) (x + 1 – x) = (x + 1 + x) × 1 = (x + 1 + x) = sum of given numbers.
623. If x is a rational number and y is an irrational number, then
- A. both x + y and xy are necessarily rational
- B. both x + y and xy are necessarily irrational
- C. xy is necessarily irrational, but x + y can be either rational or irrational
- D. x + y is necessarily irrational, but xy can be either rational or irrational
Answer: x + y is necessarily irrational, but xy can be either rational or irrational
Explanation: (a) Let x = 0 and y = 2 . Then, x is rational and y is irrational. ∴ x + y = 0 + 2 2 = which is irrational. Thus, x + y is not rational. (b) Let x = 0 and y = 2 . Then, x is rational and y is irrational. ∴ xy = 0 × 2 = 0, which is rational. Hence, xy is not irrational. (c) As shown in (b), xy is not necessarily irrational. (d) x + y is necessary irrational. But xy can be either rational or irrational. Hence, (d) is true.
Explanation: (a) Let x = 0 and y = 2 . Then, x is rational and y is irrational. ∴ x + y = 0 + 2 2 = which is irrational. Thus, x + y is not rational. (b) Let x = 0 and y = 2 . Then, x is rational and y is irrational. ∴ xy = 0 × 2 = 0, which is rational. Hence, xy is not irrational. (c) As shown in (b), xy is not necessarily irrational. (d) x + y is necessary irrational. But xy can be either rational or irrational. Hence, (d) is true.
624. If x = + 2 3, 5 y how does y change when x increases from 1 to 2?
- A. y increases from – 5 to – 5 2
- B. y increases from 2 to 5 5
- C. y increases from 5 to 5 2
- D. y decreases from – 5 to – 5 2
Answer: y increases from – 5 to – 5 2
Explanation: 1 ≤ x ≤ 2 ⇒ 1 ≤ + ≤ 2 32 5 y ⇒ ≤ ≤ ⇒ ≤≤ 2 55 (1 – 3) (2 – 3) (1 – 3) (2 – 3) 5 22 yy ⇒ ≤≤ –5 –5 . 2 y Hence, y increases from – 5 to –5 . 2
Explanation: 1 ≤ x ≤ 2 ⇒ 1 ≤ + ≤ 2 32 5 y ⇒ ≤ ≤ ⇒ ≤≤ 2 55 (1 – 3) (2 – 3) (1 – 3) (2 – 3) 5 22 yy ⇒ ≤≤ –5 –5 . 2 y Hence, y increases from – 5 to –5 . 2
625. If n = 1 + x, where x is the product of four consecu-tive positive integers, then which of the following is/are true? I. n is odd. II. n is prime. III. n is a perfect square.
- A. I only
- B. I and II only
- C. I and III only
- D. None of these
Answer: I and III only
Explanation: Let n = 1 + x = 1 + m (m + 1) (m + 2) (m + 3), where m is a positive integer. Then, clearly two of m, (m + 1), (m + 2), (m + 3) are even and so their product is even. Thus, x is even and hence n = 1 + x is odd. Also, n = 1 + m (m + 3) (m + 1) (m + 2) = 1 + (m2 + 3m) (m2 + 3m + 2) ⇒ n = 1 + y (y + 2), where m2 + 3m = y ⇒ n = 1 + y2 + 2y = (1 + y)2, which is a perfect square. Hence, I and III are true.
Explanation: Let n = 1 + x = 1 + m (m + 1) (m + 2) (m + 3), where m is a positive integer. Then, clearly two of m, (m + 1), (m + 2), (m + 3) are even and so their product is even. Thus, x is even and hence n = 1 + x is odd. Also, n = 1 + m (m + 3) (m + 1) (m + 2) = 1 + (m2 + 3m) (m2 + 3m + 2) ⇒ n = 1 + y (y + 2), where m2 + 3m = y ⇒ n = 1 + y2 + 2y = (1 + y)2, which is a perfect square. Hence, I and III are true.
626. If x and y are negative, then which of the following statements is/are always true? I. x + y is positive. II. xy is positive. III. x – y is positive.
- A. I only
- B. II only
- C. III only
- D. I and III only
Answer: II only
Explanation: If x < 0 and y < 0, then clearly xy > 0. So, whenever x and y are negative, then xy is positive. Note that (x < 0, y < 0) does not imply that (x + y) is positive. e.g. If x = – 2 and y = – 3 , then (x + y) = – 5 < 0. Note that (x < 0, y < 0) does not imply that (x – y) > 0. e.g. If x = – 5 and y = – 2, then x – y = – 5 – (– 2) = – 5 + 2 = – 3 < 0.
Explanation: If x < 0 and y < 0, then clearly xy > 0. So, whenever x and y are negative, then xy is positive. Note that (x < 0, y < 0) does not imply that (x + y) is positive. e.g. If x = – 2 and y = – 3 , then (x + y) = – 5 < 0. Note that (x < 0, y < 0) does not imply that (x – y) > 0. e.g. If x = – 5 and y = – 2, then x – y = – 5 – (– 2) = – 5 + 2 = – 3 < 0.
627. If x – y = 8, then which of the following must be true? I. Both x and y are positive. II. If x is positive, y must be positive. III. If x is negative, y must be negative.
- A. I only
- B. II only
- C. I and II
- D. III only
Answer: III only
Explanation: It is given that x – y = 8. I. We may have x = 5 and y = – 3. So, it is not necessary that both x and y are positive. II. If x is positive, then it is not necessary that y is positive, as x = 5 ⇒ y = – 3. III. If x < 0, then y = x – 8 which is clearly less than 0. So, if x is negative, then y must be negative.
Explanation: It is given that x – y = 8. I. We may have x = 5 and y = – 3. So, it is not necessary that both x and y are positive. II. If x is positive, then it is not necessary that y is positive, as x = 5 ⇒ y = – 3. III. If x < 0, then y = x – 8 which is clearly less than 0. So, if x is negative, then y must be negative.
628. If n is a negative number, then which of the following is the least?
- A. 0
- B. – n
- C. 2n
- D. n2
Answer: 2n
Explanation: n < 0 ⇒ 2n < 0, – n > 0 and n2 = (– n) 2 > 0. Thus, out of the numbers 0, – n, 2n and n2 we find that 2n is the least.
Explanation: n < 0 ⇒ 2n < 0, – n > 0 and n2 = (– n) 2 > 0. Thus, out of the numbers 0, – n, 2n and n2 we find that 2n is the least.
629. If m, n, o, p and q are integers, then m (n + o) (p – q) must be even when which of the following is even?
- A. m
- B. p
- C. m + n
- D. n + p
Answer: m
Explanation: When m is even, then m (n + 'o') (p – q) is even.
Explanation: When m is even, then m (n + 'o') (p – q) is even.
630. If A, B, C, D are numbers in increasing order and D, B, E are numbers in decreasing order, then which one of the following sequences need neither be in a decreasing nor in an increasing order?
- A. E, C, D
- B. E, B, C
- C. D, B, A
- D. A, E, C
Answer: A, E, C
631. If a and b are two numbers such that ab = 0, then
- A. a = 0 and b = 0
- B. a = 0 or b = 0 or both
- C. a = 0 and b ≠ 0
- D. b = 0 and a ≠ 0
Answer: a = 0 or b = 0 or both
632. If x is an odd integer, then which of the following is true?
- A. 5x – 2 is even
- B. 5x2 + 2 is odd
- C. 5x2 + 3 is odd
- D. None of these
Answer: 5x2 + 2 is odd
Explanation: x is odd ⇒ x2 is odd ⇒ 5x2 is odd ⇒ (5x2 + 2) is odd.
Explanation: x is odd ⇒ x2 is odd ⇒ 5x2 is odd ⇒ (5x2 + 2) is odd.
633. Which of the following is always odd?
- A. Sum of two odd numbers
- B. Difference of two odd numbers
- C. Product of two odd numbers
- D. None of these
Answer: Product of two odd numbers
Explanation: The product of two odd numbers is always odd.
Explanation: The product of two odd numbers is always odd.
634. If (n – 1) is an odd number, what are the two other odd numbers nearest to it?
- A. n, n – 1
- B. n, n – 2
- C. n – 3, n + 1
- D. n – 3, n + 5
Answer: n – 3, n + 1
Explanation: (n – 1) is odd ⇒ (n – 1) – 2 and (n – 1) + 2 are odd ⇒ (n – 3) and (n + 1) are odd.
Explanation: (n – 1) is odd ⇒ (n – 1) – 2 and (n – 1) + 2 are odd ⇒ (n – 3) and (n + 1) are odd.
635. For the integer n, if n3 is odd, then which of the following statements are true? I. n is odd. II. n2 is odd. III. n2 is even.
- A. I only
- B. II only
- C. I and II only
- D. I and III only
Answer: I and II only
Explanation: n3 is odd ⇒ n is odd and n2 is odd. ∴ I and II are true.
Explanation: n3 is odd ⇒ n is odd and n2 is odd. ∴ I and II are true.
636. If n and p are both odd numbers, which of the following is an even number?
- A. n + p
- B. n + p + 1
- C. np + 2
- D. np
Answer: n + p
Explanation: We know that the sum of two odd numbers is even. ∴ (n is odd, p is odd) ⇒ (n + p) is even.
Explanation: We know that the sum of two odd numbers is even. ∴ (n is odd, p is odd) ⇒ (n + p) is even.
637. If x, y, z and w be the digits of a number beginning from the left, the number is
- A. xyzw
- B. wzyx
- C. x + 10y + 100z + 1000w
- D. 103x + 102y + 10z + w
Answer: 103x + 102y + 10z + w
Explanation: Let the thousand’s, hundred’s, ten’s and one’s digits be x, y, z, w respectively. Then, the number is 1000 x + 100 y + 10 z + w = 103 x + 102 y + 10 z + w.
Explanation: Let the thousand’s, hundred’s, ten’s and one’s digits be x, y, z, w respectively. Then, the number is 1000 x + 100 y + 10 z + w = 103 x + 102 y + 10 z + w.
638. If x, y, z be the digits of a number beginning from the left, the number is
- A. xyz
- B. x + 10y + 100z
- C. 10x + y + 100z
- D. 100x + 10y + z
Answer: 100x + 10y + z
Explanation: Let the hundred's, ten's and one’s digits be x, y and z respectively. Then, the given number is 100 x + 10 y + z.
Explanation: Let the hundred's, ten's and one’s digits be x, y and z respectively. Then, the given number is 100 x + 10 y + z.
639. 98th term of the infinite series 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, ........ is
- A. 1
- B. 2
- C. 3
- D. 4
Answer: 2
Explanation: Given series is 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, .......... ∴ 96th term is 4. So, T97 = 1 and T98 = 2. Hence, the 98th term is 2.
Explanation: Given series is 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, .......... ∴ 96th term is 4. So, T97 = 1 and T98 = 2. Hence, the 98th term is 2.
640. 2 – 2 + 2 – 2 + .......... 101 terms =?
- A. – 2
- B. 0
- C. 2
- D. None of these
Answer: 2
Explanation: The given series is such that the sum of first hundred terms is zero, and 101st term is 2. So, the sum of 101 terms is 2.
Explanation: The given series is such that the sum of first hundred terms is zero, and 101st term is 2. So, the sum of 101 terms is 2.