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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Find the greatest common factor of 280 and 144.1). 22). 83). 64). 4 |
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Answer» 280 = 2 × 2 × 2 × 5 × 7 144 = 2 × 2 × 2 × 2 × 3 × 3 ∴ GREATEST common factor of 280 and 144 = 2 × 2 × 2 = 8 |
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| 102. |
If the sum of squares of three consecutive natural number is 2702. Then, what will be the middle number?1). 292). 303). 314). 32 |
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Answer» Solution: Let us take the there consecutive natural numbers as x ,(x +1) and (x + 2). It is given that, The sum of squares of three consecutive natural numbers= 2702 So, x² + (x + 1)² + (x + 2)² = 2702 x² + x² + 1 + 2x + x² + 4 + 4X = 2702 3x² + 6x + 5 = 2702 3x² + 6x = 2702 - 5 3x² + 6x = 2697 Dividing the whole EQUATION by three we get, x² + 2x - 899 = 0 (x - 29)(x + 31) = 0 So, x = 29 So, the middle term,(x + 1) = 30 So, the correct option is 2). 30 |
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| 103. |
The remainder when 321 is divided by 5 is1). 12). 23). 34). 4 |
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Answer» ⇒ 321 = (34)5 × 31 ⇒ last digit of 321 = 1 × 3 = 3 ∴ 3 when divided by 5 will give REMAINDER = 3 |
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| 104. |
Calculate the total number of prime factors in the expression (4)11 × (5)5 × (3)2 × (13)2.1). 302). 313). 334). 32 |
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Answer» ⇒ (2²)11 × (5)5 × (3)2 × (13)2 ⇒ (2)22 × (5)5 × (3)2 × (13)2 ∴ Total number of prime factors = SUM of the powers of the expression = 22 + 5 + 2 + 2 = 31 |
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| 105. |
A number is divided by 52, we get 27 as remainder. On dividing the same number by 13, what will be the remainder?1). 22). 73). 14). None of these |
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| 106. |
Solve:4.2 × 4.2 ÷ √17.64 × √27.041). 34.52). 21.84 3). 22.834). 4.2 |
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Answer» We can WRITE, ⇒ √17.64 = √(1764/100) = 42/10 = 4.2 ⇒ √27.04 = √(2704/100) = 52/10 = 5.2 Hence, ⇒ 4.2 × 4.2 ÷ √17.64 × √27.04 = 4.2 × 4.2 ÷ 4.2 × 5.2 = 4.2 × 1 × 5.2 = 21.84 |
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| 107. |
A number leaves remainder 197, when divided by 931. If the same number is divided by 49, the remainder will be1). 332). 253). 54). 1 |
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Answer» Here,the number LEAVES remainder 197, when DIVIDED by 931. ∴ the number can be expressed as 931k + 197 When divided by 49, 931k will be perfectly divisible by 49. ∴ Remainder will be the same as that obtained after DIVIDING 197 by 49. ∴ Remainder = 1 |
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| 108. |
The H.C.F. of 2268 and 3388 is :1). 232). 253). 364). 28 |
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Answer» H.C.F. of 2268 and 3388 is Factors of 2268 = 2 × 2 × 3 × 3 × 3 × 3 × 7 Factors of 3388 = 2 × 2 × 7 × 11 × 11 The common factors are: 2, 2, 7 H.C.F. = 2 × 2 × 7 = 28 |
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| 109. |
1). 82). 63). 44). 3 |
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Answer» Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 24, 48 Factors of 92 = 1, 2, 4, 23, 46, 92 Factors of 140 = 1, 2, 4, 5, 7, 10, 14, 20, 70, 140 ∴ HCF of 48, 92 and 140 = 4 |
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| 110. |
How many numbers less than 320 are multiples of both 5 and 3?1). 292). 213). 204). 28 |
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Answer» Multiples of both 5 and 3 means multiples of 15 ∴ Multiples of 15 that are LESS than 320 are, ⇒ 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300, 315 ∴ There are 21 numbers less than 320 which are multiples of both 5 and 3 |
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| 111. |
The number 455 has1). Four prime factors2). Three prime factors3). Two prime factors 4). No prime factors |
| Answer» ∴ THREE PRIME factors | |
| 112. |
1). 450 seconds2). 600 seconds3). 900 seconds4). 1200 seconds |
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Answer» When the four runners will MEET at the starting point, they would have completed DIFFERENT NUMBER of rounds. The time at which they will meet will be the LCM of the individual time TAKEN by them LCM of 300, 100, 150 and 450 is 900 They will meet after 900 seconds |
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| 113. |
1). 42). 33). 24). 5 |
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Answer» ⇒ Let x be the number and y be the QUOTIENT. Then, ⇒ x = 49 x y + 32 ⇒ (7 x 7 x y) + (7 x 4) + 4 ⇒ 7 x (7Y + 4) + 4 ∴ REQUIRED remainder = 4 |
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| 114. |
Find the greatest number which on dividing 3050 and 5200 leaves remainders 7 and 9 respectively.1). 1492). 1113). 1534). 179 |
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Answer» On dividing 3050 and 5200 leaves remainders 7 and 9 respectively, ⇒ Number exactly DIVIDES 3050 – 7 = 3043 and 5200 – 9 = 5191 So the number which divides 3043 and 5191 exactly, FACTORS of 3043 = 1 × 17 × 179 Factors of 5191 = 1 × 29 × 179 ∴ 179 is the greatest number. |
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| 115. |
If the unit digit of 433 × 456 × 43N is (N + 2), then what is the value of N?1). 12). 83). 34). 6 |
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Answer» If N = 1 ⇒ 433 × 456 × 431 ⇒ Last digit = 8(By multiplying last digits of all 3 number) SINCE N + 2 ≠ 8, it will not SATISFY the GIVEN condition. If N = 8 ⇒ 433 × 456 × 438 ⇒ Last digit = 4, it will also not satisfy the given condition. If N = 3 ⇒ 433 × 456 × 433 ⇒ Last digit = 4, it will also not satisfy the given condition. If N = 6 ⇒ 433 × 456 × 436 ⇒ Last digit = 8 = (N + 2) ∴ N = 6 will satisfy the given condition. |
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| 116. |
The product of all the prime numbers between 85 and 95 is1). 77432). 7200993). 82774). 89 |
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| 117. |
1). 3482). 3423). 3644). 358 |
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Answer» Quotient = X, DIVISOR = y y = 9x y = 3 × remainder = 3 × 18 = 54. Quotient = x = y/9 = 54/9 = 6. Dividend = (divisor × quotient) + remainder = (54 × 6) + 18 = 342. |
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| 118. |
Which one of the following is a multiple of the number of factors of 640?1). 602). 283). 324). 54 |
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Answer» ⇒ 640 = 27 × 51 ⇒ The number of factors of 640 = 8 × 2 = 16 ⇒ Multiple of 16 = 32 from the given options ∴ option 3 is CORRECT |
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| 119. |
1). 32). 43). 54). 2 |
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Answer» Let the NUMBER be N. As per given data, when number is divided by 3, it gives remainder 2 ⇒ N = 3A + 2 Now, again, when quotient ‘a’ is divided by 2, then it leaves a remainder of 1 ⇒ a = 2b + 1 ⇒ N = 3(2b + 1) + 2 = 6b + 3 + 2 = 6b + 5 Clearly we can see that the number when divided by 6 will LEAVE the remainder 5 ∴ The remainder when the number is divided by 6 = 5 |
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| 120. |
1). Only I2). Only II3). Both I and II4). Neither I nor II |
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Answer» We know that, 32 > 22 (i.e. 9 > 4) Also, 33 > 23 (i.e. 9 > 8) The above equations imply that if L.H.S has a number and power both greater than number and power of the R.H.S then L.H.S is always greater than R.H.S. But in the GIVEN statements we FIND that in both the statements L.H.S has lesser number and lesser power than R.H.S. So, 2600 < 7900 3400 < 9500 ∴ Neither of the statements are true |
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| 121. |
If 347P is divisible by 9, then what is the value of P?1). 22). 33). 44). 7 |
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Answer» We KNOW that if the SUM of digits in a number is divisible by 9 then the number is divisible by 9. Now 3 + 4 + 7 + P should be divisible by 9. So, 14 + P should be divisible by 9. Also P is a single DIGIT number. The number after 14 which is divisible by 9 is 18. So P is = 18 - 14 = 4. |
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| 122. |
How many numbers are there from 2000 to 7000 which are both perfect squares and perfect cubes?1). 02). 13). 24). 3 |
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Answer» If a number is both a perfect square and a perfect cube. Then, it has to be some number to the POWER of (2 × 3) = 6 ⇒ 2000 < a6 < 7000 ⇒ a = 4 satisfies the above equation ⇒ required number = 46 = 4096 ∴ the correct option is 2) |
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| 123. |
1). 212). 163). 254). 30 |
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Answer» 325 (30 + 31 + 32 + 33) = 325(1 + 3 + 9 + 27) = 325 × 40 = 324 × 3 × 40 = 324 × 120 Now check with option only 30 can divide this. |
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| 124. |
1). 292). 303). 314). 28 |
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Answer» When 1359 is DIVIDED by 48, the QUOTIENT obtained is 28 and a remainder of 15 is left ∴ The quotient is 28 |
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| 125. |
What will be the ratio of HCF and LCM of 28 and 42?1). 2 : 72). 6 : 13). 1 : 64). 1 : 14 |
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Answer» 28 = 22 × 7 42 = 2 × 3 × 7 LCM of 28 and 42 = 84 HCF of 28 and 42 = 2 × 7 = 14 ∴ Ratio of HCF and LCM = 14 : 84 = 1 : 6 |
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| 126. |
1). 02). 33). 54). 35 |
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Answer» ⇒ Let X be the number and y be the QUOTIENT. Then, ⇒ x = 387 x y + 48 ⇒ (43 x 9 x y) + (43 x 1) + 5 ⇒ 43 x (9Y + 1) + 5 ∴ REQUIRED remainder = 5 |
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| 127. |
1). 132). 11 3). 94). 4 |
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Answer» ⇒ From question 1st NUMBER = 15 ? m + 13 ⇒ 2nd number = 15 ? n + 11 ⇒ SUM of TWO numbers = 15M + 13 + 15n + 11 ⇒ 15(m + n) + 24 ⇒ 15(m + n) + 15 + 9 ⇒ 15(m + n + 1) + 9 ∴ REMAINDER will be 9 |
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| 128. |
What is the sum of the first 11 terms of an arithmetic progression if the 4th term is 11 and the 7th term is -4?1). -752). 553). 114). 100 |
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Answer» If the first term and the common DIFFERENCE of an arithmetic progression are ‘a’ and ‘d’ respectively, then, nth term of AP = a + (n - 1)d Now, 4th term = a + 3d = 11----(1) 7th term = a + 6d = -4----(2) Subtracting (1) from (2), ⇒ 6d - 3d = -4 - 11 ⇒ 3d = -15 ⇒ d = -15/3 = -5 Substituting in (1), ⇒ a = 11 - 3(-5) = 11 + 15 = 26 ? Sum of n terms of AP = n/2 [2A + (n - 1)d] ∴ Sum of 11 terms = (11/2) × [2(26) + (11 - 1)(-5)] = (11/2) × (52 - 50) = 11/2 × 2 = 11 |
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| 129. |
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:1). 752). 853). 654). 45 |
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Answer» Since the numbers are co-prime, they contain only 1 as the common factor Also, the given two products have the middle NUMBER in common So, middle number ⇒ H.C.F. of 551 and 1073 = 29 ⇒ Third number = 1073/29 = 37 ⇒ Required sum = (19 + 29 + 37) = 85 ∴ Sum of the three numbers is 85 |
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| 130. |
If (1/21) + (1/22) + (1/23) _______ (1/210) = 1/k, then what is the value of k?1). 512/5112). 1024/10233). 511/5124). 1023/1024 |
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Answer» (1/21) + (1/22) + (1/23) + ? + (1/210) = 1/k ⇒ 1/k = 1/2 × [1 – (1/210)] / (1 – 1/2) [? SUM of N terms in G.P. = a [(rn – 1) / (r – 1)] ⇒ 1/k = 1/2 × [1 – 1/1024] / (1/2) ⇒ 1/k = 1 – 1/1024 ⇒ 1/k = 1023/1024 ∴ k = 1024/1023 |
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| 131. |
1). -32). -23). 44). 5 |
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Answer» Given that (x + a)is the HCF of 2x2 + 5x - 12 and x2 + x - 12 ∴ Substitute x = - a in the any of the above equations, we get ⇒ 2 × (-a)2 + 5 × (-a) - 12 = 0 ⇒ 2a2 - 5A - 12 = 0 ⇒ 2a2 - 8a + 3A - 12 = 0 ⇒ 2a (a - 4) + 3 (a - 4) = 0 ⇒ (2a + 3) (a - 4) = 0 ⇒ 2a + 3 = 0 and a - 4 = 0 ⇒ a = - 3/2 and a = 4 ∴ The VALUE of a is 4 |
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| 132. |
1). 1/142). 1/73). 3/144). 1/28 |
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Answer» = HCF of (3, 5, 17) / LCM of (7, 14, 28) = 1/28 |
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| 133. |
Find the maximum sum of the AP 21, \(20\frac{2}{5},\;19\frac{4}{5}, \ldots .\)1). ∞2). 3783). \(431\frac{5}{7}\)4). 448 |
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Answer» Summation of AP $(= \;\frac{n}{2}\left( {{a_0} + {a_n}} \right))$ Where, n is number of TERMS, a0 is first term and an is the last term. It is a decreasing AP. Thus, maximum sum will be OBTAINED TILL the last term is positive. GIVEN AP, 21, $(20\frac{2}{5},\;19\frac{4}{5}, \ldots .)$ C.d. = 20.4 – 21 = -0.6 an = a0 + (n – 1)d = 21 – 0.6(n – 1) For it to be positive : 21 – 0.6(n – 1) > 0 ⇒ 21 > 0.6(n – 1) ⇒ n < 36 ∴ n = 35 an = 21 – 0.6 × 34 = 0.6 ∴ Maximum sum of the given AP $(= \;\frac{{35}}{2}\left( {21 + 0.6} \right) = \;378)$ |
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| 134. |
1). 102). 143). 164). 18 |
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Answer» Product of numbers = LCM × HCF ⇒ x(x + 8) = 128 ⇒ x(x + 8) = 8(8 + 8) ⇒ x = 8 ∴ Greater number = 8 + 8 = 16 |
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| 135. |
Find the unit digit of (1466/711 + 27256/24332).1). 32). 43). 54). 7 |
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| 136. |
Which among the following is true for the given numbers?1). 13/33 < 32/47 < 20/47 < 25/472). 13/33 < 20/47 < 25/47 < 32/473). 13/33 < 20/47 < 32/47 < 25/474). 20/47 < 13/33 < 32/47 < 25/47 |
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Answer» Given, 13/33 = 0.39 20/47 = 0.42 32/47 = 0.68 25/47 = 0.53 ∴ 13/33 < 20/47 < 25/47 < 32/47 |
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| 137. |
1). 512). 593). 634). 77 |
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Answer» (x + 15)/9 = 115 The number is 1020 Correct answer = (1020/15) + 9 = 77 |
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| 138. |
The LCM of two numbers is 210. lf their HCF is 35 and one of the numbers is 105, find the other number.1). 352). 703). 1054). 140 |
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Answer» HCF × LCM = n1 × n2; n1 = number1, n2 = number2 Let x be the other number. 210 × 35 = 105 × x ⇒ x = 70 |
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| 139. |
Which of the following numbers is divisible by 12?1). 1832). 1923). 1624). 128 |
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Answer» A number is divisible by 12, if it is divisible by 3 and 4 A number is divisible by 3, if the SUM of its digits is divisible by 3, while a number is divisible by 4, if the LAST two digits of the number is divisible by 4 Hence, Considering number 183, ⇒ Sum of digits = 1 + 8 + 3 = 12, which is divisible by 3 ⇒ Last two digits = 83, which is not divisible by 4 Considering number 192, ⇒ Sum of digits = 1 + 9 + 2 = 12, which is divisible by 3 ⇒ Last two digits = 92, which is divisible by 4 Considering number 162, ⇒ Sum of digits = 1 + 6 + 2 = 9, which is divisible by 3 ⇒ Last two digits = 62, which is not divisible by 4 Considering number 128, ⇒ Sum of digits = 1 + 2 + 8 = 11, which is not divisible by 3 ⇒ Last two digits = 28, which is divisible by 4 ∴ The number 192 is divisible by 12 |
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| 140. |
When a number is divided by 208, the remainder is 3. Find the remainder, when the same number is divided by 13.1). 12). 33). 54). 7 |
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Answer» Let the dividend and the QUOTIENT be ‘X’ and ‘q’ respectively ⇒ x = 208q + 3 ⇒ x = (13 × 16)q + 3 ⇒ x = 13(16Q) + 3 ∴ The remainder is 3 |
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| 141. |
How many number between 1000 and 5000 are exactly divisible by 225?1). 162). 183). 194). 12 |
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Answer» First multiple of 225 after 1000 is 1125 = 225 × 5 And last multiple of 225 before 5000 is 4950 = 225 × 22 As numbers are DIVISIBLE by 225, they will form an arithmetic PROGRESSION with first TERM 1125 and last term 4950, with common DIFFERENCE 225. ∴ 1st term of this arithmetic progression is 1125 & last term of this arithmetic progression is 4950. ∴ nth term of A.P. = 1st term + (n – 1)d Where n is number of terms & d is common difference ⇒ 4950 = 1125 + (n – 1) × 225 ⇒ 3825 = (n – 1) × 225 ⇒ n – 1 = 17 ∴ n = 18 Thus there are total 18 numbers in between 1000 & 5000 which are divisible by 225. |
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| 142. |
The sum of LCM and HCF of two numbers is 89601. If the LCM is 89600 times the HCF and one of the numbers is 175 then determine the other number.1). 1752). 11203). 5124). 80 |
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Answer» Let LCM be ‘L’ and HCF be ‘H’. ⇒ l + h = 89601-----(1) Given l is 89600 times h ⇒ l = 89600h ⇒ 89600h + h = 89601 ⇒ 89601h = 89601⇒ h = 1 ∴ from equation (1), l = 89601 – h = 89601 – 1 = 89600 ∴ l = 89600 and h = 1 We KNOW that the PRODUCT of two NUMBERS = product of their LCM and HCF ⇒ xy = lhwhere x and y are the numbers ⇒ 175x = 89600 ⇒ x = 89600/175 = 512 ∴ the other number is 512 |
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| 143. |
1). 52). 03). 34). 9 |
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Answer» As, 5! = 1 × 2 × 3 × 4 × 5 will have 0 as its unit digit because of the presence of (2 × 5). HENCE all factorials above it will have 0 as their unit digit ⇒ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 ∴ Unit digit of (1! + 2! + 3! + 4! + _________ + 20!) = Unit digit of (1! + 2! + 3! + 4!) = 3 |
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| 144. |
What is the minimum number to be subtracted from 6321, which makes it completely divisible by 14?1). 82). 123). 74). 11 |
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Answer» On dividing 6321 by 14 ∴ 7 is the MINIMUM NUMBER. |
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| 145. |
What least number must be added to 750 so that the sum is completely divisible by 59?1). 152). 183). 114). 17 |
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Answer» Check through options: 1)15 750 + 15 = 765/59 = 12.96 2) 18 750 + 18 = 768/59 = 13.01 3) 11 750 + 11 = 761/59 = 12.89 4) 17 750 + 17 = 767/59 = 13 ∴ If 17 is ADDED into 750 then number is completely DIVISIBLE by 59. |
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| 146. |
What will be the remainder when (4182 - 41) is divided by 40?1). 22). 03). 544). 50 |
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Answer» ⇒ ? = (4182 - 41) ÷ 40 ⇒ ? = 41(4181 – 1) ÷ 40 ⇒ ? = 41(4181 – 181) ÷ 40 ⇒ ? = 41(41 – 1)(4181 + …. + 181) ÷ 40 ⇒ ? = 41(4181 + ….. + 181) ∴ REMAINDER is ZERO. |
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| 147. |
1). 2225/137282). 4883/171603). 1893/102484). 4933/17160 |
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Answer» The given SERIES can be written as, $(\RIGHTARROW \frac{1}{{3 \times 5}} + \frac{1}{{1 \times 4 \times 7}} + \frac{1}{{5 \times 7}} + \frac{1}{{4 \times 7 \times 10}} + \frac{1}{{7 \times 9}} + \frac{1}{{7 \times 10 \times 13}} + \frac{1}{{9 \times 11}} + \frac{1}{{10 \times 13 \times 16}})$ ⇒ 1/2 × [1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11] + 1/6[(7 - 1)/(1 × 4 × 7) + (10 - 4)/(4 × 7 × 10) + (13 - 7)/(7 × 10 × 13) + (16 - 10)/(10 × 13 × 16)] ⇒ 1/2[1/3 - 1/11] + 1/6[1/4 - 1/208] ⇒ 4/33 + 17/416 = 2225/13728 $(\frac{1}{{{4^2} - 1}} + \frac{1}{{1 \times 4 \times 7}} + \frac{1}{{{6^2} - 1}} + \frac{1}{{4 \times 7 \times 10}} + \frac{1}{{{8^2} - 1}} + \frac{1}{{7 \times 10 \times 13}} + \frac{1}{{{{10}^2} - 1}} + \frac{1}{{10 \times 13 \times 16}})$ |
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| 148. |
When 0.090909.....is converted into a fraction, then the result is1). 1/332). 1/113). 2/334). 6/11 |
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Answer» Let 0.090909…. = x ⇒ 0.0909090 is the given DECIMAL number ⇒ Here, 09 is repeated i.e., 2 digits are repeating ∴ Multiply the decimal by 102 i.e., 10 to the power of repeating digits ⇒ 9.090909…. ⇒ 9.090909… = 9 + 0.090909… ⇒ 100x = 9 + x ⇒ 99x = 9 ∴ x = 9/99 = 1/11 ∴ The FRACTION is 1/11 |
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| 149. |
Find the sum of all two digits numbers divisible by 4.1). 11882). 11823). 11864). 1184 |
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Answer» Required NUMBERS are: 12, 16, 20, ……, 96 This is an ARITHMETIC progression series in which FIRST term, a = 12 difference, d = 4 and last term l = 96. LET no. of terms in this series is n. ⇒ l = a + (n – 1) × d ⇒ 96 = 12 + (n – 1) × 4 ⇒ (n – 1) × 4 = 84 ⇒ n – 1 = 21 ⇒ n = 22 ∴ required sum = (n/2) × (a + l) = (22/2) × (12 + 96) = 11 × 108 = 1188 |
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| 150. |
1). 792). 433). 454). 49 |
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Answer» If the first term and the common difference of an arithmetic progression are ‘a’ and ‘d’ RESPECTIVELY, then, nth term of AP = a + (n - 1)d Now, 3rd term = a + 2d = 19----(1) Subtracting (1) from (2), ⇒ 5d - 2d = 37 - 19 ⇒ 3d = 18 ⇒ d = 18/3 = 6 Substituting in (1), ⇒ a = 19 - 2(6) = 7 ∴ 13th term of AP = a + 12D = 7 + 12(6) = 7 + 72 = 79 |
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