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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. |
`{:("Across",|, "Down"),(4. " The sign of integers less than zero",|, 1. " Each of the symbols in a number system"),(6. " A number that comes just after another",|, 2. " A number that comes just before another"),(7. " The whole numbers and negative natural numbers together",|, 3. " It is another name of zero"),(10. " Expressing a number in words",|, 5. " Integers greater than zero"),(11. " An integer which is neither positive nor negative",|, 8. " Representation of a number by digits"),(12. " The order of operations",|, 9. " Hundred crores"):}` |
| Answer» `{:("Across",|, "Down"),(4. " Negative",|, 1. " Digit"),(6. " Successor",|, 2. " Predecessor "),(7. " Integers",|, 3. " Cipher"),(10. " Numeration",|, 5. " Positive"),(11. " Zero",|, 8. " Numeral"),(12. " BODMAS",|, 9. " Arab"):}` | |
| 102. |
For a point starting at a point P and travelling along a straight line, time of travel is taken as t and the distance from P as s.The relation between s and t is found to be s = 12t – 2t2, where distances to the right are taken as positive numbers and to the left as negative numbers.i. Is the position of the point to the right or left of P, till 6 seconds?ii. Where is the position at 6 seconds?iii. After 6 seconds?(Here it is convenient to write 12t – 2t2 = 2t(6 – t). |
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Answer» i. S = 12t – 2t2 Distance to the point when time is 1 second = 12t – 2t2 = 12 × 1 – 2 × 12 = 12 – 2 = 10 m Distance to the point when time is 5 second = 12t – 2t2 = 12 × 5 – 2 × 52 = 60 – 50 = 10 m Since the distance to the point till 6 seconds is positive. So the position of the point is on the right of P. ii. Distance to the point when time is 6 second = 12t – 2t2 = 12 × 6 – 2 × 62 = 72 – 72 = 0 At the 6th second, the point is at P. iii. Distance to the point when time is 7 second (after 6 sec)= 12t – 2t2 = 12 × 7 – 2 × 72 = 84 – 2 × 49 = 84 – 98 = -14 metres This is a negative number, So the position of the point is on the left of P. |
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| 103. |
m and n are integers and \(\sqrt {mn}\) =10 . Which of the following cannot be a value of m + n ? (a) 25 (b) 52 (c) 101 (d) 50 |
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Answer» (d) 50 \(\sqrt {mn}\) =10 \(\Rightarrow\) mn = 100 \(\therefore\) The possible pairs of m and n are (m, n) = (1, 100), (2, 50), (4, 25), (5, 20), (10, 10) \(\Rightarrow\) m + n can be 101, 52, 29, 25, 20 So 50 cannot be a value of m + n. |
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| 104. |
If 2.5252525........ = \(\frac{p}{q}\) (in the lowest form) then what is the value of \(\frac{q}{p}\) ? (a) 0.4 (b) 0.42525 (c) 0.0396 (d) 0.396 |
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Answer» (d) 0.396 \(\frac{p}{q}\) = 2.525252............ 2.\(\bar {52}\) = \(2\frac{52}{99}\) = \(\frac{250}{99}\) \(\therefore\) \(\frac{q}{p}\) = \(\frac{99}{250}\) = 0.396. |
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| 105. |
What number should replaced M in this multiplication problem ? (a) 0 (b) 5 (c) 7 (d) 8 |
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Answer» The answer is: (a) 0 |
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| 106. |
What is the sum of two numbers whose difference is 45, and the quotient of the greater number by the lesser number is 4 ? (a) 100 (b) 90 (c) 80 (d) 75 |
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Answer» (d) 75 Let the lesser number be x. Then, Greater number = x + 45 Given, \(\frac{X+45}{X}\) = 4 \(\implies\)x + 45 = 4x \(\implies\) 3x = 45 \(\implies\) x = 15 Then, required sum = x + x + 45 = 30 + 45 = 75 |
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| 107. |
what is Real number? |
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Answer» The totality of all rational and all irrational numbers is called real numbers which is denoted by R. The natural numbers, whole numbers, integers, non-integral rationals \((\frac{4}{7},-\frac{7}{8})\)etc., are all contained in real numbers. |
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| 108. |
Evaluate the following: (a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23 (b) 17 x 6 – 4 – 2 + 20 – (22 + 18) (c) 16 x 144 ÷ 16 ÷ 9 + 16 + 15 – 20 (d) 12 x 36 ÷ 12 ÷ 3 + 5 + 6 – 2 (e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] |
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Answer» (a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23 (Given) = 44 ÷ 2 + (7 + 8) – 14 + 23 (To complete the bracket ÷ done first) = 44 ÷ 2 + 15 – 14 + 23 (Bracket completed second) = 22 + 15 – 14 + 23 (÷ completed third) = 37 – 37 (+ completed fourth) = 0 (- completed last) ∴ 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23 = 0. (b) 17 x 6 – 4 – 2 + 20 – (22 + 18) (Given) = 17 x 6 – 4 – 2 + 20 – 40 (Bracket completed first) = 102 – 4 – 2 + 20 – 40 (x completed second) = 102 – 4 – 22 – 40 (+ completed third) = 98 – 22 – 40 (÷ completed one by one) = 76 – 40 = 36 ∴ 17 x 6 – 4 – 2 + 20 – (22 + 18) = 36 (c) 16 x 144 ÷ 16 ÷ 9 + 16 + 15 – 20 (Given) = 16 x 9 ÷ 9+16 + 15 – 20 (÷ completed first) = 16 x 1 + 16 + 15 – 20 (÷ completed second) = 16 + 16 + 15 – 20 (x completed third) = 32 + 15 – 20 (+ completed fourth) = 47 – 20 (+ completed fifth) = 27 (- completed last) ∴ 16 x 144 ÷ 16 ÷ 9 + 16 + 15 – 20 = 27 (d) 12 x 36 ÷ 12 ÷ 3 + 5 + 6 – 2 (Given) = 12 x 3 ÷ 3 + 5 + 6 – 2 (÷ completed first) = 12 x 1 + 5 + 6 – 2 (÷ completed second) = 12 + 5 + 6 – 2 (x completed third) = 17 + 6 – 2 (+ completed forth) = 23 – 2 (+ completed fifth) = 21 (- completed last) ∴ 12 x 36 ÷ 12 ÷ 3 + 5 + 6 – 2 = 21 (e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] (Given) = 15 – [17 + 30 ÷ 6 – 12 + 7] (Inner bracket completed first) = 15 – [17 + 5 – 12 + 7] (÷ completed second) = 15 – [22 – 19] (+ completed third) = 15 – 3 (bracket completed forth) = 12 (- completed last) ∴ 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] = 12. |
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| 109. |
If m and n are positive integers, then the digit in the units' place of 5n + 6m is always (a) 1 (b) 5 (c) 6 (d) n + m |
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Answer» (a) 1 The units' digit of any positive integer power of 5 is always 5 and that of any positive integral power of 6 is always 6. \(\therefore\) Units’ digit of 5m + 6n = Units' digit of (5 + 6) = Units' digit of 11 = 1. |
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| 110. |
What is the highest power of 5 that divides 90 × 80 × 70 × 60 × 50 × 40 × 30 × 20 × 10 (a) 10 (b) 12 (c) 14 (d) 15 |
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Answer» (a) 10 All the numbers that are multiplied have 0 as units' digit, so all of them are divisible by 5 once. Also 50 can be divided by 52 . So the highest power of 5 that divides the given product = (9 + 1) = 10. |
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| 111. |
Unit’s digit of any number that can be expressed as a power of any natural number between 1 and 9.Justify this statement. |
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Answer» Unit’s digit of any number that can be expressed as a power of any natural number between 1 and 9:- (i) The units’ digit of any number expressed as a power of 2 is any of the digits 2, 4, 8, 6 as 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, ....... etc. (ii) The units' digit of any number expressed as a power of 3 is any of the digits 3, 9, 7, 1 as 31= 3, 32= 9,33 = 27, 34 = 81, 35 = 243, ....... etc. (iii) The units’ digit of any number expressed as a power of 4 is 4 if the power is odd and 6 if the power is even as 41 = 4, 42 = 16, 43 = 64, 44 = 256, ......etc. (iv) The units' digit of any number expressed as a power of 5 is always 5 as 51 = 5, 52 = 25, 53 = 125, ........ etc (v) The units' digit of any number expressed as a power of 6 is always 6 as 61 = 6, 62 = 36, 63 = 216, ........ etc. (vi) The units' digit of any number expressed as a power of 7 is any of the digits 7, 9, 3, 1 as 71 = 7, 72 = 49, 73 = 343, 74 = 2401, ........ etc (vii)The units' digit of any number expressed as a power of 8 is any of the digits 8, 4, 2, 6 as 81 = 8, 82 = 64, 83 = 512, 84 = 4096, ........ etc. (viii) The units' digit of any number expressed as a power of 9 is 9 if the power is odd and 1 if the power is even as 91 = 9, 92 = 81, 93 = 729, 94 = 6561, ........ etc. |
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| 112. |
If N = 123 × 34 × 52 , then the total number of even factors of N is (a) 25 (b) 121 (c) 144 (d) 84 |
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Answer» (c) 144 N = (22 × 3)3 × 34 × 52 = 26 × 37 × 52 \(\therefore\) Total number of factors of N = (6 + 1) × (7 + 1) × (2 + 1) = 7 × 8 × 3 = 168 Some of these are odd factors and some of these are even factors. The odd factors are formed with the combination of 3s and 5s. \(\therefore\) Total number of odd factors = ( 7 + 1) × (2 +1) = 8 + 3 = 24 \(\therefore\) Number of even factors = 168 – 24 = 144. |
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| 113. |
Five digit numbers are formed such that either all the digits are even or all the digits are odd. If no digit is allowed to be repeated in one number find the difference between the maximum possible number with odd digits and the minimum possible number with even digits(a) 77063 (b) 79999 (c) 72841 (d) 86420 |
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Answer» (a) 77063 Maximum possible number with odd digits = 97531 and the minimum possible number with even digits = 20468. \(\therefore\) Required difference = 97531 – 20468 = 77063 |
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| 114. |
Total number of factors of any number. |
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Answer» The number has to be resolved into prime factors. If the prime factorisation of the given number = 2m × 3n × 5p × 11q , then the total number of factors = (m + 1) × (n + 1) × (p + 1) × (q + 1). For example, if N = 24 × 56 × 74 × 131 , then Number of factors of N = (4 + 1) × (6 + 1) × (4 + 1) × (1 + 1) = 5 × 7 × 5 × 2 = 350. 24 = 16 \(\therefore\) factors of 16 are 1, 2, 4, 8, 16, i.e., 5 in number, i.e., (4 + 1) in numbers. Similarly 74 = 2401 \(\therefore\) factors of 2401 and 1, 7, 49, 343 and 2401, i.e., 5 in number and so on. |
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| 115. |
If a number 573 xy is divisible by 90, then what is the value of x + y ? |
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Answer» 573 xy is divisible by 90, i.e., (9 × 10) \(\implies\) 573 xy is divisible by both 9 and 10. \(\implies\) y = 0 as a number is divisible by 10 if its ones' digit = 0. Also, sum of digits = 5 + 7 + 3 + x + 0 = 15 + x For divisibility by 9, 15 + x = 18 \(\Rightarrow\) x = 3 \(\therefore\) x + y = 3 + 0 = 3. |
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| 116. |
If P is a 3-digit number and Q is the number formed by reversing the digits of P, then find the difference between the place value of the digits in the tenth place.A. 1B. 9C. 0D. Cannot say |
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Answer» Correct Answer - C Let the 3-digit numbers be xyz, the 3-digit number after reversing the digits of xyz is zyx. The difference between the place values of tenth digits = 10y-10y = 0 Hence, the correct option is (c). |
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| 117. |
There is one number which is formed by writing one digit 6 times. Such number is always divisible by: (e.g., 0.111111, 0.444444 etc.) (a) 7 (b) 11 (c) 13 (d) all of these |
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Answer» (d) all of these |
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| 118. |
Find the greatest number by which the expression 72n – 32n is always exactly divisible.(a) 4 (b) 10 (c) 20 (d) 40 |
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Answer» The answer is: (d) 40 |
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| 119. |
The Arabian sea has an area of 1491000 square miles. This area lies between which numbers? (a) 1489000 and 1492540(b) 1489000 and 1490540 (c) 1490000 and 1490100 (c) 1480000 and 1490000 |
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Answer» (a) 1489000 and 1492540 1489000 < 1491000 < 1492540 |
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| 120. |
The chart below shows the number of newspapers sold as per the Indian Readership Survey in 2018. Which could be the missing number in the table?Name of the News PaperRankingSold(in Lakh)A170B250C3?D410(a) 8 (b) 52 (c) 77(d) 26 |
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Answer» (d) 26 50 > 26 > 10 |
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| 121. |
Observe the commas and write down the place value of 7. 1. 56,74,56,345 2. 567,456,345 |
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Answer» 1. 56,74,56,345 Place value of 7 is 7 x 10,00,000 = 70,00,000 = Seventy Lakhs. 2. 567,456,345 Place value of 7 is 7 x 1,000,000 = 7,000,000 = Seven Million. |
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| 122. |
How many prime factors are there in the expression (12)43 × (34)48 × (2)57? (a) 282(b) 237 (c) 142 (d) 61 |
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Answer» The answer is: (a) 282 |
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| 123. |
Calculate \(\sqrt{3}(\sqrt{48}+\sqrt{32}-\sqrt{18})\) √3(√(48) + √(32) - √(18)) |
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Answer» \(\sqrt{3}(4\sqrt{3}+4\sqrt{2}-3\sqrt{2})\) \(=\sqrt{3}(4\sqrt{3}+\sqrt{2})\) \(=12+\sqrt{6}=12+2.449=14.449\) #14.4494897428
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| 124. |
10000 – 1 equals __ |
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Answer» 10000 – 1 equals to 9999 |
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| 125. |
999 + 1 equals __ |
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Answer» 999 + 1 equals to 1000 |
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| 126. |
The Predecessor of 8970 is __ |
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Answer» The Predecessor of 8970 is 8969 |
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| 127. |
The successor of 10 million is (a) 1000001 (b) 10000001 (c) 9999999(d) 100001 |
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Answer» (b) 10000001 The successor of 10 million is 10000001 |
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| 128. |
The expanded form of the number 6,70,905 is (a) 6 x 10000 + 7 x 1000 + 9 x 100 + 5 x 1 (b) 6 x 10000 + 7 x 1000 + 0 x 100 + 9 x 100 + 0 x 10 + 5 x 1 (c) 6 x 1000000 + 7 x 10000 + 0 x 1000 + 9 x 100 + 0 x 10 + 5 x 1 (d) 6 x 100000 + 7 x 10000 + 0 x 1000 + 9 x 100 + 0 x 10 + 5 x 1 |
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Answer» (d) 6 x 100000 + 7 x 10000 + 0 x 1000 + 9 x 100 + 0 x 10 + 5 x 1 |
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| 129. |
Name the property being illustrated in each of the cases. 1. (30 + 20) + 10 = 30 + (20 + 10) 2. 10 x 35 = (10 x 30) + (10 x 5) |
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Answer» 1. Associativity 2. Distribution of multiplication over addition. |
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| 130. |
The number of employees in the Indian Railways is about 10 lakhs. Write this in the International System of numeration. |
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Answer» The number of employees in international System of numeration is 1,000,000 (one million) |
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| 131. |
Write the largest six-digit number and write the number names in words using the Indian and International system. |
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Answer» The largest six-digit number is 999999 Number names are nine lakh ninety-nine thousand nine hundred and ninety-nine Indian System
International System
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| 132. |
1 billion is equal to (a) 100 crore (b) 100 million (c) 100 lakh (d) 10000 lakh |
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Answer» (a) 100 crore 1 billion is equal to 100 crore |
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| 133. |
10 crore = __ |
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Answer» 10 crore = 100 million |
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| 134. |
Say True or False. (i) 8567 is rounded off as 8600 to the nearest 10. (ii) 139 is rounded off as 100 to the nearest 100. (iii) 1,70,51,972 is rounded off as 1,70,00,000 to the nearest lakh. |
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Answer» (i) False (ii) True (iii) False |
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| 135. |
Give 3 examples where the number of things counted by you would be a 5 digit number or more. |
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Answer» 1. Number of stars in the sky. 2. The number of people living in Tamilnadu. 3. The number of accidents in India in the year 2017. |
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| 136. |
Complete the given order. Ten crore, crore, ten lakh, __, __, __, __, __. |
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Answer» Ten crores, Crore, Ten lakh, Lakh, Ten Thousand, Thousand, Hundred, Ten, One |
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| 137. |
The estimation to the nearest hundred of 76812 is (a) 77000 (b) 76000 (c) 76800 (d) 76900 |
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Answer» Answer is (c) 76800 |
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| 138. |
The Government spends rupees 2 crores for education in a particular district every month. What would be its expenditure for over 10 months? |
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Answer» Expenditure for one month = 2 crores, Expenditure for ten months = 2,00,00,000 x 10 = 20,00,00,000 Expenditure for 10 months = twenty crores. |
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| 139. |
Evaluate : (i) 27225 `divide` 55 (ii) 44616 `divide` 156 |
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Answer» Correct Answer - (i) 495 (ii) 286 `(i) 27225 divide 55` `(27225)/(55) = (5445)/(11) = 495 = 495` `(ii) 44616 divide 156` `(44616)/(156) = (3432)/(12) = 286 = 286` |
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| 140. |
There are ten lakh people in a district. What would be the population of 10 such districts? |
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Answer» Number of people in the district = 10,00,000 ∴ Population of 10 such districts = 10,00,000 x 10 = 1,00,00,000 Total population of 10 districts would be one crore. 10 lakh = 10,000 Hundreds |
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| 141. |
The number 9785764 is rounded off to nearest lakh as(a) 9800000 (b) 9786000(c) 9795600 (d) 9795000 |
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Answer» (a) 9800000 In ten thousand places, the digit is 8 5. So 9800000 |
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| 142. |
Estimate the product 39 × 53 |
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Answer» 39 ⇒ 40 53 ⇒ 50 Product 40 × 50 = 2000 |
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| 143. |
`25 + [14 - 18 + {12 " of " 5-(16 divide 4 xx 3 -2)}]` |
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Answer» Correct Answer - 71 `25 + [14-18+{12 "of "5-(16 divide 4 xx 3-2)}]` `=25+[-4 + {60-(4 xx 3-2)}]` `=25+[-4+{60-10}] = 25+[-4+50]` ` = 25 + 46 = 71` |
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| 144. |
Simplify : `3 + 2 - 6 divide 3 xx 7` |
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Answer» Correct Answer - -9 `3 + 2 - 6 divide 3 xx 7` `= 3 + 2 - 2 xx 7 = 3+ 2 - 14 = 5 - 14 = -9` |
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| 145. |
`"Simplify" : 12 divide 4 - 3 xx 6 + 7` |
| Answer» `12 divide 4 - 3 xx 6 + 7 = 3 - 3 xx 6 + 7 = 3 - 18 + 7 = 10 - 18 = -8` | |
| 146. |
Simplify: `4 xx 3-2 + 16 divide 8` |
| Answer» `4 xx 3 - 2 + 16 divide 8 = 4 xx 3 - 2 +2 = 12 - 2 + 2 = 12` | |
| 147. |
Which integer should be placed in the box such that the result is 2? `[{(20-18-square) + 2} xx 4] divide 14` |
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Answer» Correct Answer - 3 `[{(20-bar(18-square)) + 2} xx 4] divide 14 = 2` `[{20-bar(18-square)) + 2} xx 4] = 28, " since" (28)/(14) = 2` `{(20-bar(18-square)) + 2} = 7, " since"7 xx 4 =28` `(20-bar(18-square)) = 5, " since" 5 + 2 = 7` `bar(18-square) = 15, " since" 20-15 = 5` `rArr square = 3, " since"18-15 = 3` `therefore 3 " should be there in the box." ` |
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| 148. |
Estimate the following products: (a) 578 x 161 (b) 5281 x 3491 (c) 1291 x 592 (d) 9250 x 29 |
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Answer» (a) 578 x 161 578 ⇒ 600 161 ⇒ 200 Estimated product is 600 x 200 = 1,20,000 (b) 5281 x 3491 5281 ⇒ 5000 3491 ⇒ 3500 Estimated Product = 5000 x 3500 = 1,75,00,000 (c) 1291 x 592 1291 ⇒ 1300 592 ⇒ 600 Estimated Product is = 1300 x 600 = 7,80,000 (d) 9250 x 29 9250 ⇒ 9000 29 ⇒ 30 Estimated Product is 9000 x 30 = 2,70,000 |
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| 149. |
How many hundreds are there in 10 lakh? |
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Answer»
There are four places to the left of a Hundred. |
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| 150. |
Simplify : `36-369 divide[-72 divide 24 xx 5 + 2(17-bar(7-18))]` |
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Answer» Correct Answer - 27 `36-369 divide [-72 divide 24 xx 5 +2 (17-bar(7-8))]` `=36-369 divide [-72 divide 24 xx 24 xx 5 + 2 (17-7+18)]` `=36-369 divide [-72 divide 24 xx 5 + 2(28)]` `=36-369 divide [-3 xx 5 + 56] = 36-369 divide[41]` ` = 36-9 = 27` |
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