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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.
| 1. |
Which of the following statements are true?(a) If a number is divisible by 3, it must be divisible by 9.(b) If a number is divisible by 9, it must be divisible by 3.(c) A number is divisible by 18, if it is divisible by both 3 and 6.(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.(e) If two numbers are co-primes, at least one of them must be prime.(f) All numbers which are divisible by 4 must also be divisible by 8.g) All numbers which are divisible by 8 must also be divisible by 4.(h) If a number exactly divides two numbers separately, it must exactly divide their sum.(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately. |
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Answer» b)True, If a number is divisible by 9, it must be divisible by 3. d)True,If a number is divisible by 9 and 10 both, then it must be divisible by 90. g)True, All numbers which are divisible by 8 must also be divisible by 4. h)True, If a number exactly divides two numbers separately, it must exactly divide their sum. |
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| 2. |
Write all the factors of the following numbers : (a) 24 (b) 15 (c) 21(d) 27 (e) 12 (f) 20(g) 18 (h) 23 (i) 36 |
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Answer» factors of the following are (a) `24: 1,2,3,4,6,8,12,24` (b)`15 : 1,3,5,15` (c) `21: 1,3,7,21` (d) `27: 1,3,9,27` (e) `1,2,3,4,6,12` (f) `1,2,4,5,10,20` (g)`18: 1,2,3,6,9,18` (h) `23: 1,23` (i) `36: 1,2,3,4,6,9,12,18,36` answer |
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| 3. |
Find the LCM of 20, 25 and 30 |
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Answer» `40 = 2*2*2*5` `48=2*2*2*2*5` `45= 3*3*5` LCM=`(2*2*2*2)*(3*3)*5` `=16*9*5` `=720` answer |
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| 4. |
Find the LCM of the following numbers :(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4Observe a common property in the obtained LCMs. Is LCM the product of twonumbers in each case? |
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Answer» 1)LCM=2*2*9=36 2)LCM=2*2*3*5=60 3)LCM=2*3*5=30 4)LCM=2*2*3*5=60. |
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| 5. |
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples. |
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Answer» example 1) 2*3*4=24 2)5*6*7=210 3)1*2*3=6 so, it can be seen that all the numbers are divisible by 6 |
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| 6. |
Find the LCM of 40, 48 and 45 |
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Answer» LCM(24)=2*2*2*3 LCM(90)=2*3*3*5 LCM(24,90)=`2^3*3^2*5`=360. |
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| 7. |
Express the following as the sum of two odd primes.(a) 44 (b) 36 (c) 24 (d) 18 |
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Answer» a)44=37+7 b)36=23+13 c)24=19+5 d)18=11+7. |
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| 8. |
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples. |
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Answer» `3+5 = 8` which is divisible by 4 as `8/4 = 2 ` `15+17=32` which is divisible by 4 as `32/4 = 8 ` `97+99=196` which is divisible by 4 as `196/4 = 49 ` So, above are some of the examples to verify the given statement. |
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| 9. |
Find the HCF of the following numbers :(a) 18, 48 (b) 30, 42 (c) 18, 60 (d) 27, 63 (e) 36, 84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75 |
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Answer» (a) `18 = 2*3*3` `48 = 2*2*2*2*3` HCF=`2*3 = 6` (b) `30 = 2*3*5` `42= 2*3*7` HCF=`2*3 = 6` (c)`18= 2*3*3` `60 = 2*2*3*5` HCF=`2*3=6` (d) `27=3*3*3` `63 = 3*3*7` HCF=`3*3 = 9` (e) `36= 2*2*3*3` `84= 2*2*3*7` HCF=`2*2*3 = 12` (f) `34 = 2*17` `102= 2*3*17` HCF =`2*17 = 34` (g)`70= 2*5*7` `105 = 3*5*7` `175 = 5*5*7` HCF=`5*7= 35` (h) `91= 7*13` `112 = 2*7*8` `49= 7*7` HCF=`7` (i) `18 = 2*3*3` `54= 2*3*3*3` `81 = 3*3*3*3` HCF=`3*3= 9` (j) `12=2*2*3` `45= 3*3*5` `75= 5*5*3` HCF=`3` answers |
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| 10. |
What is the HCF of two consecutive(a) numbers? (b) even numbers? (c) odd numbers? |
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Answer» 1. `(a) HCF= 1,2,3` ` 2=1*2 , 3=1*3` `:. HCF = 1` `(b) HCF = 2, 4, 6` `4=2*2 , 6=2*3` `:. HCF= 2` `(c) HCF= 1,3,5` ``3=1*3, 5=1*5 `:. HCF = 1` 2. `4=1*2*2` `15= 1*3*5` NO, `:. HCF= 1 ` |
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| 11. |
In a morning walk, three persons step off together. Their steps Measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps? |
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Answer» here we have to find the HCF of `850 & 680` `825= 2*5*5*17*1` `680=2*2*2*5*17*1` common factors are`2, 5,17` `:. HCF= 2*5*17= 170` answer |
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| 12. |
If 21y5 is a multiple of 9, where y is a digit, what is the value of y? |
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Answer» For a number to be a multiple of `9`, sum of all of its digits should be divisible by `9`. Sum of digits of gfiven number `= 2+1+y+5 = 8+y` So, `(8+y)` should be divisible by `9` if value of `y` is `1`. |
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| 13. |
Find the values of the letters and give reasons for the steps involved. `12A + 6AB =A09` |
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Answer» `12A+6AB = A09` Here, we can see that , `A+B = 9->(1)` `2+A = 0->(2)` `1+6 = A->(3)` From (2), we can see that possible values for `A` is `8` or `7`. For `A` to be `7`, `A+B` should be greater that `9` that is not possible from `(1)`. So, value of `A` is `8`. From (1), we can see that if `A` = `8`, then `B = 9 - 8= 1` So, `A = 8 and B=1` |
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| 14. |
The standard form of number 12756000000000 is |
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Answer» Standard form of a number `N` can be given by, `N = a**10^1b` In the given number `12756000000000`, there are `9` zeroes. `:. 12756000000000 = 12756**10^9` `=> 12756000000000 = 1.2756**10^4**10^9` `=> 12756000000000 = 1.2756**10^13`, which is the standard form of the given number. |
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| 15. |
Find the values of the letters and give reasons for the steps involved. `2AB + AB1=B18` |
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Answer» `2AB+AB1 = B18` If we look at the unit digits of the given equation, only value possible for `B` is `7` as `7`+`1` = `8`. So, `2A7+A71 = 718` Now, only possible value for `A` can be `4` as `4`+`7` =`11`. If we try `A=4` and `B=7`, then, it satisfies our addition. So, this is the correct answer. |
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| 16. |
Find the values of the letters and give reasons for the steps involved. `3A + 25 = B2` |
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Answer» `3A + 25 =B2` If we add only unit digits, then, `A+5 = 2` Only possible value for `A` is `7`. So, addition is `37+25 = 62`. So, `B = 2`. |
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| 17. |
Solve the cryptarithm:`B A x B3 5 7A` |
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Answer» `BAxxB3 = 57A` If we multiply any number `x` with 3 and we get a number `x` in unit digit, then `x` is either `5` or `0`. So, in the given question, `A` can be `5` or `0`. If, `A` is `5`, then, the result is `575`. Then, `3**B+1` can be `7`. So, `B` can be `2`. If we try `A=3` and `B=2`, it satisfies our multiplication. So, this is the right answer. |
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| 18. |
Write a digit in the blank space of each of the following numbers so that the numberformed is divisible by 11 :(a)92_389 (b)8_9484 |
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Answer» a)Let x be the digit to be inserted 9+3+2=14 8+x+9=17+x 17+x-14=x+3 If x+3=0 x=-3 If x+3=11 x=8 b)Let x be the digit to be inserted 4+4+x=8+x 8+9+8=25 Difference is 25-(8+x)=17-x 17-x=0 x=17 17-x=11 x=6. |
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| 19. |
Write the smallest digit and the greatest digit in the blank space of each of the followingnumbers so that the number formed is divisible by 3 :(a) (b) |
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Answer» a)6+7+2+4=19 least number to be interested is 21-19=2 largest number to be inserted is 27-19=8. b)4+7+6+5+2=24 least number to be interested is 24-24=0 largest number to be inserted is 33-24=9. |
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| 20. |
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12. |
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Answer» LCM of 6,8 and 12 is 24 now we want smallest 3 digit number which is a multiple of 24. ` 24 xx 5=120`Ans is 120. |
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| 21. |
Write the smallest 5-digit number and express it in the form of its prime factors. |
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Answer» 10000=2*2*2*2*5*5*5*5 so, 2 and 5 are the prime factors |
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| 22. |
Write all the prime numbers less than 15. |
| Answer» prime numbers less than 15=2,3,5,7,11,13. | |
| 23. |
In which of the following expressions, prime factorisation has been done?(a) 24 = 2 × 3 × 4 (b) 56 = 7 × 2 × 2 × 2(c) 70 = 2 × 5 × 7 (d) 54 = 2 × 3 × 9 |
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Answer» 1 and 4 are not prime factors 2 and 3 are prime factors. |
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| 24. |
Write seven consecutive composite numbers less than 100 so that there is no prime number between them. |
| Answer» prime number are=90,91,92,93,94,95,96. | |
| 25. |
Three sets of English, Hindi and Mathematicsbooks have to be stacked in such a way that all the books are stored topicwise and the height of each stack is the same. The number of English books is96, the number of Hindi books is 240 and the number of Mathematics books is336. Assuming that the books are of the same thickness, determine the numberof stacks of English, Hindi and Mathematics books. |
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Answer» Number of English books `=96` Number of Hindi books `=240` Number of Mathematics books `=336` Now, we have to find out the total number of stacks that can be formed when it is given that the books are of same thickness... For this we need to find out highest common factor of number of books that are 96, 240 and 336. HCF `=2*2*2*2*3 =48`Now divide the total number of books of each subject by HCF we will get the stack it will make Number of stacks that can be formed for English `=96/48=2` Number of stacks for Hindi `=240/48 = 5` Number of stacks for Mathematics `=336/48 =7` |
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