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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Robin Hood and the pedlar fought till A) they both did sweat B) blood started flowing in streams C) the evening D) Little John asked them to stop. |
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Answer» B) blood started flowing in streams |
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| 2. |
Who stopped the fight between the pedlar and Robin Hood? |
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Answer» Robin Hood himself. |
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| 3. |
Why did Robin Hood laugh when Little John lost the fight? |
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Answer» He thought it was easy to defeat the pedlar. Little John was a well-built man so he laughed at Little John for being beaten by the pedlar. |
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| 4. |
The pedlar said that he would give up the whole pack if Little John A) defeated him in the fight B) moved him one perch from the pack C) requested him D) defeated Robin Hood in a fight. |
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Answer» B) moved him one perch from the pack |
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| 5. |
Little John said …………. belonged to him. A) the pedlar’s pack B) one half of the pedlar’s pack C) one suit and one bowstring D) None of the above. |
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Answer» D) None of the above. |
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| 6. |
A pedlar with his backpack was happily walking A) in a town B) in the forest C) over the grassy land D) over the hills. |
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Answer» C) over the grassy land |
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| 7. |
The Bold Pedlar and Robin Hood by Francis J.Child About The Poet: |
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Answer» Francis James Child (1825-1896) was an American scholar and a professor at Harvard University. He was an academician, educator, and folklorist, but he is best known for his collection of folksongs known as the ‘Child Ballads’. In 1876, he was named Harvard’s first Professor of English. During that time he began his work on the ‘Child Ballads’. They are a major contribution to the study of English language folk music. |
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| 8. |
The pedlar’s name was A) Robin Hood B) Little John C) Francis J. ChildD) Gamble Gold. |
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Answer» D) Gamble Gold. |
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| 9. |
The pedlar refused to tell his name till A) Little John apologized B) Robin Hood apologized C) they told him their names first D) they defeated him. |
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Answer» C) they told him their names first |
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| 10. |
Gamble Gold was Robin Hood’s A) sister’s son B) father’s sister’s son C) mother’s sister’s son D) sister’s daughter’s son |
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Answer» C) mother’s sister’s son |
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| 11. |
The Bold Pedlar and Robin Hood Summary in English. |
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Answer» Background: Robin Hood is a heroic outlaw in English folklore who, according to legend, was a highly skilled archer and swordsman. Traditionally depicted as being dressed in Lincoln green, he is often portrayed as “robbing from the rich and giving to the poor” alongside his band of Merry Men. Robin Hood became a popular folk figure in the late medieval period and continues to be widely represented in literature, films, and television. There are many speculations as to the identity of Robin Hood. According to some historians, Robin Hood was born in the time of Henry II, perhaps Robert Fitzooth, perhaps the Earl of Huntington. His exploits centered around Bamsdale and Sherwood. Legend has it that he died in 1247 at the age of 87 at Kirkley’s Nunnery in Yorkshire. Robin Hood ballads were extremely popular with the peasantry in England for several hundred years. Thirty-seven of ‘Child Ballads’ are Robin Hood ballads. This ballad is Child Ballad 132. Summary: A pedlar meets Robin Hood and Little John and tells them what he has in his pack. Little John demands half of it. The pedlar refuses point blank to oblige Little John. They fight. The fight grows fierce. Finally it is the pedlar who wins. Robin laughs and says he has a man who could defeat him. Now it is the turn of Robin to fight with the pedlar. The pedlar wins again, and refuses to hold his hand, or tell his name, until they had told them theirs. They do, and he says his name is Gamble Gold, and he is fleeing because he killed a man in his father’s land. Robin identifies him as his mother’s sister’s son, and they go to the tavern and drink together. |
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| 12. |
For divisibility by 8, take some numbers and test the pattern in table.NumbersNumber formed by hundred, tens and unit digitDivisible by 8 Yes/No(1) 30480480/8 = 60Yes(2) 42108108/8 = ...(3) 1324324/8 = ...(4) ...... / ... = ......(5) ...... / ... = ...... |
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Answer» If the number framed by last three digits i.e., units, tens and hundred is divisible by 8 or if any number has 0 as its units, tens and nundreds digits, then the number is divisible by 8.
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| 13. |
Take some numbers for divisibility by 4 and test the pattern. |
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Answer» If a number has its last two digits divisible by 4 or if its tens and units digits are 0, then if is divisible by 4.
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| 14. |
Find out the divisibility by 6 for the numbers 336, 123, 1002, 4236. |
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Answer» Testing for divisibility by 6 for the following numbers 336, 123, 1002, 4236.
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| 15. |
Find out H.C.F. by Vedic method. (i) 8, 12 (ii) 38, 57 (iii) 117, 195 (iv) 99,165, 231 |
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Answer» Finding H.C.F. by vedic method, (i) 8, 12 First difference = 12 – 8 = 4 Thus, possible H.C.F. = 4 Second difference =8 – 4 = 4 ∵ First difference = Second difference ∴ H.C.F. of 8 and 12 = 4 (ii) 38, 57 First difference = 57 – 38 = 19 Thus, possible H.C.F. = 19 Second difference = 38 – 19 = 19 ∵ First difference = Second difference ∴ H.C.F of 38 and 57 = 19 (iii) 117,195 First difference = 195 – 117 = 78 Thus, possible H.C.F = 78 Second difference = 117 – 78 = 39 Thus, possible H.C.F = 39 Third difference = 78 – 39 = 39 ∵ Second difference = Third difference ∴ HCF 117 and 195 = 39 (iv) 99, 165, 231 Addition of two numbers = 99 + 231 = 330 First difference = 99 + 231 – 165 = 165 Thus, possible H.C.F = 165 Second difference = 231 – 165 = 66 Thus, possible H.C.F = 66 Third difference = 99 – 66 = 33 Thus, possible H.C.F = 33 ∵ Possible H.C.F is a factor of 66. ∴ HCF of 99, 165 and 231 = 33 |
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| 16. |
Raju’s cow gives 15 litres and buffalo gives 20 litres milk. Find out the maximum measurement for measuring pot for both type of milk exactly. |
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Answer» Factors of 15 = 3 × (5) Factors of 20 = 2 × 2 × (5) ∴ H.C.F. of 15 and 20 = 5 Thus, required measurement will be 5 l. |
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| 17. |
Write 3 numbers which are multiples of 4 and 6. |
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Answer» 12 = 4 × 3, 6 × 2 24 = 4 × 6, 6 × 4 36 = 4 × 9, 6 × 6 Three numbers which are multiples of 4 and 6 are 12, 24 and 36. |
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| 18. |
Which of the following numbers have 6 as a factor? 6, 10, 12, 15, 18, 25, 30, 38, 46 |
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Answer» Numbers 6, 12, 18, 30 have 6 as a factor. |
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| 19. |
Write the first five multiples of the following numbers : (i) 7 (ii) 12 (iii) 17 (iv) 15 (v) 18 |
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Answer» (i) First five multiples of 7 7 × 1 = 7 7 × 2 = 14 7 × 3 = 21 7 × 4 = 28 7 × 5 = 35 Thus, first five multiples of 7, are 7. (ii) First five multiples of 12 12 × 1 = 12 12 × 2 = 24 12 × 3 = 36 12 × 4 = 48 12 × 5 = 60 Thus first five multiples of 12 are 12. 24, 36, 48 and 60. (iii) First five multiples of 17 17 × 1 = 17 17 × 2 = 34 17 × 3 = 51 17 × 4 = 68 17 × 5 = 85 Thus, first five multiples of 17 are 17, 34, 51, 68 and 85. (iv) First five multiples of 15 15 × 1 = 15 15 × 2 = 30 15 × 3 = 45 15 × 4 = 60 15 × 5 = 75 Thus, first five multiples of 16 are 15, 30, 45, 60 and 75. (v) First five multiples of 18 18 × 1 = 18 18 × 2 = 36 18 × 3 = 54 18 × 4 = 72 18 × 5 = 90 Thus, first five multiples of 18 are 18, 36, 54, 72 and 90. |
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| 20. |
Write some more numbers in table and complete the table.NumbersDivisible by 2Divisible by 3Divisible 6216YesYesYes58YesNoNo108103......Can you see any pattern for divisibility by 6 ? |
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Answer» Completing the table
From table, we see pattern for divisibility by 6. If any number is divisible by 2 and 3 separately, then it is also divisible by 6. |
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| 21. |
Can we say that all numbers which perfectly divides 16 are factors of 16 ? |
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Answer» Yes, it is true. As 1, 2, 4, 8, 16 are factors of 16 |
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| 22. |
Write all prime numbers between 10 and 30 |
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Answer» All prime numbers between 10 and 30 are 11, 13, 17, 19, 23 and 29. |
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| 23. |
How many prime numbers did you get between 1 – 100?Write these numbers in sequence. |
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Answer» 25 prime numbers are obtained. In sequence, these numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 |
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| 24. |
Write the smallest prime number. |
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Answer» Smallest prime number is 2. |
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| 25. |
Can you say all even numbers are divisible by 2. |
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Answer» Yes, we can say all even numbers are divisible by 2. |
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| 26. |
Find out the L.C.M of 24 and 30 by Vedic method. |
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Answer» Finding L.C.M of 24 and 30 by Vedic method : Step 1. 24 and 30 is written as 24/30 in fraction form. Step 2. Prime factorising of 24 and 30 = 24/30 = (2 x 2 x 2 x 3/2 x 3 x 5) Step 3. Common in numerator and denominator = 24/30 = (2 x 2/5) = 4/5 Step 4. By cross multiplication 24 × 5 = 30 × 4 = 120 Thus required L.C.M. = 120 |
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| 27. |
Does any number not having 0 or 5 at its units place, have 5 as a factor? |
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Answer» No, if any number not having 0 or 5 at its unit place, they have not 5 as a factor |
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| 28. |
Write all factors of the following numbers. (i) 48 (ii) 36 (iii) 28 (iv) 100 (v) 125 |
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Answer» (i) Factors of 48 48 = 1 × 48 48 = 2 × 24 48 = 3 × 16 48 = 4 × 12 48 = 6 × 8 48 = 8 × 6 ∴ All factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. (ii) Factors of 36 36 =1 × 36 36 = 2 × 18 36 = 3 × 12 36 = 4 × 9 36 = 6 × 6 ∴ All factor of 36 are 1,2, 3, 4, 6, 9, 12, 18 and 36. (iii) Factors of 28 28 =1 × 28 28 = 2 × 14 28 = 4 × 7 28 = 7 × 4 ∴ All factors of 28 are 1,2, 4, 7, 14 and 28. (iv) Factors of 100 100 = 1 × 100 100 = 2 × 50 100 = 4 × 25 100 = 5 × 20 100 = 10 × 10 ∴ All factors of loo are 1, 2, 4, 5, 10, 20, 25, 50 and 100. (v) Factor of 125 125 = 1 × 125 125 = 5 × 25 125 = 25 × 5 ∴ All factors of 125 are 1, 5, 25 and 125. |
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| 29. |
Write factors of 24, 15, 26, 48,13, 11 |
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Answer» Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24 Factors of 15 = 1, 3, 5, 15 Factors of 26 = 1, 2, 13, 26 Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Factors of 13 = 1, 13 Factors of 11 = 1, 11 |
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| 30. |
Number of prime numbers between 1 to 100 are :(i) 24 (ii) 25 (iii) 26 (iv) 27 |
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Answer» (ii) The prime numbers are 25 |
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| 31. |
Factors of number 24 are : (i) 1, 2, 3 (ii) 4, 6 (iii) 12, 24 (iv) All above |
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Answer» (iv) All above |
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| 32. |
Number which is not divisible by 2 :(i) 267 (ii) 468 (iii) 192 (iv) 374 |
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Answer» (i) 267 is not divisible by 2 |
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| 33. |
Common factor of 18, 27 is : (i) 2 (ii) 4 (iii) 6 (iv) 9 |
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Answer» (iv) The common factor of 18, 27 is 9 |
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| 34. |
L.C.M. of 40, 15, 20 is : (i) 120 (ii) 100 (iii) 140 (iv) 160 |
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Answer» (i) The L.C.M. of 40, 15, 20 is 120 |
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| 35. |
Prime number in the following is : (i) 6 (ii) 7 (iii) 9 (iv) 10 |
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Answer» (ii) The Prime number in the following is : 7 |
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| 36. |
Fill in the blanks(i) Reverse process of factors is called ……………….. (ii) The numbers having more than two factors are called ……………….. (iii) ……………….. is even prime number. (iv) The numbers completely divisible by 2 are called. (v) Multiples of same multiple numbers are called ………………..(vi) Smallest odd prime number is ……………….. |
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Answer» (i) expansion (ii) composite (iii) 2 (iv) even (v) common factor (vi) 3 |
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| 37. |
Three containers contains 26 l, 65 liter 117 l milk respectively. Find maximum measurement of container milk of all three containers. |
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Answer» Milk contains in 1st container = 26 l Milk contains in 2nd container = 65 l Milk contains in 3rd container = 117 l Factors of 26 = 2 × (13) Factors of 65 = 5 × (13) Factors of 117 = 3 × 3 × (13) ∴ H.C.M. of 26, 65 and 117 = 13 Thus, required maximum measurement of container will be 13 l. |
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| 38. |
Test divisibility of 7640 by 8, without division. |
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Answer» Test divisibility by 8 In number. 7640 number formed by digits of hundred tens and ones place = 640 Dividing 640 by 8 640 ÷ 8 = 80 Which is divisible by 8. Thus, 7640 will be divisible by 8. |
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| 39. |
Test divisibility of 585 by 3, without division. |
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Answer» Test divisibility by 3 Sum of digits in number =5 + 8 + 5 = 18 ⇒ 1 + 8 = 9 Which is divisible by 3. Thus, 585 will be divisible by 3. |
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| 40. |
Write even and odd numbers separately. (i) 357 (ii) 436 (iii) 77 (iv) 1900 (v) 5001 Even numbrs …………….. Odd numbers …………….. |
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Answer» Even numbers are (ii) 436, (iv) 1900 Odd numbers are (i) 357, (iii) 77, (v) 5001 |
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| 41. |
The HCF of this pair of numbers is not 1. (A) 13,17 (B) 29,20 (C) 40, 20 (D) 14, 15 |
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Answer» (C) 40, 20 40 = 2 x 2 x 2 x 5 20 = 2 x 2 x 5 ∴ HCF of 40 and 20 = 2 x 5 = 10 |
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| 42. |
The HCF of 120 and 150 is __ (A) 30 (B) 45 (C) 20 (D) 120 |
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Answer» (A) 30 120 = 2 x 2 x 2 x 3 x 5 150 = 2 x 3 x 5 x 5 ∴ HCF of 120 and 150 = 2 x 3 x 5 = 30 |
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| 43. |
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers: i. 32, 37 ii. 46, 51 iii. 15, 60 iv. 18, 63 v. 78, 104 |
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Answer» i. 32 = 2 x 16 = 2 x 2 x 8 = 2 x 2 x 2 x 4 = 2 x 2 x 2 x 2 x 2 x 1 37 = 37 x 1 ∴ HCF of 32 and 37 =1 LCM of 32 and 37 = 2 x 2 x 2 x 2 x 2 x 37 = 1184 HCF x LCM = 1 x 1184 = 1184 Product of the given numbers = 32 x 37 = 1184 ∴ HCF x LCM = Product of the given numbers. ii. 46 = 2 x 23 x 1 51 = 3 x 17 x 1 ∴ HCF of 46 and 51 = 1 LCM of 46 and 51 = 2 x 23 x 3 x 17 = 2346 HCF x LCM = 1 x 2346 = 2346 Product of the given numbers = 46 x 51 = 2346 ∴ HCF x LCM = Product of the given numbers iii. 15 = 3 x 5 60 = 2 x 30 = 2 x 2 x 15 = 2 x 2 x 3 x 5 ∴ HCF of 15 and 60 = 3 x 5 = 15 LCM of 15 and 60 = 3 x 5 x 2 x 2 = 60 HCF x LCM = 15 x 60 = 900 Product of the given numbers = 15 x 60 = 900 ∴ HCF x LCM = Product of the given numbers. iv. 18 = 2 x 9 = 2 x 3 x 3 63 = 3 x 21 = 3 x 3 x 7 ∴ HCF of 18 and 63 = 3 x 3 = 9 LCM of 18 and 63 = 3 x 3 x 2 x 7 = 126 HCF x LCM = 9 x 126 = 1134 Product of the given numbers = 18 x 63 = 1134 ∴ HCF x LCM = Product of the given numbers. v. 78 = 2 x 39 = 2 x 3 x 13 104 = 2 x 52 = 2 x 2 x 26 = 2 x 2 x 2 x 13 ∴ HCF of 78 and 104 = 2 x 13 = 26 LCM of 78 and 104 = 2 x 13 x 3 x 2 x 2 = 312 HCF x LCM = 26 x 312 = 8112 Product of the given numbers = 78 x 104 = 8112 ∴ HCF x LCM = Product of the given numbers. |
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| 44. |
LCM × HCF = …………….. |
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Answer» Product of two numbers |
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| 45. |
Without actual division, find whether 8989794 is divisible by 9. |
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Answer» Given number is 8989794. Sum of the digits =8 + 9 + 8 + 9 + 7 + 9 + 4 = 54 – 1 If the sum of the digits of a number is divisible by 9. Then, it is divisible by 9. 54 is divisible by 9. So, 8989794 is divisible by 9. |
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| 46. |
Kaprekar’s constant is …………….. |
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Answer» Correct Answer is 6174 |
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| 47. |
Find the HCF and LCM.i. 23, 69 ii. 21, 49, 84 |
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Answer» i. 23 = 23 x 1 69 = 3 x 23 ∴ HCF of 23 and 69 = 23 ∴ LCM of 23 and 69 = 23 x 3 = 69 ii. 21 = 3 x 7 49 = 7 x 7 84 = 2 x 42 = 2 x 2 x 21 = 2 x 2 x 3 x 7 ∴ HCF of 21, 49 and 84 = 7 ∴ LCM of 21, 49 and 84 = 7 x 3 x 7 x 2 x 2 = 588 |
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| 48. |
Number divisible by 11 is A) 10934 B) 726351 C) 7138965 D) 376845 |
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Answer» Correct option is A) 10934 |
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| 49. |
In any number if the number formed by the last two digits is divisible by 4, then the number is divisible by …………………A) 2B) 4C) 6D) 8 |
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Answer» Correct option is B) 4 |
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| 50. |
HCF of 3, 5 and 7 is …………….. |
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Answer» Correct Answer is 1 |
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