Quantitative Aptitude

Number System

616 Questions
All HCF and LCM Word Problems 416 Remainder Theorems 122 Number Properties and Series 21 Successive Division 20 Divisibility Rules 19 Digit Word Problems 9 Decimal and Fraction Operations 6 Unit Digit and Last Digits 2 BODMAS Simplification 1

Number System Questions

Multiple choice

Find the greatest number that divides 880 and 1140 leaves the remainder 25 and 15 respectively. वह बड़ी से बड़ी संख्या ज्ञात कीजिए जिससे 880 तथा 1140 को भाग दिया जाता है तो क्रमशः 25 तथा 15 शेष बचता है?

  1. 35

  2. 40

  3. 45

  4. 25
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If n divides 880 and 1140 leaving remainders 25 and 15, then n divides (880-25)=855 and (1140-15)=1125 exactly. Find HCF of 855 and 1125. 855 = 3²×5×19, 1125 = 3²×5³. HCF = 3²×5 = 45.

Multiple choice

If y be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of y is

  1. 21

  2. 22

  3. 24

  4. 18
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Number is of form LCM(5,6,7,8) × k + 3 = 840k + 3, and must be divisible by 9. Testing values: k=1 gives 843 (sum=15, not ÷9), k=2 gives 1683 (sum=18, ÷9). So y = 1683. Sum of digits = 1 + 6 + 8 + 3 = 18. Option A (21) would require y with digit sum 21, but 2523 ÷ 9 = 280.33. Option C (24) would need digit sum 24, but checking values divisible by 9 doesn't yield such sum.

Multiple choice

Find the greatest 3 digit number such that when this is divided by 12, 18, 21 and 28, it leaves a remainder 3 in each case. 3 अंको की सबसे बड़ी संख्या ज्ञात कीजिए जिसे जब 12, 18, 21 और 28 से भाग दिया जाय तो प्रत्येक दशा में 3 शेषफल बचे ।

  1. 759

  2. 859

  3. 659

  4. 827
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

LCM of 12, 18, 21, 28 is 252. Numbers leaving remainder 3 are of form 252k + 3. For greatest 3-digit: 252 × 3 + 3 = 759, which is option A.

Multiple choice

The largest number that divides 248 and 1029, leaving remainder 40 and 5 respectively is – वह सबसे बड़ी संख्या जिससे कि 248 और 1029 से विभाजित करने पर क्रमशः 40 व 5 शेषफल प्राप्त हो-

  1. 16

  2. 18

  3. 17

  4. 15
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the largest number that divides 248 and 1029 leaving remainders 40 and 5, we first subtract the remainders: 248-40=208 and 1029-5=1024. Then we find HCF(208,1024). Factorizing gives 208=2^4×13 and 1024=2^10, so HCF=16. This is the largest number satisfying both conditions since any number >40 that divides 208 cannot divide 1024.

Multiple choice

Find the least number of 5 digits which when divided by 12, 15 and 18 leaves a remainder of 9, 12 and 15. 5 अंकों की छोटी से छोटी संख्या ज्ञात कीजिए जिसे 12, 15 और 18 से विभाजित करने पर क्रमशः 9, 12 तथा 15 शेषफल बचे।

  1. 10087

  2. 10077

  3. 10027

  4. 10057
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

LCM of 12,15,18 is 180. Required number = 180k - 3 (since remainders 9,12,15 are each 3 less than divisors). Smallest 5-digit: 180*56 - 3 = 10080 - 3 = 10077. Check: 10077/12=839 R9, 10077/15=671 R12, 10077/18=559 R15.

Multiple choice

The whole number (W) is divisible by 7. (W) leaves a remainder of 1 when divided by 2, 3, 4 or 5. What is the sum of the digits of the smallest value of (W)?\

  1. 4

  2. 2

  3. 7

  4. 3

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

W leaves remainder 1 when divided by 2, 3, 4, or 5, so W-1 must be divisible by LCM(2,3,4,5)=60. Thus W=60k+1. For W to be divisible by 7: 60k+1 ≡ 0 (mod 7). Since 60 ≡ 4 (mod 7), we get 4k ≡ 6 (mod 7), giving k=5 as smallest solution. W=301, sum of digits=4.

Multiple choice

Which least number should be subtracted from 427399 so that the remainder would be divisible by 15?

  1. 3

  2. 4

  3. 2

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find what must be subtracted, calculate 427399 mod 15. Divisibility by 15 requires divisibility by both 3 and 5. For divisibility by 5, the last digit must be 0 or 5. 427399 - 4 = 427395 ends in 5. Checking divisibility by 3: 4+2+7+3+9+5 = 30, which is divisible by 3. Therefore, 427395 is divisible by 15, so subtracting 4 works.

Multiple choice

Find the greatest number of six digits which when divided by 15, 20, 24 and 60 leaves remainder 12, 17, 21 and 57 respectively. छः अंको की बड़ी से बड़ी संख्या ज्ञात कीजिये जिसे जब 15, 20, 24 और 60 से विभाजित किया जाय तो शेषफल क्रमशः 12, 17, 21 तथा 57 प्राप्त हों ।

  1. 999957

  2. 999963

  3. 999953

  4. 999967
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

LCM of (15-12), (20-17), (24-21), (60-57) = LCM(3,3,3,3) = 3. Largest 6-digit multiple of 3 is 999999. Subtracting 3 gives 999996, but checking remainders: 999996 % 15 = 6 (we need 12), so add 9 to get 999957 which satisfies all conditions.