Important Number System Questions Asked in RRB NTPC 2025 with Tricky Solutions
Welcome to our detailed guide on the number system questions asked in the RRB NTPC Graduate Level 2025 CBT-1 exam. These questions, presented in both English and Hindi, cover key concepts like HCF, LCM, ratios, and division. We’ve included the exact solutions provided to help you understand the tricky approaches used. Let’s dive in!
Question 1: HCF of Three Numbers in Ratio
Three numbers are in the ratio 2:14:3, and their LCM is 4116. Find their HCF.
तीन संख्याएँ 2:14:3 के अनुपात में हैं, और उनका LCM, 4116 है। उनका HCF ज्ञात कीजिए।
Solution:
Let the numbers be \( 2x \), \( 14x \), and \( 3x \).
According to the question:
\[ 42x = 4116 \]
\[ x = \frac{4116}{42} \]
\[ x = 98 \]
Therefore, HCF = \( 98 \).
Answer: 98 (Option 1)
Question 2: Division with Same Remainder
When 1063, 2815, and 3451 are divided by the greatest number \( x \), the remainder in each case is \( y \). What is the value of \( (3x - 14y) \)?
जब 1063, 2815 और 3451 को सबसे बड़ी संख्या \( x \) से विभाजित किया जाता है, तो प्रत्येक स्थिति में शेषफल \( y \) होता है। \( (3x - 14y) \) का मान ज्ञात कीजिए।
Solution:
Numbers: (1063, 2815, 3451)
Consider differences:
\[ (2815 - 1063), (3451 - 2815), (3451 - 1063) \]
\[ (1752, 636, 2388) \]
Find HCF of (1752, 636, 2388):
\[ \text{HCF} = 12 \]
Now, divide 1063 by 12:
\[ 12 \div 1063 = 88 \text{ remainder } 7 \]
\[ 1063 = 12 \times 88 + 7 \]
Hence, \( x = 12 \), \( y = 7 \).
Calculate:
\[ 3x - 14y = 3 \times 12 - 14 \times 7 = 36 - 98 = -62 \]
Answer: -62
Question 3: Maximum Volume of Containers
Manoj has 200 litres of Oil A and 274 litres of Oil B. He fills identical containers with one type of oil per container, all completely filled. What is the maximum volume (in litres) of each container?
मनोज के पास 200 लीटर तेल A और 274 लीटर तेल B है। वह कई समान कंटेनरों को दो प्रकार के तेल से इस तरह भरता है कि प्रत्येक कंटेनर में केवल एक प्रकार का तेल होता है, और सभी कंटेनर पूरी तरह से भरे होते हैं। प्रत्येक कंटेनर का अधिकतम आयतन (लीटर में) कितना हो सकता है?
Solution:
Find HCF of (200, 274):
\[ \text{HCF} = 2 \]
So, the maximum volume of each container is 2 litres.
Answer: 2 (Option 1)
Question 4: HCF of Greater Number and 3828
Two positive numbers differ by 2858. When the greater number is divided by the smaller number, the quotient is 5 and the remainder is 178. What is the HCF of the greater number and 3828?
दो धनात्मक संख्याओं के बीच 2858 का अंतर है। जब बड़ी संख्या को छोटी संख्या से विभाजित किया जाता है, तो भागफल 5 होता है और शेषफल 178 होता है। बड़ी संख्या और 3828 का HCF कितना है?
Solution:
Let the smaller number be \( y \) and the greater number be \( x \).
According to the question:
\[ x - y = 2858 \quad (1) \]
\[ x = 5y + 178 \quad (2) \]
Substitute \( x = 5y + 178 \) in equation (1):
\[ (5y + 178) - y = 2858 \]
\[ 4y = 2680 \]
\[ y = 670 \]
Then, \( x = 5 \times 670 + 178 = 3528 \).
Find HCF of (3528, 3828):
\[ \text{HCF} = 12 \]
Answer: 12 (Option 3)
Question 5: Sum of Reciprocals of LCM and HCF
If the ratio of two numbers is 3:16 and the product of their LCM and HCF is 432, then the sum of the reciprocals of the LCM and HCF is:
यदि दो संख्याओं का अनुपात 3:16 है तथा उनके लघुत्तम समापवर्त्य (LCM) और महत्तम समापवर्तक (HCF) का गुणनफल 432 है, तो उनके लघुत्तम समापवर्त्य तथा महत्तम समापवर्तक के व्युत्क्रमों का योगफल ज्ञात कीजिए।
Solution:
Let the numbers be \( 3x \) and \( 16x \).
Product of LCM and HCF:
\[ (3x \times 16x) = 48x^2 = 432 \]
\[ x^2 = 9 \]
\[ x = \pm 3 \]
So, the numbers are \( 9 \) and \( 48 \).
HCF = 3, LCM = 48.
Sum of reciprocals:
\[ \frac{1}{3} + \frac{1}{144} = \frac{48 + 1}{144} = \frac{49}{144} \]
Answer: \(\frac{49}{144}\)
Question 6: Number of Marbles
Nandan had some marbles. When distributed equally among 32 children, 4 marbles were left. Had he distributed among 41 and 48 children, 13 and 20 marbles would remain, respectively. When distributed among 53 children, no marbles were left. The number of marbles lies between:
Solution:
Given:
\[ 32 \rightarrow 4 \text{ remainder} \ (32k + 4) \]
\[ 41 \rightarrow 13 \text{ remainder} \ (41m + 13) \]
\[ 48 \rightarrow 20 \text{ remainder} \ (48n + 20) \]
All have a common remainder pattern:
\[ (32k + 4) = (41m + 13) = (48n + 20) = 53p \]
Differences:
\[ (41m + 13) - (32k + 4) = 41m - 32k + 9 \]
\[ (48n + 20) - (41m + 13) = 48n - 41m + 7 \]
Common difference: 28.
LCM of (32, 41, 48):
\[ \text{LCM} = 3936 \]
Number of marbles: \( 3936k - 28 \).
For \( k = 2 \):
\[ 3936 \times 2 - 28 = 7872 - 28 = 7844 \]
Answer: 7844 marbles
Question 7: LCM Using Division Method
In finding the HCF of two numbers by division method, the quotients are 1, 4, and 6, respectively, and the last divisor is 29. What is the LCM of the two numbers?
विभाजन विधि द्वारा दो संख्याओं का म.स.प. (HCF) ज्ञात करने पर, भागफल क्रमशः 1, 4 और 6 प्राप्त होते हैं तथा अंतिम भाजक 29 है। दोनों संख्याओं का ल.स.प. (LCM) ज्ञात कीजिए।
Solution:
Division method steps:
\[ 725 \div 899 = 1 \text{ remainder } 174 \]
\[ 174 \div 725 = 4 \text{ remainder } 29 \]
\[ 29 \div 174 = 6 \text{ remainder } 0 \]
So, HCF = 29.
Numbers are 725 and 899.
LCM:
\[ \text{LCM} = \frac{725 \times 899}{29} = 22475 \]
Answer: 22475 (Option 1)
These solutions provide a clear and tricky approach to solving number system questions from the RRB NTPC 2025 exam. Practice these concepts to excel in your preparation!© Abhinav Sir
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