SSC CGL Tier-1 practice question
Find the smallest number that must be added to 830 so that it becomes exactly divisible by 7.
From a SSC CGL Tier-1 Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (D) 3
To find the smallest number to be added to \(830\) to make it divisible by \(7\), we first divide \(830\) by \(7\):
\[
830 = 7 \times 118 + 4
\]
The remainder is \(4\).
The number that must be added to make it divisible by \(7\) is:
\[
\text{Required number} = \text{Divisor} - \text{Remainder} = 7 - 4 = 3
\]
Check: \(830 + 3 = 833\), and \(833 \div 7 = 119\) (exactly divisible).
Hence, the smallest number to be added is **3**.
हिंदी में प्रश्न
830 में कौन सी छोटी से छोटी संख्या जोड़ी जाए कि वह 7 से पूर्णतः विभाज्य हो जाए?
उत्तर जाँचने के लिए किसी विकल्प पर टैप करें।
उत्तर और व्याख्या देखें
सही उत्तर: (D) 3
\(830\) में जोड़ी जाने वाली सबसे छोटी संख्या ज्ञात करने के लिए, \(830\) को \(7\) से भाग देते हैं:
\[
830 = 7 \times 118 + 4
\]
यहाँ शेषफल \(4\) प्राप्त होता है।
अतः जोड़ी जाने वाली अभीष्ट संख्या:
\[
\text{अभीष्ट संख्या} = \text{भाजक} - \text{शेषफल} = 7 - 4 = 3
\]
जाँच: \(830 + 3 = 833\), और \(833 \div 7 = 119\) (पूर्णतः विभाज्य)।
अतः जोड़ी जाने वाली छोटी से छोटी संख्या **3** है।
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