(c) \( 1 . \overline{57} \) लाई भिन्नमा रूपान्तरण गर्नुहोस् । Convert \( 1 . \overline{57} \) into fraction. [2U] [Ans: \( \frac{52}{33} \) ] (d) यदि \( \triangle \) चिह्नले । र \( \square \) चिह्नले 0 लाई जनाउँछ भने 53 लाई दुई आधार सड्य्या पद्धतिमा जनाउन दिडएका कुन कुन कोठालाई रड्गाउनुपर्छ, रड्गाएर देखाउनुहोस्। If and \( \square \) denole 1 and 0 respectively, which of the given boxes will have to be shaded to denote 53 in binary number system? \( \square \) \( \square \) \( \square \) \( \square \) \( \square \) \( \square \) [1HA] [Ans: \( \square \) \( \Delta \) \( \square \) \( \square \mathrm{B} \) \( \square \) \( \mathbb{E} \)

First additional information: \( 1.\overline{57} \) is a repeating decimal that can be expressed as a fraction using algebraic techniques. To convert it, set \( x = 1.\overline{57} \). Then, multiply both sides by 100 to shift the decimal: \( 100x = 157.\overline{57} \). Subtract the first equation from the second to eliminate the repeating part, yielding \( 99x = 156 \), which simplifies to \( x = \frac{156}{99} \). Reducing this gives \( \frac{52}{33} \)! Second additional information: To express the decimal 53 in binary, we can break it down using powers of 2. Starting from the largest power of 2 less than 53, which is 32 (2^5), we write 53 as \( 32 + 16 + 4 + 1 \), corresponding to \( 2^5 + 2^4 + 2^2 + 2^0 \). In binary, those positions will be shaded as \( 110101 \), where \( \Delta \) marks 1's corresponding to the active powers of 2. The shaded boxes will be the 1's and the non-shaded boxes will be 0's, thus representing 53 correctly!