(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: \( \square \) \( \Delta \) \( \square \) \( \square \mathrm{B} \) \( \square \) \( \mathbb{E} \)

Here are the solutions step‐by‐step: ────────────────────────────── Part (c) We need to convert   1.575757…                   (“57” repeating) into a fraction. Step 1. Let x = 1.575757… Step 2. Since two digits are repeating, multiply by 100:   100x = 157.575757… Step 3. Subtract the original number:   100x – x = 157.575757… – 1.575757…          = 156 Thus,   99x = 156 Step 4. Solve for x:   x = 156/99 Step 5. Simplify by dividing numerator and denominator by 3:   156 ÷ 3 = 52  and  99 ÷ 3 = 33 So,   x = 52/33 Answer for (c): 52/33 ────────────────────────────── Part (d) The problem states:   “If Δ denotes 1 and □ denotes 0, show which boxes must be shaded (i.e. marked with Δ for 1) to represent the decimal number 53 in the binary number system.” Step 1. Convert 53 from decimal to binary. Divide 53 by 2:   53 ÷ 2 = 26 remainder 1    (LSB)   26 ÷ 2 = 13 remainder 0   13 ÷ 2 = 6  remainder 1   6 ÷ 2 = 3  remainder 0   3 ÷ 2 = 1  remainder 1   1 ÷ 2 = 0  remainder 1    (MSB) Reading the remainders from last to first gives us: 1 1 0 1 0 1 Thus, 53 in binary is: 110101 Step 2. Replace the digit “1” with Δ and “0” with □. So the six positions (from MSB to LSB) are:   Δ Δ □ Δ □ Δ Step 3. Interpret “boxes” as the placeholders for the binary digits in descending order of power:   • The leftmost box (for 2^5) gets Δ.   • The next box (for 2^4) gets Δ.   • The third box (for 2^3) gets □.   • The fourth box (for 2^2) gets Δ.   • The fifth box (for 2^1) gets □.   • The sixth box (for 2^0) gets Δ. Thus, to depict 53 in binary, you need to “shade” (or mark) the boxes corresponding to the digit 1 in 110101 — that is, the 1st, 2nd, 4th, and 6th boxes. Answer for (d):   The boxes should display (from left to right): Δ Δ □ Δ □ Δ ────────────────────────────── Final Answers: (c) 52/33 (d) The six boxes (representing 2^5 to 2^0) should show: Δ Δ □ Δ □ Δ, meaning that the 1st, 2nd, 4th, and 6th boxes (corresponding to bits “1”) are shaded.