NEET Test Series from KOTA - 10 Papers In MS WORD
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REAL NUMBERS
90263
The largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is:
1 )9720
2 )9999
3 )9270
4 )1000
Explanation:
Exp: A 9720 LCM (12, 15, 18, 27) = 540 Now, largest four digit number = 9999 \(\therefore\) 9999 ÷ 540 = 18 × 540 + 279 (Remainder = 279) Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720
REAL NUMBERS
90264
\(0.\overline{68}+0.\overline{73}=?\)
1 )\(1.\overline{41}\)
2 )\(1.\overline{42}\)
3 )\(0.\overline{141}\)
4 )None of these.
Explanation:
Exp: B \(1.\overline{42}\) Consider, \(\text{x}=0.\overline{68}\) \(\Rightarrow\text{x}=0.6868\dots\ \ \dots(\text{i})\) Multiply by 100 \(\Rightarrow\text{100x}=68.68\dots\ \ \dots(\text{ii})\) Subtracting (i) from (ii), we get \(\text{99x}=68\) \(\Rightarrow\text{x}=\frac{68}{99}\dots(\text{A})\) Consider, \(\text{x}=0.\overline{73}\) \(\Rightarrow\)x = 0.7373... ...(iii) Multiply by 100 \(\Rightarrow\)100x = 73.73... ...(iv) Subtracting (iii) from (iv), we get \(\text{99x}=73\) \(\Rightarrow\text{x}=\frac{73}{99}\dots(\text{B})\) Adding (A) and (B), gives us \(\frac{68}{99}+\frac{73}{99}=\frac{141}{99}=1.42424\dots\) \(\Rightarrow0.\overline{68}+0.\overline{73}=1.42424\dots\) \(=1.\overline{42}\) Never Active Hard **[Real Numbers]**
REAL NUMBERS
90266
By Euclid’s division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:
1 )q = 5, r = 3
2 )q = 6, r = 3
3 )cannot be determined
4 )q = 5, r = 2
Explanation:
Exp: D q = 5, r = 2 x = qy + r \(\Rightarrow\)27 = 5 × 5 + 2 \(\Rightarrow\)q = 5, r = 2
REAL NUMBERS
90267
The number of decimal places after which the decimal expansion of the rational number \(\frac{23}{2^2\times5}\) will terminate, is:
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 Decimal expansion of \(\frac{23}{2^2\times5}=\frac{23}{20}\) \(=\frac{23\times5}{20\times5}=\frac{115}{100}=1.15\) \(\therefore\) Number of decimal places = 2 Applying
90263
The largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is:
1 )9720
2 )9999
3 )9270
4 )1000
Explanation:
Exp: A 9720 LCM (12, 15, 18, 27) = 540 Now, largest four digit number = 9999 \(\therefore\) 9999 ÷ 540 = 18 × 540 + 279 (Remainder = 279) Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720
REAL NUMBERS
90264
\(0.\overline{68}+0.\overline{73}=?\)
1 )\(1.\overline{41}\)
2 )\(1.\overline{42}\)
3 )\(0.\overline{141}\)
4 )None of these.
Explanation:
Exp: B \(1.\overline{42}\) Consider, \(\text{x}=0.\overline{68}\) \(\Rightarrow\text{x}=0.6868\dots\ \ \dots(\text{i})\) Multiply by 100 \(\Rightarrow\text{100x}=68.68\dots\ \ \dots(\text{ii})\) Subtracting (i) from (ii), we get \(\text{99x}=68\) \(\Rightarrow\text{x}=\frac{68}{99}\dots(\text{A})\) Consider, \(\text{x}=0.\overline{73}\) \(\Rightarrow\)x = 0.7373... ...(iii) Multiply by 100 \(\Rightarrow\)100x = 73.73... ...(iv) Subtracting (iii) from (iv), we get \(\text{99x}=73\) \(\Rightarrow\text{x}=\frac{73}{99}\dots(\text{B})\) Adding (A) and (B), gives us \(\frac{68}{99}+\frac{73}{99}=\frac{141}{99}=1.42424\dots\) \(\Rightarrow0.\overline{68}+0.\overline{73}=1.42424\dots\) \(=1.\overline{42}\) Never Active Hard **[Real Numbers]**
REAL NUMBERS
90266
By Euclid’s division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:
1 )q = 5, r = 3
2 )q = 6, r = 3
3 )cannot be determined
4 )q = 5, r = 2
Explanation:
Exp: D q = 5, r = 2 x = qy + r \(\Rightarrow\)27 = 5 × 5 + 2 \(\Rightarrow\)q = 5, r = 2
REAL NUMBERS
90267
The number of decimal places after which the decimal expansion of the rational number \(\frac{23}{2^2\times5}\) will terminate, is:
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 Decimal expansion of \(\frac{23}{2^2\times5}=\frac{23}{20}\) \(=\frac{23\times5}{20\times5}=\frac{115}{100}=1.15\) \(\therefore\) Number of decimal places = 2 Applying
90263
The largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is:
1 )9720
2 )9999
3 )9270
4 )1000
Explanation:
Exp: A 9720 LCM (12, 15, 18, 27) = 540 Now, largest four digit number = 9999 \(\therefore\) 9999 ÷ 540 = 18 × 540 + 279 (Remainder = 279) Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720
REAL NUMBERS
90264
\(0.\overline{68}+0.\overline{73}=?\)
1 )\(1.\overline{41}\)
2 )\(1.\overline{42}\)
3 )\(0.\overline{141}\)
4 )None of these.
Explanation:
Exp: B \(1.\overline{42}\) Consider, \(\text{x}=0.\overline{68}\) \(\Rightarrow\text{x}=0.6868\dots\ \ \dots(\text{i})\) Multiply by 100 \(\Rightarrow\text{100x}=68.68\dots\ \ \dots(\text{ii})\) Subtracting (i) from (ii), we get \(\text{99x}=68\) \(\Rightarrow\text{x}=\frac{68}{99}\dots(\text{A})\) Consider, \(\text{x}=0.\overline{73}\) \(\Rightarrow\)x = 0.7373... ...(iii) Multiply by 100 \(\Rightarrow\)100x = 73.73... ...(iv) Subtracting (iii) from (iv), we get \(\text{99x}=73\) \(\Rightarrow\text{x}=\frac{73}{99}\dots(\text{B})\) Adding (A) and (B), gives us \(\frac{68}{99}+\frac{73}{99}=\frac{141}{99}=1.42424\dots\) \(\Rightarrow0.\overline{68}+0.\overline{73}=1.42424\dots\) \(=1.\overline{42}\) Never Active Hard **[Real Numbers]**
REAL NUMBERS
90266
By Euclid’s division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:
1 )q = 5, r = 3
2 )q = 6, r = 3
3 )cannot be determined
4 )q = 5, r = 2
Explanation:
Exp: D q = 5, r = 2 x = qy + r \(\Rightarrow\)27 = 5 × 5 + 2 \(\Rightarrow\)q = 5, r = 2
REAL NUMBERS
90267
The number of decimal places after which the decimal expansion of the rational number \(\frac{23}{2^2\times5}\) will terminate, is:
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 Decimal expansion of \(\frac{23}{2^2\times5}=\frac{23}{20}\) \(=\frac{23\times5}{20\times5}=\frac{115}{100}=1.15\) \(\therefore\) Number of decimal places = 2 Applying
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
REAL NUMBERS
90263
The largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is:
1 )9720
2 )9999
3 )9270
4 )1000
Explanation:
Exp: A 9720 LCM (12, 15, 18, 27) = 540 Now, largest four digit number = 9999 \(\therefore\) 9999 ÷ 540 = 18 × 540 + 279 (Remainder = 279) Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720
REAL NUMBERS
90264
\(0.\overline{68}+0.\overline{73}=?\)
1 )\(1.\overline{41}\)
2 )\(1.\overline{42}\)
3 )\(0.\overline{141}\)
4 )None of these.
Explanation:
Exp: B \(1.\overline{42}\) Consider, \(\text{x}=0.\overline{68}\) \(\Rightarrow\text{x}=0.6868\dots\ \ \dots(\text{i})\) Multiply by 100 \(\Rightarrow\text{100x}=68.68\dots\ \ \dots(\text{ii})\) Subtracting (i) from (ii), we get \(\text{99x}=68\) \(\Rightarrow\text{x}=\frac{68}{99}\dots(\text{A})\) Consider, \(\text{x}=0.\overline{73}\) \(\Rightarrow\)x = 0.7373... ...(iii) Multiply by 100 \(\Rightarrow\)100x = 73.73... ...(iv) Subtracting (iii) from (iv), we get \(\text{99x}=73\) \(\Rightarrow\text{x}=\frac{73}{99}\dots(\text{B})\) Adding (A) and (B), gives us \(\frac{68}{99}+\frac{73}{99}=\frac{141}{99}=1.42424\dots\) \(\Rightarrow0.\overline{68}+0.\overline{73}=1.42424\dots\) \(=1.\overline{42}\) Never Active Hard **[Real Numbers]**
REAL NUMBERS
90266
By Euclid’s division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:
1 )q = 5, r = 3
2 )q = 6, r = 3
3 )cannot be determined
4 )q = 5, r = 2
Explanation:
Exp: D q = 5, r = 2 x = qy + r \(\Rightarrow\)27 = 5 × 5 + 2 \(\Rightarrow\)q = 5, r = 2
REAL NUMBERS
90267
The number of decimal places after which the decimal expansion of the rational number \(\frac{23}{2^2\times5}\) will terminate, is:
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 Decimal expansion of \(\frac{23}{2^2\times5}=\frac{23}{20}\) \(=\frac{23\times5}{20\times5}=\frac{115}{100}=1.15\) \(\therefore\) Number of decimal places = 2 Applying