NEET Test Series from KOTA - 10 Papers In MS WORD
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REAL NUMBERS
90428
Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion \(\frac{7}{210}\):
1 )Terminating decimal expansion
2 )Non -terminating decimal expansion
3 )Cannot be determined
4 )None
Explanation:
Exp: B Non -terminating decimal expansion Simplify it by dividing nominator and denominator both by 7 we get \(\frac{1}{30}\) Factorize the denominator we get 30 = 2 × 3 × 5 Denominator has 3 also in denominator So denominator is not in form of 2\(^{n}\) × 5\(^{n}\)
REAL NUMBERS
90180
Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
1 )\(1<\text{r}<\text{b}\)
2 )\(0<\text{r}\le\text{b}\)
3 )\(0\le\text{r}<\text{b}\)
4 )\(0<\text{r}<\text{b}\)
Explanation:
Exp: C \(0\le\text{r}<\text{b}\) Euclid's division lemma states that, For any positive integers a and b, there exist unique integers q and r such that \(\text{a}=\text{bq}+\text{r},\) where \(0\le\text{r}<\text{b}\) Never Active
REAL NUMBERS
90184
A die is thrown once. Find the probability of getting a number between 3 and 6.
1 )\(\frac{2}{3}\)
2 )\(\frac{1}{2}\)
3 )\(\frac{1}{4}\)
4 )\(\frac{1}{3}\)
Explanation:
Exp: D \(\frac{1}{3}\) Total number of outcomes ={1, 2, 3, 4, 5, 6} = 6 favourable outcomes in this case = {4, 5} = 2 \(\therefore\) p (a number between 3 and 6) \(=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{2}{6}=\frac{1}{3}\)
REAL NUMBERS
90185
If d is the HCF of 56 and 72, the values of x, y satisfying d = 56x + 72y:
1 )x = -3, y = 4
2 )x = 4, y = -3
3 )x = 3, y = -4
4 )x = -4, y = 3
Explanation:
Exp: B x = 4, y = -3 Since, HCF of 56 and 72, by Euclid’s division lemma, 72 = 56 × 1 + 16 ...(i) 56 = 16 × 3 + 8 ...(ii) 16 = 8 × 2 + 0 ...(iii) \(\therefore\) HCF of 56 and 72 is 8. \(\therefore\) 8 = 56 - 16× 3 ( from eq. (ii)) 8 = 56 - (72 - 56 × 1) × 3 [From eq. (i) : 16 = 72 - 56 × 1] 8 = 56 - 3 × 72 + 56 × 3 8 = 56 × 4 + (-3) × 72 \(\therefore\) x = 4, y = -3
90428
Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion \(\frac{7}{210}\):
1 )Terminating decimal expansion
2 )Non -terminating decimal expansion
3 )Cannot be determined
4 )None
Explanation:
Exp: B Non -terminating decimal expansion Simplify it by dividing nominator and denominator both by 7 we get \(\frac{1}{30}\) Factorize the denominator we get 30 = 2 × 3 × 5 Denominator has 3 also in denominator So denominator is not in form of 2\(^{n}\) × 5\(^{n}\)
REAL NUMBERS
90180
Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
1 )\(1<\text{r}<\text{b}\)
2 )\(0<\text{r}\le\text{b}\)
3 )\(0\le\text{r}<\text{b}\)
4 )\(0<\text{r}<\text{b}\)
Explanation:
Exp: C \(0\le\text{r}<\text{b}\) Euclid's division lemma states that, For any positive integers a and b, there exist unique integers q and r such that \(\text{a}=\text{bq}+\text{r},\) where \(0\le\text{r}<\text{b}\) Never Active
REAL NUMBERS
90184
A die is thrown once. Find the probability of getting a number between 3 and 6.
1 )\(\frac{2}{3}\)
2 )\(\frac{1}{2}\)
3 )\(\frac{1}{4}\)
4 )\(\frac{1}{3}\)
Explanation:
Exp: D \(\frac{1}{3}\) Total number of outcomes ={1, 2, 3, 4, 5, 6} = 6 favourable outcomes in this case = {4, 5} = 2 \(\therefore\) p (a number between 3 and 6) \(=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{2}{6}=\frac{1}{3}\)
REAL NUMBERS
90185
If d is the HCF of 56 and 72, the values of x, y satisfying d = 56x + 72y:
1 )x = -3, y = 4
2 )x = 4, y = -3
3 )x = 3, y = -4
4 )x = -4, y = 3
Explanation:
Exp: B x = 4, y = -3 Since, HCF of 56 and 72, by Euclid’s division lemma, 72 = 56 × 1 + 16 ...(i) 56 = 16 × 3 + 8 ...(ii) 16 = 8 × 2 + 0 ...(iii) \(\therefore\) HCF of 56 and 72 is 8. \(\therefore\) 8 = 56 - 16× 3 ( from eq. (ii)) 8 = 56 - (72 - 56 × 1) × 3 [From eq. (i) : 16 = 72 - 56 × 1] 8 = 56 - 3 × 72 + 56 × 3 8 = 56 × 4 + (-3) × 72 \(\therefore\) x = 4, y = -3
90428
Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion \(\frac{7}{210}\):
1 )Terminating decimal expansion
2 )Non -terminating decimal expansion
3 )Cannot be determined
4 )None
Explanation:
Exp: B Non -terminating decimal expansion Simplify it by dividing nominator and denominator both by 7 we get \(\frac{1}{30}\) Factorize the denominator we get 30 = 2 × 3 × 5 Denominator has 3 also in denominator So denominator is not in form of 2\(^{n}\) × 5\(^{n}\)
REAL NUMBERS
90180
Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
1 )\(1<\text{r}<\text{b}\)
2 )\(0<\text{r}\le\text{b}\)
3 )\(0\le\text{r}<\text{b}\)
4 )\(0<\text{r}<\text{b}\)
Explanation:
Exp: C \(0\le\text{r}<\text{b}\) Euclid's division lemma states that, For any positive integers a and b, there exist unique integers q and r such that \(\text{a}=\text{bq}+\text{r},\) where \(0\le\text{r}<\text{b}\) Never Active
REAL NUMBERS
90184
A die is thrown once. Find the probability of getting a number between 3 and 6.
1 )\(\frac{2}{3}\)
2 )\(\frac{1}{2}\)
3 )\(\frac{1}{4}\)
4 )\(\frac{1}{3}\)
Explanation:
Exp: D \(\frac{1}{3}\) Total number of outcomes ={1, 2, 3, 4, 5, 6} = 6 favourable outcomes in this case = {4, 5} = 2 \(\therefore\) p (a number between 3 and 6) \(=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{2}{6}=\frac{1}{3}\)
REAL NUMBERS
90185
If d is the HCF of 56 and 72, the values of x, y satisfying d = 56x + 72y:
1 )x = -3, y = 4
2 )x = 4, y = -3
3 )x = 3, y = -4
4 )x = -4, y = 3
Explanation:
Exp: B x = 4, y = -3 Since, HCF of 56 and 72, by Euclid’s division lemma, 72 = 56 × 1 + 16 ...(i) 56 = 16 × 3 + 8 ...(ii) 16 = 8 × 2 + 0 ...(iii) \(\therefore\) HCF of 56 and 72 is 8. \(\therefore\) 8 = 56 - 16× 3 ( from eq. (ii)) 8 = 56 - (72 - 56 × 1) × 3 [From eq. (i) : 16 = 72 - 56 × 1] 8 = 56 - 3 × 72 + 56 × 3 8 = 56 × 4 + (-3) × 72 \(\therefore\) x = 4, y = -3
90428
Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion \(\frac{7}{210}\):
1 )Terminating decimal expansion
2 )Non -terminating decimal expansion
3 )Cannot be determined
4 )None
Explanation:
Exp: B Non -terminating decimal expansion Simplify it by dividing nominator and denominator both by 7 we get \(\frac{1}{30}\) Factorize the denominator we get 30 = 2 × 3 × 5 Denominator has 3 also in denominator So denominator is not in form of 2\(^{n}\) × 5\(^{n}\)
REAL NUMBERS
90180
Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
1 )\(1<\text{r}<\text{b}\)
2 )\(0<\text{r}\le\text{b}\)
3 )\(0\le\text{r}<\text{b}\)
4 )\(0<\text{r}<\text{b}\)
Explanation:
Exp: C \(0\le\text{r}<\text{b}\) Euclid's division lemma states that, For any positive integers a and b, there exist unique integers q and r such that \(\text{a}=\text{bq}+\text{r},\) where \(0\le\text{r}<\text{b}\) Never Active
REAL NUMBERS
90184
A die is thrown once. Find the probability of getting a number between 3 and 6.
1 )\(\frac{2}{3}\)
2 )\(\frac{1}{2}\)
3 )\(\frac{1}{4}\)
4 )\(\frac{1}{3}\)
Explanation:
Exp: D \(\frac{1}{3}\) Total number of outcomes ={1, 2, 3, 4, 5, 6} = 6 favourable outcomes in this case = {4, 5} = 2 \(\therefore\) p (a number between 3 and 6) \(=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{2}{6}=\frac{1}{3}\)
REAL NUMBERS
90185
If d is the HCF of 56 and 72, the values of x, y satisfying d = 56x + 72y:
1 )x = -3, y = 4
2 )x = 4, y = -3
3 )x = 3, y = -4
4 )x = -4, y = 3
Explanation:
Exp: B x = 4, y = -3 Since, HCF of 56 and 72, by Euclid’s division lemma, 72 = 56 × 1 + 16 ...(i) 56 = 16 × 3 + 8 ...(ii) 16 = 8 × 2 + 0 ...(iii) \(\therefore\) HCF of 56 and 72 is 8. \(\therefore\) 8 = 56 - 16× 3 ( from eq. (ii)) 8 = 56 - (72 - 56 × 1) × 3 [From eq. (i) : 16 = 72 - 56 × 1] 8 = 56 - 3 × 72 + 56 × 3 8 = 56 × 4 + (-3) × 72 \(\therefore\) x = 4, y = -3
