Number System
- If the difference between the reciprocal of a positive proper fraction and the fraction itself be 9/20 then the fraction is
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Let us assume that the fraction is P.
Therefore Reciprocal of the given fraction P = 1 . P
According to given question,P − = 9 1 P 20 Correct Option: C
The required fraction is 4 , 5 because 5 − 4 = 25 − 16 = 9 4 5 20 20
- Six numbers are arranged in decreasing order. The average of the first five numbers is 30 and the average of the last five numbers is 25. The difference of the first and the last numbers is :
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Let us assume that six numbers are p, q, r, s, t, u.
According to the given question, Numbers are in decreasing order :
p > q > r > s > t > u
Again according to the question,
p + q + r + s + t = 5 × 30
q + r + s + t + u = 5 × 25Correct Option: B
Let us assume that six numbers are p, q, r, s, t, u.
According to the given question, Numbers are in decreasing order :
p > q > r > s > t > u
Again according to the question,
p + q + r + s + t = 5 × 30
⇒ p + q + r + s + t = 150 --- (i)
q + r + s + t + u = 5 × 25
⇒ q + r + s + t + u = 125 --- (ii)
By equation (i) – (ii)
p – u = 150 – 125 = 25
- The sum of three consecutive odd natural numbers is 147. Then, the middle number is :
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Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴ P + P + 2 + P + 4 = 147
⇒ 3P + 6 = 147
⇒ 3P = 147 – 6 = 141Correct Option: C
Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴ P + P + 2 + P + 4 = 147
⇒ 3P + 6 = 147
⇒ 3P = 147 – 6 = 141⇒ P = 141 = 47 3
∴ Middle Number = P + 2 = 47 + 2 = 49
- The sum of first 20 odd natural numbers is equal to :
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Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.
Sn = n [2a + (n − 1)d] 2
Where a : First term, d : common difference and n = Number of termsCorrect Option: C
Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.
Sn = n [2a + (n − 1)d] 2
Where a = First term, d = common difference and n = Number of terms∴ S20 = 20 × [(2 × 1) + (20 − 1) × 2] 2
∴ S20 = 10 [ 2 + 38 ] = 10 × 40 = 400
Note : Sum of first n consecutive odd numbers = n2
- The sum of all natural numbers from 75 to 97 is :
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Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
an = a + ( n – 1 )d
⇒ 97 = 75 + ( n – 1 )
⇒ n = 97 – 74 = 23Sn = n (a + 1) 2 Correct Option: D
Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
an = a + ( n – 1 )d
⇒ 97 = 75 + ( n – 1 )
⇒ n = 97 – 74 = 23Sn = n (a + 1) 2 S23 = 23 (75 + 97) 2 S23 = 23 × 172 = 1978 2
