Number System
- In a two–digit number, the digit at the unit’s place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is
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Let us assume Ten’s digit = P
Unit’s digit = 2P – 1
∴ Original number = 10P + (2P – 1) = 12P – 1
New number = 10 (2P – 1) + P
New number = 20P – 10 + x = 21P – 10
According to given question,
∴ (21P – 10) – (12P + 1) = 12P – 1 – 20Correct Option: D
Let us assume Ten’s digit = P
Unit’s digit = 2P – 1
∴ Original number = 10P + (2P – 1) = 12P – 1
New number = 10 (2P – 1) + P
New number = 20P – 10 + x = 21P – 10
According to given question,
∴ (21P – 10) – (12P + 1) = 12P – 1 – 20
⇒ 9P – 9 = 12P – 21
⇒ 3P = 12
⇒ P = 4
⇒ Original number = 12P – 1 = 12 × 4 – 1 = 47
- The sum of all natural numbers between 100 and 200, which are multiples of 3 is :
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Numbers divisible by 3 and lying between 100 and 200 are : 102, 105, 108, 111,...................... 198
Let number of terms = n
As we know the formula
an = a + ( n – 1 )d
where an = Nth Term a = First term , n = Number of terms and d = Common Difference
∴ 198 = 102 + ( n – 1 ) 3⇒ n − 1 = 198 − 102 = 32 3
⇒ n = 33∴ Sn = Number of terms = ( First term + Last term ) 2
Correct Option: B
Numbers divisible by 3 and lying between 100 and 200 are : 102, 105, 108, 111,...................... 198
Let number of terms = n
As we know the formula
an = a + ( n – 1 )d
where an = Nth Term a = First term , n = Number of terms and d = Common Difference
∴ 198 = 102 + ( n – 1 ) 3⇒ n − 1 = 198 − 102 = 32 3
⇒ n = 33∴ Sn = Number of terms = ( First term + Last term ) 2 ∴ Sn = n = ( a + l ) 2 ∴ Sn = 32 (102 + 198) = 4950 2
- By interchanging the digits of a two digit number we get a number which is four times the original number minus 24. If the unit’s digit of the original number exceeds its ten’s digit by 7, then original number is
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Let the two–digit number be 10P + Q where P < Q.
Number obtained on reversing the digits =10Q + P
According to the question, 10Q + P = 4 (10P + Q) – 24
⇒ 40P + 4Q – 10Q – P = 24
⇒ 39P – 6Q = 24
⇒ 13P – 2Q = 8 .....................(i)
Again, Q – P = 7
⇒ Q = P + 7 ....(ii)Correct Option: A
Let the two–digit number be 10P + Q where P < Q.
Number obtained on reversing the digits =10Q + P
According to the question, 10Q + P = 4 (10P + Q) – 24
⇒ 40P + 4Q – 10Q – P = 24
⇒ 39P – 6Q = 24
⇒ 13P – 2Q = 8 .....................(i)
Again, Q – P = 7
⇒ Q = P + 7 ....(ii)
Put the value of Q from equation (ii) in equation (i).
∴ 13P – 2 (P + 7) = 8
⇒ 13P – 2P – 14 = 8
⇒ 11P = 14 + 8 = 22⇒ P = 22 = 2 11
From equation (ii),
Q – 2 = 7
⇒ Q = 2 + 7 = 9
∴ Number = 10Q + P = 10 × 2 + 9 = 29
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Sum of three fractions is 2 11 24
On dividing the largest fraction by the smallest fraction 7/6 is obtained which is 1/3 greater than the middle fraction. The smallest fraction is
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Let the three fractions be p, q and r, where p < q < r.
According to the question,r = 7 ⇒ r = 7 p p 6 6
Again, middle fraction= q = 7 − 1 = 7 − 2 = 5 6 3 6 6
Correct Option: B
Let the three fractions be p, q and r, where p < q < r.
According to the question,r = 7 ⇒ r = 7 p p 6 6
Again, middle fraction= q = 7 − 1 = 7 − 2 = 5 6 3 6 6 ∴ p + q + r = 2 11 24 ⇒ p + 5 + 7 p = 59 6 6 24 ⇒ p + 7p = 59 − 5 6 24 6 ⇒ 6p + 7p = 59 − 20 = 39 6 24 24 ⇒ 13p = 39 ×6 = 39 24 4 ⇒ p = 39 = 3 4 × 13 4
- The digit in unit’s place of the number (1570)2 + (1571)2 + (1572)2 + (1573)2 is :
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Unit's digit in (1570)2 = 0
Unit's digit in (1571)2 = 1
Unit's digit in (1572)2 = 4
Unit's digit in (1573)2 = 9Correct Option: A
Unit's digit in (1570)2 = 0
Unit's digit in (1571)2 = 1
Unit's digit in (1572)2 = 4
Unit's digit in (1573)2 = 9
∴ Required unit’s digit = Unit’s digit (0 +1+4 + 9) = 4
