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Solutions
S1. Ans.(d)
Sol. Let the odd number be 2x + 1.
Then, 1(2x – 5) + (2x – 3) + (2x – 1) + (2x + 1) + (2x + 3) + (2x + 5) = 168
⇒ 12x = 168 ⇒ x = 14
∴ Largest is 2x + 5 = 28 + 5 = 33
S2. Ans.(d)
Sol. 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77
S4. Ans.(b)
Sol. Unit digit in 3⁹⁸ = Unit digit in (3⁴)²⁴ × Unit digit in (3²) = 9
∴ Dividing 3⁹⁸ by 5 or dividing 9 by 5, we get the remainder 4.
S6. Ans.(a)
Sol. ∵ Product of two whole numbers is 24.
Possible pairs = (1, 24), (2, 12), (3, 8), (4, 6)
∴ Sum of these pairs
(1, 24) = 1 + 24 = 25
(2, 12) = 2 + 12 = 14
(3, 8) = 3 + 8 = 11
(4, 6) = 4 + 6 = 10
∴ Smallest possible sum = 10
S8. Ans.(b)
Sol. (x + 1)⁴ – (x – 1)⁴
= (x⁴ + 4x³ + 6x² + 4x + 1) – (x⁴ – 4x³ + 6x² – 4x + 1)
= 2 (4x³ + 4x) = 8 (x³ + x)
∴ It is divisible by 8.
S9. Ans.(d)
Sol. Let three digit number = 111.
Since, digit at hundred place is 7 more than the digit at unit’s place.
∴ Digit at hundred place = 1 + 7 = 8
∴ Required three-digit number = 811
Now, changing the numbers new number = 118
According to the question,
811 – 118 = 693
Now, unit place digit in 693 = 3
∴ Required answer = 3
S10. Ans.(b)
Sol. Given, four digit number is 51y3.
Sum of digits = 5 + 1 + y + 3 = 9 + y
∴ We know that, the given number 51y3 will be divisible by 9 if its sum of digits is divisible by 9. So, 9 + y is divisible by 9. Which is only possible, when we take y = 0 or 9.
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