The number of integers n satisfying -n + 2 ≥ 0 and 2n ≥ 4 is
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Solution
2-n ≥ 0 ∴ n ≤ 2
2n ≥ 4 ∴ n ≤ 2
The only value that satisfies both these conditions is n = 2
273 – 272 – 271 is same as
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Solution
271(22- 2 -1) = 271(4 - 3) = 271
n3 + 2n for any natural number n is always a multiple of
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Solution
Simplest way is to put different values for number 'n' and work out with the options.
n = 1 ⇒ n3 + 2n = 3
n = 2 ⇒ 8 + 4 = 12
n = 3 ⇒ 27 + 6 = 33
n = 4 ⇒ 64 + 8 = 72
All these numbers are exactly divisible by 3
A number when divided by 5 leaves a remainder 3. What is the remainder when square of this number is divided by 5?
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Solution
Approach 1:
The number yields a remainder 3 when divided by 5. Therefore when its square is divided by 5, the remainder is square of earlier remainder i.e. 9. But 9 + 5 gives remainder 4.
Approach 2:
The number would be (5x + 3) square of it would be 25x2 + 30x + 9. Division by 5 would result in 5x2 + 6x + 14⁄5 so remainder is 4
Largest number obtained by writing number 2, three times is
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Solution
The greatest number is 222.
The sum of first 50 natural numbers is divisible by
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Solution
Sum of first n natural numbers is given by \(\frac{n(n + 1)}{2}\)
∴ \(\frac{50(51)}{2}\) = 25 x 51 is divisible by 5,25,5,17.
xyz company agreed to finish a road widening work in 150 days. It employed 75 men, each working 8 hours daily. After 90 days, it realized that only 2⁄7 of work was completed. So it employed x more men and all the men were now made to work for 10 hours daily and work was completed in time. How many more men were employed by the company?
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Solution
75 workers x 8 hours x 90 days = 54000 man hours. 2⁄7 work is done in 54000 man hours.
∴ Remaining 5/7 work will require 135000 man hours. This work is to be completed in the remaining 60 days.
∴ Number of men required =\(\frac{135000}{60\times 10}= 225\)
We already have 75 workers
∴ Additional requirement 150.
If P and q are positive integers, then which of the following is true?
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Solution
The expansion (p + q)73 will have more positive integral terms than p73 and q73.
∴ p73 + q73 < (p + q)73
Distance travelled by a particle in time (t) is given by formula s = 1+ 2t + 3t2 + 4t3, what is distance traveled in 4th second of motion?
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Solution
Distance traveled in 4th second = distance travelled in 4 seconds - distance travelled in 3 seconds
= 1+ (2 × 4) + (3 × 16)+ (4 × 64)
= 1+ (2 × 3) + (3 × 9)+ (4 × 27)
= 2 + (3 × 7)+ (4 × 37) = 171
In a circular race track of length 100 meters, 3 persons A, B, C start together. A and B start in same direction at 10 meters per second and 8 meters per second respectively. C runs in opposite direction at 15 meters per second. When will all three meet for first time after start?
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Solution
Every second a gap of 2 meter is created between A and B. They meet when a gap of 100 meter is created between them i.e. after 50 seconds.
Similarly, every second a gap of 10 + 15 = 25 meters is created between A and C. So they meet after 4 seconds \(\left (\frac{100}{25}=4 \right )\)
∴ The three of them will meet after 100 seconds. (LCM of 4 and 50)