We are aske dto find the number of numbers that are divisible by none of 2, 5 and 7. Instead, let us find the number of numbers that are divisible by atleast one of 2, 5 and 7.
Recall that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).
Therefore, Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (2 and 5) - Number of Multiples of (5 and 7) - Number of Multiples of (7 and 2) + Number of Multiples of (2 and 5 and 7)
Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (10) - Number of Multiples of (35) - Number of Multiples of (14) + Number of Multiples of (70)
Number of Multiples of (2 or 5 or 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79
Since we know the number of integers that can be divided by atleast one of 2, 5 and 7 is 79. We can arrive at the number of integers that can be divided by none of 2, 5 and 7, by subtracting from the whole of 120.
No of integers of 1, 2, … , 120, that are divisible by none of 2, 5 and 7 = 120 - 79 = 41
In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is
The difference between Asha and Bishnu marks = 23 + 34 = 57
The difference between Asha and Bishnu marks(in terms of percentage) = 64% - 52% = 12%
We need to find Geeta’s marks who secured, 84%.
Observe that 84 = 7 × 12
Therefore, Geeta’s marks = 7 × 57 = 399.
Note: You need not know or calculate the exact value of
Whatever is, we know that it is less than
So, the valuve 16 - , will be greater than 16 - or 15.5
Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15 km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?
We are given the individual speeds of Anil, Sunil and Ravi, which are 15 kmph, 10 kmph and 8 kmph respectively.
We are also told that the length of the track is 3 KM.
Let us calculate the time taken by Anil and Sunil to complete one round(which is 3KM in length):
Time taken by Anil to complete one round in minutes 12 minutes.
Time taken by Sunil to complete one round in minutes 18 minutes.
Anil completing each round in 12 minutes, reaches the starting point at 12, 24, 36, 48, 60 … minutes
Sunil completing each round in 18 minutes, reaches the starting point at 18, 36, 54, 72, 90 … minutes
Clearly, both of them reach the starting point at 36 minutes, for the first time together. Observe that 36 is teh LCM of 12 and 18
The distance covered by Ravi, whose speed is 8 kmph in 36 minutes(or = 0.6 hours) = 8 × 0.6 = 4.8 km
A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to
Let the price of Sugar per kilogram be ‘x’ rupees.
The man marks it up by 20% and sells 5 kilograms.
Marked Price = 1.2 × x = 1.2x
Therefore, the sale price of these 5 kgs totally would be = 5 × 1.2 × x = 6x
He then gives a discount of 10% on the markup and sells 15 kgs at that price. So, the price per kg now would be 0.9 × 1.2 × x = 1.08x
Therefore,
The sale price of these 15 kgs totally would be = 15 × 1.08x = 16.2x
He then looses 3 kgs of Sugar
Therefore, the sale price of these 3 kgs = 0.
There is 35 - 5 - 15 - 3 = 12 kgs of sugar remaining.
Let’s say it is sold at px price.
So, the sale price of these 12 kgs will be = 12 × px
The overall profit for the Man is 15%, So the Sale Price of the entire 35 kgs is 35 × 1.15 × x = 40.25x
Summing up and equating all the sale prices…
40.25x = 6x + 16.2x + 0x + 12 × px
40.25x = 22.2x + 12 × px
18.05x = 12 × px
Let’s approximate this to
18x = 12 × px p = = 1.5
Very importantly, px is attained after marking up the marked price.
Therefore, px = y × Marked Price
1.5x = y × 1.2x
y = = 1.25
In other words we can say that the marked price was increased by 25%.
The total number of integers from 100 to 999 is 999 - 99 = 900.
Of these 900 integers, if we could find the number of integers which do not have repetitions in them, we can also find the number of integers that have repetitions.
Let’s find the number of integers that between 100 and 999, which do not have repetitions.
The number of integers that between 100 and 999, which do not have repetitions = 9 × 9 × 8
Therefore, the number of integers that have repetitions = 900 - 9 × 9 × 8
= 9 (100 - 9 × 8)
= 9 (28)
= 180 + 72 = 200 + 52 = 252
= 0.0625
is equal to
Then m - 2n equals
12 minutes.
18 minutes.
= 0.6 hours) = 8 × 0.6 = 4.8 km
= 1.5
= 1.25
