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81. The number of two digit prime numbers which remain prime even inverting the position of its digits is:
Solution:
These numbers are 11, 13, 31, 17, 71, 37, 73, 79, 97.
82. The number of numbers from 1 to 200 which are divisible by neither 3 nor 7 is :
Number of number divisible by 3,
Number of number divisible by 7
Number of number divisible by 21,
Thus, the divisible value
Thus, number of numbers which are not divisible by 3 or 7
Solution:
The required number = Number of numbers, which are (divisible by 3 + divisible by 7 - divisible by 21)Number of number divisible by 3,
Number of number divisible by 7
Number of number divisible by 21,
Thus, the divisible value
Thus, number of numbers which are not divisible by 3 or 7
83. P is a prime number and (P2 + 3) is also a prime number . The no. of numbers that P can assume is:
P2 + 3 = 7 which is also prime
Again at P = 3, 5, 7, 11 .......
P2 + 3 = an even number which can not be prime
Solution:
Only, one value P can be assumed, which is at 2 at P = 2P2 + 3 = 7 which is also prime
Again at P = 3, 5, 7, 11 .......
P2 + 3 = an even number which can not be prime
84. A naughty boy Amrit watches a Sachin Tendulkar inning and acts according to number of runs he sees Sachin scoring. The details are:
1 run - place one orange in the basket
2 runs - place one mango in basket
3 runs - place a pear in the basket
4 runs - remove a pear and a mango from the basket
One fine day, at the start of the match, the basket is empty. The sequence of runs scored by Sachin in that inning are given as 11232411234232341121314. At the end of the above inning, how many more oranges were there compared to mangoes inside the basket ? (The Basket was empty initially).
= Number of 1 run scored = 8 oranges.
Number of mangoes in the basket,
= (number of 2 runs scored - number of 4 runs scored) = 6 - 4 = 2 mangoes.
Number of more oranges in basket than mangoes,
= 8 - 2
= 6
1 run - place one orange in the basket
2 runs - place one mango in basket
3 runs - place a pear in the basket
4 runs - remove a pear and a mango from the basket
One fine day, at the start of the match, the basket is empty. The sequence of runs scored by Sachin in that inning are given as 11232411234232341121314. At the end of the above inning, how many more oranges were there compared to mangoes inside the basket ? (The Basket was empty initially).
Solution:
Number of oranges in the basket,= Number of 1 run scored = 8 oranges.
Number of mangoes in the basket,
= (number of 2 runs scored - number of 4 runs scored) = 6 - 4 = 2 mangoes.
Number of more oranges in basket than mangoes,
= 8 - 2
= 6
85. Three mangoes, four guavas, and five watermelons cost Rs. 750. Ten watermelons, six mangoes and 9 guavas Cost Rs. 1580. What is the cost of six mangoes, ten watermelons and 4 guavas?
3M + 4G + 5W = 750 ------- (1)
6M + 9G + 10W = 1580 -------- (2)
6M + 4G + 10W = x (let) ------- (3)
On equation (1) × 2 - equation (2), we get
G = 80
On equation (2) - Equation (3), we get
5G = 1580 - x
Putting, G = 80, We get
x = 1180
Solution:
According to the question,3M + 4G + 5W = 750 ------- (1)
6M + 9G + 10W = 1580 -------- (2)
6M + 4G + 10W = x (let) ------- (3)
On equation (1) × 2 - equation (2), we get
G = 80
On equation (2) - Equation (3), we get
5G = 1580 - x
Putting, G = 80, We get
x = 1180
86. A two digit number ab is added to another number ba, which is obtained by reversing the digits then we get three digit number. Thus (a + b) equals to:
Required numbers = 64 + 46 = 110
Option D is correct
Solution:
When two two digit numbers are added and the resultant value is a three digit number. It means there must be a carry over (i.e. The sum of unit digits be greater than 9. Similarly, the sum of the tens digit is also greater than 9)Required numbers = 64 + 46 = 110
Option D is correct
87. A gardener plants his garden with 5550 trees and arranged them so that there is one plant more per row as there are rows then number of trees in a row is:
Thus, total number of trees,
n × (n +1) = 5550
Now, at this moment this problem can be solved in two ways. First by finding the roots of quadratic equation. Second, by using the values from options.
74 × 75 = 5550
i.e. (n + 1) = 75
Solution:
Let there be n rows, then number of trees in each row = (n + 1)Thus, total number of trees,
n × (n +1) = 5550
Now, at this moment this problem can be solved in two ways. First by finding the roots of quadratic equation. Second, by using the values from options.
74 × 75 = 5550
i.e. (n + 1) = 75
88. The sum of two numbers is 18. The greatest product of these two number can be:
So, maximum of (a × b) will be only when a = b
Thus, a = b = 9
Maximum of (a × b) = 9 × 9 = 81.
Solution:
a + b = 18So, maximum of (a × b) will be only when a = b
Thus, a = b = 9
Maximum of (a × b) = 9 × 9 = 81.
89. The unit digit of (316)34n + 1 is :
See the pattern, 62 = 36
63 = 216
64 = 1296
Any power of 6 will give unit digit 6. So
The unit digit of (316)34n always 6.
So, unit digit of (316)34n + 1 will be 7
Solution:
The unit digit of (316)34n, depends on the power of 6.See the pattern, 62 = 36
63 = 216
64 = 1296
Any power of 6 will give unit digit 6. So
The unit digit of (316)34n always 6.
So, unit digit of (316)34n + 1 will be 7
90. A number when divided by 14 leaves reminder of 8, but when the same number is divided by 7, it will leave the remainder :
The number = 14N + 8 (14N is divisible by 14)
When same number is divided by 7 it will give remainder 1
Solution:
When the number is divided by 14 it gives a remainder of 8,The number = 14N + 8 (14N is divisible by 14)
When same number is divided by 7 it will give remainder 1