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21. When we reverse the digits of the number 13, the increases by 18. How many other two digit numbers increases by 18 when their digits are reversed?
According to question,
(10y + x) - (10x + y) = 18
Or, 9(y - x) = 18
→ y - x = 2
So, the possible pairs of (x, y) are(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
But, we need the number other than 13.
Thus, there are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79
So, total numbers of possible numbers are 6
Solution:
Let the numbers are in the form of (10x + y), so when the digits of the number are reversed the number becomes (10y + x)According to question,
(10y + x) - (10x + y) = 18
Or, 9(y - x) = 18
→ y - x = 2
So, the possible pairs of (x, y) are(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
But, we need the number other than 13.
Thus, there are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79
So, total numbers of possible numbers are 6
22. The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?
We find that four consecutive odd numbers are 37, 39, 41 and 43
The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.
Thus, 41 is one of the odd numbers
Solution:
Using options,We find that four consecutive odd numbers are 37, 39, 41 and 43
The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.
Thus, 41 is one of the odd numbers
23. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
Now, let the required number be aabb.
Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4
Hence, 7744 is a perfect square
Solution:
Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb.
Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4
Hence, 7744 is a perfect square
24. In a 4-digit number, the sum of the first two digits is equal to that of last two digits. The sum of the first and last digits is equal to third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?
Then,
a + b = c + d ----------(i)
a + d = c ----------(ii )
b + d = 2(a + c) ----------(iii)
from eqn. (i) and (ii),
a + b = a + 2d
→b = 2d
and eqn (iii);
2d + d = 2(a + a + d)
→ 3d = 2(2a + d) → d = 4a
or, a =
→ Now, from eqn. (ii),
a + d = + d = = c
Or, c =
The value of d can be either 4 or 8.
If d = 4, then c = 5
If d = 8, then c = 10
But the value of c should be less than 10
Hence, value of c would be 5
Solution:
Let the 1st, 2nd, 3rd and 4th digits be a, b, c and d respectively.Then,
a + b = c + d ----------(i)
a + d = c ----------(ii )
b + d = 2(a + c) ----------(iii)
from eqn. (i) and (ii),
a + b = a + 2d
→b = 2d
and eqn (iii);
2d + d = 2(a + a + d)
→ 3d = 2(2a + d) → d = 4a
or, a =
→ Now, from eqn. (ii),
a + d = + d = = c
Or, c =
The value of d can be either 4 or 8.
If d = 4, then c = 5
If d = 8, then c = 10
But the value of c should be less than 10
Hence, value of c would be 5
25. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for first time?
LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3;
50 = 2 × 5 × 5
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 20 min.
They will beep together at 12:20
Solution:
They will ring together after,LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3;
50 = 2 × 5 × 5
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 20 min.
They will beep together at 12:20
26. On a road three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If the lights are first switched on at 9:00 AM sharp, at what time will they change simultaneously?
36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
72 = 2 × 2 × 2 × 3 × 3
LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds.
LCM of 36, 42 and 72 is 504
Hence, the lights will change simultaneously after 8 minutes and 24 seconds.
Solution:
LCM of 36, 42 and 72,36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
72 = 2 × 2 × 2 × 3 × 3
LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds.
LCM of 36, 42 and 72 is 504
Hence, the lights will change simultaneously after 8 minutes and 24 seconds.
27. The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 23 × 32 × 5 × 103 × 107, then the number 'N' is:
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N =
Or, N = 3 × 5
Solution:
We have,HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N =
Or, N = 3 × 5
28. Find the LCM and HCF of 2.5, 0.5 and 0.175.
Now,
LCM of two or more fractions is given by:
Solution:
Now,
LCM of two or more fractions is given by:
29. Find the HCF of (3125-1) and (335-1).
The HCF of (am - 1) and (an - 1) is given by (aHCF of m, n - 1)
Thus for this question the answer is (35 - 1)
Since, 5 is the HCF of 35 and 125
Solution:
The solution of this question is based on the rule,The HCF of (am - 1) and (an - 1) is given by (aHCF of m, n - 1)
Thus for this question the answer is (35 - 1)
Since, 5 is the HCF of 35 and 125
30. The last digit of the number obtained by multiplying the numbers 81 × 82 × 83 × 84 × 86 × 87 × 88 × 89 will be
Solution:
The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10.