Solution:
8-x / 12 â€“x = 12-x /20 â€“x
(8-x)(20-x) = (12 â€“ x)(12 â€“x )
160 â€“ 8x â€“ 20x + x2 = 144 â€“ 12x â€“ 12x + x2
4x = 16
x = 4
Solution:
Let the number is 10x + y
Given: x + y = 11 ----------(i)
According to the question,
10x + y + 45 = 10y + x
â‡’ x â€“ y = -5 -----------(ii)
From eq. (i) and (ii) , we get
x = 3 and y = 8
Solution:
Let the first number be 6x then second and third number be 3x and 2x respectively.
Given: average of these numbers = 517
Then, their sum = 517 * 3
= 1551
â‡’ 6x + 3x + 2x = 1551
â‡’11x = 1551
â‡’ x = 141
Now, the difference between first and third number = 6x â€“ 2x
= 4x
= 4 * 141
= 564
Solution:
Given fractions are 1/2, 2/3, 5/9, 6/13, and 7/9
L.C.M of their denominators ie., 2, 3, 9, 13, 9 = 234
Thus, given fractions become:
1/2 = 117/234
2/3 = (2 * 78)/234 = 156/234
5/9 = (5 * 26)/234 = 130/234
6/13 = (6 * 18)/234 = 108/234
7/9 = (7 * 26)/234 = 182/234
---> On arranging the numerators in ascending order, we get
108, 117, 130, 156, 182
Therefore Ascending order of the given fractions is:
6/13 < 1/2 < 5/9 < 2/3 < 7/9
---> Here, the fourth one is 2/3.
Solution:
Let 'x' and 'y' be the two numbers where (x > y).
Given that, Product of x and y = 9222
---> x * y = 9222 ---> eqn (1)
Given that, Difference of x and y = 19
--->. x - y = 19
---> x = 19 + y ---> eqn (2)
Substitute this 'x' value in eqn (1), we get
---> (19 + y) * y = 9222
---> 19y + y2 = 9222
---> y2 + 19y - 9222 = 0
By using the formula, y = {-b ± √[b2 - 4ac]} / 2a
We can find the value of y,
Where, b = +19
a = 1
c = -9222
---> y = {-19 ± √[(19)2 - 4 * 1 * (-9222)]} / 2 * 1
= {-19 ± √[361 + 36888]} / 2
= {-19 ± √[37249]} / 2
= {-19 ± 193} / 2
= {-19 + 193} / 2; {-19 - 193} / 2
= {174/2}; {-212/2}
= 87; -106
'y' can't be negative number (i.e, -106) as 'y' is a natural number.
Therefore, y = 87
Hence, x = 19 + y (---> from eqn (2))
--> x = 19 + 87
--> x = 106
Therefore, sum of the numbers = x + y = 106 + 87 = 193
Solution:
Multiples of 3 between 100 and 200 are 102, 105, 108,â€¦ ,198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = n
W.K.T: Arithmetic Progression Formula:
an = a1 + (n - 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 - 102 = 3
---> 198 = 102 + (n - 1) * 3
---> 198 - 102 = (n - 1) * 3
---> 96 = (n - 1) * 3
---> (n - 1) = 96/3 = 32
---> n = 32 + 1
---> n = 33
Formula:
Sum of n terms = Sn = (n/2) * (a + l)
where n = number of elements = 33
a = first term = 102
l = last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950
Solution:
W.K.T: Sum of first n terms of an Arithmetic Progression (A.P) is:
S = (n/2)[2a + (n - 1)d]
where, n = number of terms
a = first term
d = common difference
Now, sum of 1st 11 terms of an A.P = sum of 1st 19 terms of that A.P
---> (11/2)[2a + (11 - 1)d] = (19/2)[2a + (19 - 1)d]
---> 22a + 11* (10) * d = 38a + 19 *(18) *d
---> 22a + 110d = 38a + 342d
---> 38a - 22a + 342d - 110d = 0
---> 16a + 232d = 0
---> Dividing above eqn by 8, we get
---> 2a + 29d = 0 ---> eqn (1)
Then, Sum of 1st 30 terms of an A.P = (30/2)[2a + (30 - 1)d]
= (15)[2a + 29d]
= (15)[0] ----> [Since from eqn (1), 2a + 29d = 0]
= 0
Hence, Sum of 1st 30 terms of an A.P = 0
Solution:
Let the required number be 'N' which is the 'dividend'.
Given, Divisor = sum of 555 and 445 = 555 + 445 = 1000
Quotient = 2 * (555 - 445) = 2 * 110 = 220
Remainder = 30
W.K.T: Dividend = Divisor * Quotient + Remainder
---> Dividend ie., Required Number = 1000 * 220 + 30
---> Required Number = 220000â€¬ + 30
---> Required Number = 220030
Solution:
Given numbers are (9)(1/3), (20)(1/4), (25)(1/6)
L.C.M of 3, 4, and 6 = 12
Now,
(9)(12/3), (20)(12/4), (25)(12/6)
----> (9)4, (20)3, (25)2
----> (9)4 = 6561,
----> (20)3 = 8000,
----> (25)2 = 625
Therefore, right arrangement in ascending order is:
(25)(1/6) < âˆ›9 < âˆœ20
Solution:
If we write all the given numbers in a form (10a + b), b will turn out to be the unit's digit here.
So when we multiply all the given numbers, the unit's digit in the product will be nothing but the product of all unit's digits,
irrespective of what the other digits in the number are.
From the given expression,
---> unit digit of 43 * unit digit of 69 ---> 3 * 9 = 27 ---> here, unit digit = 7
7 * unit digit of 551 = 7 * 1 = 7 ---> here, unit digit =7
7 * unit digit of 9242 = 7 * 2 = 14 ---> here, unit digit = 4
Therefore, Digit at unit's place of the given expression "43 * 69 * 551 * 9242" = 4
Solution:
Total cost of the two pieces of cloth = (1.2 * 330) + (1.3 * 270) = Rs. 747
Given that he paid an amount of Rs. 1000 at the counter
Amount he get back = 1000 - 747 = Rs. 253
Solution:
Since the three numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29
Given product of the first two numbers = 551
--> First number = 551/29 = 19
Given product of the last two numbers = 1073
---> Third number = 1073/29 = 37
Thus, Required sum = (19 + 29 + 37) = 85
--> Sum of the three numbers = 85.
Solution:
The least number which must be added-to 1728 to make it a perfect square is
Reference:
----> 1728 + 36 = 1764 = (42)2
----> Hence 36 is to be added with 1728 to make it perfect square.
Hence the answer is : 36
Solution:
Find the total number of factors of 15120
If a composite no. N has been written in the form of
-----> N = ap ,b q ,c r ,d s,..........
Then the no. of total division or factors of
-----> N = (p+1)(q+1)(r+1)(s+1),.........
Hence,
-----> 15120 = 24 * 34 * 5 1 * 7 1 total no of factors,
-----> = (4+1) (3+1) (1+1) (1+1)
-----> = 80
Hence the answer is : 80
Solution:
Find the unit digit of the product of all the prime numbers between 1 and (17) 17 :
Reference:
The set of prime numbers S = { 2,3,5,7,11,13,.....}
Since there is one 5 and one 2 present in this series. If we multiply there two numbers we get â€˜0â€™ as the unit digit.
Hence the unit digit of given expression is 0.
Solution:
If (a*b) = 6a - 4b + 3ab, then ((6 * 3) + (4 * 3)) is equals to.
Reference:
----> (6 * 6) - (4 * 3) + (3 * 6 * 3 ) + (4 * 6) - (4 *3) + (3* 4* 3)
----> = 36 -12 + 54 +24 - 12 + 36
----> = 126
Hence the answer is : 126
Solution:
Which of the following cannot be the number of zeroes at the end of any factorial:
Reference:
----> We know that the number of zeroes at the end depends on the pair of (2 * 5) . We know that number of zeroes in 5! Is 1.
----> (5! to 9!) = 1
----> (10! to 14!) = 2
----> (15! to 19! ) = 3
----> (20! to 24!) = 4
----> (25! to 29!) = 6
----> Here number of zeroes between 25! to 29! is 6 because 25 = 5 2 So, 5 cannot be the number of zeroes at the end of any factorial value.
Hence the answer is : 5
Solution:
If p is prime number, then which of the following may also be a prime number:
Reference:
---> None of the prime number other than 2 is divisible by 2 and we know that 1 is not a prime.
---> p2can be factorized as (p* p) which makes it a composite number.
---> 3p is also a composite number.
---> Now, for p = 5,7 and 13 we have p -2 = 3,5 and 11 respectively, which are prime numbers.
Hence the answer is : (p -2)
Solution:
N is the largest 3-digit number, which when divided by 3, 4 and 6 leaves the remainder 1, 2 and 4 respectively. What is the remainder when N is divided by 7:
Reference :
Since,(3 -1) = (4 - 2) = (6 - 4) = 2
Least number which when divided by 3, 4 and 6 leaves 1, 2 and 4 as remainders is
----> LCM(3,4,6)
----> = 12-2
----> = 10
Largest three digits multiple of 12 which is under 1000 when we add 10 to it
----> (1000/12) = 4 (remainder)
----> 1000 - 1 = 996 (multiple of 12)
----> 996 - 12 = 984 (multiple of 12)
Required number = 984+10 = 994
Remainder = ((994)/7) = 0
Hence the answer is : 0
Solution:
What is the greatest number that will divide 1204, 3664 and 5904 leaving the same remainder:
Reference :
Required number,
----> HCF [(3664 - 1024),(5904 - 3664),(5904 - 1204)]
----> HCF(2466,2240,4700)
----> 20
Hence the answer is : 20