Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q521. 

1)  20  2)  23 
3)  169  4)  400 
5)  None of these 
Answer : 23
Explanation :
Explanation :
Let the numbers be x and y.
Then, xy = 120 and x^{2} + y^{2 }= 289.
(x + y)^{2} = x^{2} + y^{2 }+ 2xy = 289 + (2 x 120) = 529
x + y = 23.
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Q522. 

1)  101  2)  107 
3)  111  4)  185 
5)  None of these 
Answer : 111
Explanation :
Explanation :
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, coprimes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
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Q523. 

1)  40  2)  80 
3)  120  4)  200 
5)  None of these 
Answer : 40
Explanation :
Explanation :
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
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Q524. 

1)  12  2)  16 
3)  24  4)  48 
5)  None of these 
Answer : 48
Explanation :
Explanation :
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
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Q525. 

1)  37years  2)  30years 
3)  47 years  4)  57 years. 
5)  None of these 
Answer : 47 years
Explanation :
Explanation :
Total age of 30 boys = 3016 = 480
After teacher's join, total age = 3117= 527
Teacher's age = 527 480 = 47 years.
Shortcut Method :
(311) + 16 = 47 years.
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Q526.  A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is: 
1)  145  2)  253 
3)  370  4)  352 
5)  None of these 
Answer : 253
Explanation : Let the middle digit be x.
Then, 2x = 10 or x = 5. So, the number is either 253 or 352.
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253.
Explanation : Let the middle digit be x.
Then, 2x = 10 or x = 5. So, the number is either 253 or 352.
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253.
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Q527.  What is the sum of two consecutive even numbers, the difference of whose squares is 84 ? 
1)  34  2)  38 
3)  42  4)  46 
5)  None of these 
Answer : 42
Explanation : Let the numbers be x and x + 2.
Then, (x + 2)^{2}  x^{2} = 84
or 4x + 4 = 84 or 4x = 80
or x = 20. The required sum = x + (x + 2) = 2x + 2 = 42.
Explanation : Let the numbers be x and x + 2.
Then, (x + 2)^{2}  x^{2} = 84
or 4x + 4 = 84 or 4x = 80
or x = 20. The required sum = x + (x + 2) = 2x + 2 = 42.
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Q528.  The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is: 
1)  380  2)  395 
3)  400  4)  425 
5)  None of these 
Answer : 400
Explanation : Let the numbers be x and y.
Then, xy = 9375 and x/y = 15.
X = 15y
So, y x 15y = 15y^{2} = 9375
or y^{2} = 625. y = 25.
x = 15y = (15 x 25) = 375.
Thus, Sum of the numbers = x + y = 375 + 25 = 400.
Explanation : Let the numbers be x and y.
Then, xy = 9375 and x/y = 15.
X = 15y
So, y x 15y = 15y^{2} = 9375
or y^{2} = 625. y = 25.
x = 15y = (15 x 25) = 375.
Thus, Sum of the numbers = x + y = 375 + 25 = 400.
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Q529.  The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is: 
1)  20  2)  30 
3)  40  4)  24 
5)  None of these 
Answer : 20
Explanation : Let the numbers be x, y and z.
Then, x^{2} + y^{2} + z^{2} = 138 and (xy + yz + zx) = 131.
or (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 138 + 2 x 131 = 400.
or (x + y + z) = 20.
Explanation : Let the numbers be x, y and z.
Then, x^{2} + y^{2} + z^{2} = 138 and (xy + yz + zx) = 131.
or (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 138 + 2 x 131 = 400.
or (x + y + z) = 20.
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Q530.  The sum of the digits of a twodigit number is 15 and the difference between the digits is 3. What is the twodigit number ? 
1)  69  2)  78 
3)  96  4)  D.I 
5)  None of these 
Answer : D.I
Explanation : Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x  y = 3 or y  x = 3.
Solving x + y = 15 and x  y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y  x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.
Explanation : Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x  y = 3 or y  x = 3.
Solving x + y = 15 and x  y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y  x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.
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