Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q11. 

1)  Rs.8000  2)  Rs.10,000 
3)  Rs.12,000  4)  Data Inadequate 
5)  None of these 
Answer : Rs.8000
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Q12.  In a triangle ABC, the sides AB and AC have been produced to D and E. Bisectors of angle CBD and angle BCE meet at O. If angle A = 64°, then angle BOC is  
1)  54°  2)  56° 
3)  58°  4)  60° 
5)  None of these 
Answer : 58°
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Q13. 

1)  2  2)  3 
3)  4  4)  5 
5)  None of these 
Answer : 5
Explanation :
Explanation :
Since, DE parallel to BC
so, (AD / DB) = (AE / EC)
or {x / (x – 2)} = {(x + 2) / (x – 1)}
or x(x – 1) = (x + 2) (x – 1)
or x^{2} – x = x^{2 }– 4
or x = 4 .
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Q14.  If a + b + c = m and 1/a + 1/b + 1/c = 0 , then the average of a^{2}, b^{2} and c^{2} is 
1)  m2  2)  m2/3 
3)  m2/9  4)  m2/27 
5)  None of these 
Answer : m2/3
Explanation : Given, 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 0
Explanation : Given, 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 0
Or ab + bc + ca = 0
And a + b + c = m
Squaring both sides, we get, (a + b + c)^{2} = m^{2}
Or a^{2} + b^{2} + c^{2 }+2(ab + bc + ca) =m^{2}
Or a^{2} + b^{2} + c ^{2}+2(0) = m^{2}
So, the average = ( a^{2} + b^{2} + c ^{2}) /3 = m^{2}/3
^{}
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Q15.  If x + y+ z = 9 and x^{2}+y^{2}+z^{2} = 31 , then what should be the value of x^{3}+y^{3}+z^{3} 
1)  3  2)  9 
3)  27  4)  54 
5)  None of these 
Answer : 54
Explanation : Given, x + y+ z = 9 and x^{2}+y^{2}+z^{2} = 31
Explanation : Given, x + y+ z = 9 and x^{2}+y^{2}+z^{2} = 31
Then, (x + y+ z)^{2 }= 9^{2}
Or x^{2}+y^{2}+z^{2} + 2( xy + yz + zx) = 81
Or 31 + 2( xy + yz + zx) = 81
Or ( xy + yz + zx) = 25
So, x^{2}+y^{2}+z^{2 } 3xyz = (x + y + z) [x^{2}+y^{2}+z^{2} (xy + yz + zx)] = 9[3125] = 54
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Q16.  If a + b = 2c, then find the value of a/(a  c) + c/ (bc) 
1)  0  2)  1 
3)  1  4)  2 
5)  None of these 
Answer : 1
Explanation : Given, a + b = 2c
Explanation : Given, a + b = 2c
Put, a = 1, b = 2, then c = 3
So,1/(12) + 2/(32) = 1+2 = 1
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Q17.  If x^{2}+1/x^{2}=1, then find the value of x^{18}+x^{12}+x^{6}+1 
1)  0  2)  1 
3)  1  4)  2 
5)  None of these 
Answer : 0
Explanation : Given, x^{2}+1/x^{2 }=1, Or x4+1=x^{2}
Explanation : Given, x^{2}+1/x^{2 }=1, Or x4+1=x^{2}
Or x^{4}x^{2}+1=0
Multiplying both sides by (x^{2}+1), we get
(x^{2}+1)(x^{4}x^{2}+1) =0x(x^{2}+1)
Or x^{6}+1 =0
So,x^{18}+x^{12}+x^{6}+1 = x^{12}(x^{6}+1)+x^{6}+1 =x^{12}x0+0=0
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Q18.  If 1/a(a^{2}+1)=3,then what should be the value of (a^{6}+1)/a^{3} 
1)  1  2)  9 
3)  18  4)  27 
5)  None of these 
Answer : 18
Explanation : Given, 1/a(a^{2}+1) = 3
Explanation : Given, 1/a(a^{2}+1) = 3
Or a+1/a =3
Or a^{3}+1/a^{3}+3a.1/a.(a+1/a) = 33
Or a^{3}+1/a^{3 }+3x3= 27
Or a^{3}+1/a^{3} =18
Or (a^{6}+1)/a^{3}= 18^{ }
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Q19.  If x+(1/x) = 5, the find the value of 5x/(x^{2}+5x+1) 
1)  1/3  2)  1/4 
3)  1/2  4)  1/s 
5)  None of these 
Answer : 1/2
Explanation : Given, x+(1/x) = 5
Explanation : Given, x+(1/x) = 5
or (x2+x)/x = 5
or x2+x = 5x
Now,5x/(x^{2}+5x+1)= 5x/(x^{2}+1+5x) =5x/(5x+5x) =5x/10x=1/2
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Q20.  An article is marked x% above the cost price. A discount of 2x/3% is given on the marked price. If the profit is 4% of the cost price and the value of x lies between 25 and 50, then what should be the value of 50% of x ? 
1)  12  2)  14 
3)  15  4)  20 
5)  None of these 
Answer : 15
Explanation :
Explanation :
Let, the cost price of the article is Rs.100. Marked price (MP) = Rs.100+x
Selling price of the price (SP) = MP  discount = Rs.100+x  2x/3 *(Rs.100+x)/100
Also, SP = CP + 4% profit = 100+4 = Rs.104
Hence,(100+x)2/3x * (100+x)/100 = 104
or 100+x  2x/300 (100+x) = 104
or x²  50x + 600 = 0
On solving the above equation, we get possible values of x as 20 and 30. But as x lies between 25 and 50, the value should be 30.
So, x = 30.
The value of 50% of x is = 15
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