Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1.  The LCM and HCF of two numbers are 660 and 10 respectively. If the sum of the numbers is 170, the what should be difference of the numbers ? 
1)  30  2)  40 
3)  48  4)  50 
5)  None of these 
Answer : 50
Explanation : Let , two numbers are 10a and 10b ,where a and b are relatively prime numbers.
Explanation : Let , two numbers are 10a and 10b ,where a and b are relatively prime numbers.
According to the question,10a X b = 660
or ab = 66  (i)
Also given, 10a + 10b = 170
or a + b = 17  (ii)
Solving Eqs. (i) and (ii) we get a = 11 and b = 6
So, numbers are 110 and 60.
Thus, the difference is 110  60 = 50
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Q2.  Thrice the age of A is 7 times B's age. And C’s age is half of A's age. Also, nine years ago, A's age was onethird of B’s age then. What is the ratio between the age of the A,B and C age respectively ? 
1)  11:6:7  2)  11:6:3 
3)  14:6:3  4)  14:6:7 
5)  None of these 
Answer : 14:6:7
Explanation : Given, 3A = 7B
Explanation : Given, 3A = 7B
or 3A  7B = 0 (I) and A = 2C  (II)
And A9 = (B9)/3
or 3A27 = B9
or 3A  B = 18  (III)
from equations I and III , we get B = 3
A= 7 and C = 3.5
So, ratio of A,B and C i.e. A:B:C= 14:6:7
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Q3. 

1)  Rs.10,480  2)  Rs.11,240 
3)  Rs.11,520  4)  Rs.12,420 
5)  None of these 
Answer : Rs.11,520
Explanation :
Explanation :
Let the width of the rectangle = x m
Then, 250% of x = 40
x = 40 × 100/250 = 16 m
Perimeter of the rectangle = 2(length + width) = 2 × (40 + 16) = 112 cm
Perimeter of the square = (112 + 16) = 128 m
The sides of the square = 128/4 = 32 m
The area of the square field without street = (32)^{2} sq. cm = 1024 sq. m
The area of the square field with street = (32 + 8)^{2} sq. cm = 1600 sq. m
The area of the street = 1600  1024 = 576 sq. m
The total cost of constructing the street = 576 × 20 = Rs.11,520
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Q4.  A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then what was the amount (in Rs lakh) invested in the fixed deposit ? 
1)  Rs. 6 Lakhs  2)  Rs. 7 Lakhs 
3)  Rs. 8 Lakhs  4)  Rs. 9 Lakhs 
5)  None of these 
Answer : Rs. 9 Lakhs
Explanation :
Explanation :
Let Rs x be invested in FD.
So, Remaining amount = Rs. 15 Lakhs  x, which is invested in the ratio 2 : 1
So, each investment would have 1/3(15x) and 2/3(15x) investments respectively
So, Total annual interest income = 6% of x + 4% of 1/3(15x) + 3% of 2/3(15x) = 76000
Rewriting Rs.76000 as Rs. 0.76 Lakhs,
6% of x + 4% of 1/3(15x) + 3% of 2/3(15x) = 0.76
Multiply both sides by 100,
6x + 4/3(15x) + 2(15x) = 76
6x +11/3 (15x) = 76
165 + 18x 11x = 228
7x = 63
x = 9 (or) Rs. 9 Lakhs
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Q5. 

1)  60  2)  70 
3)  72  4)  80 
5)  None of these 
Answer : 80
Explanation :
Explanation :
Let, the total Maximum marks in exam = 100%
His score = 50% of 100 = 50
His marks after review = 140% of 50 = 70
His marks after post review = 120% of 70 = 84
A.T.Q 84% – 12 =70% + 30
14% = 42
or 100% = 300
So passing marks = 300(70/100) + 30 = 240
Total marks = 100% = 300
Thus, required percentage score needed for passing the examination = (240/300 ) X 100 = 80%
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Q6. 

1)  53/10 days  2)  67/10 days 
3)  63/10 days  4)  73/10 days 
5)  None of these 
Answer : 63/10 days
Explanation : A,B and C altogether work for 3 days = 3 (1/12+1/18+1/24) = 13/24 parts
Explanation : A,B and C altogether work for 3 days = 3 (1/12+1/18+1/24) = 13/24 parts
So, remaining (113/24) = 11/24 parts
B and C can do in 1 day (1/12+1/18) = 5/36 parts
Now, B & C can do this in (11/24 )/(5/36) = 33/10 days = 33/10 days
So, overall time taken to complete this workpiece = 63/10 days
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Q7. 

1)  2  2)  4 
3)  5  4)  6 
Answer : 8
Explanation :
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
Let us take number of green balls taken out from bag Y be a
Number of blue balls taken out from bag Y = a + 4
According to the question,
(10 + a)C_{2}/(30 + 2a + 4)C_{2} = 13/123
(10 + a) * (9 + a)/[(34 + 2a) * (33 + 2a)] = 13/123
Simplify the equation we get, a = 4
Number of blue balls taken out from bag Y = a + 4 = 4 + 4 = 8
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Q8. 

1)  10  2)  20 
3)  29  4)  30 
Answer : 30
Explanation :
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and
second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
This satisfies the given condition.
So, total number balls in box X should be 30.
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Q9. 

1)  2/29  2)  3/29 
3)  4/29  4)  5/29 
Answer : 3/29
Explanation :
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29) or 40/87 = 40/87
This satisfies the given condition.
Probability of both green ball (A) = 10C2/30C2
= (10 * 9)/(29 * 30) = 3/29
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Q10. 

1)  60 days  2)  80 days 
3)  90 days  4)  120 days 
5)  None of these 
Answer : 90 days
Explanation : Total workers = 0.5Z * 2 M = 0.5 * (1.2 M) * 2 M = 1.2 M2 Total work = M2 * 108 = 1.2 M2 * No of days = > No of days = (M2 * 108) / (1.2 M2 ) = 90 days
Explanation : Total workers = 0.5Z * 2 M = 0.5 * (1.2 M) * 2 M = 1.2 M2 Total work = M2 * 108 = 1.2 M2 * No of days = > No of days = (M2 * 108) / (1.2 M2 ) = 90 days
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