Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1.  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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Q2.  How many 3digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed ? 
1)  30  2)  48 
3)  60  4)  120 
5)  None of these 
Answer : 60
Explanation :
Explanation :
Let the 3digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is not allowed,
The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.
Therefore, The total number of possible 3digit numbers = 5 × 4 × 3=60
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Q3.  How many 3digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed ? 
1)  60  2)  90 
3)  120  4)  240 
5)  None of these 
Answer : 120
Explanation : Let the 3digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is allowed,
The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.
Hence, the total number possible 3digit numbers =5 × 5 × 5 =125
Explanation : Let the 3digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is allowed,
The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.
Hence, the total number possible 3digit numbers =5 × 5 × 5 =125
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Q4.  Two patterners A and B invested Rs. 1250/ and Rs.850 in a in a business respectively. Both the patterners distributes 37% of the profits equally and rest of the profit according to their ratio of capital investments. If one partners gets Rs.30 more than other, then what should be the total profit in the business. 
1)  Rs.225  2)  Rs.250 
3)  Rs.275  4)  Rs.280 
5)  None of these 
Answer : Rs.250
Explanation : The ratio of capital of A and B =1250:850 = 25:17
Explanation : The ratio of capital of A and B =1250:850 = 25:17
A and B equally share the total of 37%.
Now, rest 63% is divided in the ratio 25:17.
So, A.T.Q (63/42) x 25  (63/42)x 17 = Rs.30
Or (63/42) x (2517) = (63/42) x 8 = 30
Or (1.5 x 8 ) % = 30
So, total profit =i.e. 100% =12 x 30 =Rs.250
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Q5.  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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Q6.  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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Q7. 

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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Q8.  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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Q9. 

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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Q10. 

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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