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Mathmetics

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

Q201.
The average expenditure of a man for the first five month of a year is Rs.5,000 and for the next seven- month is Rs-5,400. He saves Rs.2,300 during the year. What his average monthly income ?
1) Rs.5,425 2) Rs.5,500
3) Rs.5,446 4) Rs.5,600
5)None of these
Answer : Rs.5,425
Explanation :
Total expenditure is = (55000 + 75400) = 62800 
His saving is Rs. 2300
Total annual income = 62800 + 2300 = 65100
Monthly income = 65100 ÷ 12 = Rs. 5425
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Q202.
Two-fifth of a number is two more than one-third another number.If the sum of the two numbers is 16,what is their product ?
1) 40 2) 48
3) 54 4) 60
5)None of these
Answer : 60
Explanation : Let, two numbers are x and y respectively. 
According to the questions, x + y = 16  ------- (I)     and 2x/5  - y/3 = 2 -------- (II)
or 6x -5y = 30
 Solving both equations , we get x = 10 and y = 6
So, their product = 10 X 6 = 60
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Q203.
The profit earned by selling an article for Rs.744 is double the loss incurred when the same article is sold for Rs.240. What would the selling price of the article if it is sold at 15% profit ? 
1) Rs.472.20 2) Rs.469.80
3) Rs.468.20 4) Rs.479.20
5)None of these
Answer : Rs.469.80
Explanation : Let, the C.P be Rs.x.
Then, according to the question, 744 - x = 2(x-240)
or 744 - x = 2x - 480 
or 3x = 744 + 480 =1224
or x = 1224/3 = Rs.408
So, required S.P = 408 X (115/100 ) = Rs.469.20 
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Q204.
The H.C.F and L.C.M of two numbers are 13 and 455 respectively. If one of the numbers lies between 75 and 125, then what is that number ?
1) 78 2) 91
3) 104 4) 117
5)None of these
Answer : 91
Explanation : Let, two numbers are 13x and 13y respectively.
So, 13x X 13y  = 13 X 455
or xy = 35
 Then, pairs  of numbers are 35 = (1 X 35) , (5 x 7).
So, second pairs is considerable i.e. 13X5 = 65 and 13 X 7 = 91 

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Q205.
The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then what is their difference ? 
1) 10 2) 12
3) 15 4) 20
5)None of these
Answer : 10
Explanation : Let, two numbers are 5x and 5y respectively. 
So, 5x X 5y = 5 X 495
or 25xy = 5 X 495 
or xy = 99
So, pairs of numbers are  i.e. 99 = (1 X 99)  and (9 x 11)
Therefore , the satisfy numbers are 9 x 5 = 45 and 11 x 5 = 55
So, the difference is 55 - 45 = 10. 

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Q206.
The L.C.M of two numbers is 45 times their H.C.F. If one of the number is 125 and the sum of H.C.F and L.C.M is 1150, the other number is - 
1) 215 2) 220
3) 225 4) 235
5)None of these
Answer : 225
Explanation : Given that, 45 H.C.F = L.C.M   and  
H.C.F + L.C.M = 1150 
or H.C.F + 45 H.C.F = 1150
or 46 H.C.F = 1150
or H.C.F = 1150/46 =25
So, L.C.M = 25 X 45 = 1125
Now, we can get 125 X x = 25 X 1125
or   x =  (25 X 1125) /125 = 225

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Q207.
What is value of the following expressions: 
 (51 + 52 + 53 + ... + 100) = ?
1) 2825 2) 2975
3) 3025 4) 3775
5)None of these
Answer : 3775
Explanation :
We know that,Sn = (1 + 2 + 3 + ... + 50 + 51 + 52 + ... + 100)/2 - (1 + 2 + 3 + ... + 50)/2
= 100 x (1 + 100) - 50 x (1 + 50) = (50 x 101) - (25 x 51)
= (5050 - 1275) = 3775.
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Q208.
A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?
1) 8 2) 9
3) 10 4) 12
5)None of these
Answer : 10
Explanation :
4 a 3  |
 9 8 4  }  ==> a + 8 = b  ==>  b - a = 8  
13 b 7  |
Also, 13 b7 is divisible by 11      (7 + 3) - (b + 1) = (9 - b)
 (9 - b) = 0
 or b = 9 (b = 9 and a = 1)     
So, (a + b) = 10.
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Q209.
When a number is divided by 5, we get 3 as remainder. What will the remainder when the square of the this number is divided by 5 ?

1) 1 2) 2
3) 3 4) 4
5)None of these
Answer : 4
Explanation :

We, know that , Dividend = (divisor x Quotient )+Reminder 
D = 5Q + 3
or    D2 = (5Q + 3)2 = (25Q2 + 30Q + 9) = 5(5Q2 + 6Q + 1) + 4
On dividing D2 by 5, we get 4 as remainder.
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Q210.
What is the total numbers of  3-digit numbers  which are completely divisible  by 6 ?
1) 140 2) 150
3) 152 4) 159
5)None of these
Answer : 150
Explanation :
The 3-digit number divisible by 6 are : 102, 108, 114,... , 996
It is an A.P. series where  a = 102, d = 6 and l = 996
Let, the number of terms be n. Then tn = 996.
or a + (n - 1) d = 996
or 102 + (n - 1) x 6 = 996
or 6 x (n - 1) = 894
or (n - 1) = 149
or n = 150
So, number of terms = 150.
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