Mathmetics

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

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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.
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 one-third 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  
or 3A - 7B = 0 ---------(I) and  A = 2C --------- (II)
And A-9 = (B-9)/3 
or 3A-27 = B-9
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.
The perimeter of a square field is 16 m more than the perimeter of a rectangle. The length of the rectangle is 40m which is 250% of its width. If a street of width 4 m surrounds from outside the square,  then find the total cost of constructing the street at the rate of Rs. 20 per sq. m?
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 :

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 :
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(15-x) and 2/3(15-x) investments respectively
So, Total annual interest income = 6% of x + 4% of 1/3(15-x) + 3% of 2/3(15-x) = 76000
Rewriting Rs.76000 as Rs. 0.76 Lakhs,
6% of x + 4% of 1/3(15-x) + 3% of 2/3(15-x) = 0.76
Multiply both sides by 100,
6x + 4/3(15-x) + 2(15-x) = 76
6x +11/3 (15-x) = 76
165 + 18x -11x = 228
7x = 63
x = 9 (or) Rs. 9 Lakhs
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Q5.
A student scores 50% in an examination and after review, even though his score is increased by 40%, he fails by 30 marks. If his post-review score is increased by 20%, he will have 12 marks more than the passing score. What should be the percentage of mark needed for passing the examination ?
1) 60 2) 70
3) 72 4) 80
5)None of these
Answer : 80
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.
A can complete a workpiece in 12 days. B requires 18 days to finish the same workpiece, whereas C can do it in 24 days. All three of them started working on the same workpiece, but due to some reason, C left work after working for 3 days with A and B. What should be the overall time taken to complete this workpiece ?
1) 5-3/10 days 2) 6-7/10 days
3) 6-3/10 days 4) 7-3/10 days
5)None of these
Answer : 6-3/10 days
Explanation : A,B and C altogether work for 3 days = 3 (1/12+1/18+1/24) = 13/24 parts
So, remaining (1-13/24) = 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 = 3-3/10 days
So, overall time taken to complete this workpiece =  6-3/10 days

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Q7.
Directions: Study the following information carefully and answer the given questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater
than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

Find the number of blue balls taken out from bag Y.

1) 2 2) 4
3) 5 4) 6
Answer : 8
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)C2/(30 + 2a + 4)C2 = 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
View Answer

Q8.
Directions: Study the following information carefully and answer the given
questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater
than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from
bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

What should be total number of balls in bag X ?

1) 10 2) 20
3) 29 4) 30
Answer : 30
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.
View Answer

Q9.
Directions : Study the following information carefully and answer the given questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

What should be the value of A ?

1) 2/29 2) 3/29
3) 4/29 4) 5/29
Answer : 3/29
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.
Direction: Data below represent the number of people required for various projects to be completed in various days:

Project                      Situation 1                                                             Situation 2
                 Number of workers       Days required          Number of workers       Days required
 A              X                                   88                                                  X+8             66
 B              Y                                  Y-1                                                  Y+6                    Y-6
 C              Z2                           75                                                   M2                    108

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
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