Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. | A mixture of 450 ml contains milk and water in the ratio 2 : 1. How much more water must be added to make the new ratio of milk and water 4 : 5 ? |
| 1) | 215 ml | 2) | 220 ml |
| 3) | 225 ml | 4) | 235 ml |
| 5) | None of these | ||
Answer : 225 ml
Explanation :
Explanation :
Old Ratio: milk: water = (2 : 1)
New Ratio:milk: water= (4 : 5)
Multiply the Old Ratio by 2 to equalise milk, so the milk parts match (4).
- Balanced Milk: (2:1) became (4:2) (Total parts = (6))
- Found the Value: (450 ml /6 = 75 ml per part
- Multiplied the Difference: Water went from (2 to 5) i.e (3parts), so 3 x 75 =225ml
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Q2. | The circumference of the back-side wheel of a vehicle is 1 m greater than that of. front-side wheel. To travel 600 m the front wheel rotates 30 times more than the back wheel. what should be the circumference of the front wheel ? |
| 1) | 5 m. | 2) | 2 m. |
| 3) | 3 m. | 4) | 4m |
| 5) | None of thes | ||
Answer : 4m
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Q3. | The circumference of the back-side wheel of a vehicle is 1 m greater than that of. front-side wheel. To travel 600 m the front wheel rotates 30 times more than the back wheel. what should be the circumference of the front wheel ? |
| 1) | 5 m. | 2) | 2 m. |
| 3) | 3 m. | 4) | 4m |
| 5) | None of thes | ||
Answer : 4m
Explanation :
Explanation :
Let the circumference of front wheel be x metres.
Then, Circumference of rear wheel = (x - 1) metres
∴600/x−600/(x+1)=30
⇒1/x(x+1)=1/20
⇒x(x−1)=20
⇒(x2+x−20)=0
⇒(x+5)(x−4)=0⇒x=4m
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Q4. | If 40 men can build a wall 300 m long in 12 days working 6; hours a day, how long will 30 men take to build a similar wall 200 m long working 8 hours a day ? |
| 1) | 4½ days | 2) | 10½ days |
| 3) | 8 days | 4) | 11 days |
| 5) | None of these | ||
Answer : 8 days
Explanation :
Explanation :
Sol. M1D1W2T1 = M2D2W1T2
or D2 = (40 x 12 x 200 x 6)/ (30 x 300 x 8 ) = 8 days
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Q5. | Three men or eight boys can do a piece of work in 17 days. How many days will two men and six boys together take to finish the same work ? |
| 1) | 9 days | 2) | 11 days |
| 3) | 14 days | 4) | 12 days |
| 5) | None of these | ||
Answer : 12 days
Explanation :
Explanation :
3 men = 8 boys
2 men =(8x2)/3 = 16/3 boys.
(2 men + 6 boys) =[(16/3 )+6] boys
= (34/3) boys
Given, 8 boys can complete the work in 17 days
(34/3 boys can complete the work in x days
34/3 : 8 = 17 : x
x =(17x8x3)/34 = 12 days
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Q6. | In a certain store, the profit is 260% of the cost. If the cost increases by 20% but the selling price remains constant, approximately what percentage of the selling price is the profit? |
| 1) | 33-1/3% | 2) | 40% |
| 3) | 54% | 4) | 66-2/3% |
| 5) | None of these | ||
Answer : 66-2/3%
Explanation :
Explanation :
Let C.P.= Rs. 100. Then, Profit = Rs. 260, S.P. = Rs. 360.
New C.P. = 120 % of Rs. 100 = Rs. 120
New S.P. = Rs. 360.
Profit = Rs. (360 - 120) = Rs.240.
Required percentage = [(240/360) x 100] = 66-2/3%
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Q7. |
|
| 1) | 8 | 2) | 10 |
| 3) | 11 | 4) | 14 |
| 5) | None of these | ||
Answer : 11
Explanation :
Explanation :
Explanation :
Let, the number of subjects = x
Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Physics and 2 more marks for his Chemistry, his average would have been 65.
=> Total marks he would have scored for all subjects = 65x
Now, we can form the equation as 65x - 63x = the additional marks of the student
= 20 + 2 = 22
=> 2x = 22
=>x=22/2=11
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Q8. |
|
| 1) | 5/12 | 2) | 7/12 |
| 3) | 11/12 | 4) | 3/11 |
| 5) | None of these | ||
Answer : 7/12
Explanation :
Explanation :
Total number of trials = 300 (Since there are 300 students all together).
Number of times a cricket player is chosen = 95 (Since 95 students play cricket).
Number of times a football player is chosen = 120.
Number of times a volleyball player is chosen = 80.
Number of times a student is chosen who plays no games = 5.
The probability of getting a player who plays either cricket or volleyball
= (95+80)/300 = 175/300 = 7/12
=
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Q9. |
|
| 1) | 18 | 2) | 20 |
| 3) | 28 | 4) | 32 |
| 5) | None of these | ||
Answer : 20
Explanation :
Explanation :
Let,the first term of an AP = a and the common difference = d
3th term of AP = A3= a+2d =-13 -------(i)
8th term = A8= a+7d =2 ------- (ii)
Subtracting equation (i) from (ii), we get :
or 7d-2d=2-(-13)
or 5d =15
or d=15/5=3
Substituting it in equation (ii),a=2-7(3)=2-21=-19
Therefore, 14th term =
A14= a + 13d = -19 + 13(3) = -19 + 39 = 20
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Q10. |
|
| 1) | 84 | 2) | 102 |
| 3) | 106 | 4) | 108 |
| 5) | None of these | ||
Answer : 102
Explanation :
Explanation :
First term of AP = a=-19 and last term = l=36
Number of terms = n=12
Sum of AP = n/2(a+l)
= 12/2(-19+36) = 17x6=102
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