Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. | 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
🔒 Click "View Answer" to reveal correct option
View Answer
Q2. | 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
🔒 Click "View Answer" to reveal correct option
View Answer
Q3. | 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%
🔒 Click "View Answer" to reveal correct option
View Answer
Q4. |
|
| 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
🔒 Click "View Answer" to reveal correct option
View Answer
Q5. |
|
| 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
=
🔒 Click "View Answer" to reveal correct option
View Answer
Q6. |
|
| 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
🔒 Click "View Answer" to reveal correct option
View Answer
Q7. |
|
| 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
🔒 Click "View Answer" to reveal correct option
View Answer
Q8. |
|
| 1) | 68 | 2) | 70 |
| 3) | 72 | 4) | 76 |
| 5) | None of these | ||
Answer : 68
Explanation :
Explanation :
Given, First term of AP = a =- 20 and last term = l = 28
Number of terms = n = 17
Sum of AP = n/2(a+l)= 17/2(-20+28)= 17x4 = 68
🔒 Click "View Answer" to reveal correct option
View Answer
Q9. |
|
| 1) | 53 | 2) | 57 |
| 3) | 63 | 4) | 72 |
| 5) | None of these | ||
Answer : 53
Explanation :
Explanation :
Given,T3 = a + 2d = 13-------(1)
T5 = a + 4d = 21-------(2)
on solving (1) and (2)
d = 4 & a = 5
T 13= a + 12d = 5 + 12(4) = 5 + 48 = 53
🔒 Click "View Answer" to reveal correct option
View Answer
Q10. |
|
| 1) | 11 | 2) | 13 |
| 3) | -11 | 4) | -13 |
| 5) | None of these | ||
Answer : 11
Explanation : Given, the 7th and the 10th terms are (a + 6d) and (a + 9d) respectively.
Explanation : Given, the 7th and the 10th terms are (a + 6d) and (a + 9d) respectively.
The ratio of these terms is -1.
Solving this ratio, we get -2a = 15d. The 16th term is (a + 15d) = -15.
Substituting for a, we get a = 15 and d = -2. Therefore, the 3rd term is (a + 2d) = 11.
🔒 Click "View Answer" to reveal correct option
View Answer
