Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q11. |
|
| 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
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
View AnswerQ12. |
|
| 1) | 196 | 2) | 630 |
| 3) | 1260 | 4) | 2520 |
| 5) | None of these | ||
Answer : 630
Explanation :
Explanation :
L.C.M. of 12, 18, 21 30
= 1260.
So, required number = (1260 ÷ 2) = 630.
View AnswerQ13. |
|
| 1) | 1677 | 2) | 1683 |
| 3) | 2523 | 4) | 3363 |
| 5) | None of these | ||
Answer : 1683
Explanation :
Explanation :
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
View AnswerQ14. |
|
| 1) | 55/601 | 2) | 12/55 |
| 3) | 11/120 | 4) | 11/112 |
| 5) | None of these | ||
Answer : 11/120
Explanation :

Explanation :

View AnswerQ15. |
|
| 1) | 279 | 2) | 283 |
| 3) | 308 | 4) | 318 |
| 5) | None of these | ||
Answer : 308
Explanation : We know that H.C.F x L.C.M = Product of two numbers
Explanation : We know that H.C.F x L.C.M = Product of two numbers
So,other number = (11 x 7700)/275 = 308.
View AnswerQ16. |
|
| 1) | 1677 | 2) | 1683 |
| 3) | 2523 | 4) | 3363 |
| 5) | None of these | ||
Answer : 1683
Explanation :
Explanation :
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
View AnswerQ17. |
|
| 1) | 196 | 2) | 630 |
| 3) | 1260 | 4) | 2520 |
| 5) | None of these | ||
Answer : 630
Explanation :
Explanation :
L.C.M. of 12, 18, 21 30 = 1260.
Required number = (1260 ÷ 2) = 630.
View AnswerQ18. |
|
| 1) | 1 | 2) | 2 |
| 3) | 3 | 4) | 4 |
| 5) | None of these | ||
Answer : 2
Explanation :
Explanation :
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
So, the required numbers are (13 x 1, 13 x 12) and
(13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
View AnswerQ19. |
|
| 1) | 20 | 2) | 23 |
| 3) | 169 | 4) | 400 |
| 5) | None of these | ||
Answer : 23
Explanation :
Explanation :
Let the numbers be x and y.
Then, xy = 120 and x2 + y2 = 289.
(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529
x + y = 23.
View AnswerQ20. |
|
| 1) | 101 | 2) | 107 |
| 3) | 111 | 4) | 185 |
| 5) | None of these | ||
Answer : 111
Explanation :
Explanation :
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
View Answer
