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- Question
The sum of a fraction and 3 times its reciprocal is 31/6. What is the fraction?
Options- A. 2/9
- B. 9/2
- C. 5/4
- D. 4/5
- Correct Answer
- 9/2
- 1. Reduce 2714/5074 to lowest terms.
Options- A. 17/23
- B. 29/43
- C. 23/43
- D. 31/37 Discuss
- 2. When 0.151515.....is converted into a fraction, then the result is
Options- A. 5/33
- B. 17/33
- C. 9/11
- D. 7/33 Discuss
- 3. The sum of a fraction and 10 times its reciprocal is 37/4. What is the fraction?
Options- A. 5/4
- B. 4/5
- C. 3/4
- D. 4/3 Discuss
- 4. When 0.090909.....is converted into a fraction, then the result is
Options- A. 1/33
- B. 1/11
- C. 2/33
- D. 6/11 Discuss
- 5. Arrange the fractions 34,512,1316,1629,38 in their ascending order of their magnitude?
Options- A. 3/4 < 3/8 < 13/16 < 16/19 < 5/12
- B. 3/8 < 5/12 < 16/29 < 3/4 < 13/16
- C. 3/8 < 5/12 < 16/29 < 13/16 < 3/4
- D. 3/8 < 5/12 < 13/16 < 16/29 < 3/4 Discuss
- 6. Sum of twice a fraction and 3 times its reciprocal is 29/3. What is the fraction?
Options- A. 2/9
- B. 5/4
- C. 4/5
- D. 9/2 Discuss
- 7. Sum of four times a fraction and 6 times its reciprocal is 11. What is the fraction?
Options- A. 3/4
- B. 4/3
- C. 4/7
- D. 7/4 Discuss
- 8. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
Options- A. 40
- B. 80
- C. 120
- D. 200 Discuss
- 9. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Options- A. 123
- B. 127
- C. 235
- D. 305 Discuss
- 10. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Decimal Fraction problems
Search Results
Correct Answer: 23/43
Correct Answer: 5/33
Correct Answer: 5/4
Correct Answer: 1/11
Correct Answer: 3/8 < 5/12 < 16/29 < 3/4 < 13/16
Correct Answer: 9/2
Correct Answer: 3/4
Correct Answer: 40
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Correct Answer: 127
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
Correct Answer: 2
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).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
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