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Question Detail
Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively
- 19
- 17
- 13
- 9
Answer: Option B
Explanation:
Answer will be HCF of (400-9, 435-10, 541-14)
HCF of (391, 425, 527) = 17
1. The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number
- 400
- 256
- 120
- 420
Answer: Option D
Explanation:
Solve this question by using below formula.
Product of 2 numbers = product of their HCF and LCM
144 * x = 12 * 5040
x = (12*5040)/144 = 420
2. Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest terms.
- \begin{aligned} \frac{11}{12} \end{aligned}
- \begin{aligned} \frac{23}{24} \end{aligned}
- \begin{aligned} \frac{26}{27} \end{aligned}
- \begin{aligned} \frac{4}{7} \end{aligned}
Answer: Option A
Explanation:
HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12
3. Find the largest number of four digits which is exactly divisible by 27,18,12,15
- 9700
- 9710
- 9720
- 9730
Answer: Option C
Explanation:
LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720
4. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.
- 15
- 16
- 30
- 31
Answer: Option D
Explanation:
LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.
Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31
5. Find the LCM of \begin{aligned} \frac{2}{3}, \frac{4}{6}, \frac{8}{27} \end{aligned}
- \begin{aligned} \frac{2}{27} \end{aligned}
- \begin{aligned} \frac{8}{3} \end{aligned}
- \begin{aligned} \frac{2}{3} \end{aligned}
- \begin{aligned} \frac{8}{27} \end{aligned}
Answer: Option B
Explanation:
Whenever we have to solve this sort of question, remember the formula.
LCM = \\begin{aligned} \\frac{HCF of Denominators}{LCM of Numerators} \\end{aligned}
So answers will be option 2,
Please also give attention to the difference in formula of HCF and LCM
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