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Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

  • 125
  • 127
  • 129
  • 131

Answer: Option B

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127

Similar Questions :

1. Find the largest number which divides 62,132,237 to leave the same reminder

  • 30
  • 32
  • 35
  • 45

Answer: Option C

Explanation:

Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35

2. Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest terms.

  • \begin{aligned} \frac{30}{25} \end{aligned}
  • \begin{aligned} \frac{28}{29} \end{aligned}
  • \begin{aligned} \frac{28}{29} \end{aligned}
  • \begin{aligned} \frac{16}{25} \end{aligned}

Answer: Option D

Explanation:

We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25

3. 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

4. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

  • 48
  • 22
  • 56
  • 27

Answer: Option A

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.

5. 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

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