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Question Detail
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
1. Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}
- 128
- 126
- 146
- 434
Answer: Option B
Explanation:
HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
2 * 3*3 * 7
2. If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.
- 2574
- 2500
- 1365
- 1574
Answer: Option A
Explanation:
HCF * LCM = 84942, because we know
Product of two numbers = Product of HCF and LCM
LCM = 84942/33 = 2574
3. 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.
4. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is
- 10:28 am
- 10:30 am
- 10:31 am
- None of above
Answer: Option C
Explanation:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.
Sometimes questions on red lights blinking comes in exam, which can be solved in the same way
5. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is
- \begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
- \begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
- \begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
- \begin{aligned} 2 \times 3 \times 5 \end{aligned}
Answer: Option C
Explanation:
As in HCF we will choose the minimum common factors among the given.. So answer will be third option
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