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Question Detail
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
1. 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
2. Find the HCF 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 A
Explanation:
Whenever we have to solve this sort of question, remember the formula.
HCF = \begin{aligned} \frac{HCF of Numerators}{LCM of Denominators} \end{aligned}
So answers will be option 1
3. 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
4. What will be the LCM of 8, 24, 36 and 54
- 54
- 108
- 216
- 432
Answer: Option C
Explanation:
LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216
5. 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
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