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Question Detail
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
1. 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
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. 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
4. 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
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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