11) 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,
- 1) 10:28 am
- 2) 10:30 am
- 3) 10:31 am
- 4) None of above
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Ans. 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
12) The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm
- 1) 16cm
- 2) 25cm
- 3) 15cm
- 4) 35cm
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Ans. D
Explanation :
So by now, you must be knowing this is a question of HCF, right.
H.C.F. of (700 cm, 385 cm, 1295 cm) = 35 cm.
13) The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M.
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Ans. 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.
14) 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 num
- 1) 400
- 2) 256
- 3) 120
- 4) 420
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Ans. 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
15) Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectiv
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Ans. B
Explanation :
Answer will be HCF of (400-9, 435-10, 541-14)
HCF of (391, 425, 527) = 17
16) Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}
- 1) 128
- 2) 126
- 3) 146
- 4) 434
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Ans. 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
17) Find the HCF of 54, 288, 36
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Ans. A
Explanation :
Lets solve this question by factorization method.
\begin{aligned}
18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
\end{aligned}
So HCF will be minimum term present in all three, i.e.
\begin{aligned}
2 \times 3^2 = 18
\end{aligned}
18) Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest term
- 1) \begin{aligned} \frac{30}{25} \end{aligned}
- 2) \begin{aligned} \frac{28}{29} \end{aligned}
- 3) \begin{aligned} \frac{28}{29} \end{aligned}
- 4) \begin{aligned} \frac{16}{25} \end{aligned}
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Ans. 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
19) Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest term
- 1) \begin{aligned} \frac{11}{12} \end{aligned}
- 2) \begin{aligned} \frac{23}{24} \end{aligned}
- 3) \begin{aligned} \frac{26}{27} \end{aligned}
- 4) \begin{aligned} \frac{4}{7} \end{aligned}
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Ans. A
Explanation :
HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12
20) HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned}
- 1) \begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
- 2) \begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
- 3) \begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
- 4) \begin{aligned} 2 \times 3 \times 5 \end{aligned}
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Ans. C
Explanation :
As in HCF we will choose the minimum common factors among the given.. So answer will be third option