1. The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is24, then the other number is
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Correct Ans:
Explanation:
First number * Second number = HCF * LCM
24 * Second number = 8 *48
Second number = (8 * 48)/24 = 16
2. HCF of 2/3, 4/5 and 6/7 is
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Correct Ans:
Explanation:
Given fractions = 2/3, 4/5 and 6/7
HCF of fractions = HCF of numerators / LCM of denominators
= HCF (2, 4, 6) / LCM (3, 5, 7)
= 2/105
3. What is the LCM of 120 and 450?
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Correct Ans:
Explanation:
LCM of 120 = 2, 2, 2, 3
LCM of 450 = 2, 3, 3, 5, 5
LCM of 120 and 450 = 2 * 2 * 2 * 3 * 3 * 5 * 5
= 1800
4. HCF of x^{2} - y^{2} and x^{3} - y^{3} is
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Correct Ans:
Explanation:
x^{2} - y^{2} = (x + y) (x - y)
x^{3} - y^{3} = (x - y) (x^{2} + xy + y^{2})
HCF = (x - y)
5. The HCF of two numbers is 24. The number which can be their LCM is
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Correct Ans:
Explanation:
HCF always divides LCM completely.
By option LCM is 120.
6. The sum of two number is 528 and their HCF is 33. How many pairs of such numbers can be?
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Correct Ans:
Explanation:
Here, the HCF = 33
Then, let the two numbers be 33a and 33b,
Respectively, where a and b are coprime.
Then, 33a + 33b = 528
a + b = 16
Hence, values of (a, b) which are coprime to each other can be
(1, 15), (3, 13), (5, 11) and (7, 9).
Hence, required number of pairs is 4.
7. Two numbers are in the ratio 3 : 4. If their LCM is 240, the smaller of the two number is
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Correct Ans:
Explanation:
Let the numbers be 3x and 4x.
LCM of 3x and 4x = 12x
Given, LCM = 240
12x = 240
x = 240/12 = 20
Smaller number = 3x = 3 * 20 = 60
8. The length, breadth and height of a room are 363m, 528m and 693m respectively. Determine the longest tape that can measure the three dimensions of the room exactly.
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Correct Ans:
Explanation:
Maximum length of tape
= HCF of 363m, 528m and 693m
363 = 3 * 11 * 11
528 = 2 * 2 * 2 * 2 * 3 * 11
693 = 3 * 3 * 7 * 11
HCF of 363m, 528m and 693m = 3 * 11
= 33 metre
9. The product of two numbers is 35828 and their HCF is 26. Find their LCM.
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Correct Ans:
Explanation:
Required LCM = Product of numbers/Their HCF
= 35828/26 = 1378
10. Find the highest common factor of 72 and 168.
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Correct Ans:
Explanation:
HCF of 72 and 168,
72 => 2 * 2 * 2 * 3 * 3
168 => 2 * 2 * 2 * 3 * 7
Thersfore, HCF = 2 * 2 * 2 * 3 = 24
11. The HCF of 252, 294 and 3#8 is 42, what is #?
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Correct Ans:
Explanation:
Multiples of 42
= 252, 294, 336, 378
So, # = 7
12. What is the least number which when divided by 7, 9 and 12 leaves the same remainder 1 in each case?
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Correct Ans:
Explanation:
LCM of 7, 9 and 12 = 84
In the given options, only 253 is the least number which when divided by 7, 9 and 12 leaves the same remainder 1 in each case, since 253 leaves the remainder 1, when 253 is divided by 84.
13. The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?
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Correct Ans:
Explanation:
HCF of two number is 8. This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers, it is divisible by the two numbers.
So, the required answer = 60
14. The LCM of two numbers is 30 and their HCF is 5. One of the numbers is 10. The other is
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Correct Ans:
Explanation:
First number * Second number = LCM * HCF
Let the second number be x.
10x = 30*5
x = 150/10 = 15
15. What is the HCF of 150, 225 and 375?
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Correct Ans:
Explanation:
150 = 2*3*5*5
225 = 3*3*5*5
375 = 3*5*5*5
Required H.C.F. = 3*5*5 = 75
16. Two numbers are in the ratio 3 : 4. Their L.C.M. is 84. The greater number is
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Correct Ans:
Explanation:
Let the numbers be 3x and 4x.
Their LCM = 12x
12x = 84
x = 84/12 = 7
Larger number = 4x = 4*7 = 28
17. If the product of two numbers is 2160 and the HCF is 6, then the ratio of HCF and LCM is
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Correct Ans:1 : 60
Explanation:
Given:
Product of two numbers = 2160; HCF = 6
WKT, LCM x HCF = Product of two numbers
LCM x 6 = 2160
LCM = 2160/6
LCM = 360
HCF : LCM = 6 : 360 = 1 : 60.
18. Find the complete remainder when a certain number divided successively by 3, 5 and 7 leaves remainders 2, 3 and 4 respectively?
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Correct Ans:71
Explanation:
Given:
Divisor, d_{1} = 3, d_{2} = 5, d_{3} = 7
Remainders, r_{1} = 2, r_{2} = 3, r_{3} = 4
Complete remainder = d_{1} d_{2} r_{3} + d_{1} r_{2} + r_{1}
= (3 x 5 x 4) + (3 x 3) + 2
= 71.
19. The product of two 2-digit numbers is 2160 and their H.C.F. is 12. The numbers are
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Correct Ans:(36, 60)
Explanation:
Given: HCF of two numbers = 12
Let numbers be 12a and 12 b where a and b are co-prime numbers.
12a x 12b = 2160
ab = 2160/(12 x 12)
ab = 15
Possible pairs of a, b are (1, 15), (3, 5).
Therefore, numbers are (12, 180), (36, 60).
From the options, (36, 60) is the correct answer.
20. Find the HCF of (4 * 54 * 3125), (16 *5 * 8) and (32 * 27 * 11 * 49).
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Correct Ans:
Explanation:
(4 * 54 * 3125) = 2^{2} * 2 * 3^{3} * 5^{5}
= 2^{3} * 3^{3} * 5^{5}
(16 *5 * 8) = 2^{4} * 5 * 2^{3}
= 2^{7} * 5
(32 * 27 * 11 * 49) = 2^{5} * 3^{3} * 11 * 7^{2}
= 2^{5} * 3^{3} * 7^{2} * 11
Therefore, required HCF = 2^{3} = 8.
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