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A Learning Portal from Recruitment India
If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.
2500
2574
1574
1365
Answer with explanation
Answer: Option BExplanation
Explanation:
HCF * LCM = 84942, because we know
Product of two numbers = Product of HCF and LCM
LCM = 84942/33 = 2574
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What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
196
1260
630
2520
Answer with explanation
Answer: Option CExplanation
12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5
L.C.M. of 12, 18, 21 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260
Required number = 1260/2
= 630
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The H.C.F and L.C.M of two numbers are 50 and 250 respectively. If the first number is divided by 2, the qotient is 50. The second number is
50
100
125
75
Answer with explanation
Answer: Option CExplanation
as we know if two numbers say a and b are multiplied the product is equal to the product of their HCF and LCM
so A.T.Q
a/2=50
a=100
therefore
hcf(100,b)*lcm(100,b)=100b
50*250=100b
12500/100=b
125=b
the second no. is 125
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The h.c.f. of two numbers is 23 and the other two factors of their l.c.m. are 13 and 14. the larger of the two numbers is
276
299
322
345
Answer with explanation
Answer: Option CExplanation
HCF of two numbers is 23.So, Let the numbers be 23x & 23y.
The LCM => 23xy.
It is given that 13 & 14 are it other factors.
So, The value of x & y will be 13 & 14.
The largest number => 23 (14)
=> 322.
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Find the greatest number of 4 digits which is divisible by 15, 25, 40 and 75
9600
9400
9800
9000
Answer with explanation
Answer: Option AExplanation
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 – 399) = 9600.
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Find the least number which when divided by 12, 15, 20 and 54 leaves 8 as remainder in each case.
582
548
584
854
Answer with explanation
Answer: Option BExplanation
LCM of 12, 15, 20, 54
12 = 2 x 2 x 3
15 = 3 x 5
20 = 2 x 2 x 5
54 = 2 x 3 x 3 x 3
LCM = 27 x 4 x 5 = 27 x 20
= 540
ADD 8
= 540 + 8 = 548
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The greatest number of four digits which is divisible by 15, 25, 40 and 75 is
9600
9400
9800
9000
Answer with explanation
Answer: Option AExplanation
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 – 399) = 9600.
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There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and of the last three is 12673. Find the fourth number?
17
23
31
29
Answer with explanation
Answer: Option DExplanation
(7429/12673) = 17/29 Fourth Number is 29
Let the four prime numbers, in the ascending order of magnitude are a, b, c and d.
Given that the product of the first three is 7429 and of the last three is 12673
So a×b×c = 7429
b×c×d = 12673
Divide them:
Thus the first number is 17
The fourth number is 29
The middle two numbers are 19 and 23.
So your answer is D. 29
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John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?
30 min
25 min
20 min
15 min
Answer with explanation
Answer: Option BExplanation
L.C.M. of 250, 300 and 150 = 1500 sec Dividing 1500 by 60 we get 25, which mean 25 minutes. John, Smith and Kate meet after
25 minutes.
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The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.
5 : 28 : 00 hrs
5 : 30 : 00 hrs
5 : 38 : 00 hrs
5 : 40 : 00 hrs
Answer with explanation
Answer: Option CExplanation
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.
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A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally he was paid only Rs. 1073. For how many days did he actually work?
39
40
37
35
Answer with explanation
Answer: Option CExplanation
HCF of 1189, 1073 = 29
1073/29 = 37
Let man s every day s wages was X
He had to work for Y days
Total wages = XY =1189
But he worked for Z days
So he got XZ =1073
X is common in both so it HCF
HCF of 1189 & 1073 is X= 29
Total Days HE worked = 1073/29
=37
Answer – 37
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On dividing a number by 4, the remainder is 2. The quotient so obtained when divided by 5, leaves the remainder 3. Now the quotient so obtained when divided by 6, leaves the remainder 5. The last quotient is 7. The number was
962
954
946
938
Answer with explanation
Answer: Option BExplanation
WKT,
Divisor x Quotient + Remainder = Number
6 x 7 + 5 = 47
5 x 47 + 3 = 238
4 x 238 + 2 = 954
Therefore, required number is 954.
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Find the side of the largest square slab which can be paved on the floor of a room 5 meters 44cm long and 3 meters 74 cm broad?
56 cm
42 cm
38 cm
34 cm
Answer with explanation
Answer: Option DExplanation
The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.
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Three different containers contain 1365 litres, 1560 litres and 1755 litres of mixtures of milk and water respectively. What biggest measure that can measure all the different quantities exactly ?
97 lit
129 lit
195 lit
37 lit
Answer with explanation
Answer: Option CExplanation
containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.
Biggest measure that can measure all these different quantities exactly can be given by
HCF of (1365, 1560, 1755)
That can be found as
HCF of 1365, 1755 = 195
Now, HCF of 195 and 1560 = 195
Hence, biggest measure that can measure all the given different quantities exactly is 195 lit
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Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?
13
17
21
78
Answer with explanation
Answer: Option AExplanation
Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 – 4 = 13; 42 – 3 = 39; 93 – 15 = 78
Now let’s find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number
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A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number ?
25 litres
15 litres
5 litres
1 litre
Answer with explanation
Answer: Option BExplanation
75 litres, 45 litres
For maximum capacity take HCF (75, 45) = 15
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The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers?
26, 78
39, 52
13, 156
36, 68
Answer with explanation
Answer: Option BExplanation
Let the numbers be 13x and 13y (∵ HCF of the numbers = 13)
13x×13y=2028⇒xy=12 co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12,
but not co-prime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers. Hence numbers are 39 and 52.
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Find the least number which when divided by 2, 3, 4 and 5 leaves a remainder 3. But when divided by 9 leaves no remainder?
63
183
153
Answer with explanation
Answer: Option BExplanation
LCM of 2, 3, 4 and 5 is 30,
let number be 30k + 3 put k = 2,
we get 63 which is divisible by 9
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The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
216
389
443
548
Answer with explanation
Answer: Option DExplanation
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
(OR)
Given numbers are 12, 15, 20, 54
To Find: least no. which is divided by given nos. and leaves remainder 8
Least no which is divisible by all given no is LCM of all nos.
LCM means the least common multiple.
First, we find LCM of 12, 15, 20, 54 by prime factorization method
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3
LCM ( 12, 15 , 20 , 54 ) = 2 × 2 × 3 × 3 × 3 × 5 = 540
To find the required no. we add 8 to LCM
⇒ Required No. = 540 + 8 = 548
Thus, 548 is the least no which when divided by 12 15 20 and 54 leaves in each case a remainder of 8.
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The LCM of two numbers is 520 and their HCF is 4. If one of the number is 52, then the other number is
42
32
50
40
Answer with explanation
Answer: Option DExplanation
First number * second number = LCM * HCF
52*Second number = 520 * 4
Second number = (520*4)/52
= 40
Then the other number is 40.
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What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?
9956
10080
10096
9240
Answer with explanation
Answer: Option BExplanation
Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.
Now, 10000/840 = 760 (remainder).
840 – 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.
So, the required number is 10080.
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If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
34
45
77
12
Answer with explanation
Answer: Option CExplanation
For any this type of question, remember
Product of two numbers = Product of their HCF and LCM
So Other number = 693×11/99 = 77
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The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
1677
1683
2523
3363
Answer with explanation
Answer: Option BExplanation
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
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