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Question 1 of 5
1. Question
What is the sum of digits of the least of multiple of 13, which when divided by 6 and 12 leaves 5 and 11, respectively, as the reminders?
Correct
Ans: d
Here, 6-5 = 1 and 12- 11 = 1
Now, LCM of 6 and 12 =12
Required number = 13(12-1)
= 13x 11 =143
So, sum of digits = 1+ 4+ 3 = 8
Incorrect
Ans: d
Here, 6-5 = 1 and 12- 11 = 1
Now, LCM of 6 and 12 =12
Required number = 13(12-1)
= 13x 11 =143
So, sum of digits = 1+ 4+ 3 = 8
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Question 2 of 5
2. Question
If (x + 1) is the HCF of Ax2 + Bx +C and Bx2+Ax + C, where A # B, then the value of C is
Correct
Ans: C
Since, (x + 1) is the HCF of Ax2 + Bx + C and Bx2+ Ax +C
So (x+1) will the factor of above equations ,now put x+1=0,x=-1
∴ A(-1)+ B(-1)+C = 0
-A+B+c=0
Þ A-B+C = 0 Þ C = B- A
or B(-1)2+ A(-1) + C = 0
Þ B – A+C=0 ÞC = A-B
Because A# B(A is not equal to B)
∴C = A-B
Incorrect
Ans: C
Since, (x + 1) is the HCF of Ax2 + Bx + C and Bx2+ Ax +C
So (x+1) will the factor of above equations ,now put x+1=0,x=-1
∴ A(-1)+ B(-1)+C = 0
-A+B+c=0
Þ A-B+C = 0 Þ C = B- A
or B(-1)2+ A(-1) + C = 0
Þ B – A+C=0 ÞC = A-B
Because A# B(A is not equal to B)
∴C = A-B
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Question 3 of 5
3. Question
The LCM of two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one of the numbers is 93, then the other number is
Correct
Ans: a
Let other number be b and HCF be x.
Because LCM = 12x
We have, x + 12x = 404 ⇒ 13x = 403⇒x =31
Because Product of two numbers = LCM ´ HCF
⇒93 x b = x x 12x ⇒ 93 x b = 12 x 31 x 31
∴b = 124
Incorrect
Ans: a
Let other number be b and HCF be x.
Because LCM = 12x
We have, x + 12x = 404 ⇒ 13x = 403⇒x =31
Because Product of two numbers = LCM ´ HCF
⇒93 x b = x x 12x ⇒ 93 x b = 12 x 31 x 31
∴b = 124
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Question 4 of 5
4. Question
The LCM of two integers is 1237. What is their HCF?
Correct
Ans: C
Given, LCM of two integers is 1237, which is a prime number.
So, their HCF is 1.
Incorrect
Ans: C
Given, LCM of two integers is 1237, which is a prime number.
So, their HCF is 1.
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Question 5 of 5
5. Question
There are 48 cricket balls, 72 hockey balls and 84 tennis balls and they have to be arranged in several rows in such a way that every row contains the same number of balls of one type. What is the minimum number of rows required for this to happen?
Correct
Ans: c
Given number of cricket balls= 48= 24 x3
Number of hockey balls = 72 = 23x 32
And number of tennis balls = 84=22x 3×7
∴HCF of 48, 72 and 84 = 22 x 3= 12
Now, minimum number of rows= (48/12)+(72/12)+(84/12)=4+6+7=17
Incorrect
Ans: c
Given number of cricket balls= 48= 24 x3
Number of hockey balls = 72 = 23x 32
And number of tennis balls = 84=22x 3×7
∴HCF of 48, 72 and 84 = 22 x 3= 12
Now, minimum number of rows= (48/12)+(72/12)+(84/12)=4+6+7=17
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