1.The least number of five digits which is exactly divisible by 12, 15 and 18, is:
10010
10051
10020
10080
Ans . D
Hint . Least number of 5 digits is 10,000 L.C.M. of 12, 15 and 18 is 180.
On dividing 10000 by 180, the remainder is 100.
∴ Required number
= 10000 + (180 – 100) = 10080.
2. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is :
9000
9400
9600
9800
Ans . C
Hint .
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.
3.The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
37
13
23
33
Ans . C
Hint .
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added
= (60 – 37) = 23.
4.Find the maximum number of students among whom 429 mangoes and 715 oranges can be equally distributed.
100
120
160
143
Ans . D
Hint . Required number
= HCF of 429 and 715 = 143
5.Find the greatest number that will divide 115, 149 and 183 leaving remainders 3, 5, 7 respectively.
14
16
18
20
Ans . B
Hint .Required number
= HCF of (115 – 3), (149 –5) and (183 – 7)
= HCF of 112, 144 and 176 = 16
6. Find the greatest number which when subtracted from 3000 is exactly divisible by 7, 11, 13.
1799
2099
1899
1999
Ans . D
Hint .
Required number
= 3000 – LCM of 7, 11, 13
= 3000 – 1001 = 1999
7.The L.C.M. of two number is 630 and their H.C.F. is 9. If the sum of numbers is 153, their difference is
17
23
27
33
Ans . C
Hint .
Let numbers be x and y.
Product of two numbers = their (LCM × HCF)
⇒ xy = 630 × 9
Also, x + y = 153 (given)
since x – y
⇒ बाकी सवाल की खुद कोशिश करे ।
8. Product of two co-prime numbers is 117. Their L.C.M. should be:
1
117
equal to their H.C.F.
cannot be calculated
Ans . B
Hint .
H.C.F of co-prime numbers is 1.
So, L.C.M. = 117/1 = 117.
10.
9. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?
264
259
269
235
Ans . B
Hint .
Required number
= LCM of ( 8, 11, 24 ) – 5
= 264 – 5 = 259
10 .If the L.C.M and H.C.F. of two numbers are 2400 and 16, one number is 480; find the second number.
40
80
60
70
Ans . B
Hint .
Product of numbers
= (LCM × HCF)
⇒ 480 × second number = 2400 × 16
⇒ second number = 80
∴ Required number
= 10000 + (180 – 100) = 10080.
2. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is :
9000
9400
9600
9800
Ans . C
Hint .
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.
3.The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
37
13
23
33
Ans . C
Hint .
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added
= (60 – 37) = 23.
4.Find the maximum number of students among whom 429 mangoes and 715 oranges can be equally distributed.
100
120
160
143
Ans . D
Hint . Required number
= HCF of 429 and 715 = 143
5.Find the greatest number that will divide 115, 149 and 183 leaving remainders 3, 5, 7 respectively.
14
16
18
20
Ans . B
Hint .Required number
= HCF of (115 – 3), (149 –5) and (183 – 7)
= HCF of 112, 144 and 176 = 16
6. Find the greatest number which when subtracted from 3000 is exactly divisible by 7, 11, 13.
1799
2099
1899
1999
Ans . D
Hint .
Required number
= 3000 – LCM of 7, 11, 13
= 3000 – 1001 = 1999
7.The L.C.M. of two number is 630 and their H.C.F. is 9. If the sum of numbers is 153, their difference is
17
23
27
33
Ans . C
Hint .
Let numbers be x and y.
Product of two numbers = their (LCM × HCF)
⇒ xy = 630 × 9
Also, x + y = 153 (given)
since x – y
⇒ बाकी सवाल की खुद कोशिश करे ।
8. Product of two co-prime numbers is 117. Their L.C.M. should be:
1
117
equal to their H.C.F.
cannot be calculated
Ans . B
Hint .
H.C.F of co-prime numbers is 1.
So, L.C.M. = 117/1 = 117.
10.
9. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?
264
259
269
235
Ans . B
Hint .
Required number
= LCM of ( 8, 11, 24 ) – 5
= 264 – 5 = 259
10 .If the L.C.M and H.C.F. of two numbers are 2400 and 16, one number is 480; find the second number.
40
80
60
70
Ans . B
Hint .
Product of numbers
= (LCM × HCF)
⇒ 480 × second number = 2400 × 16
⇒ second number = 80
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