Q.1. Two numbers are in the ratio 2 : 3.
If their L.C.M. is 48. What is sum of the numbers?
a) 28
b) 40 c) 64 d) 42
e) 32
Ans: b
Explanation:
Let the two numbers be
2x and 3x respectively.
LCM of 2x and 3x = 6x
Given, 6x = 48
So, x = 8
The required sum = 2x +
3x = 16 + 24 = 40. Ans.
Q.2. What is the greatest
number of four digits which is divisible by 15, 25, 40 and 75 ?
a) 9800 b) 9600 c) 9400 d) 9200 e) 9000
Ans:
b
Explanation:
The greatest 4- digit number = 9999
LCM of 15, 25, 40 and 75 = 600
The remainder obtained when 9999 is divided by 600 = 399
Therefore, the required greatest number = 9999 – 399 = 9600 Ans.
Q.3. Three numbers are in the
ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:
a)
40 b) 30 c) 20
d) 10 e) None of these
Ans: c
Explanation :
Let the
numbers be 2x, 3x and 4x
LCM of 2x, 3x and 4x = 12x
=> 12x = 240
=> x = 240⁄12 = 20
H.C.F of 2x, 3x and 4x = x = 20
LCM of 2x, 3x and 4x = 12x
=> 12x = 240
=> x = 240⁄12 = 20
H.C.F of 2x, 3x and 4x = x = 20
Q.4. Find the minimum number of square tiles required to
pave the floor of a room of 2m 50cm long
and 1m 50cm broad ?
a)
50 b) 750 c) 45 d) 15
e) None of these
Ans: d
Explanation:
HCF of 250 cm and 150 cm is 50 cm, which
is the side of the tile
So, the required number of tiles = (250 × 150) / (50 × 50) = 15
So, the required number of tiles = (250 × 150) / (50 × 50) = 15
Q.5. What is
the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3,
but when divided by 9 leaves no remainder?
a) 1108 b) 1683 c) 2007 d) 3363 e) None of these
Ans: b
Explanation:
Just see which of the given choices
satisfy the given conditions
Take 3363. This is not even divisible by 9. Hence this is not the answer Take 1108. This is not even divisible by 9. Hence this is not the answer Take 2007. This is divisible by 9. 2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer Take 1683. This is divisible by 9. 1683 ÷ 5 = 336, remainder = 3 1683 ÷ 6 = 280, remainder = 3 1683 ÷ 7 = 240, remainder = 3 1683 ÷ 8 = 210, remainder = 3 Hence 1683 is the answer Q.6. Find the HCF of 22×32, 2×34×7
a) 128 b) 126 c) 146 d) 434 e) 148
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 = 126
Q.7.
Find the HCF of 54, 288, 360
a) 18 b) 36 c) 54 d) 108
e) 72
Ans: a
Explanation:
Let’s
solve this question by factorization method.
54
=2×33, 288
= 25×32, 360
= 23×32×5
So HCF will be minimum term present in all three, i.e.
2×32=18
Q.8. Reduce 368/575 to the lowest terms.
a) 30/25 b) 28/29 c) 16/25 d) 11/12 e) 13/15
Ans: c
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
Q.
9.
Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers
are
a) 12, 24, 36 b) 5, 10, 15 c) 4, 8, 12 d) 6, 8, 13 e) 10, 20, 30
Ans: a
Explanation:
Numbers are 1 × 12, 2
× 12 and 3× 12
i.e. 12, 24 and 36
Q.10.
Find the greatest number that will divide 43, 91 and 183 so as to leave the
same remainder in each case.
a) 4 b) 5
c) 7 d) 9 e) 13
Ans:
Explanation:
Required number = HCF
of ( 91 – 43), (183 – 91) and (183 –
43)
= HCF of 48, 92,
140
= 4
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