Maths is an important subject in CLAT,DULLB & Other Law Exams. In any of law exam, Maths carries weightage of 20 -25 % of questions. With focused practice good marks can be fetched from this section. These questions are very important in achieving your success in CLAT, DULLB and Other Law Exams..
Q1. Which
fraction is the greatest among the following:
-
-
-
- All are same
Solution
Since
what we can say by seeing 11/32, 15/47, 37/15
11/32
will be equal to 1/3 if 11/33
15/47
will be equal to 1/3 if 15/45
37/115
will be equal to 1/3 if 37/111
So,
11/32 > 1/3 Since 32 < 33
(11/32
> 11/33 = 1/3)
15/47
< 1/3, Since 47 > 45
15/47
< 15/45 = 1/3
37/115
< 1/3, Since 115 > 111
37/115
< 37/111 = 1/3
37/115
< 1/3
Hence
11/32 Will be the greatest
Q2.which
fraction is the smallest of the following:
-
-
-
- All
are same
- Solution
,
Since
using
same principle
Alternate
Method:
When
the difference of numerator and denominator of all the numbers are
same, then choose the one for smallest numerator and denominator.
Therefore,
45/49 is the answer.
Q3. What
is the remainder if 10400 is divided by
Solution
Last
3 digits = 400 which is divisible by 8
Q4. A man gave 1/3 of his wealth to his wife, 1/2 of remaining to his first son, 1/2 of remaining to his second son & rest of Rs. 6 lacs to his youngest son. Find out his wealth.
Solution
Let
wealth be equal to w
Given
to wife = w/3
Given
to 1st son =
Given
to 2nd son =
Youngest
son = 6 lac =
= w/6
= w/6
Therefore,
w = 36 lac
Alternate
Method:
By
looking at the question, we can see remaining wealth is 6 lacs and
then by looking at the options only 36 and 42 are multiple of 6.
Then,
check both the numbers and verify the answer.
Q5.
= ?
= ?
Solution
=
Q6. The
sum of the digits of a two digit number is 8; if the digit are
reversed the number is decreased by 54. Find the number
Solution
x
+ y = 8
(10x
+ y) - (10y + x) = 54
9(x
- y) = 54,
x - y = 6
Solving
this we get, x = 7, y = 1
Alternate
Method:
Work
through the options
Q7.
= ?
= ?
Solution
Since
if a + b + c = 0, then a3
+ b3
+ c3
= 3abc
Q8.
= 32
x 32, x = ?
= 32
x 32, x = ?
Solution
= 32 x 32
= 32 x 32
(64
x 64) x x = (32)2
x (32)2
x
= 16 x 16 = 256
Q9.
= 0.0008, x = ?
= 0.0008, x = ?
Solution
= 0.0008,
=
= 5
= 0.0008,
=
= 5
x = 25
Q10. Find
the greatest number of 4 digits which when divided by 20, 30 &15
leave 18, 28 & 13 as remainders respectively.
Solution
LCM
of 20, 30 & 15 is 60 highest multiple of 60 less than 10,000 =
9960
Remainder = 18, 28, 13 or 20, 30, 15
20
– 18 = 2, 30 – 28 = 2, 15 – 13 = 2
Hence,
number = 9960 - 2 = 9958
