Here I am presenting the solution of Mathematics Questions, from the Question Paper
of Lekhpal 2015 held on 13th September for the post in the District
Lucknow on the special demand of one of our sincere visitor, Shashi. This
practice set contains 25 Questions of mathematics from the question Paper.
Q1. The area of a rhombus is 500 cm2
and the length of one of its diagonal is 25 cm. Then the length of the other
diagonal is
(a)
30 cm (b) 20 cm (c) 50 cm (d) 40 cm.
Sol. Formula:- Area of rhombus = (1/2) d1
× d2
Where
d1 and d2 are the diagonals of the rhombus.
Data
given:- Area = 500 cm2 and d1 = 25 cm and d2
=?
On
putting and solving we get:- d2 = 40 cm.
Option
(d)
Q2. The most frequently occurring value
of a data set is called as
(a)
Range (b) Mode (c) Mean (d) Median.
Sol. The most frequently occurring of a
data set is called as Mode.
Option
(b)
Q3. Raghu travelled 1200 km by air which
formed 2/5th of his trip. 1/3rd of the whole trip he
travelled by car and rest of the journey he did by train. The distance
travelled by the train is
(a)
800 km (b) 1800 km (c) 480 km (d) 1600 km
Sol. Raghu travelled 2/5th of his trip
by air = 1200 km.
Let
the total trip of Raghu = x km
According to the question:- 2/5th of trip
(x) = 1200
(2/5)
× x = 1200
x
= 3000 km.
Now
1/3rd of the whole trip by car = (1/3) × 3000 = 1000 km
Total
distance travelled by Raghu by air and car = 1200 + 1000 = 2200 km
Remaining distance travelled by train = 3000 –
2200 = 800 km.
Hence
the required distance travelled by Raghu by train = 800 km
Option
(a)
Q4. The lengths of the three sides of a
triangle are 35 cm, 40 cm and 45 cm respectively. The approximate area of the
triangle will be
(a)
762.82 cm2 (b) 678.82 cm2 (c) 760.82 cm2 (d) 670.82 cm2
Sol. Let the sides of a triangle be a =
35 cm, b = 40 cm, c = 45 cm
Formula
used:- Area of the triangle = [s (s - a) × (s - b) × (s - c)]1/2
Where
s = [(a + b + c)/2]
On
solving we get Area = [60 × 25 × 20 × 15]1/2
Area
= 300 × (5)1/2 = 672 cm2 (approx.)
Hence
the required area = 670.82 cm2
Option
(d)
Q5. The sum of three numbers is 392. The
ratio of first and second number is 2:3 and that of second and third number is
5:8. Then the first number is
(a)
80 (b) 120 (c) 60 (d) 100
Sol. 1st Number 2nd Number 3rd Number
2 3
5 8
Equating
the ratio value of 2nd Number
1st
Number : 2nd Number : 3rd
Number
10 : 15 : 24
Let
numbers be:- 1st Number = 10x,
2nd Number = 15x,
3rd Number = 24x
A.T.Q 10x + 15x + 24x = 392
x = 8
Hence
required First number = 10x = 10 × 8 = 80
Option
(a)
Q6. A trader marks his items by
increasing the price by 25%. If he gives 4% discount on the marked price, what
will be his gain percentage?
(a)
25% (b) 10% (c) 15% (d) 20%
Sol. As per concept
Percentage
Gain = (+ 25) + (- 4) + [{(+ 25) × (- 4)}/100]
Percentage
Gain = 25 – 4 – (100/100)
Percentage
Gain = 25 – 4 – 1 = 20%
Option
(d)
Q7. The average of
n,
(n + 1), (n + 2), (n + 3), (n + 4), (n + 5) is
(a)
[n + (5/2)] (b) (n + 2) (c) 3(2n + 5) (d) n (2)1/2
Sol. Average = [{n + (n + 1) + (n + 2) +
(n + 3) + (n + 4) + (n + 5)}/6]
Average
= [{6n + 15}/6]
Average
= [n + (5/2)]
Option
(a)
Q8. Suresh and Sandeep are partners in a
business. Suresh puts Rs. 5000/- for 5 months and Sandeep puts Rs. 6000/- for 6
months in the business. At the end of the year how much amount will Sandeep
receive from a total profit of Rs. 610/- ?
(a)
Rs. 360/- (b) Rs. 410/- (c) Rs. 380/- (d) Rs. 400/-
Sol. Profit Distribution Ratio = Suresh :
Sandeep
Profit
Distribution Ratio = 5000 × 5 : 6000 × 6
Profit
Distribution Ratio = 5 × 5 : 6 × 6
Profit
Distribution Ratio = 25 : 36
According
to the concept of ratio and proportion Share of Sandeep in total Profit Rs.
610/-
=
[{36/(25 + 36)} × 610]
=
Rs. 360/-
Option
(a)
Q9. The value of [15612 + {154 + (225)1/2
}1/2 ]1/2
(a)
110 (b) 115 (c) 125 (d) 99
Sol. [15612 + {154 + (225)1/2
}1/2 ]1/2
=
[15612 + {154 + 15 }1/2 ]1/2
=
[15612 + {169 }1/2 ]1/2
=
[15612 + 13 ]1/2
=
[15625 ]1/2
=
125.
Option
(c)
Q10. Two numbers are in the ratio of 3:5
and their LCM is 300. The smaller number is
(a)
45 (b) 75 (c)
60 (d) 90
Sol. Let the numbers be 3x and 5x
Then
their LCM = 15x
But
A.T. Q:- 15x = 300
→
x = 20
Hence
the smallest number = 3x = 3 × 20 = 60.
Option
(c)
Q11. If the mean and coefficient of
standard deviation of a set of data are 10 and 5 respectively, then the
standard deviation of the data set is
(a)
10 (b) 50 (c) 5 (d) 1
Sol.
Formula :- Standard Deviation = Mean of data × Coefficient of standard
deviation
So,
standard deviation = 10 × 5 = 50
Option
(b)
Q12. The largest four digit number which
can be divided by 24, 30, 36 each is
(a)
9620 (b) 9360 (c) 9960
(d) 9840
Sol. First calculate the LCM of 24, 30, 36
LCM
= 360
Now
largest four – digit number = 9999
Divide
9999 by 360 and determine the remainder
Remainder
= 279
Hence
required largest number = 9999 – 279 = 9720.
But
this is not present in given option so we try another method to get the answer
2nd
Method Hit & Trial method
In
this method we find that 9360 in the given options is the only number which is
fully divisible by 24, 30, 36
Hence
required number is 9360.
Option
(b)
Q13. If [x2 + (1/x2)]
= 23, then the value of [x4 + (1/x4)] is
(a)
526 (b) 429 (c) 629 (d) 527
Sol. Formula :- (a + b)2 = a2
+ b2 + 2ab
Now [x2 + (1/x2)] = 23
Squaring
both sides, :- [x2 + (1/x2)]2 = (23)2
x4
+ (1/x4) + 2 × x2 × (1/x2) = 529
x4
+ (1/x4) + 2 = 529
x4
+ (1/x4) = 527
Option
(d)
Q14. If the sides of a triangle are in
the ratio of 3 :4 :5 and its area is 216
cm2, then the perimeter of the triangle is
(a)
72 cm (b) 76 cm (c) 84 cm (d) 80 cm
Sol. Let the sides of the triangle 3x, 4x, 5x
According
to the concept it is clear that this triangle is a right – angled triangle with
Base
= 3x , Height = 4x and Hypotenuse = 5x
Now
Area of Right–angled Triangle = (1/2) × base × height
216
= (1/2) × (3x) × (4x)
6x2
= 216
→
x = 6
Now
the Perimeter of the triangle = 3x + 4x + 5x = 12x = 12 × 6 = 72cm
Option
(a)
Q15. A researcher has collected the
following sample data :-
5,
12, 6, 8, 5, 6, 7, 5, 12, 4, 1
The
median is
(a)
5 (b) 6
(c) 7
(d) 8.
Sol. Arrange the given data in descending
order we get the arranged data as:-
12,
12, 8, 7, 6, 6, 5, 5, 5, 4, 1
Total
number of terms = 11
Now
median of the given data = middle term of arranged data = (Total term + 1)/2
Median
= 6th term = 6
Option
(b)
Q16. The average age of 15 students in a
class is 15 years. Among this the average age of 5 students is 14 years and the
average age of 9 other students is 16 years. The age of the 15th
student is
(a)
15 years (b) 14 years (c) 11 years (d) 15 whole (2/7) years.
Sol. Total number of students = 15, Average = 15
Total
sum of ages of all 15 students = 15 × 15 = 225 years
Among
this group the average age of 5 student = 14
Then
the total sum of ages of those 5 students = 15 × 5 = 70 years
Average
of another 9 students from the same group = 16
Then
the total sum of ages of those 9 students = 16 × 9 = 144 years
Now
the total sum of ages of 14 students = 70 + 144 = 214 years
Hence
the age of the 15th student = 225 – 214 = 11 years.
Option
(c)
Q17. The LCM and HCF of two numbers are
520 and 4 respectively. If one of these numbers is 52, then the other number is
(a)
52 (b) 40 (c) 42 (d) 50.
Sol. Let the numbers be a = 52 and b = ?
Their
LCM = 520 and their HCF = 4
Formula
:- a × b = LCM × HCF
52
× b = 520 × 4
b
= 40.
Option
(b)
Q18. The height of an equilateral
triangle whose each side is 4 cm, is
(a)
4√3 cm (b) √3 cm (c) 2√3 cm (d) 3√3 cm.
Sol.
As per the conceptual formula for the height of the equilateral triangle we
know
Height
(h) = (√3/2) × a where “a” is the
side of the equilateral triangle
Given
data side of the equilateral triangle “a” = 4 cm.
Putting
the given data in the formula we get the required answer
Height
(h) = 2√3 cm.
Option
(c)
Q19. A, B and C are shareholders in a
company. In a certain year A got 1/3rd share of the profit, B
received 1/4th share and C got Rs. 5000/-. Then how much profit did
A get?
(a)
Rs. 5000/- (b) Rs. 1000/- (c) Rs. 4000/- (d) Rs. 3000/-.
Sol. Share of “A” in the profit in a
certain year = 1/3rd
Share
of “B” in the profit in a certain year = 1/4th
Share
of “C” in the profit in a certain year = [1 – {(1/3) + (1/4)}] = 5/12th
But
according to the question “C” received as his share = Rs. 5000/-
Let
the total profit in the company = x
Conceptually
we know that (5/12) of x = 5000
→
x = 12000.
Now
profit of “A” = (1/3) of 12000 = Rs. 4000/-
Option
(c).
Q20. The smallest number by which 512 is
multiplied so that the product is a perfect cube is
(a)
7 (b) 8 (c) 9 (d) 10.
Sol. The prime factors of 512 =
2×2×2×2×2×2×2×2×2
→
512 itself a perfect cube of 8. So the smallest number by which 512 is multiplied
so that the product is a perfect cube number is only and only 1, which is not
preset in the given options. In such case it is suggested to each and every
student that they leave this question number blank in their answer sheet which
will represent your concept clearance.
Option
(Do not mark any option leave it blank. This question is wrong.)
Q21. 720 sweets were distributed equally
amongst the children in such a way that the number of sweets received by each
child is 20% of the total number of children. How many sweets did each child
receive?
(a)
14 (b) 11 (c) 15 (d) 12.
Sol. Let the total number of children = x
According
to the question number of sweets received by each child = 20% of x = (x/5)
Total
number of sweets to be distributed = 720
sweets.
Now
(x) × (x/5) = 720
On
solving we get x = 60 children.
So
the number of sweets received by each child = (60/5) = 12 sweets.
Option
(d)
Q22. Three different containers contain
different quantities of a mixture of milk and water, whose measurements are 403
litres, 434 litres, and 465 litres. What biggest measure (in litres) must be
there to measure all the different quantities exactly?
(a)
51 (b) 41 (c) 31 (d) 70.
Sol. Given quantities 403 litres, 434
litres, 465 litres to be measured.
To
solve such questions we have to just find the HCF of all given quantities.
Method
(Trick) for calculating HCF :-
Prime
Factors of smallest quantities from the given quantities:-
403
= 31 × 13 (among these two factors after checking we find 31 is the only common
factor for other two remaining quantities.)
Hence
the biggest measure (in litres) to measure all the different quantities exactly
= 31 litres
Option
(c)
Q23. If 1 is added to the denominator of
a fraction, the fraction becomes ½. If 1 is added to the numerator, the
fraction becomes 1. The sum of numerator and denominator of the fraction is
(a)
14 (b) 21 (c) 5 (d) 11.
Sol. Conceptual Method :- Let the
fraction be (x/y)
Where
numerator is x and denominator is y.
According
to the question:- 1 is added to denominator the fraction becomes ½ .
[x/(y
+ 1)] = ½
On
simplifying we get the first equation 2x – y = 1 ........................ eq.no.1
According
to the question:- 1 is added to numerator the fraction becomes 1 .
[(x
+ 1)/y] = 1
On
simplifying we get the second equation x – y = -1 ........................ eq.no.2
On
(eq.no.1 – eq.no.2) we get x = 2
From
eq.no.1 (on putting x = 2) we get y = 3
Hence
the faction is (2/3) and the required sum of numerator and denominator = 2 + 3
= 5.
Option
(c)
Q24. A sum was put at simple interest at
a certain rate for 2 years. Had it been put at 3% higher rate, it would have
fetched Rs. 300/- more. The sum is
(a)
Rs. 5,500/- (b) Rs. 5,000/- (c) Rs. 5,400/- (d) Rs. 5,300/-.
Sol. Given data :- T = 2 years, R = 3%,
S.I. = 300/-, P =?
We
know the formula → S.I. = [(P×R×T)/100]
On
putting the values and solving we get P = Rs. 5000/-
Option
(b)
Q25. 36 men can complete a piece of work
in 18 days. In how many days will 27 men complete the same work?
(a)
24 days (b) 22 days (c) 12 days (d) 18 days.
Sol.
Given data M1 = 36 men, D1 = 18 days, M2 = 27
men, D2 = ?
Applying
the formula :- M1 × D1 = M2 × D2
On
putting the values and solving we get D2 = 24 days.
Option
(a)
Hope
you will be able to solve these questions by your own. Visitor’s are warm
heartedly welcomed with their related problems and queries. The solution of
these questions (practice set) will be posted soon on this blog. More
Conceptual enjoyment will be continued for you soon, just keep visiting this
blog regularly and writing me your related queries through comment or forum.
Best
of Luck for Exams. To be continued soon........................
No comments:
Post a Comment