Here
I am presenting the solution Mathematics Questions, from the Question Paper of
Lekhpal held on 13th September for the post in the District Lucknow
on the special demand of one of our sincere visitor, Vikash. Match your answer
and analyse your preparations. In case any problem you may ask for help by
sending the “solution required for question no......” in comment or through
forum.
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)
Solution continues in the next post. Check it .........................
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