The Uttar Pradesh (UP) Lekhpal Exam was held on 13th Sep 2015, Here we provides you UPSSSC Lekhpal 2015 Questions and Answers Key of all Sets with PDF download. UP Lekhpal Answer Key 2015 Download 13th Sept Exam Paper Solution. UPSSSC Chakbandi Lekhpal Exam 2015 Answer Key and Expected Cut Off Marks. Official answer key or previous question papers when searched online, cannot be found. Like all other government exams, they do not provide question papers and answer keys in the official website.

##
**UP Lekhpal Cut off marks**

Lekhpal cut off marks2015 (expected)

**General – 52 Marks**
**OBC – 50 Marks**
**SC – 45 Marks**
**ST – 40 Marks**
##
UP Lekhpal Answer Key 2015:

Visit the Official website from the link given below.

Search for UP Lekhpal Answer key 2015.

Click on the link (if declared officially)

Select set of question papers.

Download the page for answer sheet.

up rajaswa lekhpal answer key

up rajaswa lekhpal answer key 2015

**1: **If the circumference of a circle is 18.6 cm more than its diameter, then what is the diameter of the circle ?

- 8.68 cm
- 8.84 cm
- 7.54 cm
- 7.84 cm

**Explanation: **2πr = 18.6 + (D)

2πr = 18.6 + 2r

2πr - 2r = 18.6

2r (π - 1) = 18.6

2r = 186/10 X 7/15

r = 186/10 X 7/15 X 1/2

=> 1302/300

D = 2 X 1302/300 = 8.68

**2: **The HCF and LCM of two numbers are 12 and 2448 respectively. If the difference of the numbers is 60, then the sum of the numbers is ?

- 248
- 204
- 348
- 284

**Explanation: **Given data: LCM = 2448 , HCF = 12 and

difference of numbers = 60

We know that,

LCM x HCF of two numbers = Multiplication of both numbers

Let a and b are two numbers, Then

a X b = LCF X HCF

a X b = 12 X 2448 = 29376 — eqn. (1)

a – b = 60 ; given in question

as, (a – b)

^{2} = a

^{2} + b

^{2} – 2ab and

as (a + b)

^{2} = a

^{2} + b

^{2} + 2ab

(a + b)

^{2} = (a – b)

^{2} + 4ab

(a + b)

^{2} = 60X60 + 4X29376

(a + b)

^{2} = 121104

(a + b) =

√(121104)

(a + b) = 348

Sum of two numbers = 348

**3: **The prices of a scooter and a moped are in the ratio9:5. If a scooter costs Rs. 4200 more than a moped, then the price of the moped is ?

- Rs. 5700
- Rs. 6300
- Rs. 3350
- Rs. 5250

**Explanation: **If price of moped = x, then the price of scooter = x+4200.

Now,

(x+4200) : x = 9 : 5

9x = 5(x +4200)

9x = 5x + 21,000

4x = 21,000

x = 21,000/5 => 5250.

**4: **The sum of three numbers is 392. If the ratio of first number to second number is 2:3 and the ratio of second number to third number is 5:8, then the third number is ?

- 176
- 192
- 162
- 120

**Explanation: **Let the three numbers are A, B, C. Then,

A:B = 2:3 and B:C = 5:8 = ( 5 x 3/5) : ( 8 x 3/5) = 3 : 24/5

Hence, A:B:C = 2:3:24/5 => 10:15:24

C = 392 x 24/49 => 192

**5: **The perimeter of a triangular field is 540 meters and the ratio of its sides is 5:12:13. What will be the area of this field ?

- 9320 sq.metre
- 9450 sq.metre
- 9560 sq.metre
- 9720 sq.metre

**Explanation: **
**6: **The perimeter of a top surface of a rectangular table is 28 m and its area is 48 m^2. What is the length of the hypotenuse of the table ?

- 12.5 m
- 5 m
- 10 m
- 12 m

**Explanation: **The perimeter of a top surface of e rectangular table = 28 m

Formula-

The perimeter of the rectangular table = 2 (l + b)

Then, 2(l + b) = 28

l + b = 28/2

l + b = 14 ……………………………………………….. (1)

Area of surface of the rectangular table = 48 m2

Then, lb = 48

l = 48/b

Put the value of l = 48/b in equation (1),

Then, 48/b + b = 14

(48 + b

^{2}) /b = 14

48 + b

^{2} = 14b

b

^{2} – 14b + 48 = 0

b

^{2} – 8b – 6b + 48 = 0

b (b – 8) – 6 (b – 8) = 0

(b – 8) (b – 6) = 0

b – 8 = 0 & b – 6 = 0

b = 8 & b = 6

Then, l = 48/6 = 8

l = 8 cm & b = 6 m

Hypotenuse of the rectangular surface of the table = (8

^{2} + 6

^{2})1/2

= (64 + 36)1/2

= (100)1/2

= 10 m

Then, the hypotenuse of the table is 10 m.

Hence, the answer is (3) 10 m.

**7: **The perimeter of a triangle is 30 cm and its area is 30 cm

^{2}. If the length of the largest side of the triangle is 13 cm, then what is the smallest side of the triangle ?

- 5 cm
- 6 cm
- 4 cm
- 3 cm

**Explanation: **The perimeter of a triangle = 30 cm

the area of a triangle = 30 cm2

Then, ab/2 = 30

ab = 30 x 2

ab = 60 cm

a = 60/b

The a + b + 13 = 30

a + b = 30 – 13

a + b = 17 ………….. (1)

Put the value of a = 60/b in equation (1),

60/b + b = 17

60 + b2 = 17 b

b2 – 17 b + 60 = 0

b2 – 12b – 5 b+ 60

b (b – 12) – 5 (b – 12) = 0

(b – 12) (b – 5) = 0

b – 12 = 0 & b – 5 = 0

b = 12 and b = 5

Then, the sides of the triangle are 5, 7 and 13.

Then, the smallest side of the triangle is 5.

**8: **The radius of the wheel is 21 cm. How many rounds will it take to cover the distance of 792 metres ?

- 400
- 600
- 200
- 300

**Explanation: **The radius of a wheel = 21 cm

Then, the circumference of the wheel = (2 x 22 x 21)/7

= 2 x 22 x 3

= 132 cm

The distance covered by wheel in a round = 132 cm

Total distance covered by the wheel = 792 m

1 m = 100 cm

Therefore, 792 m = 792 x 100 = 79200 cm

Then, the number of rounds of the wheel to cover the distance 79200 cm = 79200/132

= 600

**9: **A man is twice as fast as a woman and a woman is twice as fast as a boy. If all of them i.e, a man, a woman and a boy can finish a work together is 4 days, in how many days will a boy do it alone ?

- 28 days
- 7 days
- 21 days
- 14 days

**Explanation: **
**10: **In a mixture of 60 litres, the ratio of acid and water is 2:1. What amount of water must be added to make the ration of acid and water as 1:2 ?

- 60 litres
- 72 litres
- 44 litres
- 52 litres

**Explanation: **The ratio of the acid and water in a mixture = 2: 1

Total quantity of mixture = 60 litres

Then, the of acid in the mixture = 60 x 2/3

= 40 litres

The, quantity of water in the mixture = 60 x 1/3

= 20 litres

If water 60 litres added, then the ratio will be 1:2

Then, the new ratio of acid and water = 40 : (20+40)

= 40 : 80

= 1:2

**11: **In ΔABC, If ∠A = 90o, a = 25 cm, b = 7 cm, then what is the value of tan B ?

- 24/7
- 24/25
- 7/25
- 7/24

**Explanation: **
C = (252 – 72)1/2

C = (625 – 49)1/2

C = (576)1/2

C = 24 cm

Then, the value of tan B = 7/24

**12: **If the interest on a sum of money be 1 paisa per rupee per month, what is the rate per cent per annum ?

Ans: b

- 10 1/2%
- 12%
- 10%
- 15%

**Explanation: **Simple Interest = 1 paisa per month

Let the rate of interest be R%.

Simple Interest = Principal x Time x Rate/100

Then, 1/100 = 1 x 1 x R/100 x 12

1 = R/12

R = 12%

Then, the rate percent of per annum is 10%.

**13: **The standard deviation of a set of 50 observations is 8. If each observation is multiplied by 2, then the value of standard deviation will be ?

- 8
- 16
- 2
- 4

**Explanation: **If we multiply all data values included in a data set by a constant k, then the standard deviation is the standard deviation of the original data set TIMES the absolute value of k.

**14: **For what value of x will the mode of the following data be 27 ?

25, 26, 27, 23, 27, 26, 24, x, 27, 26, 25, 25

- 26
- 27
- 24
- 25

**Explanation: **Mode is the value of a variable which occurs most frequently. It represents most frequent value of a series. Hence in the given set of data values if we want the mode value = 27, then 27 must be the most frequent value. Hence value of x = 27.

**15: **The sum of mean and median of 1, 6, 8, 3, 2 is ?

- 10
- 12
- 6
- 7

**Explanation: **The mean of 1, 6, 8, 3, 2 = (1 + 6 + 8 + 3 + 2)/5

= 20/5 = 4

The median if 1, 6, 8, 3, 2 = 8

Then, the sum of mean and median = 8 + 4 = 12.

**16: **A man traveled 2/11 of his journey by train, 17/22 by car and walked the remaining 1 kilometer. How far did he go ?

- 24 km
- 33 km
- 27 km
- 22 km

**Explanation: **Let the total distance of the journey be X km.

A man traveled 2/11 of his journey by train,

Then, the journey traveled by the train = (2/11)x

17/22 by car,

Then, the journey traveled by the car = (17/22)x

Total journey traveled by the train and car = X (2/11 + 17/22)

= X(4 + 17)/22

= (21/22)X

Remaining journey = X – 21X/22

= (22X – 21X)/22

= X/22

Then, X/22 = 1

X = 22 km

Then, the total distance of the journey is 22 km.

**17: **While computing mean of grouped data, we assume the frequencies are ?

- centred at the upper limit of the classes
- centred at the lower limits of the classes
- evenly distributed over all the classes
- centres at the class markes of the classes

**Explanation: **
**18: **What least fraction must be subtracted from the square root of 930.25 so that the result is a whole number ?

- 1/6
- 4/3
- 2/3
- 1/2

**Explanation: **The square root of 930.25 = 30.5

30.5 = 30 + .5

= 30 + 1/2

If 1/2 subtracted from the number 30 + 1/2, then the remaining number will be whole number is 30.

Then, 30 + 1/2 – 1/2

= 30

30 is a whole number

**19: **300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution ?

- 60 gm
- 25 gm
- 45 gm
- 40 gm

**Explanation: **Let the quantity of sugar should be added = X gm

Total quantity of the solution = 300 gm

40% sugar,

Then, the quantity of sugar in the solution = 300 x 40/100

= 120 gm

Therefore, the quantity of water in the solution = 300 – 120

= 180 gm

Then, 180 + 120 + X = (300 + X)/2 + 180

120 + X = (300 + X)/2

By cross multiplication,

2 x 120 + 2 x X = 300 + X

240 + 2 X = 300 + X

2 X – X = 300 – 240

X = 60 gm

Short Trick –

(300 + X)/2 = 180

300 + X = 180 x 2

300 + X = 360

X = 360 – 300

X = 60 gm

**20: **The difference between a number obtained by increasing a certain number by 8% and than obtained by diminishing it by 3% is 407. The original number is ?

- 3700
- 3400
- 3500
- 3600

**Explanation: **Let the original number be X

According to the question,

108 X/100 – 97 X/100 = 407

(108 X – 97 X)/100 = 407

11 X/100 = 407

11 X = 407 x 100

X = 407 x 100/11

X = 37 x 100

X = 3700

Then, the original number is 3700.

**21: **Which fraction should be added to the sum of 5 3/4, 4 4/5 and 7 3/8 to make the result a whole number ?

- 1/40
- 3/40
- 1/10
- 1/20

**Explanation: **The sum of 23/4, 24/5 and 59/8 = 23/4 + 24/5 + 59/8

= (23 x 10 + 24 x 8 + 59 x 5)/40

= (230 + 192 + 295)/40

= 717/40 = 17 + 37/40

Let the added X to make whole number.

Then, 37/40 + X = 1

(37 + 40X )/40 = 1

37 + 40 X = 40

40X = 40 – 37

X = 3/40