SAINIK SCHOOL ENTRANCE INFORMATION: November 2019

Sample maths revision papers by Defence Academy Coimbatore for Class VI sainik and RMS

Section A – Mathematics

1. Write the following digits in such a way to place 9 in ten thousands place, 4 in thousands place 3 in one’s place 7 in hindered place and 6 in tens Place

2.If a bar is placed over C, the value f C becomes____ times of the value of M

3. The greatest number that will divide 137, 182 and 422 leaving a remainder of 2 in each case is_______

M. Q.4. Find the SUCCESSOR of the answer of ( Greatest two digit number x smallest three digit number

M..Q.5.In a Right Angled tri angle ABC if Angle B is 90 degrees then the other two angles would be______ degrees each. A

B C

M.6 If A= 2510, B=2650 and C= 2512, which one of the following is greatest?

a) A x B

b) B x C

c) A x C

d) A x A

M.7.Simplify : 5/2 x ( 7/2 x 4/11) ÷ ½ = ?

M.8. Find the value of 12 ¾ dz apples in 1 ¼ of apples cost Rs.75

M.9. Find the average of the Prime numbers from 70 to 90

M.10. If 45 % of 700 is ------- Equal to 3 x 100 +10 +1+5 x1 ( is this correct?)

M.11. What is the difference XC and LXV- Write the answer is Roman Numerals
M.12. Roman numerals MCMXCIV.= ?
M.13. If 45 students can consume a stock of food in 2 months, then for how many days the same stock of food will last for 27 students?

M.14. If 24 painters working for 7 hours a day, for painting a house in 16 days. How many painters are required working for 8 hours a day will finish painting the same house in 12 days?

M.15. In a camp there is enough food for 500 soldiers for 35 days. If 200 more soldiers join the camp, how many days will the food last?

M.16. A car travels at a speed of 35 m/s. Find its speed in km/h?

M.17. Which list contains all prime numbers?

a) 1,7,13,29 and 31 (b) 2,13, 27,41, 93 c) 2,17, 19, 43, 79 d) All are prime nos

M.18,

M.19. Set -03. M.19. In a Quadrilateral, ifa the angles are in the ratio 1 :2 : 3: 4, then the greatest angle is___

M. 20. Find the Simple interest on a sum of Rs.9500 that was invested for 9 years and 9.5 %

M.21. A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

M.22. The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:

M.23. A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

M.24. Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is

M.25 A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

M.26. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

M.27. A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

M.28. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

M.29. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

M.30. How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

M.31. A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 308 m. Find the width of the track.

M.32. The area of a circle is 616 cm². Find its circumference.

M.33. The ratio of areas of two wheels is 25 : 49. Find the ratio of their radii.

M.34. The diameter of a wheel of a motorcycle is 63 cm. How many revolutions will it make to travel 99 km?

M.35. The diameter of a wheel of cycle is 21 cm. It moves slowly along a road. How far will it go in 500 revolutions?

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Set -03

94763

1000 times

Roman Numeral, Number. I, 1. V, 5. X, 10. L, 50. C, 100. D, 500. M, 1000 ... If a smaller value is placed before a larger one, we subtract instead of adding. ... Don't subtract a letter from another letter more than ten times greater. ... This system gets cumbersome quickly. When ... Using this method, 7000 would be (bar over) VII .

9901

Greatest two digit number is 99

Smallest three digit number is 100

Product of this is 99 x 100 = 9900

Predecessor of this number is 9900+1- 9901

45°

90° +45°+45°= 180°

If A= 2510, B=2650 and C= 2512, which one of the following is greatest?
Ax B = 2510 x 2650 = 6651500
B x C = 2650 x 2512 = 6656800 & A x C =2510 x 2512 = 6305120
A x A = 2510 x 2510 =6300100 & B x B = 2512 x 2512 = 6310144

5/2 x ( 7/2 x 4/11) ÷ ½ = ?
First Bracket = 7/2 x 4/11= 28/22 = 14/ 11
5/2 x 14/11 ÷ ½ or 14/11x 2/1 = 28/11
5/2 x 28/ 11 = 140/22 = 70/ 11

Rs 765.

the prime numbers. between 70 and 90 are--
71,73,79,81,83,89 476 / 6 = 79.33

It is 3 x100 +1 x 10 + 5 x1 = 315

LXV

90-65 = 25 =XXV

1994

The number 1994 is a great example of these rules. It is represented by the Roman numerals MCMXCIV. If we break it down then; M = 1,000, CM = 900, XC = 90 and IV = 4.

100 Days

Convert months into days 2 x 30 = 60 days

45----------------------60=

1 student 45 x 60 = 2700

For 27 students 2700/ 27 = 100 days

12 Days

Solution:

24 painters working for 7 hours paint a house in 16 days.

1 painter working for 7 hours paints a house in 16 × 24 days.

1 painter working for 1 hour paints a house in 16 × 24 × 7 days.

Let the required number of painters be x, then;

x painters working for 1 hour a day paint the house in (16 × 24 × 7)/x days

x painters working for 8 hours a day paint the house in (16 × 24 × 7)/(x × 8) days

But the number of days given = 12

According to the problem;

(16 × 24 × 7)/(x × 8) = 12 or 2688/8x = 12

8x × 12 = 2688 or 96x = 2688 or x = 2688/96

x = 28

Therefore, 28 painters working for 8 hours a day will finish the same work in 12 days.

25 Days

Solution:

This is a situation of inverse variation, now we solve using unitary method.

For 500 soldiers, food lasts for 35 days.

For 1 soldier, food lasts for (35 × 500) days.

Since 200 more join. So, now the number of soldiers is (500 + 200) = 700.

For 700 soldiers, food lasts for (35 × 500)/700 days

Therefore, for 700 soldiers, food lasts for = 25 days.

126 km/hr

35 x 18/5 = 7x 18 = 126 km/hr

ANS only C column are prime numbers

104

(84+90+665)/ 105 = 7 A/105
or 838/105 = 7A/105 = (735+a) / 105
or 7 104/105 = 7 A/`105
here compare left side and right side we get A= 104
= 839 = 735+1 A= 839-735 = 104

144 Degrees

Ans E
A Quadrilateral 4 side 4 angles 1+2+3+4 = 10
Full measure is 360 degrees
so 1 part is 36
the greatest angle given is 4 parts
hence 4 x 36 = 144 degrees

Ans B
(9500 x 9.5 x 9) /(100 x 10)
95 x95 x 9 / 10 =81225/ 10 = Rs. 8122.5
SI = Rs. 8122.50

150 m

Speed =		60 x	5	m/sec	=		50	m/sec.
			18				3

Length of the train = (Speed x Time).

Length of the train =		50	x 9	m = 150 m.
		3

245 m

Explanation:

Speed = 45 x 5 m/sec = 25 m/sec.

18 2

Time = 30 sec.

Let the length of bridge be x metres.

Then, 130 + x = 25

30 2

2(130 + x) = 750

x = 245 m.

240 m

Explanation:

Speed = 54 x 5 m/sec = 15 m/sec.

Length of the train = (15 x 20)m = 300 m.

Let the length of the platform be x metres.

Then, x + 300 = 15 or 36 x + 300 = 540

x = 240 m.

7 days

Explanation:

1 woman's 1 day's work =	1
	70

1 child's 1 day's work =	1
	140

(5 women + 10 children)'s day's work =		5	+	10		=		1	+	1		=	1
		70		140				14		14			7

5 women and 10 children will complete the work in 7 days.

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs.	6
	5

For Rs.	6	, toffees sold = 6.
	5

For Re. 1, toffees sold =	6 x	5	= 5.

26. 85 kg

27. 700 apples
28. Rs. 698

29.

Principal

= Rs.		100 x 4016.25
		9 x 5

= Rs.		401625
		45

= Rs. 8925.

30 4. Years Answer: Option B

Explanation:

Time =		100 x 81	years	= 4 years.
		450 x 4.5

M.31 Width 14 m

Answer is E

Solution:

Let r₁ and r₂ be the outer and inner radii of ring.

Then 2πr₁ = 308 or 2 × 22/7 r₁ = 308

⇒ r₁ = (308 × 7)/(2 × 22) or ⇒ r₁ = 49 m

2πr₂ = 220 or ⇒ 2 × 22/7 × r₂ = 220

⇒ r₂ = (220 × 7)/(2 × 22)

⇒ r₂ = 35 m

Therefore, width of the track = (49 - 35) m = 14 m

32. Solution:

We know area of circle = πr²

⇒ 22/7 × r² = 616

⇒ r² = (616 × 7)/22

⇒ r² = 28 × 7

⇒ r = √(28 × 7)

⇒ r = √(2 × 2 × 7 × 7)

⇒ r = 2 × 7

⇒ r = 14 cm

Therefore, circumference of circle = 2πr

= 2 × 22/7 × 14

= 88 cm

33.

The diameter of a wheel of a motorcycle is 63 cm. How many revolutions will it make to travel 99 km?

Solution:

The diameter of the wheel of a motorcycle = 63 cm

Therefore, circumference of the wheel of motorcycle = πd

= 22/7 × 63

= 198 cm

Total distance travelled by motorcycle = 99 km

= 99 × 1000

= 99 × 1000 × 100 cm

Therefore, number of revolutions = (99 × 1000 × 100)/198 = 50000

M,35, Solution:

In revolution, distance that wheel covers = circumference of wheel Diameter of wheel = 21 cm

Therefore, circumference of wheel = πd

= 22/7 × 21

= 66 cm

So, in 1 revolution distance covered = 66 cm

In 500 revolution distance covered = 66 × 500 cm

= 33000 cm

= 33000/100 m

= 330 m

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Sample maths paper class VI Sainik and RMS

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