Maths revision papers fully solved for sainik and RMS class VI
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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
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?
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?
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a)
A x B
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b) B x C
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c) A x C
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d) A x A
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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 NumeralsM.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
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1
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94763
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2
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1000 times
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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 .
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3
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15
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4
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9901
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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
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5
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45°
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90°
+45°+45°= 180°
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6
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B
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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 |
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7
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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 |
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8
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Rs 765.
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9
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the prime numbers. between 70 and 90 are-- 71,73,79,81,83,89 476 / 6 = 79.33 |
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10
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No
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It is 3 x100 +1 x 10 + 5 x1 = 315 |
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11
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12
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1994
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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.
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13
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100 Days
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Convert months into days 2 x 30 = 60 days
45----------------------60=
1 student 45 x 60 =
2700
For 27 students 2700/ 27 = 100 days
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14
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12 Days
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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.
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15
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25 Days
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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.
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16
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126 km/hr
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35 x 18/5 = 7x 18 = 126 km/hr
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17
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ANS only C column are prime numbers
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18
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104
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(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 |
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19
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144 Degrees
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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 |
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20
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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 |
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21
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150 m
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Length of the train = (Speed x Time).
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22
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245 m
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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.
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23
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240 m
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Explanation:
Speed
= 54 x 5 m/sec
= 15 m/sec.
18
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.
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24
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7 days
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Explanation:
 5
women and 10 children will complete the work in 7 days. |
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25
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5
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C.P. of 6 toffees = Re. 1
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26. 85 kg
27. 700 apples
28. Rs. 698
28. Rs. 698
29.
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Principal
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=
Rs. 8925.
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30 4. Years Answer: Option B
Explanation:
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Time =
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 |
100 x 81
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 years |
= 4 years.
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450 x 4.5
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M.31 Width 14 m
Answer is E
Solution:
Let r₁ and r₂ be the outer and inner radii of ring.
Let r₁ and r₂ be the outer and inner radii of ring.
Then 2Ï€r₁ = 308 or 2 × 22/7 r₁ = 308
⇒ r₁ = (308 × 7)/(2 × 22) or
⇒ r₁ = 49 m
2Ï€r₂ = 220 or ⇒ 2 × 22/7 × r₂ = 220
2Ï€r₂ = 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 = Ï€r²
We know area of circle = Ï€r²
⇒ 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 = 2Ï€r
Therefore, circumference of circle = 2Ï€r
= 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
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
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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