Q1.
ABCD is a parallelogram, E and F are mid points of sides AB and
CD respectively.
Prove that the line AF and CE trisect the diagonal BD.
Q2.
Reconstruct the division problem
**** *
* * * * * *9 *
**
***
**
** *
** *
**
**
Q2.
The seven consecutive squares are 81, 100, 121, 144, 169,
196, 225 with its digits sums to a square numbers : Let us find
another set of seven consecutive squares with same property.
Q2.
Solve
x….
x
x
x
x
x = 2
Q3.
In trapezium the parallel sides are 4 cm and 16 cm. The lower
base angle is 30o and 60o. What is the distance between midpoints
of two parallel sides.
Q4.
Prove that 11n+2 + 12
2n+1
is divisible by 133 for any natural
number n.
Q5.
There are 7 girls and 2 boys. A team of 4 persons must be chosen
with at least 1 boy on the team. In how many ways can this be
done.
Q6.
Find a thousand natural numbers such that their sum equals their
product.
Q7.
In the figure place the number 0 and 9 in the circles without
repetition so that all the sums of the numbers in the vertices of the
shaded triangles are equal.
Q8.
Four small circles of radius 1 are tangent to each other and to a
large circle containing them, as shown in the diagram. What is the areas
of the region inside the larger circle, but outside all the smaller circles ?
Q9.
3 persons A, B and C with help of a monkey collected many
cocoanuts, got tired and fell asleep. At night A woke up and
decided to have his shares. He divided cocoanuts into 3 equal
shares giving the left out single cocoanut to monkey for it’s hard
labour and fell aleep again. In the same way in order B and C
wake up. Not knowing whether anybody woke up and each of
them divided the cocoanuts into three shares, every time giving
the left out single cocoanut to the monkey. Early in the morning
all of them woke up together, divided the remaining cocoanuts
into 3 equal shares and left gave out single cocoanut to the
monkey. What is the minimum number of cocoanut they
collected?
Q10. How many diagonals are there if the regular convex polygon has
n sides ?
Q11. Resolve into factors
(a+b)2 (b+c) (c+a)2 + abc {2(a+b) (b+c) (c+a) + abc}
Q12. Solve
x
4x − 3
1
2
=3
x+
1
2
− 22 x −1
Q13. Solve
3 − x +1 = x
Q14. A machine is sold by a shop for Rs. 19200 cash or 4800 cash
down payment together with five equal monthly installments. If
the rate of interest charged by the shop is 12% per annum, find
each installment.
Q15. Prove
1
1
1
1
n −1
+
+
+ .... +
=
1x 2 2x3 3x 4
(n − 1) xn
n
Q16. A natural number ends in 2. If we move this digit 2 to the
beginning of the number, then the number will be doubled. Find
the smallest number with this property.
Q17. ABCD is rectangle chose point G and H (G on BC and H on CD)
such that triangle AGH is equilateral.
Q18. ABC is an isosceles with angle A = 20o E and F are points on AC
and AB such that ∠EBC = 60o and ∠FCB = 50o. Calculate
∠BEF.
Q19. Reconstruct the division problem.
*****
* * * * * *2 *
**
***
**
***
***
**
**
20.
Reconstruct the division problem
* * * * * ** * * * *
***
** * *
***
** * *
** *
** * *
** * *
Q21. Let N be the sum of the digits of a natural number A, let B =
A+N, Let A’ be the sum of the digits of the number B, and let C =
B+A’. Find A if the digits of C are those of A in reverse order.
Q22. Prove that 22225555 + 55552222 is divisible by 7.
Q23. A number of bacteria are placed in a utensil. One second later
each bacterium divides in two, the next second each of the
resulting bacteria divides in two again and so on. After a minute
the utensil is full. When was utensil half full ?
Q24. Prove that a square can be divided into 1989 squares.
Q25. Find the smallest natural number which is 4 times smaller than the
number written with the same digit but in the reverse order.
Q26. Prove that
3999991 is not prime
Q27. Draw four straight lines without lifting the pen passing through all
9 points in figure.
.
.
.
.
.
.
.
.
.
Q28. Draw six straight lines through the 16 points shown without
lifting your pencil from the page.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Q29. A square is inscribed in an equilateral triangle as depicted. Find
the ratio of area of square to area of triangle.
Q30. ABCD is a square. W, X, Y and Z are mid points of side AB, BC,
CD and DA respectively. Find the ratio of area of shaded square
to square ABCD.
Q. 31. In a vertical cylindrical container water is being poured for first
three minutes at the rate of one life per minute. During second three
minutes at the rate of two litres per minute. During third three minutes at
the rate of three liters per minute and so on. Find
(a) After 50 minutes how deep the water is ?
(b) How long it will take to fill the cylinder ?
Q. 32 Which is greater 300 or 2300
Q. 33 Which is greater :
240 or 328
Q. 34 Prove that
1 1 1
1
1
1
− + − ............. −
+
>
2 3 4
99 100 5
Q. 35 Find the last two digits of (rightmust) of 3999.
Q. 1 Diagram
∆ADF ≅ ∆CEB
So, AF = EC
And AE = FC
So, AECF is a parallelogram
FL || CM and DF = FC
⇒ DL = LM
Similarly, LM = BM
So DL = LM = MB
Q. 2 The solution is
49 59 04 99
Q. 2 The seven numbers starting with
9999.
Q. 2
X
X
X
X= 2
⇒ x
2
=2
⇒ x =± 2
Let value of one side of trapezium as in figure = x
So, AH
x
3
x
x , CH =
= DG; So, GB =
2
2
2 3
So, AH =
3
x
2
x
, HG = 4 cm, GB = 2 3
From the question,
⇒ x = 6 3 cm
So, AH =
3
x6 3
2
x
3
x +4+
= 16cm
2
2 3
= 9 cm
⇒ IH = AH – AI
= 9-8 = 1 cm
So, IF =
(3 3 ) + (3)
2
2
= 9 cm
Q. 4 11n+2 +1222n+1 =
121.11n+12.122n
=
133.11n-12.112n+12.122n
=
12
(122n –11n)+ 133.11n
=
12
(144n-11n)+133.11n
=
12 x (144-11) x some number + 133.11n
So it is divisible by 133.
Q.5
There must be 1 or 2 boys in the team. In the later case 2 girls can
be added to the team in
7 C 3 ways.
If there is only 1 boy in the team. There are 2 ways to choose him.
So the team can be formed by adding 3 girls in
Therefore total ways =
=
∠7
+2
∠7.2∠2
∠7
+2
∠ 5∠ 2
7 c + 2.7 c
2
3
∠7
∠7.3 ∠3
∠7
∠4∠3
7C3
different ways.
=
=
7.6
+2
2
=
21+70
=
91 ways
.
7.6.5
3.2
Q.6 Solution is
1x1x1x….. 1 x 334 x 4 = 1336
998 times
and 998 + 334+ 4 = 1336
Q. 7 The crux sum = 13. The solution is
Q. 8
From figure the radius of bigger circle is
2+2 2
=1+ 2
2
e
So, area bound = π 1+ 2
j
2
Area inside four smaller circle = 4 π
So, area of the region inside the larger circle, but outside all smaller
circle
=
=
Q.9
e
π 1+ 2
j
e
2
− 4π
j
π 2 2 −1
Let total cocoanuts =x
So, A gets
x −1
3
and remaining = x −
LM x −1 +1OP
N3 Q
2x − 2
3
=
Similarly, B gets
=
=
and new remaining =
=
FG 2x − 2 − 1IJ x 1
H 3 K3
2x − 5
9
LM
N
OP
Q
2x − 5
2x − 2
−
+1
3
9
6x − 6 − 2x + 5 − 9
9
4x − 10
=
So, C gets
=
9
FG 4x − 10 − 1IJ x 1
H 9 K3
4 x − 10 − 9
27
=
4 x − 19
=
So, New remaining =
=
27
4 x − 10
9
−
FG 4x − 19 + 1IJ
H 27 K
12x − 30 − 4 x + 19 − 27
27
8x − 38
27
=
This is divided amongst A,B, C and one for monkey
FG
H
IJ
K
8 x 38 − 27
1 8x − 38
−1 =
81
3
27
So A gets :
8x − 65
81
As, x is whole number and
is also whole number,
8x − 65
= 7 ⇒ x = 79 minimum).
Hence,
81
Q.10
The number of diagonals
b
n n −3
=
g
2
ba + b g bb + cg ( c + a)
2
Q.11
=
=
=
=
2
2
l
q
+ abc 2( a + b ) ( b + c) ( c + a) + abc
ca + b h2 cb + ch2 ( c + a)2 + 2(a + b) ( b + c) ( c + a)abc + (abc)2
m(a + b ) (b + c) ( c + a) + abcr2
RS(a + b ) (ac + a2 + bc + ab) + abcUV2
T
W
RS( b + c)ac + ( b + c)a2 + ( b − c)bc + ( b + c)ab + abcUV2
T
W
RSabc + ac2 + a2 b + a2c + b 2c + bc2 + ab2 + abc + abcUV2
T
W
=
=
=
=
Q.12
RSabc + b 2c + bc2 + a2c + abc + ac2 + a2b + ab2 + abcUV2
T
W
mbc(a + b + c) + ca(a + b + c) + ab(a + b + c)r2
m(a + b + c) (ab + bc + ca)r2
x1
x +1
x
2
4 −3
= 3 2 − 2 2 x −1
⇒2 2 x + 2 2 x −1 =3
FG
H
⇒2 2 x 1 +
x+
1
1
x
2 +3 2
1
IJ =3x .2.3 2
2K
1
2 2x 2 3
x2
⇒ x =
3
3
22x
4
=
3x 31 / 2
So, taking log on both side
1
2x log 2 − x log 3 = 2 log 2 − log 3
2
1
2
⇒x( log 4 − log 3 ) = log 4 − log 3
1
log 4 − log 3
2
⇒x =
log 4 − log 3
Q.13
3 − x +1= x
⇒3 − x = x 2 − 2x +1
⇒x 2 − x − 2 = 0
⇒x 2 − 2 x + x − 2 = 0
⇒x ( x − 2 ) +1( x − 2) = 0
⇒( x − 2) ( x +1) = 0
⇒x −2 =0
∴x = 2
or
x+1 = 0
⇒x =−1
So solution is x=-1 or 2.
Q. 14 Cash price = Rs. 19200
Cash down payment = Rs 4800
Price to be paid in installments = Rs. 14400
Number of equal installments = 5
Rate of interest = 12% per annum
Let amount of one installment = x
So interest paid in installment scheme = (5x-14400)
Principal for first installment
= Rs. 14400
Principal second installment
= Rs. (14400-x)
Principal second installment
= Rs. (14400-2x)
For 3rd installment Principal
= Rs. (14400-3x)
For 4th installment Principal
= Rs. (14400-4x)
Total
= Rs (72000-10x)
i.e. Principal for one month
Balance of the last principal (Rs. 14400-4x)
+ the interest charged Rs. (5x-14400)
= monthly installment (Rs x)
( 72000 −10 x )x12 x1
100 x12
So interest =
(ii)
From (i) and (ii)
( 72000 −10 x )x12
= 5x −14400
1200
⇒x =2964.70
So amount of each (equal) monthly installment = Rs. 2964.70
(i)
1
1
1
+
+
+..
1x 2 2x 3 3x 4
Q. 15
=
=
=
+
1
( n −1)n
FG1− 1IJ + FG 1 − 1IJ + FG 1 − 1IJ +... +FG 1 − 1 IJ
H1 2K H 2 3K H 3 4K
H (n −1) n K
1−
1
n
n −1
n
Q. 16 The Solution is
105263157894736842
Q. 17
Let AB =DC =a,
BC = AD = b,
And DQ = x and BP = y
a2 + y 2 = b2 + x 2
(try for a better solution !)
So,
ba−x g2 + cb −y h2
⇒2cax + by h=a2 + y 2 = b 2 + x 2
=
and we have
FH x 2 −2ax +b IK 2 =4b 2 FH b 2 +x 2 IK −4ab2
FH
⇒ x 2 + b2
IK RSFH x 2 + b 2 IK − 4ax − 4b 2 UV = 0
T
W
⇒ x = 2a − 3 b
And
y = 2 b − 3a
Q. 18 It is a difficult problem :
To Solution :
∆ABC is isosceles, so ∠ABC = ∠ACB
Q ∠ BAC = 20 so ∠ABC = ∠ACB = 80
0
0
∴ ∠EBD = 200 ∠ECB = 500 and ∠BDC=400
in ∆ BEC, ∠BEC = ∠ECD = 50o, so, BE = BC
using sine Law, in both triangles.
In ∆ BED,
Sin (160 − x )
=
BD
in ∆DBC,
sin 40
BC
sin(160 − x )
sin x
So,
∴
sin x
BE
=
si n 80
BD
=
BD
BE &
BD
BC =
Sin 80
Sin 40
sin(160 − x ) sin 80
=
sin x
sin 40
⇒ sin(160 − x ) =
sin 80
.sin x
Sin 40
⇒ sin160 . cos x - sin x . cos 160 = 1.532 sin x
⇒ 0.34 2 . Cos x + 0.93969 sin x = 1.532 sin x
⇒ tanx = 0.5774
⇒ tan 300
so x = 30o
19
Solution is : 49 591528
Q.20 The solution is
333 43301322
Q.21 if A has x digits,
Then, C can not exceed A by more than 18x. On the other hand, if
C is the reverse of A, then C exceeds A by at least
and by at least
So,
9.10
A = 12 or 69.
x −3
2
9.10
if x is odd. This implies X ≤ 2
x −2
2
if x is even,
Q.22 22225555 + 55552222
= 22225555 + 45555 + 55552222 – 42222 – 45555 + 42222
= (22225555 +45555) + (55552222 –42222) – 42222 (43333-1)
As, 22225555 +45555 is
divisible by 2222+4
i.e. 2226
= 7x318
and
55552222 –42222 is divisible by 5555-4
i.e. 5551
= 7x793
c4 h
3
also 43333 –1 =
=
1111−1
641111-1 is divisible by
64-1 = 63
=7x9
So, 22225555 + 55552222 can be split into 3 terms each divisible by 7.
Q.23 On 59th Second.
Q.24 1989 = 3x3
=
x 17x13
39 x 51
39 can be put in forum of squares.
9 squares
9 squares
9
12 sq.
As, 12 =
1
1
1
9
1
1
1
4
9
9
9
12
51 can be put in forum
4
7
9
31
7 =
31 =
Q.25 2178 x 4 = 8712
Q.26 39 99 991
= 1997 x 2003
So 39 99 991 is not prime.
Q. 27
Q. 28
Q. 29
Let side of square =a
So, AG =
3a
2
, GD=a
So, AD = AG+GD
 3 
+ 1a =
= 
2


But attitude of equilateral triangle =
[if x is side of equilateral triangle]
 3 
3
+ 1a =
x
So, 
2
2


3
2
⇒a =
x
3 +2
(
)
2
3x
2
So ratio of area = 25 3 − 48 .
Q.30 From the question :
Let AB=a, ∠CWB = θ
So,
CB
S in θ =
CW
a
=
5
2
a
So, RB =
5
RW =
So, ∆ RWB =
=
a
2
5
a
20
a2
20
So, are RBxQ = (area ∆CWB) – 2x area (∆CQX)
1 2 1 2
a − a
10
= 4
=
3 2
a
20
So area (PQRS) = area (ABCD) – 4 x area (RBxQ) – 4x (RWB)
a2 −
=
=
12 2
4a 2
a −
20
20
FG1 − 16 IJ a
H 20 K
2
=
1 2
a
5
1
So, ratio =
5
Q. 31
Answer are
(a)
56.277 cm
(b)
67 minutes and 9 seconds
Solution
(a) 3x(1+2+3+..+16) +2x (17) = 442 lines
i.e.
502 π d = 442000 [d is depth]
⇒ d = 56.277
(b)
for total fill up
required volume of water = 50 2 x π x 100
250 π litter
b g
3n n + 1
Here 3(1+2+3+…+n)
=
2
≤ 250 π
⇒ n = 22
So, This leaves
250 π − 3 x 22 x
23
liter
2
The water is now flowing at the rate of 23 liter per minute.

So, it will take  250 π − 3x 22 x

23 
 / 23
2 
= approximately 1 minute 9 seconds.
So total time = 67 minutes and 9 seconds
Q. 32 We have
32 = 9
8 = 23
and
9>8
⇒ 3 2 > 23
100
100
2I
3I
F
F
⇒ 3
H K > H2 K
⇒ 3 200 > 2300
Q 33.
210
=
1024
37
=
2187
2187 > 1024
⇒ 3 7 > 210
F I 4 > F 210 I 4
H K H K
⇒ 37
⇒ 328 > 2 40
1 1 1
− + −...
2 3 4
Q. 34
=
=
-
1
1
+
99 100
3−2
5−4
7−6
+
+
+...
6
20
42
1 1
1
+ + +...
6 20 42
Let us calculate
+
+
1
9900
1 1
+
6 20
=
10 + 3
60
=
13
60
=
12 + 1
60
=
1 1
+
5 60
>
1
5
Hence
1 1 1
− + −...
2 3 4
>
1
5
-
1
1
+
99 100
100 - 99
9900
Q. 35 The last digit for power of 3
Powers
Last two digit
31
3
32
9
33
27
34
81
35
43
36
29
37
87
38
31
39
93
310
79
911
37
312
11
313
33
314
99
315
97
316
91
317
73
318
19
319
57
320
71
321
13
322
39
323
17
324
51
325
53
326
59
327
47
328
41
329
23
330
69
331
07
332
21
333
63
334
89
335
67
336
01
337
03
338
09
339
27
& So on
For
3999 = 33x333
= 327x37
The last two digit
at 337 = 03
Same rotates
at 32x37 = 03
So last two digit = 03