Download 10th National ISMO Class 5 Question Paper With Solutions

PAKTURK 10th National Interschool Maths Olympiad
Q1:
A)
11 3
 ?
7 7
1
B)
Class 5 Solutions
Q4:
A)
2
C)
3
D)
10  10  10  10  ?
11
B)
111
C)
101
D)
20
4
Solution:
Solution:
10  10  10  10
11 3 11  3 14
 

2
7 7
7
7
 100  1
Answer: B
 101
Answer: C
Q2:
How many zeros are needed to write the
numerical for one million?
Q5:
A)
3
B)
4
C)
5
D)
6
A)
Solution:
1,000,000 is the numerical for one million.
So we need six zeros.
1 1 1
 
?
6 3 18
1
9
B)
1
3
C)
1
2
D)
2
Solution:
Answer: D
1 1 1
1
1
2
1
 




6 3 18 18 18 18 9
Answer: A
Q3:
Which of the following is the simplest fraction
form of 75 percent?
Q6:
A)
2
5
B)
1
4
C)
3
5
D)
What is the H.C.F of 480 and 516?
3
4
A)
4
B)
12
C)
16
D)
24
Solution:
Solution:
75% 
75
100
After simplification we get
480  25  3  5 and 516  22  3  43
The H.C.F
75
3

100 4
22  3  12
Answer: B
Answer: D
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Page 1 of 12
PAKTURK 10th National Interschool Maths Olympiad
Q7:
Class 5 Solutions
How many Rs. 23 balls can you buy for Rs.713?
Q10:
A)
25
B)
27
C)
29
What is 20% of
31
D)
A)
Solution:
70
B)
7
350 
1
?
7
C)
1
D)
10
Solution:
The total number of balls you can buy is
Rs.713
Rs.23
=31
Answer: D
1  20
1

 20%   350   
 350 
7  100
7

1 350 1
 
  50  10
5
7
5
Answer: D
Q8:
A)
48  12  3  ?
108
B)
96
C)
12
D)
15
Solution:
Q11: What is the quotient when
divided by 2014 ?
A)
2014
B)
1
C)
2014  2014 is
0
D)
2
48  12  3
 48  36
Solution:
 12
Answer: C
2014  2014
 2014
2014
Answer: A
Q9:
A)
0.2  0.4  0.8  ?
0.064
B)
0.64
C)
64
D)
640
Q12: Which one of the following fractions is the
largest one?
A)
Solution:
7
8
B)
13
16
C)
31
40
D)
63
80
 0.2  0.4  0.8
2 4 8
 
10 10 10
64

1000
 0.064
Solution:

Answer: A
7
 0.875
8
31
 0.775
40
13
 0.8125
16
63
 0.7875
80
It is clear that the largest one is
7
 0.875
8
Answer: A
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Page 2 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Q16:
Q13:
What is the average of first 5 multiples of 5?
B)
37.5
C)
15
D)
35
131
B)
139
in
C) 141
D)
143
Solution:
Solution:
2014 - 12 = 14 
5  10  15  20  25 75

 15
5
5
 2002  14 
Answer: C

2014 - 12 = 14 ?
A)
A) 7.5
What number will come instead of

2002
 
14
Answer: D
Q14: Which of the following numbers can be the
factors of a prime number?
A) 1 and 13
C) 1 and 15
Q17: The square of 4 has the same value as the
square root of a number.
B) 1,2 and 7
D) 1 and 21
Which of the following is the number?
Solution:
A prime number has only two factors and the first one
is 1 and the other one is itself.
A) 256
In option A:
1 and 13 are the only factors of 13. As
you see one is itself and the other one is 1.
Solution:
Answer: A
B) 32
C) 16
D) 64
42  ?
 16  ?
The question mark represent the square of 16 which is
equal to 256.
Q15:
What percent of 0.2 is equal to 0.4?
A) 400 %
B) 200 %
C) 100 %
Answer: A
D) 150 %
Q18:
Solution:
3  2014  2013  2012  ........
Find the missing number.
100 4 100 2
10
0.4 
 
  100 
0.2 10 2
5
2
10
2
  100  5  2  100  200
5
A)
2007
B)
2006
C)
2017
D)
2008
Solution:
Answer: B
3  2014  2013  2012  ........
 6042  2013  2012
 2017
Answer: C
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Page 3 of 12
PAKTURK 10th National Interschool Maths Olympiad
Q19: The price of a table is Rs. 1600 and the price of
a chair is Rs. 1200.
Class 5 Solutions
Q22: Adeel ate 5 slices of cake and Atif ate 7 slices. If
there were initially 15 slices of equal size, what fraction
of the cake was eaten?
How much money will be left if you buy 2 tables and 4
chairs with Rs. 9600?
A) Rs. 1200
C) Rs. 1600
A)
B) Rs. 1800
D) Rs. 1400
5
4
B)
3
4
C)
3
5
4
5
D)
Solution:
Solution:
If you buy two tables, it will cost Rs. 32000 and if you
buy 4 chairs that will cost Rs. 4800
The total slices of cake were eaten is by Atif and Adeel
is 7+5=12.
The fraction will be
Total cost will be Rs. 3200 + Rs. 4800= Rs. 8000
12 4

15 5
Answer: D
The money that will be left is:
Rs. 96000-Rs. 8000=Rs. 1600
Answer: C
Q23:
A)
Which of the following is equal to 4.25?
5
2
B)
7
4
C)
19
3
D)
17
4
Q20: What is the highest common factor of 24 and a
number that is a multiple of 24?
Solution:
A)
48
B)
36
C)
24
D)
12
Solution:
425
425
25  17

4.25 
100 100
4
25
The multiples of 24 are 24, 48, 72, ....
Answer: D
As you see above the highest common factor of 24 and
any of its multiple is equal to 24.
Answer: C
Q24: Which of the following number should be
multiplied by 3.125 to get a natural number?
A) 4
Q21:
B) 5
C) 6
D) 8
The price of a book is Rs. 2400.
Solution:
What percent of the price of the book can Usman pay
with Rs. 1800?
A)
75%
B)
60%
C)
80%
D)
72%
Solution:
3125
125
3125
125
25
5  25
3.125 



40
1000 1000
40
8
25
5
So if we multiply the final fraction by 8, it will be a
natural number.
Rs.1800 1800 3

  75%
Rs. 2400 2400 4
Answer: D
Answer: A
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Page 4 of 12
PAKTURK 10th National Interschool Maths Olympiad
Q25: ISMO is a four digit number and
O are 4 digits.
I , S , M and
A) 18
Q28:
What is the least common multiple (L.C.M) of
denominators of
2014  2013  2012  ISMO
What is the value of
Class 5 Solutions
A) 280
1 1 1
1
, ,
, and ?
3 5 4
7
B) 420
C) 105
D) 390
I+S+M+O ?
B) 14
C) 8
D) 24
Solution:
Solution:
As any two of the denominators has no common factor,
then the L.C.M will be the product of all.
2014  2013  2012  ISMO
3  5  4  7  420
 ISMO=6039
Answer: B
 I=6, S=0, M=3, O=9
 I+S+M+O  6  0  3  9  18
Answer: A
Q29: Naseera and Minal have 64 toffees in total, but
Minal has 12 more toffees than Naseera.
How many toffees does Naseera have?
Q26:
Round off 283 to the nearest ten.
A) 276
B) 278
C) 280
D) 290
Solution:
If we keep 12 toffees for Minal and share the remaining
toffes among Minal and Naseera, they will have equal
number of toffees.
Solution:
We will round off the last digit only. As the last digit is
3, it will be round off to 0 but as 3 is less than 5, 8 will
remain same.
The toffes which we kept for Minal makes the number
of toffes of Minal's 12 more than the number of toffees
of Naseera's.
After rounding off, the number will be 280
Answer: C
 64  12  52
52
 26
2
Now both have 26 toffees.
Q27: What is the least common multiple (L.C.M) of
first 5 prime numbers?
A) 44
B) 2
C) 2310
Naseera has 26 toffees and Minal has 38 toffees.
Answer: B
D) 4110
A) 40
B) 26
C) 44
D) 28
Solution:
The first prime numbers are 2,3,5,7,11. As they do not
have any common, we can multiply all to find the L.C.M.
That will be
2  3  5  7  11  2310
Answer: C
We believe what is taught with love lasts forever
Page 5 of 12
PAKTURK 10th National Interschool Maths Olympiad
Q30:
A father bought 22 toys and paid Rs. 1100.
Class 5 Solutions
Q33: A student in class eight scored 72 marks in
maths test. How much should she score in order to
increase her average to 80?
What is the price of one toy?
A) 88
A) Rs. 50
B) Rs. 60
C) Rs. 75
B) 90
C) 80
D) 84
D) Rs. 80
Solution:
Solution:
He should get 88 in order to increase her average to 80.
The price of one toy is
1100
 50
22
Answer: A
72  88 160

 80
2
2
She scored 8 less than 80, now she need to get 8 more
than 80.
Answer: A
Q31: Which one of the following fractions has 7 in
denominator when it is reduced to its lowest form?
A)
35
28
B)
28
35
C)
3
14
D)
60
84
Q34: The sum of any two adjacent numbers is equal
to the number on them.
Solution:
For example: a+b=c
60
60
12 5


84 84
7
12
5
is the lowest form and it has 7 in its denominator.
7
Answer: D
Q32:
What is the value of "a" if the sum of numbers in 3rd
line is equal to 17?
What is 20% of 40% of 48000?
A) 3972
B) 3840
C) 3800
D) 3755
A) 4
B) 5
C) 6
D) 7
Solution:
20%  40%  48000
20
40

 48 000
1 00 1 00
 80  48  3840

Answer: B
We believe what is taught with love lasts forever
Page 6 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Q36: A boy selected five numbers as 78, 66, 64, 96,86
and noticed that one of the number is the average of
other four numbers.
Solution:
Sum of the number is 3rd line is 17. So
a  b  4  17  a  b  13
What is that number?
a  b  c  13
A) 64
B) 66
C) 78
D) 86
 c  13
Solution:
Arrange numbers in ascending order as 64,66,78,86,96
If one of the number is the average of remaining
numbers, the average won't change when the average
number is also included in the numbers.
So, that number will be
c  b  4  25
Let's check:
13  b  4  25
64  66  78  86  96
 78
5
64  66  86  96
 78
4
b  25  17
Answer: C
b8
Now we can find the value of a
a  b  13
Q37: ABCD is a parallelogram and the length of AE is
half of the length of CD.
a  8  13
a5
Answer: B
Q35: The price of an item is Rs. 1200. The
shopkeeper wants to sell 5 items for Rs. 9000. How
much should the shopkeeper increase the price of the
item?
A) Rs. 600
C) Rs. 800
B) Rs. 700
D) Rs. 900
What is area of parallelogram if the length of AB is 12
cm?
2
2
A) 96 cm
2
C) 64 cm
B) 72 cm
2
D) 84 cm
Solution:
Solution:
If he sells 5 items for Rs. 1200, he will get Rs. 6000
The area of parallelogram is height times base.
The line segment AE is the height which is half length
of CD. The lengths of opposite sides in a parallelogram
He should sell all the 5 items for
Rs.9000
=Rs.1800
5
are equal. ( CD 
AB )
to get Rs. 90000
1
1
AE  CD  12cm=6cm
2
2
So he needs to increase the price of an item from Rs.
1200 to Rs. 1800 to get Rs. 9000
Area of the parallelogram: AE  CD  6 cm  12 cm  72 cm2
Answer: B
Incensement is Rs. 1800 - Rs. 1200 = Rs. 600
Answer: A
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Q38: Which four operation sign should be used in
order to get 10?
Page 7 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Solution:
7 square on the left, 7 on the rigt, 7 on the bottom and
7x7=49 in middle. Total number of squares will be 70.
A)
C)
, , , 
, , , 
B)
D)
, , , 
, , , 
Solution:
7  5  12
See the figure above.
12  4  3
Answer: B
3  8  11
11  1  10
So the sign should be used are
12600 is found
wrong by a student as 2  3  5  7 .
Q40:
, , , 
The prime factorization of
3
Answer: D
2
What is the missing number in above prime
factorization?
A) 2
B) 3
C) 5
D) 7
Solution:
Let's find the correct prime factorization of 12600.
Q39:
How many squares will be used in 7th figure?
1st Figure
2nd Figure
12600  23  32  52  7
As you see above in the correct factorization, one 5 is
missing in the wrong factorization.
3rd Figure
Answer: C
Q41:
A) 40
B) 70
C) 80
How many squares are there in the figure?
D) 100
A) 15
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B) 17
C) 19
D) 22
Page 8 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
A number multiplied by itself, added 1, the
result multiplied by 10, added 3, and the result
multiplied by 4. What is the number if the final
answer is 2012?
Q43:
Solution:
A) 9
B) 8
C) 7
D) 6
Solution:
Let's find the number by starting with 2012 and apply
reverse operations.
Answer: D
2012
 503
4
503  3  500
500
 50
10
50  1  49
Mohseen has 1600 marbles. Half of them
are blue, one quarter are red, and one eighth are
green.
Q42:
49  7  7
So the number is
7
What percent of the marbles are of some other
colour?
A) %10
B) %12
C) %12.5
Answer: C
D) %15
Solution:
Q44: The average of twelve numbers is 45. The
average of ten numbers when two numbers removed is
48.
1600
 800
2
1600
Red:
 400
4
1600
Green:
 200
8
Blue:
What is the average of those two numbers which are
removed?
A) 30
B) 32
C) 44
D) 48
Solution:
The number of marbles in other colour:
1600   800  400  200  200
The sum of twelve numbers is
The sum of ten numbers is
12  45  540
10  48  480
200
Percentage will be
16  12.5  12.5%
1600
100
16
Answer: C
That means the sum of those two numbers which are
removed is 540  480  60
Average of the two number is
60
 30
2
Answer: A
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Page 9 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Q46:
Q45: In the following diagram AF=BF, BG=GC,
AK=KE and BL=LG.
ABCD is a rectangle and EFGH is a square.
What is the total area of shaded regions?
What is the perimeter of shaded region?
A) 58 cm
C) 62 cm
2
A) 100 cm
2
C) 108 cm
B) 60 cm
D) 64 cm
2
B) 112 cm
2
D) 148 cm
Solution:
Solution:
Lets find the all shaded region by finding the area of
four parts.
We have two rectangles and two triangles. The lengths
of sides are mentioned by red text.
The area of the first rectangle is 4 cm  8 cm  32 cm
The area of the second rectangle is
2
4 cm  8 cm  32 cm2
The area of first triangle is
8 cm  8 cm
 32 cm 2
2
4 cm  8 cm
The area of second triangle is
 16 cm 2
2
The perimeter of the shaded regions is the lengths
which are shown by red colour.
The sum of the lengths of two red part is 6 cm. (
AB  FG )
The perimeter is
10cm+16cm  10cm  6cm  6cm  10cm  6cm=64 cm
Answer: D
The total area of the shaded region is
32 cm2  32 cm2  32 cm2  16 cm2  112 cm2
Answer: B
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Page 10 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Q47: The rent of a building increases in a sequence
as follow.
Year
2012
2013
2014
2015
Q49: The tank of a car has 23.4 litre petrol when the
pointer (arrow) is as in the figure. The tank is empty
when the pointer shows 0. What is the capacity of the
tank?
Rent
Rs. 12000
Rs. 15000
Rs. 19000
Rs. 24000
What will be the rent in 2023?
A) Rs. 90000
C) Rs. 105000
B) Rs. 100000
D) Rs. 120000
Solution:
A) 62
B) 62.4
C) 64.8
D) 72.4
Solution:
There are total 8 equal parts from 0 to 1 and the
pointers at the end of third part.
The capacity of three parts is 23.4.
It means that the capacity of one part is
The incensement in 2013 is Rs. 3000, in 2014 is Rs.
4000, in 2015 is Rs. 5000. The incensement increases
by Rs. 1000 every year.
23.4
 7.8
3
There are total five parts, so the capacity of the tank is
8  7.8  62.4
Answer: B
So the rent will be Rs. 100000 in 2023
Answer: B
Q48: What is the sum of digits of
according the rule below?
111111  111111
Q50: In how many ways the ant can rich the wheat by
moving up and right only?
11  1
11  11  121
111  111  12321
1111  1111  1234321
A) 12
B) 18
C) 24
D) 36
Solution:
According the rule above
equal to 12345654321
Sum of the digits will be
111111  111111 will be
A) 11
1  2  3  4  5  6  5  4  3  2  1  36
B) 12
C) 13
D) 14
Answer: D
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Page 11 of 12
PAKTURK 10th National Interschool Maths Olympiad
Class 5 Solutions
Solution:
There are total 14 different ways. See the figure below.
Answer: D
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Page 12 of 12