Q1. The largest natural number, by which the
product of three consecutive even numbers is always divisible, is _____.
Solution. First of all list down different
set of three consecutive even numbers
(2, 4, 6)
(4, 6, 8) (6, 8, 10) (8, 10, 12) etc
Product
of numbers in first set = 2 × 4 × 6
Product
of numbers in second set = 4 × 6 × 8
Product
of numbers in third set = 6 × 8 × 10
Product
of numbers in fourth set = 8 × 10 × 12
NOTE: The product of three consecutive
natural numbers is always divisible by 6.
Now, the
largest natural number which divides the product is 12.
As each
set can written as
2 × 4 × 6
= 2 × {1 × 2 × 3}
4 × 6 × 8
= 2 × {2 × 3 × 4}
The
product of numbers given in brackets will always be divisible by 6.
Therefore,
when 2 is multiplied by 6, the product will be divided by 12.
(Hell of
a question for a Class V student, no matter how much brilliant he/she is!)
Q2. To the product of smallest twin prime
numbers, add the HCF of two co-primes. The difference of the face value and the
place value of the digit in Tens place of the sum, is _________.
Solution. Twin prime
numbers: A pair of prime numbers which
differ by 2. The smallest twin prime numbers are (3, 5) etc.
CO-prime: Two integers a and b are said to be coprime if
the only positive integer that evenly divides both of them is 1.
Product
of smallest twin prime numbers = 3 × 5 = 15
HCF of
two co-prime numbers = 1
After
sum, the result = 15 + 1 = 16
Place
value of digit in Tens place of sum = 10
Face
value of digit in Tens place of sum = 1
Difference
= 10 ~ 1 = 9
Q3. 0.2% of a number is 0.3, the number is
________.
Solution. 0.2% of a number = 0.3
(Observe
carefully there is only single after decimal on both sides)
Þ
2% of a number = 3
Þ
100% of a number = 150 (multiply both sides by 50)
(Remember
the rule that every quantity is 100% in itself)
Q4. In a school, there are four boys to every
three girls. If there are 304 boys, the number of girls in the school is ________.
Solution. 4 Boys for every 3 girls
Let’s
understand this as
The above
figure represents a group of 4 boys for every 3 girls.
Now, to
count 304 boys how many group of such 4 boys will be required = 304 ÷ 4 = 76
In each
of these 76 groups, there will be 3 girls.
So, total
number of girls = 76 × 3 = 228
Q5. A snack bar sells five items with an
average price of Rs. 0.60. Which two items from the following items can be
added to the menu without changing the average price?
Solution. Average price changes with number
of items. If we do not want to change it, then we should try keeping the
average price constant.
So, the
average price of each item is Rs. 0.60. Therefore, for two items it will become
0.60 + 0.60 = Rs. 1.20.
Now,
check for options only the price of Cake
slice and Chewing Gum together
makes Rs. 1.20 possible.
Q6. If 40 nails are used in making one shoe,
the number of nails needed to make 20 pairs of shoes are _______
Solution. Total number of shoes in 20 pairs
= 2 × 20 = 40 shoes
One shoe
requires = 40 nails
40 shoes
will require = 40 × 40 = 1600 nails
Q7. Students of a class took a maths test, 1/3
of the class got B grade, ¼ got B+ grade, 1/6 got C grade and 1/8 failed. The
remaining students got A grade. The number of students who got an A grade is
________. (Number of students in the class is less than 30)
Solution. Total number of students in class
= Students who got A grade + students who got B grade + students who got (B+)
grade + Students who got C grade + student who failed
= Students
who got A grade + 1/3 + ¼ + 1/6 + 1/8
= student
who got A grade + 7/8
Students
who got A grade will be 1/8.
Q8. Manya is 5 year 8 months old. Her sister
Sanya is three-quarters her age. When Manya is 9 year 8 months old, the age of
her sister would be _______.
Solution. Age of Manya = 5 yrs 8 months or
68 months
Age of
her sister Sanya = ¾ × 68 = 51 months
Now,
Manya will be 9 yrs 8 months or 116 months old.
Manya’s
age increased by = 116 months – 68 months = 48 months
Therefore,
Manya’s sister will also become older by 48 months
Sanya age
= 51 + 48 = 109 months or 9 yrs 1 month.
Q9. Half of a pie is divided into 3 equal
pieces. Each piece is _______of the whole pie.
Q10. In a division sum, the divisor is 12 times
the quotient and 5 times the remainder. If the remainder is 48, then the
dividend is _________.
Solution. Divisor is 5 times the remainder.
Þ Divisor = 48 × 5 = 240
Divisor
is 12 times the quotient.
Þ Quotient = 240 ÷ 12 = 20
Dividend
= Divisor × Quotient + Remainder
= 240 ×
20 + 48 = 4800 + 48 = 4848
Q11. Bananas cost Rs. 36 per dozen, and an apple
costs Rs. 7 each. A plate of fruit salad consists of 3 bananas and 1/4th
of an apple. Find the cost of 12 such plates.
Solution. Cost of each banana = 36 ÷ 12 =
Rs. 3
Cost of
each apple = Rs. 7
Cost of a
plate consisting of 3 banana and 1/4th apple
= 3 × 3 +
7 × ¼ = 9 + 7/4 = 43/4
Cost of
12 such plates = 12 × 43/4 = 3 × 43 = 129
Q12. 4 boxes contain a total of 96
sweets. If each sweet costs 45p, the cost of each box of sweet is ________.
Solution. Number of sweets per box = 96 ÷ 4
= 24
So, each
box contains 24 sweets.
Cost of
each box = 24 × 0.45 = Rs. 10.80
Q13.
Solution. The pattern to find middle number
is:
58 = 9 ×
8 – 10 – 4
Similarly,
15 × 8 – 10 – 9 = 101
(In such
type of questions generally arrange given digits using four basic mathematical
operators i.e. addition, subtraction, multiplication and subtraction)
Q14. The least number which when
divided by 5, 6, 7 and 8 leaves remainder 3, but when divided by 9 leaves no
remainder is _______.
Solution. Least number divided by 5, 6, 7
and 8 = LCM of (5, 6, 7 and 8)
= 840
But they
leave a remainder of 3. Thus, actual number which when divided by 5, 6, 7 and 8
leaving remainder 3 in each case will be 840 + 3 = 843.
843 when
divided 9 leave a remainder of 6.
Series of
such numbers will be (1 × 843), (2 × 843), (3 × 843) etc
Out of
these numbers, 3 × 843 is exactly divisible by 9 and leaves a remainder of 3
with 5, 6, 7, and 8.
Solution. Size of each paper = ¾ m
Total
length of the paper = 17(1/2) m or 35/2 m.
Number of
pieces cut = Total length of the paper ÷ Size of each paper
= 35/2 ÷
¾ = 70/3 or 23(1/3)
This
shows that exactly 23 pieces can be cut and some paper would be left.
Total
length paper used in 23 pieces = 23 × ¾ = 69/4 m
Paper
left = 35/2 – 69/4 = ¼ m
Q16. Puneet cut 85 cm of Brass
from one end of a piece of Brass rod 2.3 m long. What decimal fraction of 5 m
is the length of the remaining piece?
Solution. Length of the remaining piece of
rod = 2.3 m – 0.85 m = 1.45 m
Required
decimal fraction = 1.45/ 5 = 0.29
Q17. Jay bought 3 packets of pins,
2 packets of tattoos, and 1 packet of glitter for Rs. 19.10. Each packet of pin
costs Rs. 1.90 and the packet of glitter costs twice as much as a packet of
tattoos. The packet of tattoos costs him _________.
Solution. Total cost = 3 packets of pins + 2
packets of tattoos + 1 packet of glitter
19.10 = 3
× 1.90 + 2 packets of tattoos + 2 packets of tattoos
(Since, a
packet of glitter costs twice the packet of tattoos)
Cost of 4
packets of tattoos = 19.10 – 5.70 = Rs. 13.40
Cost of a
packet of tattoo = 13.40 ÷ 4 = Rs. 3.35
Q18. Using the digits 1, 2, 3, 4,
5, 6, 7, 8, 9 each exactly once, write any three 3-digit numbers so that the
second number is twice the first and third number is thrice the first number.
1) __________ 2) __________ 3) ___________
Solution. Highly time consuming question.
Even not at par with the knowledge level of a class V student.
Q19. In a party, 52% of the guests
liked the decorations and 40% liked the food, 27% liked both. The percentage of
guests who did not like the food as well as the decoration is __________.
Solution. Answer to this question will be
(52 + 40 – 27 = 35%). But to teach this concept to a class V student is a
tedious task. So, I am leaving the solution part of the question.
Q20. A super fast running at the
speed of 94 km/hr develops a snag and stops. It had covered (1/17)th
of the total distance in the 4 hors it had been running. The distance yet to be
covered is ___________.
Solution. In 4 hours the train has traveled
some distance. Let’s calculate it first.
In 1
hour, train covers = 94 km
In 4
hours, train covers = 4 × 94 = 376 km
Now, this
376 km is (1/17)th part of the total distance to be covered.
Therefore,
the total distance will be = 376 × 17 =6392 km
CDI =
(500 – 100) + 1 = 401
XLIV =
(50 – 10) + (5 – 1) = 44
XVIII
CMXII = 10 + 8 + (1000 – 100) + 10 + 2 = 930
CXCVII =
100 + (100 – 10) + 7 = 197
Þ CDI + XLIV × XVIII CMXII ÷ CXCVII
= 401 +
44 × 930 ÷ 197
Q22. The average weight of 6
students increase by 3 kg when one of the students whose weight is 58 kg is
replaced by a new student. The weight of the new student is
Solution. As Class V students are not
exposed to ‘Average’, answering this question to explain them will be bit
difficult.
Q23. Mohit had 390 marbles. He
shared it with 8 boys and 13 girls so that each boy got 2 marbles more than
each girl. The number of marbles left with him is _______.
Solution. Each boy got 2 marbles more than
each girl. As there are 8 boys, then in all 8 × 2 = 16 extra marbles are distributed
between boys and girls beside the number of equally divided marbles.
Number of
marbles left to be divided equally between boys and girls = 390 – 16 = 374
Now,
these 374 marbles are divided between boys and girls in such a way that they
get equal number of marbles.
Total
number of person = 8 boys + 13 girls = 21
Number of
marbles each of them got = 374 ÷ 21 = 17 quotient and 17 remainder.
This
implies that each boy and girl got 17 marbles.
Total
number of marbles distributed = 17 × 21 = 357
Number of
marbles left to be distributed = 374 – 357 = 17
CDI =
(500 – 100) + 1 = 401
XLIV =
(50 – 10) + (5 – 1) = 44
XVIII
CMXII = 10 + 8 + (1000 – 100) + 10 + 2 = 930
CXCVII =
100 + (100 – 10) + 7 = 197
Þ CDI + XLIV × XVIII CMXII ÷ CXCVII
= 401 +
44 × 930 ÷ 197
Q22. The average weight of 6
students increase by 3 kg when one of the students whose weight is 58 kg is
replaced by a new student. The weight of the new student is
Solution. As Class V students are not
exposed to ‘Average’, answering this question to explain them will be bit
difficult.
Q23. Mohit had 390 marbles. He
shared it with 8 boys and 13 girls so that each boy got 2 marbles more than
each girl. The number of marbles left with him is _______.
Solution. Each boy got 2 marbles more than
each girl. As there are 8 boys, then in all 8 × 2 = 16 extra marbles are distributed
between boys and girls beside the number of equally divided marbles.
Number of
marbles left to be divided equally between boys and girls = 390 – 16 = 374
Now,
these 374 marbles are divided between boys and girls in such a way that they
get equal number of marbles.
Total
number of person = 8 boys + 13 girls = 21
Number of
marbles each of them got = 374 ÷ 21 = 17 quotient and 17 remainder.
This
implies that each boy and girl got 17 marbles.
Total
number of marbles distributed = 17 × 21 = 357
Number of
marbles left to be distributed = 374 – 357 = 17
Solution. i) To find the bus which takes
least time, we have to find the time taken by bus after departing from Noodle
Pond till it arrives at Lake
Cola.
Time
taken by Bus 1 = 02:35 – 20:10 = 6:25 hours
Time
taken by Bus 2 = 16:35 – 11:25 = 5:10 hours
Time
taken by Bus 3 = 13:10 – 08:10 = 5:00 hours
Thus, we
can conclude that Bus 3 takes the least time to reach Lake Cola
from Noodle Pond.
ii) In
the same manner, we will find which bus takes least time (to be fast) to arrive
Salad Town after departing from Burger
Junction.
Time
taken by Bus 1 = 01:05 – 21:55 = 03:10 hours
Time
taken by Bus 2 = 15:30 – 12:20 = 02:50 hours
Time
taken by Bus 3 = 11:15 – 09:05 = 02:10 hours
Thus, we
can conclude that Bus 3 takes least time. Hence, it is fastest.
iii) As
the person has missed Bus 2 at Chocopur, he has to wait till the next Bus
arrives. It implies that we have found the time difference between the
departures from Chocopur to arrival at Salad Town
for different buses.
Time
difference for Bus 1 = 01:05 – 23:45 = 01:20 hours
Time
difference for Bus 2 = 15:30 – 14:05 = 01:25 hours
Time
difference for Bus 3 = 11:15 – 10:55 = 20 minutes
iv) TO
find which is the fastest Bus from Burger Junction to Chocopur, we have to find
which Bus takes the least time.
Time
difference from Bus 1 = 23:20 – 21:55 = 01:25 hours
Time
difference from Bus 2 = 13:50 – 12:20 = 01:30 hours
Time
difference from Bus 3 = 10:40 – 09:05 = 01:35 hours
Thus, it
is clear that Bus 1 takes the least time and hence, the fastest.
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