Question 13 Marks
A vessel has $13L$ $200mL$ of fruit juice. In how many glasses each of capacity $60mL$, can it be filled?
AnswerWe have, total fruit juice $=13L 200mL= 13200mL [1L = 1000mL]$ and capacity of $1$ glass $= 60mL$
$\therefore$ Number of glasses that can be filled $=\frac{\text{Total fruit juice}}{\text{Capacity of 1 glass}}$
$=\frac{13200}{60}$ $=220$
Hence, $220$ glasses each of 60mL can be filled with $13L$ $200mL$ of fruit juice.
View full question & answer→Question 23 Marks
Three brands $A, B$ and $C$ of biscuits are available in packets of $12, 15$ and $21$ biscuits, respectively. If a shopkeeper wants to buy an equal number of biscuits of each brand, what is the minimum number of packets of each brand, he should buy?
AnswerHere, The minimum number of packets of each brand that the shopkeeper should buy, is equal to the $LCM$ of $12,15$ and $21.$
So, $LCM$ of $12, 15$ and $21$ by division method.
$\begin{array}{c|c}2&12,15,21\\ \hline2&6,15,21\\ \hline3&3,15,21\\ \hline5&1,5,7\\ \hline7&1,1,7\\ \hline&1,1,1\end{array}$
$LCM$ of $12, 15$ and $21 = 2^2\times 3 \times 5 \times 7 = 420$
Now, minimum number of packets of brand $A$ $=\frac{420}{12}=35$
Minimum number of packets of brand $S$ $=\frac{420}{15}=28$
Minimum number of packets of brand $C$ $=\frac{420}{21}=20$
View full question & answer→Question 33 Marks
In a five-digit number, digit at ten’s place is $4$, digit at unit’s place is one-fourth of ten’s place digit, digit at hundred’s place is $0$, digit at thousand’s place is $5$ times of the digit at unit’s place and ten thousand’s place digit is double the digit at ten’s place. Write the number.
AnswerA number consists of $5$ digits.
Now, the digit at ten’s place $= 4$,
the digit at unit’s place the digit at hundred’s place $= 0$,
the digit at thousand’s place $= 5 \times 1 = 5$
the digit at ten thousand’s place $= 2 \times 4 = 8$
Therefore, the number is $85041.$
View full question & answer→Question 43 Marks
In the marriage of her daughter, Leela spent $Rs. 216766$ on food and decoration, $Rs. 122322$ on jewellery, $Rs. 88234$ on furniture and $Rs. 26780$ on kitchen items. Find the total amount spent by her on the above items.
AnswerGiven, amount spent on food and decoration $= Rs. 216766$
Amount spent on jewellery $= Rs. 122322$
Amount spent on furniture $= Rs. 88234$ and amount spent on kitchen items $= Rs. 26780$
Total amount spent $= 216766 + 122322 + 88234 + 26780 = Rs. 454102$
Hence, the total amount spent by Leela on her daughter’s marriage is $Rs. 454102.$
View full question & answer→Question 53 Marks
Find the $LCM$ of $160, 170$ and $90.$
Answer$LCM$ of $160, 170$ and $90$ by division method.
$\begin{array}{c|c}2&160,170,90\\ \hline 2&80,85,45\\ \hline2&40,85,45\\ \hline2&20,85,45\\ \hline2&10,85,45\\ \hline3&5,85,45\\ \hline3&5,85,15\\ \hline5&5,85,5\\ \hline17&1,17,1\\ \hline&1,1,1\end{array}$
$LCM$ of $160,170$ and $90 = 25 \times 32 \times 5 \times 17 = 24480$
View full question & answer→Question 63 Marks
Determine the least number which when divided by $3, 4$ and $5$, leaves remainder $2$ in each case.
AnswerFirstly, we have to find the $LCM$ of $3, 4$ and $5$.
$\begin{array}{c|c}2&3,4,5\\ \hline2&3,2,5\\ \hline3&3,1,5\\ \hline3&1,1,5\\ \hline&1,1,1\end{array}$
$LCM$ of $3, 4$ and $5 = 2 \times 2 \times 3 \times 5 = 60$
Now, required number $= LCM$ of $3, 4$ and $5$ + Remainder $= 60 + 2 = 62$
Hence, the required number is $62.$
View full question & answer→Question 73 Marks
India’s population has been steadily increasing from $439$ million in $1961$ to $1028$ million in $2001$. Find the total increase in population from $1961$ to $2001$. Write the increase in population in Indian System of Numeration, using commas suitably.
AnswerGiven, population of India in $1961 = 439$ million.
$= 439 \times 1000000 = 439000000 [ 1$ million $= 1000000]$
population of India in $2001 = 1028$ million. $= 1028 \times 1000000 = 1028000000 [ 1$ million $= 1000000]$
Total increase in population from $1961$ to $2001 =$ (Population in $2001$ - Population in $1961$)
$= 1028000000 - 439000000 = 589000000 $
$= 589 \times 1000000$
$ = 589$ million
So, the increase population in Indian System of Numeration $= 58,90,00,000$
View full question & answer→Question 83 Marks
Using each of the digits $1, 2, 3$ and $4$ only once, determine the smallest $4$-digit number divisible by $4$.
AnswerGiven digits are: $1,2,3$ and $4$. For smallest number, we will arrange the given digits in ascending order.
But the number should also be divisible by $4$.
so we will arrange the digits in such a way that it’s last two digits should be divisible by $4$. Required number $= 1324$
View full question & answer→Question 93 Marks
Find the $LCM$ of $80, 96, 125$ and $160.$
Answer$LCM$ of $80, 96,125$ and $160$ by division method.
$\begin{array}{c|c} 2 & 80,96,125,160 \\ \hline 2& 40,48,125,80\\ \hline2&20,24,125,40\\ \hline2&10,12,125,20\\ \hline2&5,6,125,10\\ \hline3&5,3,125,5\\ \hline5&5,1,125,5\\ \hline5&1,1,25,1\\ \hline5&1,1,5,1\\ \hline&1,1,1,1 \end{array}$
The $LCM$ of $80, 96, 125$ and $160$ is:$ = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 5 = 12000$
View full question & answer→Question 103 Marks
In a school library, there are $780$ books of English and $364$ books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves, such that each shelf should have the same number of books of each subject. What should be the minimum number of books in each shelf?
AnswerGiven, number of English books $= 780$ and Number of Science books $= 364$ To find the minimum number of books in each shelf, we have to find the $HCF$ of $780$ and $364.$
New, $HCF$ of $780$ and $364 = 52$
Hence, the minimum number of books in each shelf $= HCF$ of $780$ and $364 = 52.$ View full question & answer→Question 113 Marks
Using divisibility test, determine which of the following numbers are divisible by $4$?
$a.\ 4096$
$b.\ 21084$
$c.\ 31795012$
AnswerWe know that, the number is divisible by $4,$ if its last two digits or digits at ten’s and one’s place is divisible by $4.$
$a.\ $We have, $4096 .$
$4096$ is divisible by $4$.
Since, its last two digits i.e. $96$ is divisible by $4.$
$b.\ $We have, $21084$
$21084$ is divisible by $4.$
Since, its last two digits i.e. $84$ is divisible by $4.$
$c.\ $We have, $31795012$
$31795012$ is divisible by $4.$
Since, its last two digits i.e. $12$ is divisible by $4.$
View full question & answer→Question 123 Marks
Using divisibility test, determine which of the following numbers are divisible by $9$?
$a.\ 672$
$b.\ 5652$
AnswerWe know that, a number is divisible by $9$, if the sum of its digits is divisible by $9$.
$a.\ $We have, $672$
The sum of digits =$ 6 + 7 + 2 = 15$
which is not divisible by $9.$
$b.\ $We have, $5652$
The sum of digits $= 5 + 6 + 5 + 2 = 18$
which is divisible by $9$.
View full question & answer→Question 133 Marks
A loading tempo can carry $482$ boxes of biscuits weighing $15\ kg$ each. Whereas a van can carry $518$ boxes each of the same weight. Find the total weight that can be carried by both the vehicles.
AnswerGiven, number of boxes in a loading tempo $= 482$ Weight of each box $ = 15\ kg$
Weight of $482$ boxes $= 482 × 15 = 7230kg$
and number of boxes in a van $= 518$
Weight of each box $= 15kg$
Weight of $518$ boxes $= 518 \times 15 = 7770kg$
Now, total weight $= 7230 + 7770 = 15000kg$
Hence, the total weight that can be carried by both the vehicles is $15000\ kg.$
View full question & answer→Question 143 Marks
Write the following in expanded form.
$a.\ 74836$
$b.\ 74021$
$c.\ 8907010$
AnswerExpanded form of given numbers are:
$a.\ 74836 = 7 \times 10000 + 4 \times 1000 + 8 \times 100 + 3 \times 10 + 6 \times 1$
$b.\ 574021 = 5 \times 100000 + 7 \times 10000 + 4 \times 1000 + 0 \times 100 + 2 \times 10 + 1 \times 1$
$c.\ 8907010 = 8 \times 1000000 + 9 \times 100000 + 0 \times 10000 + 7 \times 1000 + 0 \times 100 + 1 \times 10 + 0 \times 1$
View full question & answer→Question 153 Marks
A box contains $5$ strips having $12$ capsules of 500mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.
AnswerHere,
number of strips in each box $= 5$
Number of capsules in $1 $strip $= 12$
Weight of medicine in each capsule $= 500mg$
So, total weight of medicine in one box $= 5 \times 12 \times 500$
$= 30000mg$
Now, total weight of medicine in $32$ such boxes
$= 32 \times 30000$
$= 960000mg$
View full question & answer→Question 163 Marks
Chinmay had $Rs. 610000$. He gave $Rs. 87500$ to Jyoti, $Rs. 126380 $to Javed and $Rs. 350000$ to John. How much money was left with him?
AnswerGiven,
Chinmay’s total money $= Rs. 610000$
Money given to Jyoti by Chinmay $= Rs. 87500$
Money given to Javed by Chinmay $= 1126380$
and money given to John by Chinmay $= Rs. 350000$
Money left with Chinmay = Total money - Distributed money
$= 610000 - (87500 + 126380 + 350000)$
$= 610000 - 563880 = Rs. 46120$
Hence, $Rs. 46120$ was left with him.
View full question & answer→Question 173 Marks
Determine the sum of the four numbers as given below:
$a.\ $Successor of $32$
$b.\ $Predecessor of 49
$c.\ $Predecessor of the predecessor of $56$
$d.\ $Successor of successor of $67$
AnswerWe know that, for getting the successor of a number, we add $1$ in that number and for getting the predecessor, we subtract $1$ from that number.
$a.\ $Successor of $32 = 32 + 1 = 33$
$b.\ $Predecessor of $49 = 49 - 1 = 48$
$c.\ $Predecessor of the predecessor of $56 = 56 - 1 - 1 = 54$
$d.\ $Successor ofsuccessor of $67 = 67 + 1 + 1 = 69$
View full question & answer→Question 183 Marks
A factory has a container filled with $35874 $litres of cold drink. In how many bottles of $200mL$ capacity each, can it be filled?
AnswerGiven,
total cold drink in the container $= 35874$ litres
$= 35874000mL [1$ litres $= 1000ml]$
and capacity of one bottle $= 200ml$
Now, number of bottles required = $=\frac{\text{Total cold drink in the container}}{\text{Capacity of one bottle }}$
$=\frac{35874000\text{ml}}{200\text{ml}}$
$=179370\text{ml}$
Hence, $179370$ bottles of $200ml$ are required to fill $35874$ litres of cold drink.
View full question & answer→Question 193 Marks
Radius of the Earth is $6400\ km$ and that of Mars is $4300000m$. Whose radius is bigger and by how much?
AnswerGiven, radius of the Earth $= 6400km$
$= 6400000m [1km = 1000m]$
radius of Mars $= 4300000m [1km = 1000m]$
On comparing both the radius, we get
Radius of Earth > Radius of Mars
Difference between the two radius $= 6400000 - 4300000 = 210000$0m
Hence, the radius of Earth is bigger and by $2100000m.$
View full question & answer→Question 203 Marks
A mobile number consists of ten digits. First four digits are $9, 9, 7$ and $9$. Make the smallest mobile number by using only one digit twice from $8, 3, 5, 6, 0.$
AnswerGiven, total number of digits $= 10$ and first four digits of the mobile number $= 9, 9, 7, 9$
Also, given digits $= 8, 3, 5, 6, 0$ Now, to make the smallest mobile number,
we will use the smallest digit twice, i.e. $0$.
Required mobile number $= 9979003568$
View full question & answer→Question 213 Marks
A mobile number consists of ten digits. The first four digits of the number are $9, 9, 8$ and $7$. The last three digits are $3, 5$ and $5$. The remaining digits are distinct and make the mobile number which is the greatest possible number. What are these digits?
AnswerGiven, total number of digits $= 10$ First four digits of the mobile number $= 9, 9, 8,7$ and last three digits of the mobile number$ = 3,5,5$ Now, to make the mobile number, the possible distinct digits are $6, 4, 2,1$ and $0$.
For making the greatest mobile, we can select the digits $6, 4$ and $2$.
Required mobile number $= 9987642355$
Hence, the required digits are $6, 4$ and $2.$
View full question & answer→Question 223 Marks
As per the census of $2001$, the population of four states are given below:
| $(a)$ |
Maharashta |
$96878627$ |
| $(b)$ |
Andhra Pradesh |
$76210007$ |
| $(c)$ |
Bihar |
$82998509$ |
| $(d)$ |
Uttar Pradesh |
$166197921$ |
Arrange the states in ascending and descending order of their populations. AnswerOn arranging the population of four states in ascending order,
we get: $76210007 < 82998509 < 96878627 < 166197921$ (Andhra Pradesh) < (Bihar) < (Maharashtra) < (Uttar Pradesh) Again, rearranging the population of four states in descending order,
we get: $166197921 > 96878627> 82998509 > 76210007$ (Uttar Pradesh) > (Maharashtra) > (Bihar) > (Andhra Pradesh)
View full question & answer→Question 233 Marks
Fatima wants to mail three parcels to three village schools. She finds that the postal charges are $Rs. 20, Rs. 28$ and $Rs. 36,$ respectively. If she wants to buy stamps only of one denomination, what is the greatest denomination of stamps she must buy to mail the three parcels?
AnswerHere,
The greatest denomination of stamps that Fatima must buy to mail the three parcels.
The $HCF$ of $Rs. 20.$
$\begin{array}{c|c}2&20\\ \hline2&10\\ \hline 5&5\\ \hline&1\end{array}$
Prime factorization of $20 = 2 \times 2 \times 5$
The $HCF$ of $Rs. 28.$
$\begin{array}{c|c}2&28\\ \hline2&14\\ \hline 7&7\\ \hline&1\end{array}$
Prime factorization of $28 = 2 \times 2 \times 7$
The $HCF$ of $Rs. 36.$
$\begin{array}{c|c}2&36\\ \hline2&18\\ \hline3&9\\ \hline 3&3\\ \hline&1\end{array}$
Prime factorization of $36 = 2 \times 2 \times 3 \times 3$
Common factors of $20, 28$ and $36$ are 2(occurring twice).
$HCF$ of $20, 28$ and $36 = 2 \times 2 = 4$
Thus, the greatest denomination of stamps that Fatima must buy is $Rs. 24.$
View full question & answer→MCQ 243 Marks
Estimate each of the following by rounding off each number to nearest hundreds.
- A
$874 + 478$
- B
$793 + 397$
- C
$11244 + 3507$
- ✓
Answer
$a.\ $We have,
$874 + 478$
Rounded off $874$ to nearest hundreds $= 900$ and
rounded off $478$ to nearest hundreds $= 500$
So, estimated sum $= 900 + 500 = 1400$
$b.\ $We have,
$793 + 397$
Rounded off $793$ to nearest hundreds $= 800$ and
rounded off $397$ to nearest hundreds $= 400$
So, estimated sum $= 800 + 400 = 1200$
$c.\ $We have,
$11244 + 3507$
Rounded off $11244$ to nearest hundreds $= 11200$ and
rounded off $3507$ to nearest hundreds $= 3500$
So, estimated sum $= 11200 + 3500 = 14700$
$d.\ $We have,
$17677 + 13589$
Rounded off $17677$ to nearest hundreds $= 17700$ and
rounded off $13589$ to nearest hundreds $= 13600$
So, estimated sum $= 17700 + 13600 = 31300$
View full question & answer→Question 253 Marks
A merchant has $120L$ of oil of one kind, $180L$ of another kind and $240L$ of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such in tin?
AnswerHere, the greatest capacity of tin will be equal to the$ HCF$ of $120$.
$\begin{array}{c|c}2&120\\ \hline2&60\\\hline2&30\\ \hline3&15\\ \hline5&5\\ \hline&1\end{array}$
Prime factorization Of 1$20 = 2 \times 2 \times 2 \times 3 \times 5$ the greatest capacity of tin will be equal to the $HCF$ of $180$.
$\begin{array}{c|c}2&180\\ \hline2&90\\\hline3&45\\ \hline3&15\\ \hline5&5\\ \hline&1\end{array}$
Prime factorization Of $180 = 2 \times 2 \times 3 \times 5 \times 3$ the greatest capacity of tin will be equal to the $HCF$ of $240$. $\begin{array}{c|c}2&240\\ \hline2&120\\ \hline2&60\\\hline2&30\\ \hline3&15\\ \hline5&5\\ \hline&1\end{array}$
Prime factorization of $240 = 2 \times 2 \times 2 \times 3 \times 5 \times 2$ Common factors of $120,180$ and $240$ are $2$ (occurring twice), $3$ and $5$.
Thus, the $HCF$ of $120, 180$ and $240 = 2 \times 2 \times 3 \times 5 = 60$ Greatest capacity of tin $= HCF$ of $120, 180$ and $240 = 60L$
View full question & answer→Question 263 Marks
Estimate each of the following by rounding off each number to nearest tens.
$a.\ 11963 - 9369$
$b.\ 76877 - 7783$
$c.\ 10732 - 4354$
$d.\ 78203 - 16407$
Answer$a.\ $We have,
$11963 - 9369$
Rounded off $11963$ to nearest tens $= 11960$ and
rounded off $9369$ to nearest tens $= 9370$
So, estimated difference $= 11960 - 9370 = 2590$
$b.\ $We have,
$76877 - 7783$
Rounded off $76877$ to nearest tens $= 76880$ and
rounded off $7783$ to nearest tens $= 7780$
So, estimated difference $= 76880 - 7780 = 69100$
$c.\ $We have,
$10732 - 4354$
Rounded off $10732$ to nearest tens $= 10730$ and
rounded off $4354$ to nearest tens $= 4350$
So, estimated difference $= 10730 - 4350 = 6380$
$d.\ $We have,
$78203 - 16407$
Rounded off $78203$ to nearest tens $= 78200$ and
rounded off $16407$ to nearest tens $= 16410$
So, estimated difference $= 78200 - 16410 = 61790$
View full question & answer→Question 273 Marks
Find a $4$-digit odd number using each of the digits $1, 2, 4$ and $5$ only once, such that when the first and the last digits are interchanged, it is divisible by $4.$
AnswerWe know that, $4$-digit number is said to be an odd number, if unit place digit is an odd number (i.e. $1$ or $5$).
Given digits are $1, 2, 4$ and $5$.
Total such odd numbers are: $4125, 4215, 1245, 1425, 2145, 2415, 4251, 4521, 5241, 5421, 2451$ and $2541.$
Also, we know that, any $4$-digit number can be divisible by $4$,
if the last two digits of that number is divisible by $4$.
Consider a number $4521$.
If we interchange the first and the last digits, then the new number $= 1524$
Here, we see that the last two digits (i.e. $24$) which is divisible by $4$.
So, $1524$ is divisible by $4$. Required $4$-digit number $= 4521$.
There are three more numbers which is divisible by $4$ such that $2415, 2451$ and $4125.$
View full question & answer→Question 283 Marks
Estimate each of the following products by rounding off each number to nearest tens.
$a.\ 87 \times 32$
$b.\ 311 \times 113$
$c.\ 3239 \times 28$
$d.\ 1385 \times 789$
Answer
$a.\ $We have,
$87 \times 32$
Rounded off $87$ to nearest tens $= 90$ and
rounded off $32$ to nearest tens $= 30$
So, estimated product $= 90 \times 30 = 2700$
$b.\ $ We have,
$311 \times 113$
Rounded off $311$ to nearest tens $= 310$ and
rounded off $113$ to nearest tens $= 110$
So, estimated product $= 310 \times 110 = 34100$
$c.\ $We have,
$3239 \times 28$
Rounded off $3239$ to nearest tens $= 3240$ and
rounded off $28$ to nearest tens $= 30$
So, estimated product $= 3240 \times 30 = 97200$
$d.\ $We have,
$1385 \times 789$
Rounded off $1385$ to nearest tens $= 1390$ and
rounded off $789$ to nearest tens $= 790$
So, estimated product $= 1390 \times 790 = 1098100$
View full question & answer→Question 293 Marks
Test the divisibility of following numbers by $11.$
$a.\ 5335$
$b.\ 9020814$
AnswerWe know that, a number is divisible by $11$, if the difference of sum of digits at even and odd places is either zero or divisible by $11.$
$a.\ $We have, $5335$
Sum of digits at even places from the right $= 3 + 5 = 8$
Sum of digits at odd places from the right $= 5 + 3 = 8$
So, difference $= 8 - 8 = 0$
Hence, it is divisible by $11.$
$b.\ $We have, $9020814$
Sum of digits at even places $= 1 + 0 + 0 = 1$
Sum of digits at odd places from the right $= 4 + 8 + 2 + 9 = 23$
So, difference $= 23 - 1 = 22 ($divisible by $11)$
Hence, it is divisible by $11.$
View full question & answer→Question 303 Marks
In a colony of $100$ blocks of flats numbering $1$ to $100$, a school van stops at every sixth block, while a school bus stops at every tenth block. On which stops will both of them stop, if they start from the entrance of the colony?
AnswerGiven, in a colony of $100$ blocks, there are $100$ flats numbering from $1$ to $100$.
Here, school van and bus stop at the same stoppage means it is the $LCM$ of both stoppages.
Now, $LCM$ of $6$ and $10$
$\begin{array}{c|c}2&6,10\\ \hline3&3,5\\ \hline5&1,5\\ \hline&1,1\end{array}$
$LCM$ of $6$ and $10 = 2 \times 3 \times 5 = 30$
It shows that first time they both meets at $30th$ stoppage and the next time they again meet at a multiple of $30$.
Multiples of $30$, which are less than $100$, are $30, 60$ and $90$.
Hence, both will stop on the stop blocks $30th, 60th$ and $90th$, if they start from the entrance of the colony.
View full question & answer→