Quiz Level-2

Hint

The required number leaves the remainder 1 and 4 on dividing 70 and 50 respectively.
This means that the number exactly divides 69 and 46.
So, we need to find the HCF of 69 is \((3 \times 23)\) and 46 is \((2 \times 23)\).
Now, the HCF of \(69,46=23\)
Hence, the required number is 23.

Hint

In this type of question, we need to find the LCM of the divisors and add the common remainder (i.e., 3 ) to it.
Let us find the LCM of \(5,7,9,12\).
Thus,
\(
\begin{array}{l|l}
3 & 5,7,9,12 \\
\hline & 5,7,3,4
\end{array}
\)
The LCM of \(5,7,9,12=3 \times 5 \times 7 \times 3 \times 4=1260\)
Now, the required number \(=1260+3=1263\)
Hence, the least number is 1263.

Hint

Since the remainder is same in each case, hence we will use the following formula
\(
\text { H.C.F }(x, y, z)=\text { H.C.F of }(x-y),(y-z),(z-x)
\)
\(N=\) H.C.F. of (6905 – 4665), \((4665\) – 1305), and (6905 – 1305)
\(=\) H.C.F. of 2240,3360 and 5600
HCF of 5600 and 3360 :
\(5600=3360 \times 1+2240\)
\(3360=2240 \times 1+1120\)
\(2240=1120 \times 2+0\)
\(\therefore\) HCF of 5600 and \(3360=1120\)
Now, HCF of 2240 and \(1120=1120\)
So, the HCF of \((3360,2240\) and 5600\()=N=1120\)
Sum of digits in \(\mathrm{N}=(1+1+2+\mathrm{0})=4\)

Hint

Let the numbers be \(37 \mathrm{a}\) and \(37 \mathrm{~b}\).
Then, \(37 a \times 37 b=4107\)
\(
a b=3
\)
Now, co-primes with product 3 are \((1,3)\).
So, the required numbers are \((37 \times 1,37 \times 3)\) i.e., \((37,111)\).
Greater number \(=111\).

Hint

Let the numbers be \(3 x, 4 x\) and \(5 x\).
Then, their L.C.M. \(=60 \mathrm{X}\).
So, \(60 x=2400\) or \(x=40\).
\(\therefore\) The numbers are \((3 \times 40),(4 \times 40)\) and \((5 \times 40)\).
Hence, required H.C.F. \(=40\).

Hint

Let the numbers \(13 \mathrm{a}\) and \(13 \mathrm{~b}\).
Then, \(13 \mathrm{a} \times 13 \mathrm{~b}=2028\)
\(
\Rightarrow \mathrm{ab}=12 \text {. }
\)
Now, the co-primes with product 12 are \((1,12)\) and \((3,4)\).
[Note: Two integers a and b are said to be coprime or relatively prime of they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are \((13 \times 1,13 \times 12)\) and \((13 \times 3,13 \times 4)\).
Clearly, there are 2 such pairs.

Hint

We first find the L.C.M of \(6,9,15\) and 18 , which is 90 .
For getting remainder 4 , we need to add 4 to it.
i.e., \(90+4=94\).
But, 94 is not divisible by 7 . So, we proceed with the next multiple of 90 , which is 180.
Adding 4 to 180 , we get 184 , which is also not divisible by 7 .
The next multiple of 90 is 270 .
Adding 4 to 270 , we get 274 , again not divisible by 7 .
The next multiple of 90 is 360 .
Adding 4 to it gives us 364 , divisible by 7 ( as \(\frac{364}{7}=52\) ).
Hence, the answer is 364 .

Hint

L.C.M. of 5, 6, 4, and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

 Number to be added = (60 – 37) = 23.

Hint

L.C.M. of \(5,6,7,8=840\).
\(\therefore\) Required number is of the form \(840 k+3\)
Least value of \(k\) for which \((840 k+3)\) is divisible by 9 is \(k=2\).
\(\therefore\) Required number \(=(840 \times 2+3)=1683\)

Hint

L.C.M. of 252, 308 and 198 = 2772.

So, A, B, and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Hint

\(
\text { Other number }=\left(\frac{11 \times 7700}{275}\right)=308 \text {. }
\)

Hint

The least number which is exactly divisible by \(12,18,21,90\) will be LCM of \(12,18,21,90\). So our required answer will be half of LCM
\(
\operatorname{LCM}(12,18,21,90)=1260
\)
So our required answer will be half of \(1260=630\)

Hint

Let the numbers be \(3 x\) and \(4 x\). Then, their H.C.F. \(=x\). So, \(x=4\).
So, the numbers 12 and 16.
L.C.M. of 12 and \(16=48\).

Hint

\(
\begin{aligned}
& \text { Required number }=(\text { L.C.M. of } 12,16,18,21,28)+7 \\
& =1008+7 \\
& =1015
\end{aligned}
\)

Hint

\(
\text { Required length }=\text { H.C.F. of } 700 \mathrm{~cm}, 385 \mathrm{~cm} \text { and } 1295 \mathrm{~cm}=35 \mathrm{~cm} \text {. }
\)

Hint

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number \(=\) H.C.F. of 551 and \(1073=29\);
First number \(=\left(\frac{551}{29}\right)=19 ; \quad\) Third number \(=\left(\frac{1073}{29}\right)=37\).
\(\therefore\) Required sum \(=(19+29+37)=85\).

Hint

Required number \(=(\) L.C.M. of \(12,15,20,54)+8\)
\(
\begin{aligned}
& =540+8 \\
& =548 .
\end{aligned}
\)

Hint

Required number = H.C.F. of (1657 – 6) and (2037 – 5)  = H.C.F. of 1651 and 2032 = 127.

Hint

Let the numbers be \(2 x\) and \(3 x\).
Then, their L.C.M. \(=6 x\).
So, \(6 x=48\) or \(x=8\).
\(\therefore\) The numbers are 16 and 24 .
Hence, required sum \(=(16+24)=40\).

Hint

Let the numbers be \(a\) and \(b\).
Then, \(a+b=55\) and \(a b=5 \times 120=600\).
\(\therefore\) The required sum \(=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{55}{600}=\frac{11}{120}\)

Hint

Given numbers are 1.08, 0.36 and 0.90

G.C.D. of 108, 36, and 90 is 18.

Thus, G.C.D. of given numbers = 0.18

Hint

From the given information, the numbers can be assumed as \(6 x\) and \(7 x\).
We can find LCM of \(6 x\) and \(7 x\).
Therefore, LCM of \((6 x, 7 x)\) is
\(
\begin{aligned}
& =x \cdot 6 \cdot 7 \\
& =42 x
\end{aligned}
\)
Given : LCM of the two numbers is 84 .
Then,
\(
42 x=84
\)
Divide each side by 42 .
\(
x=2
\)
Substitute 2 for \(x\) in \(6 x\) and \(7 x\).
\(
\begin{aligned}
& 6 x=6 \cdot 2=12 \\
& 7 x=7 \cdot 2=14
\end{aligned}
\)
So, the two numbers are 12 and 14.

Hint

Sara obviously wants to use all the flowers so she does not any left over. She needs to find a number which divides into 16 and 24,
This is just an indirect way of using the HCF of 16 and 24, which is 8.
16=2×8
24=3×8
She will be able to make 8 bouquets:
Each bouquet will have 2 red flowers and 3 yellow flowers.

Hint

Think about multiples of the two numbers 9 and 15. And we will have to assume that they both started blinking at the same time (at time zero)
For 9 the lights will blink at seconds: 0, 9, 18, 27, 36, 45, 54, 63,
For 15 the lights will blink at seconds: 0, 15, 30, 45, 60, 75
They will blink together at 45 seconds (LCM of both)

Hint

Given:
One neon light blinks every 4 seconds
The other blinks every 6 seconds.
To find In 60 seconds, how many times will they blink at the same time

One neon light blinks every 4 seconds
The other blinks every 6 seconds.
Therefore we find the LCM of 6 abd 4.
LCM(6,4)=12.
Therefore to find the no of times it will blink at the same time in 60s
=60/12
=5
In 60 seconds they blink 5 times at the same time.

Hint

Given : 12 coloring books, 28 markers, and 36 crayons.
To Find : greatest number of baskets she can make if each type of toy is equally distributed among the baskets
How many of each supply will go into the baskets

Find HCF of 12 , 28 and 36
12 = 2 x 2 x 3
28 = 2 x 2 x 7
36 = 2 x 2 x 3 x 3
HCF = 2 x 2 = 4
greatest number of baskets she can make = 4
each supply will go into 1 basket
12/4 = 3 coloring books,
28/4 = 7 markers,
36/4 = 9 crayons.

Hint

If we consider today as Day 0, then
Days with tacos: \(6,12,18,24, \ldots\)
Days with cheeseburgers: \(8,16,24, \ldots\)
It can be seen that after 24 days, both will be on the menu again.
In fact, this utilizes LCM (lowest common multiple) in calculations. By prime factorization,
\(
\begin{aligned}
& 6=2 \cdot 3 \\
& 8=2 \cdot 2 \cdot 2
\end{aligned}
\)
As both of them have a 2 , we can take the two out and count it once. Therefore,
\(\operatorname{LCM}(6,8)=2 \cdot 3 \cdot 2 \cdot 2=24\),
Where the first 2 is the common factor, 3 comes from the factor of 6, and the \(2 \times 2\) from 8.
In this way, we can find the number of days, which is 24.

Hint

3 baskets
Step-by-step explanation: GCF(18, 27,12) = 3.
18= 6 x 3
27 = 9 x 3
12 = 4 x 3
Therefore she will have 3 baskets each with 19 fruits .6 oranges,9 pears, and 4 bananas in each basket.

Hint

\(
\begin{array}{ll}
10-10,20,30,40,50, \underline{60} & \\
12-12,24,36,48, \underline{\mathbf{6 0}} & \text { LCM }=\mathbf{6 0} \\
60 \div 10=6 \text { packages of pencils } & \\
60 \div 12=5 \text { packages of erasers } 
\end{array}
\)

Hint

\(
\begin{aligned}
& 30-1,2,3,5, \underline{\mathbf{6}}, 10,15,30 \\
& 48-1,2,3,4, \underline{\mathbf{6}}, 8,12,16,24,48 \quad \text { GCF }=\mathbf{6}
\end{aligned}
\)
Kiara needs \(\mathbf{6}\) plastic containers for her cookies.