NUMBER SYSTEMS EXAMPLE QUESTIONS PART 3
1. Find the LCM of following three fractions:
Explanation:
Numerators = 36, 48 and 72.
72 is largest number among them. 72 is not divisible by 36 or 48
Start with table of 72.
72 x 2 = 144 = divisible by 72, 36 and 48
∴ LCM of numerators = 144
Denominators = 225, 150 and 65
We can see that they can be divided by 5.
On dividing by 5 we get 45, 30 and 13
We cannot divide further.
So, HCF = GCD = 5
2. Find the HCF of following three fractions:
Explanation:
Numerators = 36, 48 and 72.
We can see that they can be divided by 12.
On dividing by 12 we get 3, 4 and 6.
We cannot divide further.
∴ HCF = GCD of numerators = 12
150 is largest number among them. 75 can divide 150, so neglect 75
Let's find LCM of 150 and 135
----------------------------------
5 150 135
----------------------------------
3 30 27
3 10 9
10 3
----------------------------------
∴ LCM of denominators = 5 x 3 x 3 x 3 x 10 = 1350
3. Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?
a. 13
b. 17
c. 78
d. 89
4. Find the smallest number which leaves the remainders 13, 41 and 29 at the end when divided by 20, 48 and 36 respectively.
a. 187
b. 713
c. 720
d. 727
5. A number when divided by 36, 24 and 16, leaves the remainder 11 in each case. Find the smallest value of this number.
a. 36
b. 133
c. 144
d. 155
| 36 | , | 48 | , | 72 |
| 225 | 150 | 65 |
| a. | 72 |
| 225 |
| b. | 36 |
| 65 |
| c. | 144 |
| 5 |
| d. | 288 |
| 5 |
| Answer: c. | 144 |
| 5 |
Explanation:
Tip:
| LCM of fraction = | LCM of numerators |
| HCF of denominators |
Numerators = 36, 48 and 72.
72 is largest number among them. 72 is not divisible by 36 or 48
Start with table of 72.
72 x 2 = 144 = divisible by 72, 36 and 48
∴ LCM of numerators = 144
Denominators = 225, 150 and 65
We can see that they can be divided by 5.
On dividing by 5 we get 45, 30 and 13
We cannot divide further.
So, HCF = GCD = 5
| LCM of fraction = | 144 |
| 5 |
2. Find the HCF of following three fractions:
| 36 | , | 48 | , | 72 |
| 75 | 150 | 135 |
| a. | 12 |
| 1350 |
| b. | 72 |
| 225 |
| c. | 150 |
| 36 |
| d. | 1350 |
| 36 |
| Answer: a. | 12 |
| 1350 |
Explanation:
Tip:
| LCM of fraction = | LCM of numerators |
| HCF of denominators |
Numerators = 36, 48 and 72.
We can see that they can be divided by 12.
On dividing by 12 we get 3, 4 and 6.
We cannot divide further.
∴ HCF = GCD of numerators = 12
150 is largest number among them. 75 can divide 150, so neglect 75
Let's find LCM of 150 and 135
----------------------------------
5 150 135
----------------------------------
3 30 27
3 10 9
10 3
----------------------------------
∴ LCM of denominators = 5 x 3 x 3 x 3 x 10 = 1350
| HCF of fraction = | 12 |
| 1350 |
3. Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?
a. 13
b. 17
c. 78
d. 89
Answer: a. 13
Explanation:
Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 - 4 = 13; 42 - 3 = 39; 93 - 15 = 78
Now let's find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number
Explanation:
Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 - 4 = 13; 42 - 3 = 39; 93 - 15 = 78
Now let's find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number
4. Find the smallest number which leaves the remainders 13, 41 and 29 at the end when divided by 20, 48 and 36 respectively.
a. 187
b. 713
c. 720
d. 727
Answer: b. 713
Explanation:
Here least number is needed that means we need the Least Common Multiple i.e. LCM
We must now first find LCM of 20, 36 and 48
-----------------------------------------
4 20 36 48
-----------------------------------------
3 5 9 12
5 3 4
-----------------------------------------
∴ LCM = 4 x 3 x 5 x 3 x 4 = 720
But this is not the answer because there are remainders as per the given condition.
If we observe closely, the difference between the given numbers and remainders is same
20 - 13 = 7; 48 - 1 = 7; 36 - 29 = 7
Difference is same = 7
So simply subtract this difference from LCM.
Number = 720 - 7 = 713
Explanation:
Here least number is needed that means we need the Least Common Multiple i.e. LCM
We must now first find LCM of 20, 36 and 48
-----------------------------------------
4 20 36 48
-----------------------------------------
3 5 9 12
5 3 4
-----------------------------------------
∴ LCM = 4 x 3 x 5 x 3 x 4 = 720
But this is not the answer because there are remainders as per the given condition.
If we observe closely, the difference between the given numbers and remainders is same
20 - 13 = 7; 48 - 1 = 7; 36 - 29 = 7
Difference is same = 7
So simply subtract this difference from LCM.
Number = 720 - 7 = 713
5. A number when divided by 36, 24 and 16, leaves the remainder 11 in each case. Find the smallest value of this number.
a. 36
b. 133
c. 144
d. 155
Answer: d. 155
Explanation:
Here smallest (least) number is needed that means we need the LCM
We must first find LCM of 36, 24 and 16
--------------------------------------
4 16 24 36
--------------------------------------
3 4 6 9
2 4 2 3
2 1 3
--------------------------------------
∴ LCM = 4 x 3 x 2 x 2 x 1 x 3 = 144
Since remainder is same just add it to this LCM
Number = 144 + 11 = 155
6. A 4 digit number which when divided by 12, 16 and 18 gives out a common remainder i.e. 21. Find the smallest/ least such number.
a. 36
b. 133
c. 144
d. 1461
7. The two given numbers A and B are in the ratio 5:6 such that their LCM is 480. Find their HCF.
a. 12
b. 16
c. 96
d. 240
8. The sum of two given numbers P and Q is 56. Their LCM and HCF is 96 and 8 respectively.
Explanation:
Let numbers be P and Q
Also, P + Q = 56 and PQ = HCF x LCM = 8 x 96
9. Find the largest number which when divides 47, 35 and 27 leaves the same remainder. Also find the value of the remainder.
a. 3, 4
b. 4, 3
c. 1, 9
d. 9, 1
10. Find the largest scale size to measure accurately, the three equilateral triangles with sides measuring 76 cm, 114 cm and 152 cm.
a. 19 cm
b. 21 cm
c. 38 cm
d. None of the above
a. 90
b. 180
c. 183
d. 900
17. A pattern is formed with six type of LEDs. They start lighting together and then light at intervals of 2, 4, 6, 8, 10 and 12 seconds, respectively. In half an hour, how many times do they light together?
a. 4
b. 10
c. 15
d. 16
18. Find the largest of following fractions.
a. 63/80
b. 31/40
c. 13/16
d. 7/8
19. Find the smallest number which when reduced by 7, can be perfectly divided by 12, 16, 18, 21 and 28 each.
a. 1008
b. 1015
c. 1022
d. 1032
20. A and B are two primes numbers, A > B. Their LCM is 161. Find the value of 3B - A :
a. -2
b. -1
c. 1
d. 2
Explanation:
Here smallest (least) number is needed that means we need the LCM
We must first find LCM of 36, 24 and 16
--------------------------------------
4 16 24 36
--------------------------------------
3 4 6 9
2 4 2 3
2 1 3
--------------------------------------
∴ LCM = 4 x 3 x 2 x 2 x 1 x 3 = 144
Since remainder is same just add it to this LCM
Number = 144 + 11 = 155
6. A 4 digit number which when divided by 12, 16 and 18 gives out a common remainder i.e. 21. Find the smallest/ least such number.
a. 36
b. 133
c. 144
d. 1461
Answer: d. 1461
Explanation:
Here least possible 4 digit number is needed that means we need the LCM
We must first find LCM of 12, 16, 18 and 20
--------------------------------------------------------
2 12 16 18 20
--------------------------------------------------------
3 6 8 9 10
2 2 8 3 10
1 4 3 5
--------------------------------------------------------
∴ LCM = 2 x 3 x 2 x 1 x 4 x 3 x 5 = 720
Now this is a 3 digit number.
If we multiply it by 2 we get (720 x 2) = 1440 → (4 digit number)
But 1440 is divisible by 12, 16, 18 and 20
We must have 21 as remainder, as per given condition.
So, Number = 1440 + 21 = 1461
Explanation:
Here least possible 4 digit number is needed that means we need the LCM
We must first find LCM of 12, 16, 18 and 20
--------------------------------------------------------
2 12 16 18 20
--------------------------------------------------------
3 6 8 9 10
2 2 8 3 10
1 4 3 5
--------------------------------------------------------
∴ LCM = 2 x 3 x 2 x 1 x 4 x 3 x 5 = 720
Now this is a 3 digit number.
If we multiply it by 2 we get (720 x 2) = 1440 → (4 digit number)
But 1440 is divisible by 12, 16, 18 and 20
We must have 21 as remainder, as per given condition.
So, Number = 1440 + 21 = 1461
7. The two given numbers A and B are in the ratio 5:6 such that their LCM is 480. Find their HCF.
a. 12
b. 16
c. 96
d. 240
Answer: b. 16
Explanation:
Let K be common factor. So 2 numbers are 5K and 6K
Also K is the greatest common factor (HCF) as 5 and 6 have no other common factor
∴ 5K x 6K = 480 x K
K = 16 = HCF
Explanation:
Tip:
If A and B are two numbers,
A x B = HCF of A and B x LCM of A and B
If A and B are two numbers,
A x B = HCF of A and B x LCM of A and B
Let K be common factor. So 2 numbers are 5K and 6K
Also K is the greatest common factor (HCF) as 5 and 6 have no other common factor
∴ 5K x 6K = 480 x K
K = 16 = HCF
8. The sum of two given numbers P and Q is 56. Their LCM and HCF is 96 and 8 respectively.
| Find the sum of | 1 | + | 1 | . |
| P | Q |
| a. | 1 |
| 96 |
| b. | 1 |
| 56 |
| c. | 7 |
| 96 |
| d. | 1 |
| 8 |
| Answer: c. | 7 |
| 96 |
Explanation:
Tip:
If A and B are two numbers,
A x B = HCF of A and B x LCM of A and B
If A and B are two numbers,
A x B = HCF of A and B x LCM of A and B
Let numbers be P and Q
Also, P + Q = 56 and PQ = HCF x LCM = 8 x 96
| 1 | + | 1 | = | P + Q | = | 56 | = | 7 |
| P | Q | PQ | 8 x 96 | 96 |
9. Find the largest number which when divides 47, 35 and 27 leaves the same remainder. Also find the value of the remainder.
a. 3, 4
b. 4, 3
c. 1, 9
d. 9, 1
Answer: b. 4, 3
Explanation:
What to do when we don't know the remainder and the largest number which divides?
In such cases subtract the 3 numbers from each other
47 - 35 = 12
35 - 27 = 8
27 - 47 = -20 (Ignore minus sign)
Largest number is needed that means we need HCF of 12, 8 and 20
------------------------------------
4 12 8 20
------------------------------------
3 2 5
------------------------------------
Cannot divide further
∴ HCF = 4 = Largest number that can divide leaving common remainder
∴ Common remainder = 3
Explanation:
What to do when we don't know the remainder and the largest number which divides?
In such cases subtract the 3 numbers from each other
47 - 35 = 12
35 - 27 = 8
27 - 47 = -20 (Ignore minus sign)
Largest number is needed that means we need HCF of 12, 8 and 20
------------------------------------
4 12 8 20
------------------------------------
3 2 5
------------------------------------
Cannot divide further
∴ HCF = 4 = Largest number that can divide leaving common remainder
| 47 | = 11 | 3 |
| 4 | 4 |
10. Find the largest scale size to measure accurately, the three equilateral triangles with sides measuring 76 cm, 114 cm and 152 cm.
a. 19 cm
b. 21 cm
c. 38 cm
d. None of the above
Answer: c. 38 cm
Explanation:
Here the answer is HCF of 114, 76 and 152
------------------------------------------
2 76 114 152
------------------------------------------
19 38 57 76
1 3 2
------------------------------------------
∴ HCF = 2 x 19 = 38 = Maximum scale size needed to measure all 3 exactly
11. Given:
Eq 1: x2 - 8x + 15
Eq 2: x2 - px +1
HCF of Eq1 & Eq 2 = (x - 4)
Find the value of p.
c. 4
d. 5
12. The given five signals light up automatically at intervals of 3 min, 4 minutes, 8 min, 10 min, and 12 min respectively. How many times in 8 hours will they light up together from the time they start?
a. 3 times
b. 4 times
c. 5 times
d. 12 times
13. In a race on a circular track, the three athletes complete one round in 27 minutes, 45 minutes and 63 minutes respectively. Find the time after which they meet again at the starting point, since the time they started.
a. 9 hours
b. 126 minutes
c. 135 minutes
d. 945 minutes
14. A child goes to play with some pebbles in his bag. The number of pebbles is such that he can arrange them in rows of 18, 10 and 15 & form a square in each case. How many minimum number of pebbles are there in his bag?
a. 43
b. 90
c. 133
d. 900
15. A wall is 4.5 meters long and 3.5 meters high. Find the number of maximum sized wallpaper squares, if the wall has to be covered with only the square wall paper pieces of same size.
a. 8
b. 12
c. 15.75
d. 63
Answer: d. 63
Explanation:
Wall can be covered only by using square sized wallpaper pieces.
Different sized squares are not allowed.
Length = 4.5 m = 450 cm;
Height = 3.5 m = 350 cm
Maximum square size possible means HCF of 350 and 450
We can see that 350 and 450 can be divided by 50.
On dividing by 50, we get 7 and 9.
Since we cannot divide further, HCF = 50 = size of side of square
16. I want to stack the wooden blocks left after the carpentry work
at home. I stack them in groups of 4 but 3 blocks get left out. So, I
attempt with stacks of 5 blocks each but again 3 blocks remain left out.
The same thing happens even when I try with 9 blocks and 10 blocks in
each stack. Find the total number of blocks I have.Explanation:
Here the answer is HCF of 114, 76 and 152
------------------------------------------
2 76 114 152
------------------------------------------
19 38 57 76
1 3 2
------------------------------------------
∴ HCF = 2 x 19 = 38 = Maximum scale size needed to measure all 3 exactly
11. Given:
Eq 1: x2 - 8x + 15
Eq 2: x2 - px +1
HCF of Eq1 & Eq 2 = (x - 4)
Find the value of p.
| a. | 8 |
| 15 |
| b. | 15 |
| 8 |
d. 5
Answer: c. 4
Explanation:
(x - 4) is HCF i.e. a factor of both equations
∴ The equations must get satisfied for x = 4
Also, when x = 4, both equations are equal in value.
Putting x = 4 in both equations
42 - 8(4) + 15 = 42 - p(4) +1
∴ 31 - 32 = 17 - 4p
∴ p = 4
Explanation:
(x - 4) is HCF i.e. a factor of both equations
∴ The equations must get satisfied for x = 4
Also, when x = 4, both equations are equal in value.
Putting x = 4 in both equations
42 - 8(4) + 15 = 42 - p(4) +1
∴ 31 - 32 = 17 - 4p
∴ p = 4
12. The given five signals light up automatically at intervals of 3 min, 4 minutes, 8 min, 10 min, and 12 min respectively. How many times in 8 hours will they light up together from the time they start?
a. 3 times
b. 4 times
c. 5 times
d. 12 times
Answer: b. 4 times
Explanation:
We need the next instances when the signals light up.
That means the Least Common Multiple (LCM) of 3, 4, 8, 10, 12
3, 4 divide 12 so neglect them.
LCM of 8, 10 and 12
-----------------------------------
4 8 10 12
-----------------------------------
2 2 10 3
1 5 3
-----------------------------------
∴ LCM = 4 x 2 x 5 x 3 = 120 = They light up together after 2 hours
After starting, they light up together 1st time in 2 hours.
Then 2nd time in 2 + 2 = 4 hours.
Then 3rd time in 6 hours.
And 4th time in 8 hours.
Explanation:
We need the next instances when the signals light up.
That means the Least Common Multiple (LCM) of 3, 4, 8, 10, 12
3, 4 divide 12 so neglect them.
LCM of 8, 10 and 12
-----------------------------------
4 8 10 12
-----------------------------------
2 2 10 3
1 5 3
-----------------------------------
∴ LCM = 4 x 2 x 5 x 3 = 120 = They light up together after 2 hours
After starting, they light up together 1st time in 2 hours.
Then 2nd time in 2 + 2 = 4 hours.
Then 3rd time in 6 hours.
And 4th time in 8 hours.
13. In a race on a circular track, the three athletes complete one round in 27 minutes, 45 minutes and 63 minutes respectively. Find the time after which they meet again at the starting point, since the time they started.
a. 9 hours
b. 126 minutes
c. 135 minutes
d. 945 minutes
Answer: d. 945 minutes
Explanation:
We need the next instance that means the LCM of times of all 3 athletes.
--------------------------------------------
9 27 45 63
--------------------------------------------
3 5 7
-------------------------------------------
∴ LCM = 9 x 3 x 5 x 7 = 945 = They meet after 945 minutes
Explanation:
We need the next instance that means the LCM of times of all 3 athletes.
--------------------------------------------
9 27 45 63
--------------------------------------------
3 5 7
-------------------------------------------
∴ LCM = 9 x 3 x 5 x 7 = 945 = They meet after 945 minutes
14. A child goes to play with some pebbles in his bag. The number of pebbles is such that he can arrange them in rows of 18, 10 and 15 & form a square in each case. How many minimum number of pebbles are there in his bag?
a. 43
b. 90
c. 133
d. 900
Answer: d. 900
Explanation:
We need the least number, which means we must first find LCM of 18, 10 and 15
-------------------------------------------------
3 18 10 15
---------------------------------------------
2 6 10 5
5 3 5 5
3 1 1
-----------------------------------------
∴ LCM = 3 x 2 x 5 x 3 = 90
90 is not a perfect square.
90 = 3 x 3 x 2 x 5
Here there are two 3's (this gives square of 3) but only one 2 and one 5
If we multiply by 2 and 5 then we get, 3 x 3 x 2 x 2 x 5 x 5 = 900
900 is perfect square = number of marbles
Explanation:
We need the least number, which means we must first find LCM of 18, 10 and 15
-------------------------------------------------
3 18 10 15
---------------------------------------------
2 6 10 5
5 3 5 5
3 1 1
-----------------------------------------
∴ LCM = 3 x 2 x 5 x 3 = 90
90 is not a perfect square.
90 = 3 x 3 x 2 x 5
Here there are two 3's (this gives square of 3) but only one 2 and one 5
If we multiply by 2 and 5 then we get, 3 x 3 x 2 x 2 x 5 x 5 = 900
900 is perfect square = number of marbles
15. A wall is 4.5 meters long and 3.5 meters high. Find the number of maximum sized wallpaper squares, if the wall has to be covered with only the square wall paper pieces of same size.
a. 8
b. 12
c. 15.75
d. 63
Answer: d. 63
Explanation:
Wall can be covered only by using square sized wallpaper pieces.
Different sized squares are not allowed.
Length = 4.5 m = 450 cm;
Height = 3.5 m = 350 cm
Maximum square size possible means HCF of 350 and 450
We can see that 350 and 450 can be divided by 50.
On dividing by 50, we get 7 and 9.
Since we cannot divide further, HCF = 50 = size of side of square
| Number of squares = | Wall area | = | 450 x 350 | = 63 |
| Square area | 50 x 50 |
a. 90
b. 180
c. 183
d. 900
Answer: c. 183
Explanation:
Since we need total number of blocks, we must first find LCM of 4, 5, 9, 10
-------------------------------------------------
2 4 5 9 10
-------------------------------------------------
5 2 5 9 5
2 1 9 1
-------------------------------------------------
∴ LCM = 2 x 5 x 2 x 1 x 9 x 1 = 180
Since remainder is same i.e. 3 blocks in each case, just add it to this LCM
Number of blocks = 180 + 3 = 183
Explanation:
Since we need total number of blocks, we must first find LCM of 4, 5, 9, 10
-------------------------------------------------
2 4 5 9 10
-------------------------------------------------
5 2 5 9 5
2 1 9 1
-------------------------------------------------
∴ LCM = 2 x 5 x 2 x 1 x 9 x 1 = 180
Since remainder is same i.e. 3 blocks in each case, just add it to this LCM
Number of blocks = 180 + 3 = 183
17. A pattern is formed with six type of LEDs. They start lighting together and then light at intervals of 2, 4, 6, 8, 10 and 12 seconds, respectively. In half an hour, how many times do they light together?
a. 4
b. 10
c. 15
d. 16
Answer: d. 16
18. Find the largest of following fractions.
a. 63/80
b. 31/40
c. 13/16
d. 7/8
Answer: d. 7/8
19. Find the smallest number which when reduced by 7, can be perfectly divided by 12, 16, 18, 21 and 28 each.
a. 1008
b. 1015
c. 1022
d. 1032
Answer: a. 1008
20. A and B are two primes numbers, A > B. Their LCM is 161. Find the value of 3B - A :
a. -2
b. -1
c. 1
d. 2
Answer: a. -2
21. Consider the following:
A and B are two numbers such that (A + B) = 528
HCF of A and B = 33.
How many pairs of such numbers exist?
a. 4
b. 6
c. 8
d. 12
22. Given:
i. X, Y, Z are three numbers in the ratio - 1 : 2 : 3.
ii. Their HCF = 12.
Find the numbers.
a. 4, 8, 12
b. 5, 10, 15
c. 10, 20, 30
d. 12, 24, 36
23. How can I represent 252 as a product of just the prime numbers?
a. 2 * 2 * 2 * 3 * 7
b. 2 * 2 * 3 * 3 * 7
c. 2 * 3 *3 * 3 * 7
d. 3 * 3 *3 * 3 * 7
24. Which of the following numbers has the maximum number of divisors?
a. 99
b. 101
c. 176
d. 182
25. Given:
Three number: 1.75, 5.6 and 7
Find their HCF.
a. 0.07
b. 0.7
c. 0.35
d. 3.5
21. Consider the following:
A and B are two numbers such that (A + B) = 528
HCF of A and B = 33.
How many pairs of such numbers exist?
a. 4
b. 6
c. 8
d. 12
Answer: a. 4
22. Given:
i. X, Y, Z are three numbers in the ratio - 1 : 2 : 3.
ii. Their HCF = 12.
Find the numbers.
a. 4, 8, 12
b. 5, 10, 15
c. 10, 20, 30
d. 12, 24, 36
Answer: d. 12, 24, 36
23. How can I represent 252 as a product of just the prime numbers?
a. 2 * 2 * 2 * 3 * 7
b. 2 * 2 * 3 * 3 * 7
c. 2 * 3 *3 * 3 * 7
d. 3 * 3 *3 * 3 * 7
Answer: b. 2 * 2 * 3 * 3 * 7
24. Which of the following numbers has the maximum number of divisors?
a. 99
b. 101
c. 176
d. 182
Answer: c. 176
25. Given:
Three number: 1.75, 5.6 and 7
Find their HCF.
a. 0.07
b. 0.7
c. 0.35
d. 3.5
Answer: c. 0.35
Comments
Post a Comment
COMMENT