Quant - Random Questions
Folks,
This post is for all CAT aspirants. You might have already covered these areas by now - this is just an attempt to help you recollect those topics in a random fashion. So here we go:
Quadratic Equations:
This post is for all CAT aspirants. You might have already covered these areas by now - this is just an attempt to help you recollect those topics in a random fashion. So here we go:
Quadratic Equations:
Consider the quadratic equation ax2+bx+c=0
If A and B are roots of the equation then (x-A)(x-B)=0
A+B=(-b/a)
AB=(c/a)
Sum of the roots=-coefficient of x/coefficient of x2
Product of the roots=constant term/coefficient of x2
Cubic Equation:
For any cubic equation of the form ax3+bx2+cx+d=0.
If roots are A,B & C then,
A+B+C=(-b/a)
AB+BC+CA=(c/a)
ABC=(-d/a)
Q) :f(x) is a cubic polynomial with f(2a) = f(2b) = f(2c) = 0 and f(1) = 114
It is known that a + b + c = 2, a2 + b2 + c2 = 6 and abc = −8
What is the value of f(2) ?
It is known that a + b + c = 2, a2 + b2 + c2 = 6 and abc = −8
What is the value of f(2) ?
So: f(x) is a cubic polynomial with roots 2a, 2b and 2c.
Thus,
x3 – (2a + 2b + 2c)x2 + (4ab + 4bc + 4ac)x – 8abc = 0 ...(i)
We have 2(ab + bc + ac) = (a + b + c)2 – (a2 + b2 + c2) = 4 – 6 = −2
Substituting this in equation (i), we get
x3 – 4x2 – 4x + 64 = 0 ...(ii)
We know that f(x) must be a real number multiple of (ii)
∴ f(x) = k(x3 – 4x2 – 4x + 64)
f(1) = 114
∴ 57k = 114
∴ k = 2
Thus,
x3 – (2a + 2b + 2c)x2 + (4ab + 4bc + 4ac)x – 8abc = 0 ...(i)
We have 2(ab + bc + ac) = (a + b + c)2 – (a2 + b2 + c2) = 4 – 6 = −2
Substituting this in equation (i), we get
x3 – 4x2 – 4x + 64 = 0 ...(ii)
We know that f(x) must be a real number multiple of (ii)
∴ f(x) = k(x3 – 4x2 – 4x + 64)
f(1) = 114
∴ 57k = 114
∴ k = 2
∴ f(x) = 2(x3 – 4x2 – 4x + 64)
Putting x = 2, we get f(2) = 2(8 – 16 – 8 + 64) = 96.
In this given series all the positive terms form a series 1,4,7,11.... and all negative terms form a series -5, -9, -11...
It has 11 terms of first AP and 10 terms of second AP.
Thought process:
As the question is about sum of the terms in AP, you should immediately recollect the formula for sum of the terms in AP. It is (n/2)[2a+(n-1)d].
So (11/2)[2(1)+10(3)] + (10/2)[2(-5)+9(-4)] = 176-230 = -54
Putting x = 2, we get f(2) = 2(8 – 16 – 8 + 64) = 96.
Q) What is the sum of first 21 terms of the series given below
1-5+4-9+7....In this given series all the positive terms form a series 1,4,7,11.... and all negative terms form a series -5, -9, -11...
It has 11 terms of first AP and 10 terms of second AP.
Thought process:
As the question is about sum of the terms in AP, you should immediately recollect the formula for sum of the terms in AP. It is (n/2)[2a+(n-1)d].
So (11/2)[2(1)+10(3)] + (10/2)[2(-5)+9(-4)] = 176-230 = -54
Q) 1 × 31 + 2 × 32 + 3 × 33 + … + n × 3n
What is the value of n?
What is the value of n?
Thought Process:
This is not in the form of GP a+ar+ar2+...arn
We should get it into that form by considering
Sn = 1*3 + 2 *32 +......n*3n
3Sn = 1*3 +..........(n-1)*3n + n*3n+1
Subtracting we get
-2Sn = 1*3+ 1*32 +....+1*3n-n*3n+1
-2Sn=3*(3n - 1)/2 - n*3n+1
4Sn= 3*(1-3n) +2*n*3n+1
Sn= (3/4)(1-3n + 2n*3n)
Sn = (3/4)[(2n-1)*3n+1]
Therefore n = 18
Q) A trader sells vegetables at a profit of 20%. Further, he uses a false weight and weighs only 800 gm instead of 1000 gm. What is his profit percentage?
Thought process:
Let us assume the cost price of veg be 1000Rs. So 20% profit means 200Rs. Hence the selling price=1200Rs. But the cost is only for 800
Profit=((1200-800)/800) * 100=50%.
Assumption is key here - when the cost price is not given we have to assume some value - normally 100 or 1000 as it will be easy to calculate. Here it was for 1000 gm so we have assumed CP as 1000.
Prime Numbers:
Some of the key points to remember regarding prime numbers are
1. All the prime numbers greater than 3 will be in the form of 6k+1 or 6k-1
2. There are 25 prime numbers between 1-100
3. All the prime numbers except 2 are odd
4.The number of ways in which a number X can be as expressed as a product of two factors which are relatively prime to each other is given by the formula:- 2n-1
Percentages
If C=A*B, and if A is increased by a%, and c is constant, then b will decrease by (a*100)/(a+100)
If C=A*B, and if A is decreased by b%, and c is constant, then b will increase by (a*100)/(a-100)
This formula can be used in host of other topics like
Distance= Speed*Time
Revenue= Sales units*Price
Expenditure= Price*Volume
Work Done= Men* No of Days
Infinite Equilateral Triangle
The middle points of the sides of the triangle are joined to form a second triangle . Again a third triangle is formed by joining the middle points of this second triangle and the process is repeated infinitely. If the perimeter and area of the outer triangle are P and A respectively, what will be the sum of the perimeters of triangles thus formed?
Sol:
When the mid points are joined in an equilateral, the total perimeter of the triangle also becomes half of the original perimeter.
Sum = P+P/2+P/4+P/8+P/16+P/32+P/64+P/128+.....= 2P.
Using the formula to sum to infinity of GP,
S= a/(1-r)
S=P/(1-(1/2)) = 2P
Q) Michael and Amanda are swimming with constant speed from two different ends of a swimming pool starting at the same time. When they meet for the first time, they are 750 m from Michael’s side of pool. They meet again for the second time at a distance of 500 m from Amanda’s side of the pool. If Amanda is yet to complete swimming the length of the pool, find the length of the swimming pool in metres?
Sol:
Let ‘t’ be the time taken by both when they meet for the first time.
When they meet for the second time, they will take another ‘t’ time since their speeds are constant and total distance travelled by them will be also same. Let 'd' be the length of the pool. When they meet first, Michael has covered 750 m and Amanda has covered (d – 750) m in time ‘t’.
When they meet for the second time, Amanda is yet to complete a round.
From the time they have started till they meet for second time, Amanda has completed 500 m in time ‘2t’ .
This means that Amanda must have completed 250 m in time ‘t’.
Total length of the pool = Distance covered by Amanda in time 't' + Distance covered by Michael in time 't' = 750 + 250 = 1000 m
Q) N' is the smallest natural number such that (25! / N) is a perfect square. Find the number of factors of N.
Sol:
The prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.
Let us find the highest power of all the above prime numbers in 25!. 25! = 25 x 24 x 23 x 22 ................................................x 3 x 2 x 1
25! = 222 × 310 × 56 × 73 × 112 × 13 × 17 × 19 × 23
For, (25! / N) to be a perfect square, all prime factors should have even power.
In the factorization, the numbers having odd power are 7, 13, 17, 19 and 23.
Since ‘N’ is the smallest number, we have
N = 7 × 13 × 17 × 19 × 23
Now, for a number A = axbycz, the number of factors = (x + 1)(y + 1)(z + 1) Hence, the number of factors of N = (1+1) (1+1) (1+1) (1+1) (1+1) = 32
Q) Four numbers are selected randomly from the first ten natural numbers and product of these four natural numbers is equal to 'M'? What is the probability that 'M' is a multiple of 8?
Sol:
We know that first ten natural numbers are, 1, 2, 3 ................, 10. Now we can choose 4 numbers out of 10 numbers in 10C4 ways = 10! / 6! 4! = 10 x 9 x 8 x 7 / 4 x 3 x 2 x 1 = 210 ways. Hence, 'M' can take 210 different values.
Now let us find out the number of values of 'M' which are divisible by 8:
Case1: - 8 is one of the number which is selected for sure. Now we can choose balance 3 numbers out of 9 numbers in 9C3 ways = 9! / 6! 3! = 9 x 8 x 7 / 3 x 2 x 1 = 84 ways. Hence, 'M' can take 84 values in this situation.
Case2: - 2 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 7 numbers (except 8, we are not counting 8 as all the cases pertaing to 8 have already been counted in Case1) in 7C2 ways = 7! / 5! 2! = 7 x 6 / 2 x 1 = 21 ways. Hence, 'M' can take 21 values in this situation.
Case3: - 6 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 6 numbers (except 8 and 2, we are not counting 8 and 2 as all the cases pertaing to 8 and 2 have already been counted in Case1 and Case2) in 6C2 ways = 6! / 4! 2! = 6 x 5 / 2 x 1 = 15 ways. Hence, 'M' can take 15 values in this situation.
Case4: - 10 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 5 numbers (except 8, 6 and 2, we are not counting 8, 6 and 2 as all the cases pertaing to 8, 6 and 2 have already been counted in Case1, Case2 and Case3) in 5C2 ways = 5! / 3! 2! = 5 x 4 / 2 x 1 = 10 ways. Hence, 'M' can take 10 values in this situation.
Case5: - 2, 6 and 10 are choosen for sure but 4 and 8 are not choosen. Now we can choose balance 1 numbers out of 5 numbers in 5 ways. Hence, 'M' can take 5 values in this situation.
Hence, total number of values 'M' can have which are divisible by 8 = 84+21+15+10+5 = 135
or, the required Probability = Desirable number of cases / Total number of cases = 135/210 = 9/14
Q) There are three taps ‘A’, ‘B’, ‘C’ in a tank filled with water and that of capacity 60 m3. Tap ‘A’ can empty the tank in 20 min., tap ‘B’ can empty it in 15 min. whereas tap ‘C’ can empty it in 12 min. All three taps are opened simultaneously when tank is full with water. In the process at some point of time one of the taps got blocked and it took exactly 7 min. to empty the tank. Which of the tap got blocked and when?
a) 'A' after 1 min b) 'B' after 1 min c) 'C' after 1 min d) 'A' after 30 secs.
Sol:
‘A’ can empty the tank in 20 min. and hence water pouring out = 60/20 = 3 m3
‘B’ can empty the tank in 15 min. and hence water pouring out = 60/15 = 4 m3
‘C’ can empty the tank in 12 min. and hence water pouring out = 60/12 = 5 m3
Hence total water coming out if all three taps are open = 3+4+5 = 12 m3 Total time to be taken to empty the tank = 60/12 = 5 minutes, but it is taking 7, minutes If tap ‘A’ got blocked after x min. then, 3(x) + 4(7) + 5(7) = 60, which is not possible. If tap ‘B’ got blocked after x min. then, 3(7) + 4(x) + 5(7) = 60, which gives the value of ‘x’ as 1 min. if tap ‘C’ got blocked after x min. then, 3(7) + 4(7) + 5(x) = 60, which gives the value of ‘x’ as 2.2 min. Given the available options, tap ‘B’ got blocked after 1 min. is the correct option. Hence, correct option is (b).
Q) In a 10-team NBA baskete-ball league, each team plays each of the other teams 18 times. No game ends in a tie, and, at the end of the season, each team is ahead of the next best team by the same positive number of games. What is the greatest number of games that the last place team could have won?
Sol:
The number of games played is (18 x (9) x 10)/2 = 810
If 'n' is the number of wins of the last-place team, and d is the common difference of wins between successive teams, then n + (n + d) + (n + 2d) + • • • + (n + 9d) = 10n + 45d = 810 or, 2n + 9d = 162 Now 'n' is maximum when 'd' is minimum (but not zero, because there are no ties). Hence, the smallest integral value of 'd' for which 'n' is integral is d = 2, substituting this in the equation, we will get, n = 72
Q)Find the remainder when the LCM of 13471345 - 1 and 13471345 + 1, is divided by 10.
Sol:
The given numbers are two even consecutive numbers.
For any two even consecutive even number, the HCF = 2
According to the property, HCF x LCM = Product of the two numbers
As last digit of 13471345-1 is 6 and that of 13471345+1 is 8, the last digit of the product = 6 x 8 = 48
Since, the product ends in 8 and HCF is 2, we get the last digit of LCM as 4.
The remainder of any number when divided by 10 is always the last/unit digit of the number. Since the LCM of the two numbers end in 4, when divided by 10 will give a remainder 4.
Q) Rohit wants to propose to his girlfriends, Sonia and Priyanka. The probability of him proposing to Sonia is 0.3 and proposing to Priyanka is 0.4. The probability of Sonia accepting the proposal is 0.7 and Priyanka accepting the same proposal is 0.6. If Rohit‘s proposal is being accepted, then what is the probability that it is accepted by Priyanka?
Sol:
This is the case of Conditional probability.
Required Probability = (Priyanka accepting proposal)/(Priyanka accepting proposal+Sonia accepting proposal)
Probability of Priyanka accepting proposal = 0.4 * 0.6
Probability of Sonia accepting proposal = 0.3 * 0.7
Required probability = ( 0.4 * 0.6 ) / ( 0.3 * 0.7 + 0.4 * 0.6)
= 24 / 45
= 8 / 15
CUBES:
If we paint a cube of the dimension n * n * n in any one color and cut it to get n3 symmetric cubelets; then the number of cubelets with color on different faces can be categorized as follows:
Cubelets with only one face painted - 6 * (n-2)2
Cubelets with only two faces painted - (n-2) * 12
Cubelets with three faces painted - 8
Cubelets with no face painted - (n-2)3
3Sn = 1*3 +..........(n-1)*3n + n*3n+1
Subtracting we get
-2Sn = 1*3+ 1*32 +....+1*3n-n*3n+1
-2Sn=3*(3n - 1)/2 - n*3n+1
4Sn= 3*(1-3n) +2*n*3n+1
Sn= (3/4)(1-3n + 2n*3n)
Sn = (3/4)[(2n-1)*3n+1]
Therefore n = 18
Q) A trader sells vegetables at a profit of 20%. Further, he uses a false weight and weighs only 800 gm instead of 1000 gm. What is his profit percentage?
Thought process:
Let us assume the cost price of veg be 1000Rs. So 20% profit means 200Rs. Hence the selling price=1200Rs. But the cost is only for 800
Profit=((1200-800)/800) * 100=50%.
Assumption is key here - when the cost price is not given we have to assume some value - normally 100 or 1000 as it will be easy to calculate. Here it was for 1000 gm so we have assumed CP as 1000.
Prime Numbers:
Some of the key points to remember regarding prime numbers are
1. All the prime numbers greater than 3 will be in the form of 6k+1 or 6k-1
2. There are 25 prime numbers between 1-100
3. All the prime numbers except 2 are odd
4.The number of ways in which a number X can be as expressed as a product of two factors which are relatively prime to each other is given by the formula:- 2n-1
Percentages
If C=A*B, and if A is increased by a%, and c is constant, then b will decrease by (a*100)/(a+100)
If C=A*B, and if A is decreased by b%, and c is constant, then b will increase by (a*100)/(a-100)
This formula can be used in host of other topics like
Distance= Speed*Time
Revenue= Sales units*Price
Expenditure= Price*Volume
Work Done= Men* No of Days
Infinite Equilateral Triangle
The middle points of the sides of the triangle are joined to form a second triangle . Again a third triangle is formed by joining the middle points of this second triangle and the process is repeated infinitely. If the perimeter and area of the outer triangle are P and A respectively, what will be the sum of the perimeters of triangles thus formed?
Sol:
When the mid points are joined in an equilateral, the total perimeter of the triangle also becomes half of the original perimeter.
Sum = P+P/2+P/4+P/8+P/16+P/32+P/64+P/128+.....= 2P.
Using the formula to sum to infinity of GP,
S= a/(1-r)
S=P/(1-(1/2)) = 2P
Q) Michael and Amanda are swimming with constant speed from two different ends of a swimming pool starting at the same time. When they meet for the first time, they are 750 m from Michael’s side of pool. They meet again for the second time at a distance of 500 m from Amanda’s side of the pool. If Amanda is yet to complete swimming the length of the pool, find the length of the swimming pool in metres?
Sol:
Let ‘t’ be the time taken by both when they meet for the first time.
When they meet for the second time, they will take another ‘t’ time since their speeds are constant and total distance travelled by them will be also same. Let 'd' be the length of the pool. When they meet first, Michael has covered 750 m and Amanda has covered (d – 750) m in time ‘t’.
When they meet for the second time, Amanda is yet to complete a round.
From the time they have started till they meet for second time, Amanda has completed 500 m in time ‘2t’ .
This means that Amanda must have completed 250 m in time ‘t’.
Total length of the pool = Distance covered by Amanda in time 't' + Distance covered by Michael in time 't' = 750 + 250 = 1000 m
Q) N' is the smallest natural number such that (25! / N) is a perfect square. Find the number of factors of N.
Sol:
The prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.
Let us find the highest power of all the above prime numbers in 25!. 25! = 25 x 24 x 23 x 22 ................................................x 3 x 2 x 1
25! = 222 × 310 × 56 × 73 × 112 × 13 × 17 × 19 × 23
For, (25! / N) to be a perfect square, all prime factors should have even power.
In the factorization, the numbers having odd power are 7, 13, 17, 19 and 23.
Since ‘N’ is the smallest number, we have
N = 7 × 13 × 17 × 19 × 23
Now, for a number A = axbycz, the number of factors = (x + 1)(y + 1)(z + 1) Hence, the number of factors of N = (1+1) (1+1) (1+1) (1+1) (1+1) = 32
Q) Four numbers are selected randomly from the first ten natural numbers and product of these four natural numbers is equal to 'M'? What is the probability that 'M' is a multiple of 8?
Sol:
We know that first ten natural numbers are, 1, 2, 3 ................, 10. Now we can choose 4 numbers out of 10 numbers in 10C4 ways = 10! / 6! 4! = 10 x 9 x 8 x 7 / 4 x 3 x 2 x 1 = 210 ways. Hence, 'M' can take 210 different values.
Now let us find out the number of values of 'M' which are divisible by 8:
Case1: - 8 is one of the number which is selected for sure. Now we can choose balance 3 numbers out of 9 numbers in 9C3 ways = 9! / 6! 3! = 9 x 8 x 7 / 3 x 2 x 1 = 84 ways. Hence, 'M' can take 84 values in this situation.
Case2: - 2 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 7 numbers (except 8, we are not counting 8 as all the cases pertaing to 8 have already been counted in Case1) in 7C2 ways = 7! / 5! 2! = 7 x 6 / 2 x 1 = 21 ways. Hence, 'M' can take 21 values in this situation.
Case3: - 6 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 6 numbers (except 8 and 2, we are not counting 8 and 2 as all the cases pertaing to 8 and 2 have already been counted in Case1 and Case2) in 6C2 ways = 6! / 4! 2! = 6 x 5 / 2 x 1 = 15 ways. Hence, 'M' can take 15 values in this situation.
Case4: - 10 and 4 are choosen for sure. Now we can choose balance 2 numbers out of 5 numbers (except 8, 6 and 2, we are not counting 8, 6 and 2 as all the cases pertaing to 8, 6 and 2 have already been counted in Case1, Case2 and Case3) in 5C2 ways = 5! / 3! 2! = 5 x 4 / 2 x 1 = 10 ways. Hence, 'M' can take 10 values in this situation.
Case5: - 2, 6 and 10 are choosen for sure but 4 and 8 are not choosen. Now we can choose balance 1 numbers out of 5 numbers in 5 ways. Hence, 'M' can take 5 values in this situation.
Hence, total number of values 'M' can have which are divisible by 8 = 84+21+15+10+5 = 135
or, the required Probability = Desirable number of cases / Total number of cases = 135/210 = 9/14
Q) There are three taps ‘A’, ‘B’, ‘C’ in a tank filled with water and that of capacity 60 m3. Tap ‘A’ can empty the tank in 20 min., tap ‘B’ can empty it in 15 min. whereas tap ‘C’ can empty it in 12 min. All three taps are opened simultaneously when tank is full with water. In the process at some point of time one of the taps got blocked and it took exactly 7 min. to empty the tank. Which of the tap got blocked and when?
a) 'A' after 1 min b) 'B' after 1 min c) 'C' after 1 min d) 'A' after 30 secs.
Sol:
‘A’ can empty the tank in 20 min. and hence water pouring out = 60/20 = 3 m3
‘B’ can empty the tank in 15 min. and hence water pouring out = 60/15 = 4 m3
‘C’ can empty the tank in 12 min. and hence water pouring out = 60/12 = 5 m3
Hence total water coming out if all three taps are open = 3+4+5 = 12 m3 Total time to be taken to empty the tank = 60/12 = 5 minutes, but it is taking 7, minutes If tap ‘A’ got blocked after x min. then, 3(x) + 4(7) + 5(7) = 60, which is not possible. If tap ‘B’ got blocked after x min. then, 3(7) + 4(x) + 5(7) = 60, which gives the value of ‘x’ as 1 min. if tap ‘C’ got blocked after x min. then, 3(7) + 4(7) + 5(x) = 60, which gives the value of ‘x’ as 2.2 min. Given the available options, tap ‘B’ got blocked after 1 min. is the correct option. Hence, correct option is (b).
Q) In a 10-team NBA baskete-ball league, each team plays each of the other teams 18 times. No game ends in a tie, and, at the end of the season, each team is ahead of the next best team by the same positive number of games. What is the greatest number of games that the last place team could have won?
Sol:
The number of games played is (18 x (9) x 10)/2 = 810
If 'n' is the number of wins of the last-place team, and d is the common difference of wins between successive teams, then n + (n + d) + (n + 2d) + • • • + (n + 9d) = 10n + 45d = 810 or, 2n + 9d = 162 Now 'n' is maximum when 'd' is minimum (but not zero, because there are no ties). Hence, the smallest integral value of 'd' for which 'n' is integral is d = 2, substituting this in the equation, we will get, n = 72
Q)Find the remainder when the LCM of 13471345 - 1 and 13471345 + 1, is divided by 10.
Sol:
The given numbers are two even consecutive numbers.
For any two even consecutive even number, the HCF = 2
According to the property, HCF x LCM = Product of the two numbers
As last digit of 13471345-1 is 6 and that of 13471345+1 is 8, the last digit of the product = 6 x 8 = 48
Since, the product ends in 8 and HCF is 2, we get the last digit of LCM as 4.
The remainder of any number when divided by 10 is always the last/unit digit of the number. Since the LCM of the two numbers end in 4, when divided by 10 will give a remainder 4.
Q) Rohit wants to propose to his girlfriends, Sonia and Priyanka. The probability of him proposing to Sonia is 0.3 and proposing to Priyanka is 0.4. The probability of Sonia accepting the proposal is 0.7 and Priyanka accepting the same proposal is 0.6. If Rohit‘s proposal is being accepted, then what is the probability that it is accepted by Priyanka?
Sol:
This is the case of Conditional probability.
Required Probability = (Priyanka accepting proposal)/(Priyanka accepting proposal+Sonia accepting proposal)
Probability of Priyanka accepting proposal = 0.4 * 0.6
Probability of Sonia accepting proposal = 0.3 * 0.7
Required probability = ( 0.4 * 0.6 ) / ( 0.3 * 0.7 + 0.4 * 0.6)
= 24 / 45
= 8 / 15
CUBES:
If we paint a cube of the dimension n * n * n in any one color and cut it to get n3 symmetric cubelets; then the number of cubelets with color on different faces can be categorized as follows:
Cubelets with only one face painted - 6 * (n-2)2
Cubelets with only two faces painted - (n-2) * 12
Cubelets with three faces painted - 8
Cubelets with no face painted - (n-2)3
Q)125 small but identical cubes have been put together to form a large cube.how many such small cubes would be required to cover this large cube completely?
Sol:
The size of the cube made using 125 small cubes is 5 x 5 x 5. After covering this the size of the cube will become (n+2) x (n+2) x (n+2) i.e. 7 x 7 x 7.
So total numbers of cubes in 7 x 7 x 7 = 343.
Extra number of cubes required = 343 - 125 = 218
So total numbers of cubes in 7 x 7 x 7 = 343.
Extra number of cubes required = 343 - 125 = 218
2)what is the maximum number of identical pieces of cube that can be cut into by 3 cuts?
Sol:
I think its clear that the 3 cuts are to be put in the three different axis.
Next putting 1 cut, splits an edge into 2 equal halves, 2 cuts will split into 3 equal parts and so on.
Here 1 cut is out in each axis ... So new cube dimensions are 2 x 2 x 2. Total number of smaller cubes = 8
Next putting 1 cut, splits an edge into 2 equal halves, 2 cuts will split into 3 equal parts and so on.
Here 1 cut is out in each axis ... So new cube dimensions are 2 x 2 x 2. Total number of smaller cubes = 8
Formula: 2n Where n is the number of cuts
3) 125 small but identical cubes are cut to form a large cube.This large cube is now painted on all six faces. How many of the smaller cubes have no faces painted at all / have exactly one face painted?
Sol:
Number of cubes with no face painted = 3 x 3 x 3 = 27
One face painted = 6 x (n-2) x (n-2). So it is = 6 x 3 x 3 = 54
4)
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