1. The wages of 24 men and 16 women amounts to Rs.11600 per day. Half the number of men and 37 women earn the same amount per day. What is the daily wage of a man?
Let the wage of a man is m and woman be w.
24m+16w=11600
12m+37w = 11600
Solving we get m = 350
2. The sum of three digits a number is 17. The sum of square of the digits is 109. If we substract 495 from the number, the number is reversed. Find the number.
Let the number be abc.
Then a + b + c= 17 .....(1)
${a^2} + {b^2} + {c^2} = 109$ .....(2)
100a+10b+c -495 = 100c+10b+a ......(3)
From 3, we get a - c = 5
So the possibilities for (a, c, b) are (6,1,10), (7,2,8), (8,3,6), (9,4,4)
From the above, (8,3,6) satisfies the condition.
3. A calculator has a key for squaring and another key for inverting. So if x is the displayed number, then pressing the square key will replace x by x^2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternatively presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be
Even number of inverse key has no effect on the number. For example, Initially the given number is 6. Square key makes it ${{6^2}}$ and invert key makes it $\dfrac{1}{{{6^2}}}$. Now again square key makes it ${\left( {\dfrac{1}{{{6^2}}}} \right)^2} = \dfrac{1}{{{6^4}}}$ and invert key makes it ${6^4}$. Now observe clearly, after pressing square key 2 times, the power of 6 became 4.
By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of ${2^n}$. So after the 16 pressings the power becomes ${2^{16}}$
So the final number will be ${6^{{2^{16}}}} = {6^{65536}}$
4. How many two digit numbers are there which when subtracted from the number formed by reversing it's digits as well as when added to the number formed by reversing its digits, result in a perfect square.
Let the number xy = 10x + y
Given that, 10x+y - (10y - x) = 9(x-y) is a perfect square
So x-y can be 1, 4, 9. -------- (1)
So given that 10x+y +(10y +x) = 11(x+y) is a perfect square.
So x+y be 11. Possible options are (9,2), (8,3),(7,4),(6,5) ---------(2)
From the above two conditions only (6,5) satisfies the condition
Only 1 number 56 satisfies.
5. Find the 55th word of SHUVANK in dictionary
Sol: Arranging the letters in alphabetical order we get : A H K N S U V
Now Total words start with A are 6!
Total words start with AH are 5! = 120
Now
Total words start with AHK are 4! = 24
Total words start with AHN are 4! = 24
Total words start with AHSK are 3! = 6
Now AHSNKUV will be the last word required.
6. Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph
Relative speed = 60 - 40 = 20 kmph
Initial gap as car B leaves after 2 hours = 40 x 2 = 80 kms
Car B should be 9 km ahead of the A at a required time so it must be 89 km away
Time = 89 / 20 = 4.45 hrs or 267 mins
7. Find the average of the terms in the series 1-2+3-4+5....+199-200
Sol:(1-2) +(3-4) + (5-6) +........(199-200) = -100
Average = 100 / 200 = -0.5
8. n is a natural number and n^3 has 16 factors. Then how many factors can n^4 have?
Total factors of a number N=${a^p}.{b^q}.{c^r}...$ is (p+1)(q+1)(r+1)...
As ${n^3}$ has 16 factors ${n^3}$ can be one of the two formats given below
${n^3}$ =${a^{15}}$
${n^3}$ = ${a^3}.{b^3}$
If ${n^3}$ =${a^{15}}$ then n = ${a^5}$ and number of factors of ${n^4}$ = 21
${n^3}$ = ${a^3}.{b^3}$ then n = ab and number of factors ${n^4}$ = 25
9. Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged?
Let the total time taken by the cars be a and b
Let the time after which the speed is interchanged be t
For car A, 60t+90(a-t) = 420, 90a - 30t = 420 .......(1)
For car B, 90t + 60(b-t) = 420, 60b + 30t = 420 ....(2)
Using both (1) and (2), we get 90a + 60b = 840
But as a - b =1, 90a + 60(a-1) = 840.
Solving a = 6.
Substituting in equation 1, we get t = 4
Let the wage of a man is m and woman be w.
24m+16w=11600
12m+37w = 11600
Solving we get m = 350
2. The sum of three digits a number is 17. The sum of square of the digits is 109. If we substract 495 from the number, the number is reversed. Find the number.
Let the number be abc.
Then a + b + c= 17 .....(1)
${a^2} + {b^2} + {c^2} = 109$ .....(2)
100a+10b+c -495 = 100c+10b+a ......(3)
From 3, we get a - c = 5
So the possibilities for (a, c, b) are (6,1,10), (7,2,8), (8,3,6), (9,4,4)
From the above, (8,3,6) satisfies the condition.
3. A calculator has a key for squaring and another key for inverting. So if x is the displayed number, then pressing the square key will replace x by x^2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternatively presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be
Even number of inverse key has no effect on the number. For example, Initially the given number is 6. Square key makes it ${{6^2}}$ and invert key makes it $\dfrac{1}{{{6^2}}}$. Now again square key makes it ${\left( {\dfrac{1}{{{6^2}}}} \right)^2} = \dfrac{1}{{{6^4}}}$ and invert key makes it ${6^4}$. Now observe clearly, after pressing square key 2 times, the power of 6 became 4.
By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of ${2^n}$. So after the 16 pressings the power becomes ${2^{16}}$
So the final number will be ${6^{{2^{16}}}} = {6^{65536}}$
4. How many two digit numbers are there which when subtracted from the number formed by reversing it's digits as well as when added to the number formed by reversing its digits, result in a perfect square.
Let the number xy = 10x + y
Given that, 10x+y - (10y - x) = 9(x-y) is a perfect square
So x-y can be 1, 4, 9. -------- (1)
So given that 10x+y +(10y +x) = 11(x+y) is a perfect square.
So x+y be 11. Possible options are (9,2), (8,3),(7,4),(6,5) ---------(2)
From the above two conditions only (6,5) satisfies the condition
Only 1 number 56 satisfies.
5. Find the 55th word of SHUVANK in dictionary
Sol: Arranging the letters in alphabetical order we get : A H K N S U V
Now Total words start with A are 6!
Total words start with AH are 5! = 120
Now
Total words start with AHK are 4! = 24
Total words start with AHN are 4! = 24
Total words start with AHSK are 3! = 6
Now AHSNKUV will be the last word required.
6. Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph
Relative speed = 60 - 40 = 20 kmph
Initial gap as car B leaves after 2 hours = 40 x 2 = 80 kms
Car B should be 9 km ahead of the A at a required time so it must be 89 km away
Time = 89 / 20 = 4.45 hrs or 267 mins
7. Find the average of the terms in the series 1-2+3-4+5....+199-200
Sol:(1-2) +(3-4) + (5-6) +........(199-200) = -100
Average = 100 / 200 = -0.5
8. n is a natural number and n^3 has 16 factors. Then how many factors can n^4 have?
Total factors of a number N=${a^p}.{b^q}.{c^r}...$ is (p+1)(q+1)(r+1)...
As ${n^3}$ has 16 factors ${n^3}$ can be one of the two formats given below
${n^3}$ =${a^{15}}$
${n^3}$ = ${a^3}.{b^3}$
If ${n^3}$ =${a^{15}}$ then n = ${a^5}$ and number of factors of ${n^4}$ = 21
${n^3}$ = ${a^3}.{b^3}$ then n = ab and number of factors ${n^4}$ = 25
9. Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged?
Let the total time taken by the cars be a and b
Let the time after which the speed is interchanged be t
For car A, 60t+90(a-t) = 420, 90a - 30t = 420 .......(1)
For car B, 90t + 60(b-t) = 420, 60b + 30t = 420 ....(2)
Using both (1) and (2), we get 90a + 60b = 840
But as a - b =1, 90a + 60(a-1) = 840.
Solving a = 6.
Substituting in equation 1, we get t = 4
