Answer :
The CP of profitable cow = 9900/1.1 = 9000
and profit = Rs. 900
The CP of loss yielding cow = 9900/0.8 = 12375
and loss = Rs. 2475
so, the net loss = 2475 - 900 = 1575.
Answer :
Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420) * 100
= 70%(approx).
Answer :
Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80x(15/100) = Rs. 12 = 80 + 12 = Rs. 92
Present Savings = 100 – 92 = Rs. 8
100 ------ 8
? --------- 400 => 5000
His salary = Rs. 5000.
Answer :
Taking the 2 investments to be 3x and 5x respectively
Total income of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25
Therefore, Gain% = 0.25/8 x 100 = 3.125 %.
Answer :
Let 1kg of Rs. 100 then 840gm is of Rs. 84.
Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.]
So now S.P = Rs. 96
But actually it contains 840 gm so C.P for shopkeeper = Rs. 84
S.P = Rs. 96
C.P = Rs. 84
Profit% = {(S.P-C.P)/C.P}x100
{(96-84)/84} x 100 = 14.28571429% PROFIT.
Answer :
Given that SP = Rs. 12000 - 10% = Rs. 10,800
Loss% = 4
We know that, C.P = 100/(100 - Loss%) x 100
=> 100/100-4 x 10800
=> 1080000/96
C.P = Rs. 11,250.
Answer :
Let the cost price of a ball is Rs.x
Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls
The equation is :
17x - 720 = 5x
Solving the equation
we get x = 60
Therefore, cost price of a ball is Rs. 60.
Answer :
Let S be the sample space.
Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.
P(E) = n(E)/n(S) = 169/1326 = 13/102.
Question 9. The Table Is Bought For Rs. 1950 And Sold At Rs. 2340. Find The Profit Percent?
Answer :
Cost Price = Rs. 1950
Selling Price = Rs. 2340
Profit = S.P – C.P
Profit = Rs. 2340 – 1950 = 390
Profit % = (Profit/C.P) x 100
Profit % = (390/1950) x 100
Profit % = 20 %.
Answer :
P(first letter is not vowel) = 2/4
P(second letter is not vowel) = 1/3
So, probability that none of letters would be vowels is = 2/4×1/3=1/6.
Answer :
The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
= 6*6*6*6=64
n(S) = 64
Let X be the event that all dice show the same face.
X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}
n(X) = 6
Hence required probability = n(X)n(S)=6/64=1216.
Question 12. Three Unbiased Coins Are Tossed. What Is The Probability Of Getting At Most Two Heads?
Answer :
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
Question 13. Three Unbiased Coins Are Tossed.what Is The Probability Of Getting At Least 2 Heads?
Answer :
Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
P(E) = n(E) / n(S)
= 4/8= 1/2.
Answer :
Here n(S) = (6 x 6) = 36
Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Answer :
Amount of Rs. 100 for 1 year
when compounded half-yearly = Rs.[100*(1+3/100)^2]=Rs.106.09
Effective rate=(106.09-100)%=6.09%.
Answer :
C.I.= Rs.[4000*(1+10/100)^2-4000]
=Rs.840
sum=Rs.(420 * 100)/3*8=Rs.1750.
Answer :
Amount
= Rs.(25000x(1+12/100)³
= Rs.(25000x28/25x28/25x28/25)
= Rs. 35123.20.
C.I = Rs(35123.20 -25000)
= Rs.10123.20.
Answer :
Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = P(1 + (r/100))n
(P+C.I) = P[1 + (10/100)]2
(P+420) = P[11/10][11/10]
P-1.21P = -420
0.21P = 420
Hence, P = 420/0.21 = Rs. 2000.
Answer :
S.I. = Rs.(1200*10*1)/100=rs.120
C.I. =rs[1200*(1+5/100)2-1200]=rs.123
Difference = Rs.(123-120) =Rs.3
Answer :
Principal = Rs.16,000;
Time=9 months = 3 quarters;
Amount
=Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20]
= Rs.18522.
C.I
= Rs.(18522 - 16000)
= Rs.2522.
Question 21. In How Many Ways Can The Letters Of The Word 'leader' Be Arranged ?
Answer :
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360.
Answer :
Clearly, Rate = 5% p.a .,
Time = 3 years
S.I =Rs.1200.
So,Principal
=Rs.(100 x 1200/3x5)
=Rs.8000.
Amount
=Rs.[8000 x (1+5/100)³]
=Rs(8000x21/20x21/20x21/20)
= Rs.9261
C.I
=Rs.(9261-8000)
=Rs.1261.
Answer :
Let the required no of working hours per day be x.
More pumps , Less working hours per day (Indirect Proportion)
Less days, More working hours per day (Indirect Proportion)
Pumps4 : 3Days1 : 2}?? 8:x
=> (4 * 1 * x) = (3 * 2 * 8)
=> x=12
Answer :
Divisor = 2/3 x dividend
and Divisor = 2 x remainder
or 2/3 x dividend = 2 x 5
Dividend = 2 x 5 x 3 / 2 = 15.
Question 25. How Many Figures (digits) Are Required To Number A Book Containing 200 Pages ?
Answer :
Number of one digit pages from
1 to 9 = 9
Number of two digit pages from
10 to 99 = 90
Number of three digit pages from
100 to 200 = 101.
Answer :
Let 1/2 of the no. = 10x + y
and the no. = 10v + w
From the given conditions,
w= x and v = y-1
Thus the no. = 10 (y-1) + x
? 2(10x + y ) = 10 (y-1) + x
? 8y - 19x = 10 ...(i)
v + w = 7
? y-1 + x = 7
? x + y = 8
Solving equations (i) and (ii) , we get
x = 2 and y = 6.
Answer :
Let the two-digit number be 10x + y
10x + y = 7(x + y)
? x = 2y ...(i)
10(x +2 ) + (y + 2) = 6(x + y + 4) + 4
or 10x + y + 22 = 6x + 6y + 28
? 4x - 5y = 6 ...(ii)
Solving equations (i) and (ii)
We get x = 4 and y = 2.
Answer :
(x - 4) / 6 = 9
Multiply both sides by 6:
x - 4 = 54
Add 4 to both sides:
x = 58
(58 - 3) / 5 = 55 / 5 = 11.
Answer :
xy = 96050 ...(i)
and xz = 95625 ...(ii)
and y - z = 1 ... (iii)
Dividing (i) by (ii) we get
y/z = 96050 / 95625
= 3842 / 3825
= 226 / 225 ... (iv)
Combining (iii) and (iv) we get z = 225.
Answer :
Initial speed = 80km/hr
Total distance = 80 x 10 = 800km
New speed = 800/4 =200km/hr
Increase in speed = 200 - 80 = 120km/hr.
Answer :
Let the distance traveled be x km.
Then, x/10 - x/15 = 2
3x - 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.
Answer :
Speed = 44 kmph x 5/18 = 110/9 m/s
We know that, Time = distance/speed
Time = (360 + 140) / (110/9)
= 500 x 9/110 = 41 sec.
Answer :
Suppose they meet after 'h' hours
Then
3h + 4h = 17.5
7h = 17.5
h = 2.5 hours
So they meet at => 4 + 2.5 = 6:30 pm.
Answer :
Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.
Answer :
Total running distance in four weeks = (24 x 240) + (4 x 400)
= 5760 + 1600
= 7360 meters
= 7360/1000
=> 7.36 kms.
Answer :
Using pythagarous theorem,
distance travelled by first train = 36x5/18x30 = 300m
distance travelled by second train = 48x5/18x30 = 400m
so distance between them =v( 90000 + 160000) = v250000 = 500mts.
Answer :
To find the minimum distance, we have to get the LCM of 75, 80, 85
Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400
Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.
Question 38. Running 3/4th Of His Usual Rate, A Man Is 15min Late. Find His Usual Time In Hours ?
Answer :
Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time
so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.
Answer :
Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.
Answer :
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
Question 41. Which Of The Following Has Most Number Of Divisors?
Answer :
99 = 1 x 3 x 3 x 11
101= 1 x 101
176= 1 x 2x 2 x 2 x 2 x 11
182= 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, 99
divisors of 101 are 1,101
divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176
divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182
Hence , 176 hasthe most number of divisors.
Answer :
Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120.
Answer :
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4.
Question 44. Three Numbers Are In The Ratio Of 3:4:5 And Their L.c.m Is 3600.their Hcf Is?
Answer :
Let the numbers be 3x, 4x, 5x.
Then, their L.C.M = 60x.
So, 60x=3600 or x=60.
Therefore, The numbers are (3 x 60), (4 x 60), (5 x 60).
Hence,required H.C.F=60.
Question 45. Find The Lowest Common Multiple Of 24, 36 And 40?
Answer :
To find the LCM of 24, 36 and 40
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
40 = 2 x 2 x 2 x 5
Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5
= 8 x 9 x 5
= 72 x 5
= 360.
Answer :
Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4
Total value of 25 paise coins =(1 / 7 ) x 105 = 15
Total value of 50 paise coins = (2 / 7) x 105 = 30
Total value of 100 paise coins = (4 / 7) x 105 = 60
No. of 25 paise coins = 15 x 4 = 60 coins
No. of 50 paise coins = 30 x 2 = 60 coins
No. of 1 rupee coins = 60 x 1 = 60 coins.
Answer :
Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.
No. of 1 rupee coins = (11x / 1) =11x
No. of 50 paise coins = (9x / 0.5) = 18x
No. of 25 paise coins = (5x / 0.25) = 20x
11x + 18x + 9x = 342
38x = 342
x = 9
Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins.
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