Cost price of 25 kg = Rs. (15 x 14.50 + 10 x 13) = Rs. 347.50.
Sell price of 25 kg = Rs. (25 x 15) = Rs. 375.
profit = Rs. (375 — 347.50) = Rs. 27.50.
Toatal bill paid by Amit, Raju and Ram = ( 50 + 55 +75 ) = Rs. 180
Let amount paid by Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively.
Therefore, (3x + 4x + 5x ) = 180
12x = 180
x = 15
Therefore, amount paid by,
Amit = Rs. 45
Raju = Rs. 60
Ram = Rs. 75
But actually as given in the question, Amit pays Rs. 50, Raju pays Rs. 55 and Ram pays Rs. 8@Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he needs to pay Rs. 5 to Raju settle the amount.
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
Gain =(80/1600*100) % = 5%.
Here always remember, when ever x% loss,
it me S.P. = (100 - x)% of C.P
when ever x% profit,
it me S.P. = (100 + x)% of C.P
So here will be (100 - x)% of C.P.
= 80% of 1200
= (80/100) * 1200
= 960.
Let the cost price = Rs 100
then, Marked price = Rs 135
Required gain = 8%,
So Selling price = Rs 108
Discount = 135 - 108 = 27
Discount% = (27/135)*100 = 20%.
Assume original salaries of Ram and Sham as 4x and 5x respectively.
Therefore,
(4x + 5000)/= 50
(5x + 5000) 60
60 (4x + 5000) = 50 (5x + 5000)
10 x = 50,000
5x = 25, 000
Sham's present salary = 5x + 5000 = 25,000 + 5000
Sham's present salary = Rs. 30,000.
Required largest number must be divisible by the L.C.M. of 12, 15 and 18
L.C.M. of 12, 15 and 18
12 = 2 × 2 × 3
15 =5 × 3
18 = 2 × 3 × 3
L.C.M. = 180
Now divide 9999 by 180, we get remainder as 99
The required largest number = (9999 – 99) =9900
Number 9900 is exactly divisible by 180.
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9.
Sell price of 2 dozen oranges = Rs. 11.
If profit is Rs 2, oranges bought = 2 dozen.
If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.
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 me 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.
Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.
Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x.
[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105
[15x + 8x + 12x] = 105
35x = 105
x = 3
Hence, man's daily wage = 5x = 5 x 3 = Rs. 15
Wife's daily wage = 4x = 4 x 3 = Rs. 12
Daughter's daily wage = 3x = 3 x 3 = Rs. 9.
All 3 digit numbers divisible by 3 are :
102, 105, 108, 111, ..., 999.
This is an A.P. with first element 'a' as
102 and difference 'd' as 3.
Let it contains n terms. Then,
102 + (n - 1) x3 = 999
102 + 3n-3 = 999
3n = 900 or n = 300
Sum of AP = n/2 [2*a + (n-1)*d]
Required sum = 300/2[2*102 + 299*3] = 165150.
SP2 = 2/3 SP1
CP = 100
SP2 = 80
2/3 SP1 = 80
SP1 = 120
100 --- 20 => 20%.
Here greatest number that can divide me 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.
Let the article costs 'X' to A
Cost price of B = 1.2X
Cost price of C = 0.75(1.2X) = 0.9X
Cost price of D = 1.4(0.9X) = 1.26X = 252
Amount paid by A for the article = Rs. 200.
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.
Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.
160@A Man Walking At The Rate Of 5 Km/hr Crosses A Bridge In 15 Minutes. The Length Of The Bridge (in Metres) Is?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.
Let C.P. = Rs. C.
Then, 832 - C = C - 448
2C = 1280 => C = 640
Required S.P. = 150% of Rs. 640 = 150/100 x 640 = Rs. 960.
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = (75/25) * 100 = 300%.
Let S.P. of 45 lemons be Rs. x.
Then, 80 : 40 = 120 : x or x = 40×120/80= 60
For Rs.60, lemons sold = 45
For Rs.24, lemons sold =4560×24= 18.
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.