Answer :
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.
Answer :
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. 80. 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.
Answer :
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.
Question 4. Two Dice Are Tossed.the Probability That The Total Score Is A Prime Number Is?
Answer :
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.
Question 5. Find The Lcm Of Following Three Fractions:36/,48/,72/?
Answer :
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.
Answer :
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.
Answer :
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.
Answer :
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.
Question 9. Find The Largest Number Of 4-digits Divisible By 12, 15 And 18?
Answer :
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.
Answer :
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%.
Answer :
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.
Answer :
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
Question 13. If The Cost Price Is 25% Of Selling Price. Then What Is The Profit Percent?
Answer :
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = (75/25) * 100 = 300%.
Answer :
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.
Question 15. The Sum Of All 3 Digit Numbers Divisible By 3 Is?
Answer :
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.
Answer :
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.
Answer :
speed= (5x5/18)m/sec=25/18 m/sec.
Distance covered in 15 minutes= (25/18 x 15 x 60)m= 1250 m.
Answer :
Total time taken = (160/64 + 160/8)hrs
= 9/2 hrs.
Average speed = (320 x 2/9) km.hr
= 71.11 km/hr.
Answer :
4107 is the square of 37.
So let two numbers be 37x and 37y.
37x × 37y = 4107
xy = 3
3 is the product of (1 and 3)
x = 1 and y = 3
37x = 37 × 1 =37
37y = 37 × 3 = 111
Greater number = 111.
Answer :
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.
Answer :
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.
Question 22. The Product Of Two Numbers Is 2028 And Their H.c.f. Is 13. The Number Of Such Pairs Is?
Answer :
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
Question 23. Three Number Are In The Ratio Of 3 : 4 : 5 And Their L.c.m. Is 2400. Their H.c.f. Is?
Answer :
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Answer :
Let the required numbers be 33a and 33b.
Then 33a +33b= 528 => a+b = 16.
Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).
Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)
The number of such pairs is 4.
Answer :
Let the numbers be x and (100-x).
Then,x(100-x)=5*495
=> x2-100x+2475=0
=> (x-55) (x-45) = 0
=> x = 55 or x = 45
The numbers are 45 and 55
Required difference = (55-45) = 10.
Answer :
3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7
5.25 meters=525 cm = 5 × 5 × 3 × 7
Hence common factors are 3 and 7
Hence LCM = 3 × 7 = 21
Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.
Question 27. What Is The Probability Of Getting 53 Mondays In A Leap Year?
Answer :
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364days
366 – 364 = 2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days = 2/7.
Answer :
Let original Cost price is x
Its Selling price = (105/100) * x = 21x/20
New Cost price = (95/100) * x = 19x/20
New Selling price = (110/100 )* (19x/20 )= 209x/200
[(21x/20) - (209x/200)] = 1
=> x = 200.
Answer :
Let cp= 100,
35 % increase in sp=135
10 % discount in 135((135*10)/100)=13.5
so 1st sp=(135-13.5)=121.5, again 15 % discount in 1st sp((121.5*15)/100)=18.225
2nd sp=(121.5-18.225)=103.275,
so finally cp=100,sp=103.275 ,gain by 3.27%.
Answer :
Given M.P=45,S.P=42, Profit = 0.05
Let C.P=x , Then
Profit = (42-x)/x = 0.05
=> x = 40.
Answer :
Let the shopkeeper buy 300g for Rs.300. Now he sells 100g for Rs.110, another 100g for Rs120, and the rest 100g for Rs94. sir
Therefore, the total amount he receives = Rs.110 + Rs.120 + Rs.94 = 324.
Therefore, the shopkeeper spends Rs.300 and gets back Rs.324.
Therefore, his profit percentage = 24/300x100 % = 8%.
Answer :
Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).
Sell price of 5 kg = Rs. (22x5) = Rs. 110.
[{110 - (28 + 3x)}/(28 + 3x) ]* 100 =10
[82-3x/28 + 3x]= 1 / 10
820 - 30x = 28 +3x
33x = 792
x = 24.
Answer :
SP2 = 2/3 SP1
CP = 100
SP2 = 80
2/3 SP1 = 80
SP1 = 120
100 --- 20 => 20%.
Answer :
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.
Answer :
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.
Answer :
Here always remember, when ever x% loss,
it means S.P. = (100 - x)% of C.P
when ever x% profit,
it means S.P. = (100 + x)% of C.P
So here will be (100 - x)% of C.P.
= 80% of 1200
= (80/100) * 1200
= 960.
Answer :
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.
Answer :
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%.
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