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Principal = 60000
Rate = 12% per annum
As the interest rate is to be compounded half yearly so, r = 12/2 = 6%
(i) Amount after 6 months = 60000*[1 + 6/100]^1 = 60000*106/100 = Rs 63,600
(ii) Amount after 1 year(which can be counted as two 6 month period) = 60000*[1 + 6/100]^2 = (60000*106*106)/100*100 = Rs 67,416
Incomplete Question. Please mention time taken to get back to original point.
Principal = 20,000
Amount after 1 year = 25,500
(i) Let the rate be r
t = 1
Amount = P(1 + r/100)^t
25500 = 20000(1 + r/100)^1
1 + r/100 = 25500/20000
100 + r = 127.5
r = 27.5%
(ii) t = 3
Amount after 3 years = 20000 (1 + 27.5/100)^3
Amount = 20000 (1 + 275/1000)^3
Amount = 20000 (1275/1000)^3
Amount = 20000*1275*1275*1275/ (1000*1000*1000)
Amount = Rs 41,453.44
Answer is option (A).
Circumference of circle = Perimeter of Square
Perimeter of square = 4*a = 44
a = 11 cm
Circumference of circle = 2*pi*r = 44
2*(22/7)*r = 44
r = 7 cm
Area of square = a*a = 11*11 = 121 cm^2
Area of circle = pi*r*r = (22/7)*7*7 = 154 cm^2
Difference between Area of circle & Area of square = 154 - 121 = 33 cm^2
Total pens = 18
Yellow pens = 7
Blue pens = 6
Green Pens = 5
P(Yellow pen) = 7/18
P(Blue pen) = 6/18
P(Green pen) = 5/18
(i) P(neither green nor blue) = 1 - P(either blue or green) = 1 - (6/18 + 5/18) = 1 - 11/18 = 7/18
(ii) P(not blue) = 1 - P(blue) = 1 - 6/18 = 12/18 = 2/3
Total Cards = 52
P(Card) = 1/52
(i) There are 4 ace in a deck of cards. So, the probability of drawing an ace is
P(Ace) = 4/52 = 1/13
(ii) Thera are total 26 red cards in a deck.
13 Hearts + 13 Diamonds = 26 Red cards
So, probability of drawing a red card is
P(Red) = 26/52 = 1/2
p(Not Red) = 1 - P(Red) = 1 - 1/2 = 1/2
Speed = Distance / Time
Speed = 14 / 25 km/min = 14*60/25 km/hr = 33.6 km/hr
Travelling in 5 hours will be
Distance = Speed * Time
D = 33.6 * 5 = 168 km
Answer is option (C).
Let AB be building and CD be tower.
Let BD = AE = b & CE = h.
AB = DE = 30m.
Angle CAE = 45
Angle DAE = 60
So, in ∆ADE,
tan 60 = DE / AE
√3 = 30 / b
b = 30 / √3 = 10*√3
Now in ∆ACE,
tan 45 = CE / AE
1 = h / b
h = 10√3
Height of tower = CD = CE + DE = h + 30 = 10√3 + 30 = 10*1.732 +30 = 17.32 +30 = 47.32m
2kg sugar = 9 * 10^6 crystals
1kg sugar = (9 * 10^6)/2 crystals
(i) 5kg sugar = (5 * 9 * 10^6)/2 = 22.5 * 10^6 crystals
(ii) 1.2kg sugar = (1.2 * 9 * 10^6)/2 = 5.4 * 10^6 crystals
In 6 hrs = 840 bottles
In 1 hr = 840/6 = 140 bottles
S, in 5 hrs = 5 * 140 = 700 bottles.
Graph or data is not given.Problem can not be solved.
5^m / 5^(-3) = 5^5
5^m = 5^5 * 5^(-3)
As the base of multiplying numbers is same then power will add. So now equating the powers,
m = 5 + (-3)
m = 5 – 3
m = 2
Given the shape of room is Rectangle.
Let ABCD is rectangular room. Then,
BC = AD = 4.5 feet
AC = 7.5 feet
Let AB = CD = l
In triangle ABC, by applying Pythagoras theorem
AC^2 = AB^2 + BC^^2
7.5^2 = l^2 + 4.5^2
l^2 = 7.5^2 - 4.5^2
l^2 = 56.25 - 20.25
l^2 = 36
l = 6 feet
Area of room = 6 * 4.5 = 27 square feet
Let length & breadth of rectangle be l & b respectively.
Perimeter of rectangle = 2(l+b)
So, 2(l+b)/b = 5/1
2l + 2b = 5b
3b = 2l
b = 2l/3 ---------(1)
Now given,
Area of rectangle = 216
l * b = 216
From (1) put the value of 'b'
l * 2l/3 = 216
2 * l^2 = 3 * 216
l^2 = 3 * 216/2
l^2 = 324
l = 18
Length of rectangle = 18 cm
Perimeter of park = Distance covered in 88 minutes
Distance = 12 * 88 / 60 = 88/5 km = 88*1000/5 = 17600 m
Given, l/b = 3/2
l = 3b/2 -----------(1)
Perimeter = 17600 m
2(l+b) = 17600
3b/2 + b = 8800
5b/2 = 8800
b = 3520 m
By putting the value of 'b' in (1)
l = 3*3520/2 = 3 * 1760 = 5280 m
Area of park = l * b = 3520 * 5280 = 18,585,600 sq. m
Let the length & breadth of plot be 'l' & 'b' respectively.
l = b +20 -----(1)
Cost of fencing per meter = Rs. 26.50
Total cost of fencing = Rs. 5300
Total cost of fencing = Perimeter * Cost per meter = 5300
2(l+b) * 26.50 = 5300
From (1)
(b +20 +b)*53 = 5300
2b +20 = 5300/53
2b = 100 - 20
2b = 80
b = 80/2 = 40 m
So,
l = b + 20 = 40 + 20 = 60 m
Either the question or given options are incorrect.
l = 2b
A = l*b
(l-55)*(b+55) = A + 75
l*b + 55l - 55b -3025 = A + 75
A + 55*2b - 55b = A + 3100
110b - 55b = 3100
55b = 3100
b = 3100/55 = 620/11
So,
l = 2b = 2 * 620/11 = 112.73 cm
Let the age of A, B & C are A yrs, B yrs & C yrs. So,
A + B = B + C + 12
A = C + 12
C = A - 12
So, C is 12 years younger than A.
Let the monthly salary of surf be 'x'. Then,
Amount given to mother = x*25/100 = x/4
Now the amount left = x - x/4 = 3x/4
Total percentage spent on rent and monthly expenses = 15 + 25 = 40%
So, now out of left amount she spent 40% on rent & monthly expenses.
Amount to be spent on rent & monthly expenses = (3x/4)*(40/100) = 3x/10
So, now the amount she kept in bank = (3x/4) - (3x/10) = 9x/20
So, now
Amount in bank + Amount given to mother = 42000
9x/20 + x/4 = 42000
14x/20 = 42000
x = 60,000
So, monthly salary of surf is Rs. 60,00
Let the speed of the boat in still water be x kmph. Then,
Speed downstream = (x + 3) kmph,
Speed upstream = (x - 3) kmph.
Distance = Speed * Time
Since the Distance is same in both upstream & downstream. So,
(x + 3) * 1 = (x - 3) * 4/3
3x + 9 = 4x - 12
x = 21
So, speed of boat in still water is 21 kmph.
Speed = Distance / Time
Speed against the current = 2/2 = 1 kmph
Speed along the current = 1*60/20 = 3 kmph
Speed in still water = (1 + 3)/2 = 2 kmph
Time taken to go 5km in stationary water = 5/2 = 2.5 hrs = 2 hours 30 minutes
Speed = Distance / Time
Let man's speed against current be 'x kmph'
Man's speed with current = 15 kmph
Speed of current = 2,5 kmph
Then,
2.5 = (15 - x)/2
5 = 15 - x
x = 10
Man's speed against the current is 10 kmph.
Answer is option (A)
Let there are 'n' other workers & sum of their salary is N.
Let the sum of all technician's salary be S.
(S + N) / (7 + n) = 8000 ----------(1)
S / 7 = 12000
S = 84000 -----------(2)
N / n = 6000
N = 6000n -------------(3)
Now, put the value of S & N from (2) & (3) in (1). We get,
(84000 + 6000n) / (n + 7) = 8000
84000 + 6000n = 8000n + 56000
2000n = 28000
n = 14
So total workers = n+7 = 14+7 = 21
Sum of 36 students' age = 36*14 = 504
When we add teacher's age then
Average age = 15
Teacher's age = 37*15 - 504 = 555 - 504 = 51 years
Mean = Sum of the numbers / Total count of numbers
12 = (3+11+7+9+15+13+8+19+17+21+14+x) / 12
12*12 = 137 + x
144 = 137 + x
x = 144 - 137
x = 7
Time = Distance / Time
Time taken to reach the place = 150 / 50 = 3 hr
Time taken to return = 150 / 30 = 5 hr
Total distance = 150 + 150 = 300 km
Total time = 3 + 5 = 8 hr
Average speed = Total distance / Total time
Avg speed = 300 / 8 = 37.5 km/hr
Total runs in 10 matches = 38.9 * 10 = 389
Total runs in first 6 matches = 42 * 6 = 252
So, total runs in last 4 matches = 389 - 252 = 137
Average for last 4 matches = 137/4 = 34.25
Average weight of whole class = (36*40 + 44*35) / 36+44 = (1440 + 1540) / 80 = 2980 / 80 = 37.25
Avg marks of all students = (55*50 + 60*55 + 45*60) / (55+60+45)
= (2750 + 3300 + 2700) / 160 = 8750 / 160 = 54.6875
Answer is option (A)
Price of 357 apples = Rs. 1517.25
Price of 1 apple = 1517.25 / 357
Price of 49 dozen apples = (1517.25*49*12) / 357 = Rs. 2499
Let there are 'x' persons.
'x' person can finish work in = 100 days
(x-10) person can finish work in = 110 days
100 * (x-10) / x = 110
100x - 100*10 = 110x
110x - 100x = 1000
10x = 1000
x = 1000/10 = 100
So, there were originally 100 persons.
Answer is option (D)
Let PQ be tower & A & B are two points.
Let AQ = a & BQ = b
PQ = 600
In triangle APQ,
Angle APQ = 30
tan 30 = AQ / PQ
1/sqrt 3 = a/600
a = 600 / sqrt 3 = 346.41
In Triangle BPQ,
angle BPQ = 45
tan 45 = BQ/PQ
1 = b/ 600
b = 600
Difference between points = b-a = 600 - 346.41 = 254m (approx)
Let PQ be tower & bus at point b.
Let BQ be 'b'
PQ = 80
In Triangle BPQ,
angle PBQ = 30
tan 30 = PQ / BQ
1/ sqrt 3 = 80 / b
b = 80 * sqrt3 = 138.56
So, bus is 138.56 m from the tower.
Let the distance between foot of hill & pole be 'a' and height of pole be 'h'.
tan 30 = (100 - h) / a
1 / sqrt3 = (100 - h) /a
a = sqrt3 * (100-h) ----------------- (1)
tan 60 = 100 / a
sqrt3 = 100 / a
a = 100 / sqrt3 ---------------------- (2)
Now equating (1) & (2),
100 / sqrt3 = sqrt3 * (100-h)
100 = 3*(100-h)
100 = 300 - 3h
3h = 300 - 100
h = 200/3
Height of pole = 200/3m
Let 'x' be the distance between foot of ladder from wall.
sin 60 = x / 10
sqrt3 / 2 = x / 10
2*x = 10* sqrt3
x = 5 * sqrt3 = 5 * 1.73 = 8.65
So, distance of foot of ladder from wall is 8.65 m.
Thickness of 1 book = 20 mm
Thickness of all books = 5 * 20 = 100 mm
Thickness of 1 paper sheet = 0.016 mm
Thickness of all paper sheets = 5 * 0.016 = 0.08 m
Total thickness = 100 + 0.08 = 100.08 mm
18 km = 1 cm
1 km = 1/18 cm
72 km = 72/18 cm = 4 cm
People going on Sunday = 845
People going on Monday = 169
Difference = 845 - 169 = 676
Percentage decrease = 676 * 100 / 845 = 80%
CP of 1 article = 2400 / 80 = Rs 30
Profit on 1 article = 16*30 / 100 = 4.8
SP of 1 article = CP + Profit = 30 + 4.8 = Rs. 34.80
Amount = P[1 + r/100]^n
P = 26400
r = 15
n = 2 + 4/ 12 = 7/3
Amount = 26400 * [1 + 15/100]^ (7/3) = Rs 36,579
Price before VAT = 5400 * 100/108 = Rs 5,000
Price to be paid for jeans = 1450 - 1450*10/100 = 1450 - 145 = Rs 1305
Price to be paid for shirts = 850 - 850*10/100 = 850 - 85 = Rs 765 each
Total price paid = 1305 + 2*765 = Rs 2835
Depreciated value of scooter after 1 year = 8% of 42,000 = 8*42000 / 100 = 3360
Scooter's value after 1 year = 42000 - 3360 = Rs 38,640
Population in 2001 = 54000 / (1 + 5/100)^2 = 48,980 ; As it is 2 years ago
Population in 2005 = 54000 * (1 + 5/100)^2 = 59,535 ; As it is 2 years after
Let marked price be 'x'
Marked Price = Selling price + discounted price
x = 1600 + 20x/100
x - x/5 = 1600
4x/5 = 1600
x = 1600*5/4 = 2000
Marked Price = Rs 2000
Let no of females be x & no of males be y.
x + y = 750 -------- (1)
x/2 + y/4 = 750/3
2x + y = 1000 --------- (2)
From (1)& (2),
x = 250
No. female democrats = x/2 = 250/2 = 125
Total no. of people surveyed = (n+100)
People in town of Eros who prefer brand A = n/2
People in town of Angie who prefer brand A = 100*60/100 = 60
All people who prefer brand A = (n + 100)*55/100
n/2 + 60 = (n+100)*55/100
n +120 = (n+100)*11/10
10n + 1200 = 11n + 1100
n = 100
Total number of people surveyed = n+100 = 100+100 = 200
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