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STD 7 Scholarship Exam Maths (2003) (7 solved)
(1/5)x 2 = (3/10)x 3
(Bring x terms and non-x terms on same sides, signs will be changed while changing sides)
(1/5)x (3/10)x = 3 + 2
(Make denominators common to solve x-terms)
(2/10)x (3/10)x = -1
-(1/10)x = -1
x = -1 * (-10) = 10
|x| > 0 (as modulus value is always positive)
So x > 0
So x is a positive value, hence x is negative, hence x < 0
Let us assume cost price of an article = Re1
So Total cost price = Rs 50
His profit = cost price of 6 articles = Rs. 6
So % profit = 12%
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57.03/0.0001901 = (5703 * 10^-2)/(1901*10^-7) = 3*10^5(Measure number of digits after the decimal point in numerator and denominator)
By Pythagoras theorem,
l(AP) = SqRoot(l(AD)^2 - l(DP)^2) = SqRoot(100 36) = SqRoot(64) = 8 cm
A(Qadr ABCD) = A(Tri ADP) + A( Rect APQB) + A(Tri BQC)
= (1/2) * 6 * 8 + 14* 8 + (1/2) * 10 * 8
= 24 + 112 + 40
= 176 Sq Cm
19.9^2 = (199/10)^2 = 39601/100 = 396.01
Interest rate = 16 pcpa = 16/12 pcpm (per month)
On Rs 100, you will get Rs 16/12 each month
For the amount to be double, one should get interest of Rs 100.
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This will happen in 100/(16/12) = 1200/16 = 75 months = 6 years, 3 months
Each unit between 0 and -1/3 is equally divided into 5 parts.
So each small dot represents (0 ( 1/3) / 5 = 1/15
So Coordinate of K = -4/3 1/15 = -21/15 = -1(6/15) = -1(2/5) = -1.4
=[ (5^(0+5))/(3^(-7+0))]^(3/5)^(10/3)
= ((5^5)/(3^-7))^((3/5)*(10/3))
=( (5^5*3^7))^2
= 5^10 * 3^14
Fat in 120 L milk = 120 * 0.1 = 12 units
Fat in 200 L milk = 200 * 0.08 = 16 units
Total fat = 12 + 16 = 28 units
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Total Milk = 120 + 200 = 320 L
% fat = (28/320)*100 = 8.75%
Assume first number is x
so other number is (4/3) x
Product of two numbers =their LCM * their GCD
x * (4/3)x = 6 * 72
So x^2 = 6 * 72 * (3/4) = 18 * 18
x = 18 so other number = (4/3)*18 = 24
Bus travels 126 Km in 4 H, 5 M i.e. in 4 *60 + 5 = 245 minutes.
So in 1 min bus travels 126/245 km
in 35 min it will travel 35*(126/245) = 18 Km
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STD 7 Scholarship Exam Maths (2005)
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