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Monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenses are in the ratio 7 : 9. If each saves Rs. 50 per month find their monthly incomes.
Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are incresed by 12.
The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age.
Let the age of father be x and that of his son be a,b
We have x=2(a+b) and we have been given that 5 years ago we had the following situation which is given by the equation x-5=3(a+b-10)
On solving the above equations we obtain
a+b=25 thus the age of father is 50
A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number.
The wages of 8 men and 6 boys amount to Rs. 33. If 4 men earn Rs. 4.50 more than 5 boys determine the wages of each man and boy.
A number consisting of two digits is four times the sum of its digits and if 27 be added to it the digits are reversed. The number is :
Of two numbers, 1/5th of the greater is equal to 1/3rd of the smaller and their sum is 16.The numbers are:
Y is older than x by 7 years 15 years back X’s age was 3/4 of Y’s age. Their present ages are:
The sum of the digits in a three digit number is 12. if the digits are reversedthe number is increased by 495 but reversing only of ten’s and the unit digits increases the number by 36
Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3 of the smaller and 1/5 of the greater number are together 21. The numbers are :
The demand and supply equations for a certain commodity are 4q + 7p = 17 and respectively where p is the market price and q is the quantity then the equilibrium price and quantity are:
If the roots of the equation 2x2 + 8x – m3 = 0 are equal then value of m is
The cost of 10 kg of apples is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour?
Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = Rs.877.40
If œ ß be the roots of the equation 2x2 – 4x – 3 = 0 the value of α2 + β2 is
If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals then
The equation x2 –(p+4)x + 2p + 5 = 0 has equal roots the values of p will be.
If x = m is one of the solutions of the equation 2x2 + 5x – m = 0 the possible values of m are
If p and q are the roots of x2 + x + 1 = 0 then the values of p3 + q3 becomes
If L + M + N = 0 and L M N are rationals the roots of the equation (M+N–L) x2 + (N+L–M)x + (L+M–N) = 0 are
If p ≠ q and p2 = 5p – 3 and q2 = 5q – 3 the equation having roots as
If one rot of 5x2 + 13x + p = 0 be reciprocal of the other then the value of p is
A solution of the quadratic equation (a+b–2c)x2 + (2a–b–c)x + (c+a–2b) = 0 is
If the root of the equation x2–8x+m = 0 exceeds the other by 4 then the value of m is
The values of x in the equation 7(x+2p)2 + 5p2 = 35xp + 117p2 are
Time taken by a train to slow down from 80 kmh-1 to 20 kmh-1 with a uniform deceleration of 2 ms-2 is
The equation has got two values of x to satisfy the equation given as
Te sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are
The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are.
Five times of a positive whole number is 3 less than twice the square of the number. The number is
The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solve it to find the length and breadth of the field. The length and breadth are
Let the length be l and the breadth be b. now area = 2000 sq.m so length * breadth = 2000; l*b=2000.
now perimeter = 180m
so 2 ( l + b ) =180;
l + b = 90;
l = 90 - b.
substituting this value of l in l * b = 2000,
( 90 - b ) * b = 2000;
90b - b^2 = 2000;
b^2 - 90b + 2000 = 0;
b^2 - 40b - 50b - 2000 = 0;
b ( b - 40 ) - 50( b - 40 ) = 0;
( b - 40 )( b - 50 ) = 0;
b = 40 or b = 50;
so l = 40 or l = 50 but the longer side is considered the length and the shorter one the breadth.
so length = 50 m and breadth = 40 m
Two squares have sides p cm and (p + 5) cms. The sum of their squares is 625 sq. cm. The sides of the squares are